# Logarithms: a bit of history (and the main rules)

This post will probably be of little or no interest to you. I wrote it to get somewhat more acquainted with logarithms myself. Indeed, I struggle with them. I think they come across as difficult because we don’t learn about the logarithmic function when we learn about the exponential function: we only learn logarithms later – much later. And we don’t use them a lot: exponential functions pop up everywhere, but logarithms not so much. Therefore, we are not as familiar with them as we should be.

The second point issue is notation: x = loga(y) looks more terrifying than y = ax because… Well… Too many letters. It would be more logical to apply the same economy of symbols. We could just write x = ay instead of loga(y), for example, using a subscript in front of the variable–as opposed to a superscript behind the variable, as we do for the exponential function. Or, else, we could be equally verbose for the exponential function and write y = expa(x) instead of y = ax. In fact, you’ll find such more explicit expressions in spreadsheets and other software, because these don’t take subscripts or superscripts. And then, of course, we also have the use of the Euler number e in eand ln(x). While it’s just a real number, is not as familiar to us as π, and that’s again because we learned trigonometry before we learned advanced calculus.

Historically, however, the exponential and logarithmic functions were ‘invented’, so to say, around the same time and by the same people: they are associated with John Napier, a Scot (1550–1617), and Henry Briggs, an Englishman (1561–1630). Briggs is best known for the so-called common (i.e. base 10) logarithm tables, which he published in 1624 as the Arithmetica Logarithmica. It is logical that the mathematical formalism needed to deal with both was invented around the same time, because they are each other’s inverse: if y = ax, then x = loga(y).

These Briggs tables were used, in their original format more or less, until computers took over. Indeed, it’s funny to read what Feynman writes about these tables in 1965: “We are all familiar with the way to multiply numbers if we have a table of logarithms.” (Feynman’s Lectures, p. 22-4). Well… Not any more. And those slide rules, or slipsticks as they were called in the US, have disappeared as well, although you can still find circular slide rules on some expensive watches, like the one below.

It’s a watch for aviators, and it allows them to rapidly multiply numbers indeed: the time multiplied by the speed will give a pilot the distance covered. Of course, there’s all kinds of intricacies here because we’ll measure time in minutes (or even seconds), and speed in knots or miles per hour, and so that explains all the other fancy markings on it. 🙂 In case you have one, now you know what you’re paying for! A real aviator watch! 🙂

How does it work? Well… These slide rules can be used for a number of things but their most basic function is to multiply numbers indeed, and that function is based on the logb(ac) = logb(a) + logb(c). In fact, this works for any base so we can just write log(ac) = log(a) + log(c). So the numbers on the slide rule below are the a, b and c. Note that the slides start with 1 because we’re working with positive numbers only and log(1) = 0, so that corresponds with the zero point indeed. The example below is simple (2 times 3 is six, obviously): it would have been better to demonstrate 1.55×2.35 or something. But you see how it goes: we add log(2) and log(3) to get log(6) = log(2×3). For 1.55×2.35, the slider would show a position between 3.6 and 3.7. The calculator on my \$30 Nokia phone gives me 3.6425. So, yes, it’s not far off. However, it’s hard to imagine that engineers and scientists actually used these slide rules over the past 300 years or so, if not longer.

Of course, Briggs’ tables are more accurate. It’s quite amazing really: he calculated the logarithms of 30,000 (natural) numbers to to fourteen decimal places. It’s quite instructive to check how he did that: all he did, basically, was to calculate successive square roots of 10.

Huh?

Yes. The secret behind is the basic rule of exponentiation: exponentiation is repeated multiplication, and so we can write: am+n =aman and, more importantly, am–n = ama–n = am/an. Because Briggs used the common base 10, we should write 10m–n = 10m/10n. Now Briggs had a table with the successive square roots of 10, like the one below (it’s only six significant digits behind the decimal point, not fourteen, but I just want to demonstrate the principle here), and so that’s basically what he used to calculate the logarithm (to base 10) of 30,000 numbers! Talking patience ! Can you imagine him doing that, day after day, week after week, month after month, year after year? Waw !

So how did he do it? Well… Let’s do it for x = log10(2) = log(2). So we need to find some x for which 10x = 2. From the table above, it’s obvious that log(2) cannot be 1/2 (= 0.5), because 101/2 = 3.162278, so that’s too big (bigger than 2). Hence, x = log(2) must be smaller than 0.5 = 1/2. On the other hand, we can see that x will be bigger than 1/4 = 0.25 because 101/4 = 1.778279, and so that’s less than 2.

In short, x = log(2) will be between 0.25 (= 1/4) and 0.5. What Briggs did then, is to take that 101/4 factor out using the 10m–n = 10m/10n formula indeed:

10x–0.25 = 10x/100.25 = 2/1.778279 = 1.124683

If you’re panicking already, relax. Just sit back. What we’re doing here, in this first step, is to write 2 as

2 = 10x = 10[0.25 + (x–0.25)] = 101/410x–0.25 = (1.778279)(1.124683)

[If you’re in doubt, just check using your calculator.] We now need log(10x–0.25) = log(1.124683). Now, 1.124683 is between 1.154782 and 1.074608 in the table. So we’ll use the lowest value (101/32) to take another factor out. Hence, we do another division: 1.124683/1.074608 = 1.046598. So now we have 2 = 10x = 10[1/4 + 1/32 + (x – 1/4 – 1/32)] = (1.778279)(1.074608)(1.046598).

We now need log(10x–1/4–1/32) = log(1.046598). We check the table once again, and see that 1.046598 is bigger than the value for 101/64, so now we can take that 101/64 value out by doing another division. (10x–1/4–1/32)/101/64 = 1.046598/1.036633 = 1.009613. Waw, this is getting small! However, we can still take an additional factor out because it’s larger than the 1.009035 value in the table. So we can do another division: 1.009613/1.009035 = 1.000573. So now we have 2 = 10x = 10[1/4 + 1/32 + 1/64 + 1/256 + (x – 1/4 –1/32 – 1/64 –1/256)] = 101/4101/32101/64101/25610x–1/4–1/32–1/64–1/256 = (1.778279)(1.074608)(1.036633)(1.009035)(1.000573).

Now, the last factor is outside of the range of our table: it’s too small to find a fraction. However, we had a linear approximation based on the gradient for very small fractions x: 10= 1 + 2.302585·r. So, in this case, we have 1.000573 = 1 + 2.302585·r and, hence, we can calculate r as 0.000248. [I can shown where this approximation comes from: just check my previous posts if you want to know. It’s not difficult.] So, now, we can finally write the result of our iterations:

2 = 10x ≈ 10(1/4 + 1/32 + 1/64 + 1/256 + 0.000248)

So log(2) is approximated by 0.25 + 0.03125 + 0.015625 + 0.00390625 + 0.000248 = 0.30103. Now, you can check this easily: it’s essentially correct, to an accuracy of six digits that is!

Hmm… But how did Briggs calculate these square roots of 10? Well… That was done ‘by cut and try’ apparently! Pf-ff ! Talk of patience indeed ! I think it’s amazing ! And I am sure he must have kept this table with the square roots of 10 in a very safe place ! 🙂

So, why did I show this? Well… I don’t know. Just to pay homage to those 17th century mathematicians, I guess. 🙂 But there’s another point as well. While the argument above basically demonstrated the am+n = amaformula or, to be more precise, the am–n = am/an formula, it also shows the so-called product rule for logarithms:

logb(ac) = logb(a) + logb(c)

Indeed, we wrote 2 as a product of individual factors 10and then we could see the exponents r in all of these individual factors add up to 2. However, the more formal proof is interesting, and much shorter too: 🙂

1. Let m = loga(x) and n = loga(y)
2. Write in exponent form: x = aand y = an
3. Multiply x and y: xy = aman = am+n
4. Now take loga of both sides: loga(xy) = loga(am+n) = (m+n)loga(a) = m+n = loga(x) + loga(y)

You’ll notice that we used another rule in this proof, and that’s the so-called power rule for logarithms:

loga(xn)= nloga(x)

This power rule is proved as follows:

1. Let m = loga(x)
2. Write in exponent form: x = am
3. Raise both sides to the power of n: xn = (am)n
4. Convert back to a logarithmic equation: loga(xn)= mn
5. Substitute for m = loga(x): loga(xn)= n loga(x)

Are there any other rules?

Yes. Of course, we have the quotient rule:

loga(x/y) = loga(x) – loga(y)

The proof of this follows the proof of the product rule, and so I’ll let you work on that.

Finally, we have the ‘change-of-base’ rule, which shows us how we can easily switch from one base to another indeed:

The proof is as follows:

1. Let x = loga b
2. Write in exponent form: a= b
3. Take log c of both sides and evaluate:

log c ax = log c b
x
log c a = log c b

[I copied these rules and proofs from onlinemathlearning.com, so let me acknowledge that here. :-)]

Is that it? Well… Yes. Or no. Let me add a few more lines on these logarithmic scales that you often encounter in various graphs. It the same scale as those logarithmic scales used for that slide that we showed above but it covers several orders of magnitude, all equally spaced: 1, 10, 100, 1000, etcetera, instead of 0, 1, 2, 3, etcetera. So each unit increase on the scale corresponds to a unit increase of the exponent for a given base (base 10 in this case): 101, 102, 103, etcetera. The illustration below (which I took from Wikipedia) compares logarithmic scales to linear ones, for one or both axes.

So, on a logarithmic scale, the distance from 1 to 100 is the same as the distance from 10 to 1000, or the distance from 0.1 to 10, or the distance between any point that’s 100 (= 102) times another point. This is easily explained by the product rule, or the quotient rule rather:

log(10) – log(0.1) = log(101/1–1) = log(102) = 2

= log(1000) – log(10) = log(103/11) = log(102/) = 2

= log(100) – log(1) = log(102/100) = log(102) = 2

And, of course, we could say the same for the distance between 1 and 1000, and 0.1 and 100. The distance on the scale is 3 units here, while the point is 1000 = 10the other point.

Why would we use logarithmic scales? Well… Large quantities are often better expressed like that. For example, the Richter scale used to measure the magnitude of an earthquake is just a base–10 logarithmic scale. With magnitude, we mean the amplitude of the seismic waves here. So an earthquake that registers 5.0 units on the Richter scale has a ‘shaking amplitude’ that is 10 times greater than that of an earthquake that registers 4.0. Both are fairly light earthquakes, however: magnitude 7, 8 or 9 are the big killers. Note that, theoretically, we could have earthquakes of a magnitude higher than 10 on the Richter scale: scientists think that the asteroid that created the Chicxulub crater created a cataclysm that would have measured 13 on Richter’s scale, and they associate it with the extinction of the dinosaurs.

The decibel, measuring the level of sound, is another logarithmic unit, so the power associated with 40 decibel is not two times but one hundred times that of 20 decibel!

Now that we’re talking sound, it seems that logarithmic scales are more ‘natural’ when it comes to human perception in general, but I’ll let you have fun googling some more stuff on that! 🙂

# Real exponentials and double roots: a post for my kids

There is one loose end related to exponentials that I want to tie up here. It’s the issue of multiple roots (or multiple-valuedness as it’s called in the context of inverse functions).

Introduction

You’ll remember that, for integer exponents n, we had two inverse operations for an:

1. The logarithm: the instruction here is to find n (i.e. the exponent) given the value an and given a (i.e. the base).
2. The ‘nth root’ function: the instruction here is find a (i.e. the base) given the value an and given n (i.e. the exponent).

We have two inverse operations because the exponentiation operation is not commutative: while a + b = b + a (and, therefore, a×b = b×a, so multiplication is commutative as well), an is surely not the same as na (except if a = n, of course).

Having two inverse operations is somewhat confusing, of course. However, when we expand the domain of the exponential function to also include rational exponents, the ‘nth root’ function becomes an exponential function itself: a1/n. That’s nice, because it tidies things up. We only have one inverse operation now: the logarithm.

Now, my kids understand exponentials, but they find logarithms weird. There are two reasons for that. The most important one is that we don’t learn about the logarithm function when we learn about the exponential function. We only learn logarithms later – much later. Therefore, we are not as familiar with them as we should be. There is no good reason for that but that’s what it is. [I guess I am like Euler here: I’d suggest logarithms and complex numbers should be taught earlier in life. Then we would have less trouble understanding them.]

The second one is notation, I think. Indeed, x = loga(y) looks much more frightening than y = abecause… Well… Too many letters. It would be more logical to apply the same economy of symbols. We could just write x = ay instead of loga(y), for example, using a subscript in front of the variable–as opposed to a superscript behind the variable, as we do for the exponential function. Or, else, we could be equally verbose for the exponential function and write y = expa(x) instead of y = ax. In fact, you’ll find such more explicit expressions in spreadsheets and other software, because these don’t take subscripts or superscripts.

In any case, that’s not the point here. I will come back to the logarithmic function later. The point that I want to discuss here is that, while we sort of merged our ‘nth root function’ with our exponential function as we allowed for rational exponents as well (as opposed to integers only), we’re actually still taking roots, so to say, and then we note another problem: the square root function yields not one but two numbers when the base (a) is real and positive: ± a1/2.

In fact, that’s a more general problem.

Odd and even rational exponents

You’ll remember the following rules for exponentiation:

1. For a positive real number a, we have always have two real nth roots when n is even: a1/n: ± a1/n. That’s obviously a consequence of having two real square roots ± a1/2, because the definition of even parity is that n can be written as n = 2k with k any integer, i.e. k ∈ Z (so k can be negative). Hence, a1/n can then be written as a1/2k = a1/2k = (a1/k)1/2. Hence, whatever the value of a1/k (if k is even, then we have two kth roots once again, but that doesn’t matter), we will have two real roots: plus (a1/k)1/2 and minus (a1/k)1/2

2. If n is uneven (or odd I should say), so n ∈ {2k+1: k ∈ Z}, we have only one real root a1/2k+1: that root is positive when a is positive and negative when a is negative.

3. For the sake of completeness, let me add the third case: a is negative and n is even. We know there’s no real nth root of a in that case. That’s why mathematicians invented i: we’ll associate an even root of a negative real number with two complex-valued roots: a1/n: ± ia1/n.

The first and second case are illustrated below for n = 2 and n = 3 respectively. The complex roots of the third case cannot be visualized because y is a real axis. Of course, we could imagine the complex-roots ± ia1/n if we would flip or mirror the blue and red graph (i.e. the graphs for n = 2) along the vertical axis and re-label that axis as the iy-axis, i.e. the imaginary axis. But so I’ll leave that to your imagination indeed.

How does this parity business turn out for rational exponents?

If r is a rational number r = m/n, we’ll have to express it as an irreducible fraction first, so the numerator m and denominator n have no other common divisors than 1, or –1 when considering negative numbers. But let’s look at positive numbers first. If we write r as an irreducible fraction m/n, then m and n cannot both be even. Why not? Because m and n can then both be divided by 2 and m/n is not an irreducible fraction in that case. Let’s assume m is even. Hence, n must be odd in that case. We can then write a2k/n as (ak/n)2. This number will always be positive, because we are squaring something. So it doesn’t matter if ak/n has one or two roots: we’ll square them and so the result will always be positive.

Now let’s assume the second possibility: m is odd. We can then write am/n as (a1/n)m. So now it will depend on whether or not n is even. If n is even, we have two real roots, if n is uneven, then we have only one. Let’s work a few examples:

• 82/3 = (81/3)= 2= 4
• 43/2 = (41/2)= (±2)=±2=±23 = ±8
• 161/4 = (161/2)1/2 = (±4)1/2 =±41/2 =±2 = ±2
• (–8)5/3 = (–81/3)= (–2)= 32

So we have two roots if m is odd and n is even, and only one root in all other cases. However, we said that m and n cannot both be even, hence, if n is even, m must be odd. In short, we can say that a rational exponent m/n is even (i.e. there will be two roots), if n is even. Does that work for complex roots as well? Let’s work that out with an example:

(–4)3/2 = (–41/2)= (±2i)=(±2)3i=±8i

So, yes! It works for complex roots as well. 🙂

OK. But let’s ask the obvious question now: where are these even numbers on the real line?

Well… They are everywhere: we can start from 1/2 and then change the numerator: 3/2, 5/2, etcetera. It’s all fine, as long as we use an odd number. However, we can also go down and change the denominator: 1/4, 1/6, 1/8 etcetera. And then we can, of course, take odd multiples of these fractions once again, such as 1025/1024 = 1.0009765625, for example, or on the other side, 1023/1024 = 0.9990234375. So we have two even numbers here right next to the odd number 1. We may increase the precision: we could take ± 1/3588 for example. 🙂

Of course, you may have noticed something here. The first thing, of course, is that we’ve defined these two even numbers 1.0009765625 and 0.9990234375 with a precision of 10 digits behind the decimal point, i.e. 1/1024 = 1/210 = 0.0009765625. The second point to note is that the last digit of these two rational coefficients, when expressed as a decimal, was 5. Now, you may think that should always be the case because of that 1/2 factor. But it’s not true: 1/6, for example, is a rational number that, written in decimal form, will yield 0.166666… This is an expression with a recurring decimal. And 1/10, of course, just yields 0.1. So there’s no easy rule here. You need to look at the fraction itself, and rational numbers are either as a finite decimal or an infinite repeating decimal. Of course, there are rules for that, but this is not a post on number theory, so I won’t write anything more on this: you can Google some more stuff yourself if you’re interested in this.

Irrational exponents

How does the business of parity work for irrational exponents? The gist of the rather long story above can be summarized easily. We can write am/n as am/n = am·(1/n) =(a1/n)m = a1/n·a1/n·a1/n·a1/n =·… (m times) and so whether or not we have multiple roots (two instead of one) depends on whether or not n is even. Indeed, remember – once again – that exponentiation is repeated multiplication, and so for the sign of the result, what matters is whether or not the number of times that we do that multiplication is even or odd, not only for integer but for rational exponents as well.

For irrational exponents, we also have repeated multiplication, but now we have an infinite expression, not a finite one:

a= ar(1/Δ + 1/Δ + 1/Δ + 1/Δ +…) = ar/Δ·ar/Δ·ar/Δ·ar/Δ

I explained this expression in my previous post: 1/Δ is an infinitesimally small fraction. In fact, I calculated rational powers of using the fraction 1/Δ = 1/1024 = 1/210. I used that fraction because I had started backwards, taking successive square roots of e, so e1/2, and then  e1/4, e1/8e1/16, etcetera.

However, as I mentioned when I started doing that, there was no compelling reason to cut things up by dividing them in 2. We could use 1/3 as the fraction to start with and, then, of course, or fraction 1/Δ would have been equal to 1/310 = 1/59049, so we have an odd number in the denominator here. So that’s one problem: we cannot say if Δ is even or odd. And the the second problem, of course, is that it’s an infinite expression and, hence, we cannot say if we multiplied 1/Δ an even or an odd number of times.

That leads to the third problem: we cannot say if r itself is even or uneven, which is basically what we were looking at: can we define irrational exponents as even or odd?

In short, the answer is no. In practice, that means that we will associate awith one ‘rth root’ only.

Hmm… That obviously makes a lot of sense but how do we ‘justify’ it from a more formal point of view? Where do these negative roots (for even powers) go? I am not sure. I guess there must be some more formal argument but I’ll leave that to you to look it up. I am fairly happy with what Wikipedia writes on that:

“[Real] Powers of a positive real number are always positive real numbers. […]  If the definition of exponentiation of real numbers is extended to allow negative results then the result is no longer well behaved.”

In fact, the article actually does give a somewhat more formal argument, as it writes:

• Neither the logarithm method nor the rational exponent method can be used to define br as a real number for a negative real number b and an arbitrary real number r. Indeed, eris positive for every real number r, so ln(b) is not defined as a real number for b ≤ 0.
• As for the rational exponent method, that cannot be used for negative values of b because it relies on continuity. The function f(r) = br has a unique continuous extension from the rational numbers to the real numbers for each b > 0. But when b < 0, the function f is not even continuous on the set of rational numbers r for which it is defined.

I am not quite sure I fully understand the last line, but I guess this refers to what I pointed out above: all these even and odd numbers that are so close to each other. When we go from rational to irrational exponents, we can no longer define odd or even.

The bottom line

The bottom line is that, in practice, we will only work with positive real bases. Hence, if b is negative, then we will define br as –(–b)r. Huh?

Yes. Think about it. If b is negative, we’ll just multiply it with –1 to ensure that the base is a positive real number. And then we just put a minus in front to get a graph such as, for example, that x1/3 function for the negative side of the x-axis as well.

You should also note that most applications, like the one I use to draw simple graphs like the ones above (rechneronline.de/function-graphs) are not capable of showing you both roots. They do check whether the exponent is even or odd though, because it plots the function x1/3 on both sides of the zero point, and the  x1/2 graph on the positive side only: it’s just not capable to associate more than one y value with one x value indeed. [In case you’re curious to see what it does with an irrational exponent, go and check it yourself: you can put in x^pi or x^e. Will it give function values for negative values of x as well? What’s your guess? :-)]

You’ll wonder why I am emphasizing this point. Well… I just wanted to note that we should be aware of the fact that, as we go from rational to irrational exponents, we sort of deliberately ‘forget’ about the second (negative) root. The point to note is that the issue of multiple-valued functions – such as discussed in the context of, for example, Riemann surfaces – is not necessarily related to complex-valued functions. We have it here (double roots), and we also have it, in general, for periodic functions.

But that’s for a next post. And there we’ll use our ‘natural’ exponential ex, and its inverse function, ln(x), an awful lot. So I’ll just conclude here with their graphs, noting, as Wikipedia does, that, nowadays, the term ‘exponential function’ is almost exclusively used as a shortcut for describing the natural exponential function ex. But, to my kids, I say: it’s good that you know where it comes from. 🙂

Post scriptum:

When thinking about such minor things, it’s always to good to think about why we are manipulating all these symbols. Exponentiation is repeated multiplication. What does it mean to multiply something with a negative number? A minus sign is an instruction to reverse direction, to turn around, 180 degrees. So we multiply the magnitudes of both numbers a and b, but we change the direction: if we’re walking down the positive real axis, then now we’re walking down the negative axis.

So repeated multiplication with a negative real number means we’re switching back and forth, wildly jumping from the positive to the negative side of the zero point and then back again. You’ll admit you would appreciate being told in advance how many times we need to do the multiplication if the multiplier is negative: if n is even, then we’ll end up going in the same direction:  (–1)= 1. No sign reversal. If n is uneven, then we know that, besides the ‘booster’ effect (i.e. the exponentiation operation), we’re expected to speed in the opposite direction: (–1)= –1.

Hence, if b would happen to be a negative real number, then defining br as –(–b)r, or assuming that, in general, our base will be a positive real number makes sense. Of course, the math has to keep track of the theoretical possibility that, if the exponent would happen to be even, b might be a negative number, but you can see it’s more of a theoretical possibility indeed. Not something we’d associate with something happening in real life.

In that sense, I should note that multiplication with a complex multiplier is much more ‘real-life’, so to say. Multiplying something with a complex number does the same to the magnitude of both numbers as real multiplication: it multiplies the magnitudes, thereby changing the scale. So the product of a vector that’s 2 units long and a vector that’s 3 units long will still be 6 units long. However, complex numbers also allow for a more gradual change of direction. Instead of just a gear to move forward and backward, we also get a steering wheel so to say: multiplying two complex numbers also adds their angles (as measured from some kind of zero direction obviously), besides multiplying their magnitudes. For example, suppose that the zero direction is east, and we have a vector pointing east indeed (that means its imaginary part is zero) that we need to multiply with a vector pointing north (so that’s a vector with a zero real part, along the imaginary axis), then the final vector will be pointing north.

However, with that subtlety comes complexity as well. With real numbers, you can go in the same direction by reversing direction two times, and so that’s why we have two 2nd roots (i.e. two square roots) of 1: (a) +1, so then we just stay where we are, and (b) –1, so then we rotate two times a full 180 degrees around the zero point: indeed, (–1)(–1) corresponds to two successive rotations by 180 degrees (or π in radians)–clockwise or counterclockwise, it doesn’t matter: one full loop around the zero point will get us back to square one, or point 1, I should say. 🙂

With complex numbers, it all depends. The 3rd root (i.e. the cube root) of 1 was only 1 in the real space but, in the complex space, we have three 3 cube roots of unity. The first one (W= W3) is the root we’re used to: unity itself, so the angle here is zero, i.e. straight ahead. In fact, with 1, we just stay where we are: 1×1×1 = 1= 1 indeed. But that’s not the only way. The illustration below shows two other ways to end up where we are (i.e. at point 1):

• The second cube root is W2: 120 degrees. You can see we get back at 1 by making three successive turns of 120 degrees indeed, so that’s one full loop around the or<igin. Using complex numbers (in polar notation), we write e2π/3×e2π/3×e2π/3 = e6π/3 = e2π e= 1.
• The third cube root is W1: that’s 240 degrees ! Indeed, here we get back at square one by making three successive turns of 4π/3 radians, i.e .by making two loops, in total, around the origin: e4π/3×e4π/3×e4π/3 = e12π/3 = e4π e= 1.

In short, we gain flexibility (of course, we have four 4th roots (with which we make 0, 1, 2 and 3 loops around the origin respectively), 5th roots, and so on), and the great Leonhard Euler was obviously fully right: complex numbers are more ‘natural’ numbers as they allow us to model real-life situations much better.

However, if you think that double roots are a problem… Well… Think again ! With complex numbers, the problem of multiple-valuedness is much more ‘real’, I’d say. 🙂

P.S: As mentioned in my previous post, I talk about that problem of multiple-valuedness when talking about Riemann surfaces in my October-November 2013 posts, so I won’t repeat what I wrote there. It’s about time I get back to both Feynman as well Penrose. 🙂

Just one last (philosophical) question to test your understanding. Negative real numbers have no real square root. That includes –1 obviously. Why is that? Why do we have two square real roots for +1 and no (none!) (real) square roots for –1?

[…] No? Come on!

[…] OK. Let me tell you: it’s all a question of definition. What’s implicit here is that we have only one real direction: from zero to infinity along the positive axis, and then –1 is nothing but a reversal of direction. So it’s an operation really, not a ‘real’ number. In a philosophical sense, of course: negative numbers don’t exist, so to say! Indeed, ask yourself: what is a negative number? It’s an operation: we subtract things when we use the minus sign, and we reverse direction when multiplying numbers with –1. So, if we multiply something with –1 two times in succession, we are back where we are.

Of course, we could say that the negative direction is the ‘real’ direction and, hence, that it’s the positive numbers that don’t ‘really’ exist. Indeed, math doesn’t care about what we say, so let’s say that the negative axis is the ‘real’ one, in a physical sense. What happens then? Well… Let’s see… Let’s do what we did before. We still define –1 as a reversal of direction, or a rotation by 180 degrees and, hence, doing that two times should bring us back where we want to be, so that’s –1 now. OK. So we have (–1)(–1)(–1) = (–1). But so that means that (–1)(–1) = 1, and… How can we write something like that for –1? What number a gives us the result that a×a = –1. Hmm… Only this imaginary number: i×i = i2 = –1. So, no matter how hard you try: the way we use symbols is pretty consequent, and so you will find that (–1)(–1) = 1×1 = 1 (so we have two square roots of 1), but we will not find that 1×1 = –1. If you would want to do that, you’d have to define +1 as a reversal of direction, so that basically means that the + sign would take the function of the – sign. Huh?

🙂 You must think I’ve gone crazy. I don’t think so. The idea I want to convey here is that, no matter how abstract math may seem to be – when everything is said and done – it’s intimately connected to our most basic notions of space, and our motion in that space. We go from here to there, or backwards, we change direction, we count things, we measure lengths or distances,… All that math does is to capture that in a non-ambiguous and consistent way. That also results in terse ‘truths’ such as: 1 has two real square roots, +1 and –1, but the square roots of –1 are only imaginary: ± i.

However, that terse statement hides another fun ‘truth’: +i and −i are as real as –1. Indeed, they are a rotation by 90 degrees, counterclockwise (+i) or clockwise (−i), as opposed to, for example, a rotation by 180 degrees (–1), or a full loop (1). 🙂

# Euler’s formula revisited

This post intends to take some of the magic out of Euler’s formula. In fact, I started doing that in my previous post but I think that, in this post, I’ve done a better job at organizing the chain of thought. [Just to make sure: with ‘Euler’s formula’, I mean ei= cos(x) + isin(x). Euler produced a lot of formulas, indeed, but this one is, for math, what E = mcis for physics. :-)]

The grand idea is to start with an initial linear approximation for the value of the complex exponential eis near s = 0 (to be precise, we’ll use the eiε = 1 + iε formula) and then show how the ‘magic’ of i – through the i= –1 factor – gives us the sine and cosine functions. What we are going to do, basically, is to construct the sine and cosine functions algebraically.

Let us, as a starting point – just to get us focused – graph (i) the real exponential function ex, i.e. the blue graph, and (ii) the real and imaginary part of the complex exponential function ei= cos(x) + isin(x), i.e. the red and green graph—the cosine and sine function.   From these graphs, it’s clear that ex and eiare two very different beasts.

1. eis just a real-valued function of x, so it ‘maps’ the real number x to some other real number y = ex. That y value ‘rockets’ away, thereby demonstrating the power of exponential growth. There’s nothing really ‘special’ about ex. Indeed, writing einstead of 10obviously looks better when you’re doing a blog on math or physics but, frankly, there’s no real reason to use that strange number e ≈ 2.718 when all you need is just a standard real exponential. In fact, if you’re a high school student and you want to attract attention with some paper involving something that grows or shrinks, I’d recommend the use of πx. 🙂

2. eiis something that’s very different. It’s a complex-valued function of x and it’s not about exponential growth (though it obviously is about exponentiation, i.e. repeated multiplication): y = eidoes not ‘explode’. On the contrary: y is just a periodic ‘thing’ with two components: a sine and a cosine. [Note that we could also change the base, to 10, for example: then we write 10ix. We’d also get something periodic, but let’s not get lost before we even start.]

Two different beasts, indeed. How can the addition of one tiny symbol – the little i in ei– can make such big difference?

The two beasts have one thing in common: the value of the function near x = 0 can be approximated by the same linear formula:

In case you wonder where this comes from, it’s basically the definition of the derivative of a function, as illustrated below. This is nothing special. It’s a so-called first-order approximation of a function. The point to note is that we have a similar-looking formula for the complex-valued eifunction. Indeed, its derivative is d(eix)/dx = ieiand when we evaluate that derivative at x = 0, then we get ie= i. So… Yes, the grand result is that we can effectively write:

eiε ≈ 1 + iε for small ε

Of course, 1 + iε is also a different ‘beast’ than 1 +  ε. Indeed, 1 + ε is just a continuation of our usual walk along the real axis, but 1 +  iε points in a different direction (see below). This post will show you where it’s headed.

Let’s first work with eagain, and think about a value for ε. We could take any value, of course, like 0.1 or some fraction 1/n. We’ll use a fraction—for reasons that will become clear in a moment. So the question now is: what value should we use for n in that 1/n fraction? Well… Because we are going to use this approximation as the initial value in a series of calculations—be patient: I’ll explain in a moment—we’d like to have a sufficiently small fraction, so our subsequent calculations based on that initial value are not too far off. But what’s sufficiently small? Is it 1/10, or 1/100,000, or 1/10100? What gives us ‘good enough’ results? In fact, how do we define ‘good enough’?

Good question! In order to try to define what’s ‘good enough’, I’ll turn the whole thing on its head. In the table below, I calculate backwards from e= e by taking successive square roots of eHuh? What? Patience, please! Just go along with me for a while. First, I calculate e1/2, so our fraction ε, which I’ll just write as  x, is equal to 1/2 here, so the approximation for e1/2 is 1 + 1/2 = 1.5. That’s off. How much? Well… The actual value of e1/2 is about 1.648721 (see the table below (or use a calculator or spreadsheet yourself): note that, because I copied the table from Excel, ex is shown as e^x). Now, 1.648721 is 1.5 + 0.148721, so our approximation (1.5) is about 9% off (as compared to the actual value). Not all that much, but let’s see how we can improve. Let’s take the square root once again: (e1/2)1/2 e1/4, so x = 1/4. And then I do that again, so I get e1/8, and so on and so on. All the way down to x = 1/1024 = 1/210, so that’s ten iterations. Our approximation 1 + x (see the fifth/last column in the table below is then equal to 1 + 1/1024 = 1 + 0.0009765625, which we rounded to 1.000977 in the table.

The actual value of e1/1024 is also about 1.000977, as you can see in the third column of the table. Not exactly, of course, but… Well… The accuracy of our approximation here is six digits behind the decimal point, so that’s equivalent to one part in a millionth. That’s not bad, but is it ‘good enough’? Hmm… Let’s think about it, but let’s first calculate some other things. The fourth column in the table above calculates the slope of that AB line in the illustration above: its value converges to one, as we would expect, because that’s the slope of the tangent line at x = 0. [So that’s the value of the derivative of eat x = 0. Just check it: dex/dx = ex, obviously, and e= 1.] Note that our 1 + x approximation also converges to 1—as it should!

So… Well… Let’s now just assume we’re happy with with that approximation that’s accurate to one part in a million, so let’s just continue to work with this fraction 1/1024 for x. Hence, we will write that e1/1024 ≈ 1 + 1/1024 and now we will use that value also for the complex exponentialHuh? What? Why? Just hang in here for a while. Be patient. 🙂 So we’ll just add the again and, using that eiε ≈ 1 + iε expression, we write:

ei/1024 ≈ 1 + i/1024

It’s quite obvious that 1 + i/1024 is a complex number: its real part is 1, and its imaginary part is 1/1024 = 0.0009765625.

Let’s now work our way up again by using that complex number 1 + i/1024 = 1 + i·0.0009765625 to calculate ei/512, ei/256, ei/128 etcetera. All the way back up to x = 1, i.e. ei. I’ll just use a different symbol for x: in the table below, I’ll substitute x for s because I’ll refer to the real part of our complex numbers as ‘x’ from time to time (even if I write a and b in the table below), and so I can’t use the symbol x to denote the fraction. [I could have started with s, but then… Well… Real numbers are usually denoted by x, and so it was easier to start that way.] In any case…

The thing to note is how I calculate those values ei/512, ei/256, ei/128 etcetera. I am doing it by squaring, i.e. I just multiply the (complex) number by itself. To be very explicit, note that ei/512 = (ei/1024)= ei·2/1024 = (ei/1024)(ei/1024). So all that I am doing in the table below is multiply the complex number that I have with itself, and then I have a new result, and then I square that once again, and then again, and again, and again etcetera. In other words, when going back up, I am just taking the square of a (complex) number. Of course, you know how to multiply a number with itself but, because we’re talking complex numbers here, we should actually write it out:

(a + i·b)= a– b2 + i·2ab = a– b2 + 2abi

[It would be good to always separate the imaginary unit from real numbers like a, b, or ab, but then I am lazy and so I hope you’ll always recognize that is the imaginary unit.] In any case… When we’re going back up (by squaring), the real part of the next number (i.e. the ‘x’ in x + iy) is a– b2 and the complex part (the ‘y’) is 2abi. So that’s what’s shown below—in the fourth and fifth column, that is.

Look at what happens. The x goes to zero and then becomes negative, and the y increases to one. Now, we went down from e1/n = e1 = e1/1 to e1/n = e1/1024, but we could have started with e2, or e4/n, or whatever. Hence, I should actually continue the calculations above so you can see what happens when s goes to 2, and then to 3, and then to 4, and so on and so on. What you’d see is that the value of the real and imaginary part of this complex exponential goes up and down between –1 and +1. You’d see both are periodic functions, like the sine and cosine functions, which I added in the last two columns of the table above. Now compare those a and b values (i.e. the second and third column) with the cosine and sine values (i.e. the last two columns). […] Do you see it? Do you see how close they are? Only a few parts in a million, indeed.

You need to let this sink it for a while. And I’d recommend you make a spreadsheet yourself, so you really ‘get’ what’s going on here. It’s all there is to the so-called ‘magic’ of Euler’s formula. That simple (a + ib)= a– b2 + 2abformula shows us why (and how) the real and imaginary part oscillate between –1 and +1, just like the cosine and sine function. In fact, the values are so close that it’s easy to understand what follows. They are the same—in the limit, of course

Indeed, these values a– b2 and 2ab, i.e. the real and imaginary part of the next complex number in our series, are what Feynman refers to as the algebraic cosine and sine functions, because we calculate them as (a + ib)= a– b2 + 2abi. These algebraic cosine and sine values are close to the real cosine and sine values, especially for small fractions s. Of course, there is a discrepancy becomes – when everything is said and done – we do carry a little error with us from the start, because we stopped at 1/n = 1/1024, before going back up.

There’s actually a much more obvious way to appreciate the error: we know that e1/1024 should be some point on the unit circle itself. Therefore, we should not equate a with 1 if we have some value b > 0. Or – what amounts to saying the same – if if b is slightly bigger than 0, then a should be slightly smaller than 1. So the eiε ≈ 1 + iε is an approximation only. It cannot be exact for positive values of ε. It’s only exact when ε = 0.

So we’re off—but not far off as you can see. In addition, you should note that the error becomes bigger and bigger for larger s. For example, in the line for s = 1, we calculated the values of the algebraic cosine and sine for s = 2 (see the a^2 – b^2 and 2ab column) as –0.416553 and 0.910186, but the actual values are cos(2) = –0.416146 and sin(2) = 0.909297, which shows our algebraic cosine and sine function is gradually losing accuracy indeed (we’re off like one part in a thousand here, instead of one part in a million). That’s what we’d expect, of course, as we’re multiplying the errors as we move ‘back up’.

The graph below plots the values of the table.

This graph also shows that, as we’re doubling our ratio r all the time, the data points are being spaced out more and more. This ‘spacing out’ gets a lot worse when further increasing s: from s = 1 (that’s the ‘highest’ point in the graph above), we’d go to s = 2, and then to s = 4, s = 8, etcetera. Now, these values are not shown above but you can imagine where they are: for s = 2, we’re somewhere in the second quadrant, for s = 4, we’re in the third, etcetera. So that does not make for a smooth graph. We need points in-between. So let’s ‘fix’ this problem by taking just one value for s out of the table (s = 1/4, for example) and we’ll continue to use that value as a multiplier.

That’s what’s done in the table below. It looks somewhat daunting at first but it’s simple really. First, we multiply the value we got for e1/4 with itself once again, so that gives us a real and an imaginary part for e1/8 (we had that already in the table above and you can check: we get the same here). We then take that value (i.e. e1/8) not to multiply it with itself but with e1/4 once again. Of course, because the complex numbers are not the same, we cannot use the (a + ib)= a– b2 + 2abi rule any more. We must now use the more general rule for multiplying different complex numbers: (a + ib)(c + id) = (ac – bd) + i(ad + bc). So that’s why I have an a, b, c and d column in this table: a and b are the components of the first number, and c and d of the second (i.e. e1/4 = 0.969031 + 0.247434i)

In the table above, I let s range from zero (0) to seven (7) in steps of 0.25 (= 1/4). Once again, I’ve added the real cosine and sine values for these angles (they are, of course, expressed in radians), because that’s what s is here: an angle, aka as the phase of the complex number. So you can compare.

The table confirms, once again, that we’re slowly losing accuracy (we’re now 3 to 4 parts in a thousand off), but it is very slowly only indeed: we’d need to do many ‘loops’ around the center before we could actually see the difference on a graph. Hey! Let’s do a graph. [Excel is such a great tool, isn’t it?] Here we are: the thick black line describing a circle on the graph below connects the actual cosine and sine values associated with an angle of 1/4, 1/2, 3/8 etcetera, all the way up to 7 (7 is about 2.3π, so we’re some 40 degrees past our original point after the ‘loop’), while the little ‘+‘ marks are the data points for the algebraic cosine and sine. They match perfectly because our eye cannot see the little discrepancy.

So… That’s it. End of story.

What?

Yes. That’s it. End of story. I’ve done what I promised to do. I constructed the sine and cosine functions algebraically. No compass. 🙂 Just plain arithmetic, including one extra rule only: i= –1. That’s it.

So I hope I succeeded. The goal was to take some of the magic out of Euler’s formula by showing how that eiε = 1 + iε approximation and the definition of i= –1 gives us the cosine and sine function itself as we move around the unit circle starting from the unity point on the real axis, as shown in that little graph:

Of course, the ε we were working with was much smaller than the size of the arrow suggests (it was equal to 1/1024 ≈ 0.000977 to be precise) but that’s just to show how differentials work. 🙂 Pretty good, isn’t it? 🙂

Post scriptum:

I. If anything, all this post did was to demonstrate multiplication of complex numbers. Indeed, when everything is said and done, exponentiation is repeated multiplication–both for real as well as for complex exponents. The only difference is–well… Complex exponents give us these oscillating things, because a complex exponent effectively throws a sine and cosine function in.

Now, we can do all kinds of things with that. In this post, we constructed a circle without a compass. Now, that’s not as good as squaring the circle 🙂 but, still, it would have awed Pythagoras. Below, I construct a spiral doing the same kind of math: I start off with a complex number again but now it’s somewhat more off the unit circle (1 + 0.247434i). In fact, I took the same sine value as the one we had for ei/4 but I replaced the cosine value (0.969031) with 1 exactly). In other words, my ε is a lot bigger here.

Then I multiply that complex number 1 + 0.247434with itself to get the next number (0.938776 + 0.494868i), and then I multiply that result once again with my first number (1 + 0.247434i), just like we did when we were constructing the circle. And then it goes on and on and on. So the only difference is the initial value: that’s a bit more off the unit circle. [When we constructed the circle, our initial value was also a bit off but much less. Here we go for a much larger difference.]

So you can see what happens: multiplying complex numbers amounts to adding angles and multiplying magnitudes: αeiβ·γeiδ = αγei(β+δ) =|αeiβ|·|γeiδ|ei(β+δ)| = |α||γ|ei(β+δ). So, because we started off with a complex number with magnitude slightly bigger than 1 (you calculate it using Pythagoras’ theorem: it’s 1.03, more or less, which is 3% off, as opposed less than one part in a million for the 1 + 0.000977i number), the next point is, of course, slightly off the unit circle too, and some more than 3% actually. And so that goes on and on and on and the ‘some more’ becomes bigger and bigger in the process.

Constructing a graph like this one is like doing the kind of silly stuff I did when programming little games with our Commodore 64 in the 1980s, so I shouldn’t dwell too much on this. In fact, now that I think of it: I should have started near –i, then my spiral would have resembled an e. 🙂 And, yes – for family reading this – this is also like the favorite hobby of our dad: calculating a better value for π. 🙂

However… The only thing I should note, perhaps, is that this kind of iterative process resembles – to some extent – the kind of process that iterative function systems (IFSs) use to create fractals. So… Well… It’s just nice, I guess. [OK. That’s just an excuse. Sorry.]

II. The other thing that I demonstrated in this post may seem to be trivial but I’ll emphasize it here because it helped me (not sure about you though) to understand the essence of real exponentials much better than I did before. So, what is it?

Well… It’s that rather remarkable fact that calculating (real) irrational powers amounts to doing some infinite iteration. What do I mean with that?

Well… Remember that we kept on taking the square root of e, so we calculated e1/2, and then (e1/2)1/2 = e1/4, and then (e1/4)1/2 e1/8, and then we went on: e1/16e1/32e1/64, all the way down to e1/1024, where we stopped. That was 10 iterations only. However, it was clear we could go on and on and on, to find that limit we know so well: e1/Δ tends to 1 (not to zero (0), and not to either!) for Δ → ∞.

Now, e = e1 is an exponential itself and so we can switch to another base, base-10 for example, using the general a= (bk)= bks = bt formula, with k = logb(a). Let’s do base-10: we get e1 = [10log10(e)]=  100.434294…etcetera. Now, because is an irrational number, log10(e) is irrational too, so we indeed have an infinite number of decimals behind the decimal point in 0.434294…etcetera. In fact, e is not only irrational but transcendental: we can’t calculate it algebraically, i.e. as the root of some polynomial with rational coefficients. Most irrational numbers are like that, by the way, so don’t think that being ‘transcendental’ is very special. In any case… That’s a finer point that doesn’t matter much here. You get the idea, I hope. It’s the following:

1. When we have a rational power am/n , it helps to think of it as a product of m factors a1/n (and surely if we would want to calculate am/n without using a calculator, which, I admit, is not very fashionable anymore and so nobody ever does that: too bad, because the manual work involved does help to better understand things). Let’s write it down: am/n = am·(1/n) =(a1/n)m = a1/n·a1/n·a1/n·a1/n =·… (m times). That’s simple indeed: exponentiation is repeated multiplication. [Of course, if m is negative, then we just write am/n as 1/(am/n), but so that doesn’t change the general idea of exponentiation.]
2. However, it is much more difficult to see why, and how, exponentiation with irrational powers amounts to repeated multiplication too. The rather lengthy exposé above shows… Well, perhaps not why, but surely how. [And in math, if we can show how, that usually amounts to showing why also, isn’t it? :-)] Indeed, when we think of ar (i.e. an irrational power of some (real) number a), we can think of it as a product of an infinite number of factors ar/Δ. Indeed, we can write aas:

a= ar(1/Δ + 1/Δ + 1/Δ + 1/Δ +…) = ar/Δ·ar/Δ·ar/Δ·ar/Δ

Not convinced? Let’s work an example: 10π = [eln10]π = [eln10]π = eln10·π = eln10·π = e7.233784… Of course, if you take your calculator, you’ll find something like 1385.455731, both for 10π  and e7.233784 (hopefully!), but so that’s not the point here. We’ve shown that is an infinite product e1/Δ·e1/Δ·e1/Δ·e1/Δ·… =e(1/Δ+1/Δ+1/Δ+1/Δ+…) eΔ/Δ with Δ some infinitely large (but integer) number. In our example, we stopped the calculation at Δ = 1024, but you see the logic: we could have gone on forever. Therefore, we can write e7.233784… as

e7.233784… = e7.233784…(1/Δ+1/Δ+1/Δ+1/Δ+…) = e7.233784…/Δ·e7.233784…/Δ·e7.233784…/Δ

Still not convinced? Let’s revert back to base 10. We can write the factors e7.233784…/Δ as e(ln10·π)/Δ = [eln10]π/Δ = 10π/Δ. So our original power 10π is equal to: 10π = 10π/Δ·10π/Δ·10π/Δ·10π/Δ·10π/Δ·10π/Δ… = 10π(Δ/Δ), and of course, 101/Δ also tends to 1 as Δ goes to infinity (not to zero, and not to 10 either). 🙂 So, yes, we can do this for any real number a and for any r really.

Again, this may look very trivial to the trained mathematical eye but, as a novice in Mathematical Wonderland, I felt I had to go through this to truly understand irrational powers. So it may or may not help you, depending on where you are in MW.

[Proving that the limit for Δ/Δ goes to 1 as Δ goes to ∞ should not be necessary, I hope? 🙂 But, just in case you wonder how the formula for rational and irrational powers could possibly be related, we can just write am/n = a(m/n)(1/Δ + 1/Δ + 1/Δ + 1/Δ +…) = am/nΔ·am/nΔ·am/nΔ·am/nΔ·…= (a1/Δ + 1/Δ + 1/Δ + 1/Δ +…)m/n = am/n, as we would expect. :-)]

III. So how does that a= ar/Δ·ar/Δ·ar/Δ·ar/Δ… formula work for complex exponentials? We just add the i, so we write air but we know what effect that has: we have a different beast now. A complex-valued function of r, or… Well… If we keep the exponent fixed, then it’s a complex-valued function of a! Indeed, do remember we have a choice here (and two inverse functions as well!).

However, note that we can write air in two slightly different ways. We have two interpretations here really:

A. The first interpretation is the easiest one: we write air as air =  (ar)i = (ar/Δ + r/Δ + r/Δ + r/Δ +…)i.

So we have a real power here, ar, and so that’s some real number, and then we raise it to the power i to create that new beast: a complex-valued function with two components, one imaginary and one real. And then we know how to relate these to the sine and cosine function: we just change the base to e and then we’re done.

In fact, now that we’re here, let’s go all the way and do it. As mentioned in my previous post  – it follows out of that a= (ek)= eks = eformula, with k = ln(a) – the only effect of a change of base is a change of scale of the horizontal axis: the graph of as is fully identical to the graph of et indeed: we just we need to substitute s by t = ks = ln(a)·s. That’s all. So we actually have our ‘Euler formula for aihere. For example, for base 10, we have 10i= cos[ln(a)·s] + isin[ln(a)·s].

But let’s not get lost in the nitty-gritty here. The idea here is that we let ‘act’ on ar, so to say. And then, of course, we can write ar as we want, but that doesn’t change the essence of what we’re dealing with.

B. The second interpretation is somewhat more tricky: we write air as air = air/Δ·air/Δ·air/Δ·air/Δ·…

So that’s a product of an (infinite) number of complex factors air/Δ. Now, that is a very different interpretation than the one above, even if the mathematical result when putting real numbers in for a and r will – obviously – have to be the same. If the result is the same, then what am I saying really? Well… Nothing much, I guess. Just that the interpretation of an exponentiation as repeated multiplication makes sense for complex exponentials as well:

• For rational r, we’ll have a finite number of complex factors: aim/n = ai/n·ai/n·ai/n·ai/n·… (m times).
• For irrational r, we’ll have an infinite number of complex factors air = air/Δ·air/Δ·air/Δ·air/Δ… etcetera.

So the difference with the first interpretation is that, instead of looking at aias a real number ar that’s being raised to the complex power i, we’re looking at aias a complex number ai that’s being raised to the real power r. As said, the mathematical result when putting real numbers in for a and r will – obviously – have to be the same. [Otherwise we’d be in serious trouble of course: math is math. We can’t have the same thing being associated with two different results.] But, as said, we can effectively interpret air in two ways.

[…]

What I am doing here, of course, is contemplating all kinds of mathematical operations here – including exponentiation – on the complex space, rather on the real space. So the first step is to raise a complex number to a real power (as opposed to raising a real number to a complex power). The next step will be to raise a complex number to a complex power. So then we’re talking complex-valued functions of complex variables.

Now, that’s what complex analysis is all about, and I’ve written very extensively about that in my October-November 2013 post. So I would encourage you to re-read those, now that you’ve got, hopefully, a bit more of an ‘intuitive’ understanding of complex numbers with the background given in this and my previous post.

Complex analysis involves mapping (i.e. mapping from one complex space to another) and that, in turn, involves the concept of so-called analytic and/or holomorphic functions. Understanding those advanced concepts is, in turn, essential to understanding the kind of things that Penrose is writing about in Chapter 9 to 12 of his Road to Reality. […] I’ll probably re-visit these chapters myself in the coming weeks, as I realize I might understand them somewhat better now. If I could get through these, I’d be at page 250 or so, so that’s only one quarter of the total volume. Just an indication of how long that Road to Reality really is. 🙂

And then I am still not sure if it really leads to ‘reality’ because, when everything is said and done, those new theories (supersymmetry, M-theory, or string theory in general) are quite speculative, aren’t they? 🙂

# Reflecting on complex numbers (again)

This will surely be not my most readable post – if only because it’s soooooo long and – at times – quite ‘philosophical’. Indeed, it’s not very rigorous or formal, unlike those posts on complex analysis I wrote last year. At the same time, I think this post digs ‘deeper’, in a sense. Indeed, I really wanted to get to the heart of the ‘magic’ behind complex numbers. I’ll let you judge if I achieved that goal.

Complex numbers: why are they useful?

The previous post demonstrated the power of complex numbers (i.e. why they are used for), but it didn’t say much about what they are really. Indeed, we had a simple differential equation–an expression modeling an oscillator (read: a spring with a mass on it), with two terms only: d2x/dt2 = –ω2x–but so we could not solve it because of the minus sign in front of the term with the x.

Indeed, the so-called characteristic equation for this differential equation is r2 = –ω2 and so we’re in trouble here because there is no real-valued r that solves this. However, allowing complex-valued roots (r = ±iω) to solve the characteristic equation does the trick. Let’s analyze what we did (and don’t worry if you don’t ‘get’ this: it’s not essential to understand what follows):

• Using those complex roots, we wrote the general solution for the differential equation as Aeiωt+ Beiωt. Now, note that everything is complex in this general solution, not only the eiωt and eiωt  ‘components’ but also the (random) coefficients A and B.
• However, because we wanted to find a real-valued function in the end (remember: x is a vertical displacement from an equilibrium position x = 0, so that’s ‘real’ indeed), we imposed the condition that Aeiωtand Beiωt had to be each other’s complex conjugate. Hence, B must beequal to A* and our ‘general’ (real-valued) solution was Aeiωt+ A*eiωt. So we only have one complex (but equally random) coefficient now – A – and we get the other one (A*) for free, so to say.
• Writing A in polar notation, i.e. substituting A for A = x0eiΔ, which implies that A* = x0e–iΔ, yields A0eiΔeiωt + A0e-iΔeiω = A0[ei(ωt + Δ) + ei(ωt + Δ)].
• Expanding this, using Euler’s formula (and the fact that cos(-α) = cosα but sin(-α) = sinα) then gives us, finally, the following (real-valued) functional form for x:

A0[cos(ωt + Δ) + isin(ωt + Δ) + cos(ωt + Δ) – isin(ωt + Δ)]

= 2A0cos(ωt + Δ) = x0cos(ωt + Δ)

That’s easy enough to follow, I guess (everything is relative of course), but do we really understand what we’re doing here? Let me rephrase what’s going on here:

• In the initial problem, our dependent variable x(t) was the vertical displacement, so that was a real-valued function of a real-valued (independent) variable (time).
• Now, we kept the independent variable t real – time is always real, never imaginary 🙂 – but so we made x = x(t) a complex (dependent) variable by equating x(t) with the complex-valued exponential ert. So we’re doing a substitution here really.
• Now, if ert is complex-valued, it means, of course, that r is complex and so that allows us to equate r with the square root of a negative number (r = ±iω).
• We then plug these imaginary roots back in and get a general complex-valued solution (as expected).
• However, we then impose the condition that the imaginary part of our solution should be zero.

In other words, we had a family of complex-valued functions as a general solution for the differential equation, but we limited the solution set to a somewhat less general solution including real-valued functions only.

OK. We all get this. But it doesn’t mean we ‘understand’ complex numbers. Let’s try to take the magic out of those complex numbers.

Complex numbers: what are they?

I’ve devoted two or three posts to this already (October-November 2013) but let’s go back to basics. Let’s start with that imaginary unit i. The essence of– and, yes, I am using the term ‘essence’ in a very ‘philosophical’ sense here I guess: i‘s intrinsic nature, so to speak – is that its square is equal to minus one: i2= –1.

That’s it really. We don’t need more. Of course, we can associate i with lots of other things if we would want to (and we will, of course!), such as Euler’s formula for example, but these associations are not essential – or not as essential as this definition I should say. Indeed, while that ‘rule’ or ‘definition’ is totally weird and – at first sight – totally random, it’s the only one we need: all other arithmetic rules do not change and, in fact, it’s just that one extra rule that allows us to deal with any algebraic equation – so that’s literally every equation involving addition, multiplication and exponentiation (so that’s every polynomial basically). However, stating that i2= –1 still doesn’t answer the question: what is a complex number really?

In order to not get too confused, I’ve started to think we should just take complex numbers at face value: it’s the sum of (i) some real number and (ii) a so-called imaginary part, which consists of another real number multiplied with i. [So the only ‘imaginary’ bit is, once again, i: all the rest is real! ] Now, when I say the ‘sum’, then that’s not some kind of ‘new’ sum. Well… Let me qualify that. It’s not some kind of ‘new’ sum because we’re just adding two things the way we’re used to: two and two apples are four apples, and one orange plus two more is three. However, it is true that we’re adding two separate beasts now, so to say, and so we do keep the things with an i in them separate from the real bits. In short, we do keep the apples and the oranges separate.

Now, I would like to be able to say that multiplication of complex numbers is just as straightforward as adding them, but that’s not true. When we multiply complex numbers, that i2= –1 rule kicks in and produces some ‘effects’ that are logical but not all that ‘straightforward’ I’d say.

Let’s take a simple example–but a significant one (if only because we’ll use the result later): let’s multiply a complex number with itself, i.e. let’s take the square of a complex number. We get (a + bi)2= (a + bi)(a + bi) = a·a + a·(bi) + (bi)·a + (bi)·(bi) = a+ 2abi + b2i= a2 + 2abi – b2. That’s very different as compared to the square of a real sum a + b: (a + b)= a+ 2ab + b2. How? Just look at it: we’ve got a real bit (a2 – b2) and then an imaginary bit (2abi). So what?

Well… The thumbnail graph below illustrates the difference for a = b: it maps x to (a) 4x[i.e. (x + x)2] and to (b) 2x2 [i.e. (x + ix)2] respectively. Indeed, when we’re squaring real numbers, we get (a + b)= 4a2–i.e. a ‘real bit’ only, of course!–but when we’re squaring complex numbers, we need to keep track of two components: the real part and the imaginary part. However, the real part (a2 – b2) is zero in this case (a = b), and so it’s only the imaginary part 2abi = 2a2i that counts!

That’s kids stuff, you’ll say… In fact, when you’re a mathematician, you’ll say it’s a nonsensical graph. Why? Because it compares an apple and an orange really: we want to show 2ixreally, not 2x2.

That’s true. However, that’s why the graph is actually useful. The red graph introduces a new idea, and with a ‘new’ idea I mean something that’s not inherent in the i2= –1 identity: it associates i with the vertical axis in the two-dimensional plane.

Hmm… This is an idea that is ‘nice’ – very nice actually – but, once again, I should note that it’s not part of i‘s essence. Indeed, the Italian mathematicians who first ‘invented’ complex numbers in the early 16th century (Tartaglia (‘the Stammerer’) and da Vinci’s friend Cardano) introduced roots of –1 because they needed them to solve algebraic equations. That’s it. Full stop. It was only much later (some hundred years later that is!) that Euler and Descartes associated imaginary numbers (like 2ix2) with the vertical coordinate axis. To my readers who have managed not to fall asleep while reading this: please continue till the end, and you will understand why I am saying the idea of a geometrical interpretation is ‘not essential’.

To the same readers, I’ll also say the following, however: if we do associate complex numbers with a second dimension, then we can associate the algebraic operations with things we can visualize in space. Most of you–all of you I should say–know that already, obviously, but let’s just have a look at that to make sure we’re on the same page.

A very basic thing in physical mathematics is reversing the direction of something. Things go in one direction, but we should be able to visualize them going in the opposite direction. We may associate this with a variable going from 0 to infinity (+∞): it may be time (t), or a time-dependent variable x, y or z. Of course, we know what we have here: we think of the positive real axis. So, what we do when we multiply with –1 is reversing its direction, and so then we’re talking the negative real axis: a variable going from 0 to minus infinity (-∞). Therefore, we can associate multiplication by –1 with a full rotation around the center (i.e. around the zero point) by 180 degrees (i.e. by π, in radians).

You may think that’s a weird way of looking at multiplication by minus one. Well… Yes and no. But think of it: the concept of negative numbers is actually as ‘weird’ as the concept of the imaginary unit in a way. I mean… Think about it: we’re used to use negative numbers because we learned about them when we were very small kids but what are they really? What does it mean to have minus three apples? You know the answer of course: it probably means that you owe someone three apples but that you don’t have any right now. 🙂 […] But that’s not the point here. I hope you see what I mean: negative numbers are weird too, in a sense. Indeed, we should be aware of the fact that we often look at concepts as being ‘weird’ because we weren’t exposed to them early enough: the great mathematician Leonhard Euler thought complex numbers were so ‘essential’ to math and, hence, so ‘natural’ that he thought kids should learn complex numbers as soon as they started learning ‘real’ numbers. In fact, he probably thought we should only be using complex numbers because… Well… They make the arithmetic space complete, so to say. […] But then I guess that’s because Euler understood complex numbers in a way we don’t, which is why I am writing about them here. 🙂

OK. Back to the main story line. In order to understand complex numbers somewhat better, it is actually useful – but, again, not necessarily essential – to think of i as a halfway rotation, i.e. a rotation by 90 degrees only, clockwise or counterclockwise, as illustrated above: multiplication with i means a counterclockwise rotation by 90 degrees (or π/2 radians) and multiplication with –i means a clockwise rotation by the same amount. Again, the minus sign gives the direction here: clockwise or counterclockwise. It works indeed: i·i =(-i)·(-i) = –1.

OK. Let’s wrap this up: we might say that

• a positive real number is associated with some (absolute) quantity (i.e. a magnitude);
• a minus sign says: “Go the opposite way! Go back! Subtract!”– so it’s associated with the opposite direction or the opposite of something in general; and, finally,
• the imaginary unit adds a second dimension: instead of moving on a line only, we can now walk around on a plane.

Once we understand that, it’s easy to understand why, in most applications of complex numbers, you’ll see the polar notation for complex numbers. Indeed, instead of writing a complex number z as z = a+ ib, we’ll usually see it written as:

z = reiθ with eiθ = cosθ + isinθ

Huh? Well… Yes. Let me throw it in here straight away. You know this formula: it’s Euler’s formula. The so-called ‘magical’ formula! Indeed, Feynman calls it ‘our jewel’: the ‘most remarkable formula in mathematics’ as he puts it. Waw ! If he says so, it must be right. 🙂 So let’s try to understand it.

Is it magical really? Well… I guess the answer is ‘Yes’ and ‘No’ at the same time:

• No. There is no ‘magic’ here. Associating the real part a and the imaginary part b with a magnitude r and an angle θ (a = rcosθ and b = rcosθ) is actually just an application of the Pythagorean theorem, so that’s ‘magic’ you learnt when you were very little and, hence, it does not look like magic anymore. [Although you should try to appreciate its ‘magic’ once again, I feel. Remember that you heard about the Pythagorean theorem because your teacher wanted to tell you what the square root of 2 actually is: a so-called irrational number that we get by taking the ‘one-half power’ of 2, i.e. 21/2 = 20.5, or, what amounts to the same, the square root of 2. Of course, you and I are both used to irrational numbers now, like 21/2, but they are also ‘weird’. As weird as i. In fact, it is said that the Greek mathematician who claimed their existence was exiled, because these irrational numbers did not fit into the (early) Pythagorean school of thought. Indeed, that school of thought wanted to reduce geometry to whole numbers and their ratios only. So there was no place for irrational numbers there!]
• Yes. It is ‘magical’. Associating eiθ – so that’s a complex exponential function really! – with the unit circle is something you learnt much later in life only, if ever. It’s a strange thing indeed: we have a real (but, I admit, irrational) number here – e is 2.718 followed by an infinite number of decimals as you know, just like π – and then we raise to the power iθ, so that’s i once again multiplied by a real number θ (i.e. the so-called phase or – to put it simply – the angle). By now, we know what it means to multiply something with i, and–of course–we also know what exponentiation is (it’s just a shorthand for repeated multiplication), but we haven’t defined complex exponentials yet.

In fact… That’s what we’re going to do here. But in a rather ‘weird’ way as you will see: we won’t define them really but we’ll calculate them. For the moment, however, we’ll leave it at this and just note that, through Euler’s relation, we can see how a fraction or a multiple of i, e.g. 0.1i or 2.3i, corresponds to a fraction or a multiple of the angle associated with i, i.e. 0.1 times π/2 or 2.3 times π/2. In other words, Euler’s formula shows how the second (spatial) dimension is associated with the concept of the angle.

[…] And then the third (spatial) dimension is, of course, easy to add: it’s just an angle in another direction. What direction? Well… An angle away from the plane that we just formed by introducing that first angle. 🙂 […] So, from our zero point (here and now), we use a ruler to draw lines, and then a compass to measure angles away from that line, and then we create a plane, and then we can just add dimensions as we please by adding more ‘angles’ away from what we already have (a line, or a plane, and any higher-dimensional thing really).

Dimensions

I feel I need to digress briefly here, just to make sure we’re on the same page. Dimensions. What is a dimension in physics or in math? What do we mean if we say that spacetime is a four-dimensional continuum? From what we wrote above, the concept of a spatial dimension should be obvious: we have three dimensions in space (the x, y and z direction), and so we need three numbers indeed to describe the position of an object, from our point of view that is (i.e. in our reference frame).

But so we also have a fourth number: time. By now, you also know that, just like position and/or motion in space, time is relative too: that is relative to some frame of reference indeed. So, yes, we need four numbers, i.e. four dimensions, to describe an event in spacetime. That being said, time is obviously still something different (I mean different than space), despite the fact that Einstein’s relativity theory relates it to space: indeed, we showed in our post on (special) relativity that there’s no such thing as absolute time. However, that actually reinforces the point: a point in time is something fundamentally different than a point in space. Despite the fact that

1. Time is just like a space dimension in the physical-mathematical meaning of the term ‘dimension’ (a dimension of a space or an object is one of the coordinates that is needed to specify a point within that space, or to ‘locate’ the object – both in time and space that is); and that,
2. We can express distance and time in the same units because the speed of light is absolute (so that allows us to express time in meter, despite the fact that time is relative or “local”, as Hendrik Lorentz called it); and that, finally,
3. If we do that (i.e. if we express time and distance in equivalent units), the equations for space and time in the Lorentz transformation equations mirror each other nicely – ‘mixing’ the space and time variables in the same way, so to say – and, therefore, space and time do form a ‘kind of union’, as Minkowski famously said;

Despite all that, time and space are fundamentally different things. Perhaps not for God – because He (or She, or It?) is said to be Everywhere Always – but surely for us, humans. For us, humans, always busy constructing that mental space with our ruler and our compass, time is and remains the one and only truly independent variable. Indeed, for us, mortal beings, the clocks just tick (locally indeed – that’s why I am using a plural: clocks – but that doesn’t change the fact they’re ticking, and in one direction only).

And so things happen and equations such as the one we started with – i.e. the differential equation modeling the behavior of an oscillator – show us how they happen. In one of my previous posts, I also showed why the laws of physics do not allow us to reverse time, but I won’t talk about that here. Let’s get back to complex numbers. Indeed, I am only talking about dimensions here because, despite all I wrote above about the imaginary axis in the complex plane, the thing to note here is that we did not use complex numbers in the physical-mathematical problem above to bring in an extra spatial dimension.

We just did it because we could not solve the equation with one-dimensional numbers only: we needed to take the square root of a negative number and we couldn’t. That was it basically. So there was no intention of bringing in a y- or z-dimension, and we didn’t. If we would have wanted to do that, we would have had to insert another dependent variable in the differential equation, and so it would have become a so-called partial differential equation in two or three dependent variables (x, y and z), with time – once again – as the independent variable (t). [A differential equation in one variable only (real- or complex-valued), like the ones we’re used to now, are referred to as ordinary differential equations, as opposed to… no, not extraordinary, but partial differential equations.]

In fact, if we would have generalized to two- or three-dimensional space, we would have run into the same type of problem (roots of negative numbers) when trying to solve the partial differential equation and so we would have needed complex-valued variables to solve it analytically in this case too. So we would have three ‘dimensions’ but each ‘dimension’ would be associated with complex (i.e. ‘two-dimensional) numbers. Is this getting complicated? I guess so.

The point is that, when studying physics or math, we will have to get used to the fact that these ‘two-dimensional numbers’ which we introduced, i.e. complex numbers, are actually more ‘natural’ ‘numbers’ to work with from a purely analytic point of view (as for the meaning of ‘analytic’, just read it as ‘logical problem-solving’), especially when we write them in their polar form, i.e. as complex exponentials. We can then take advantage of that wonderful property that they already are a functional form (z =reiθ), so to speak, and that their first, second etcetera derivative is easy to calculate because that ‘functional form’ is an exponential, and exponentials come back to themselves when taking the derivative (with the coefficient in the exponent in front). That makes the differential equation a simple algebraic equation (i.e. without derivatives involved), which is easy to solve.

In short, we should just look at complex numbers here (i.e. in the context of my three previous posts, or in the context of differential equations in general) as a computational device, not as an attempt to add an extra spatial dimension to the analysis.

Now, that’s probably the reason why Feynman inserts a chapter on ‘algebra’ that, at first, does not seem to make much sense. As usual, however, I worked through it and then found it to be both instructive as well as intriguing because it makes the point that complex exponentials are, first and foremost, an algebraic thing, not a geometrical thing.

I’ll try to present his argument here but don’t worry if you can’t or don’t want to follow it all the way through because… Well… It’s a bit ‘weird’ indeed, and I must admit I haven’t quite come to terms with it myself. On the other hand, if you’re ready for some thinking ‘outside of the box’, I assure you that I haven’t found anything like this in a math textbook or on the Web. This proves the fact that Feynman was a bit of a maverick… Well… In any case, I’ll let you judge. Now that you’re here, I would really encourage you to read the whole thing, as loooooooong as it is.

Complex exponentials from an algebraic point of view: introduction

Exponentiation is nothing but repeated multiplication. That’s easy to understand when the exponents are integers: a to the power n (an) is a×a×a×a×… etcetera – repeated n times, so we have n factors (all equal to a) in the product. That’s very straightforward.

Now, to understand rational exponents (so that’s an m/n exponent, with m and n integers), we just need to understand one thing more, and that is the inverse operation of exponentiation, i.e. the nth root. We then get am/n = (am)1/n. So, that’s easy too. […] Well… No. Not that easy. In fact, our problems starts right here:

• If n is even, and a is a positive real number, we have two (real) nth roots a1/n: ± a1/n.
• However, if a is negative (and n is still even obviously), then we have a problem. There’s no real nth root of a in that case. That’s why Cardano invented i: we’ll associate an even root of a negative real number with two complex-valued roots.
• What if n is uneven? Then we have only one real root: it’s positive when a is positive, and negative when a is negative. Done.

But let’s not complicate matters from the start. The point here is to do some algebra that should help us to understand complex exponentials. However, I will make one small digression, and that’s on logarithmic functions. It’s not essential but, again, useful. […] Well… Maybe. 🙂 I hope so. 🙂

We know that exponentials are actually associated with two inverse operations:

1. Given some value y and some number n, we can take the nth root of y (y1/n) to find the original base x for which y = xn.
2. Given some value y and some number a, we can take the logarithm (to base a) of y to find the original exponent x for which y = ax.

In the first case, the problem is: given n, find x for which y = xn. In the second case, the problem is: given a, find x for which y = ax. Is that complicated? Probably. In order to further confuse you, I’ve inserted a thumbnail graph with y = 2x (so that’s the exponential function with base 2) and y = log2x (so that’s the logarithmic function with base 2). You can see these two functions mirror each other, with the x = y line as the mirror axis.

We usually find logarithms more ‘difficult’ than roots (I do, for sure), but that’s just because we usually learn about them much later in life–like in a senior high school class, for example, as opposed to a junior high school class (I am just guessing, but you know what I mean).

In addition, we have these extra symbols ‘log‘–L-O-G :-)–to express the function. Indeed, we use just two symbols to write the y = 2function: 2 and x – and then the meaning is clear from where we write these: we write 2 in normal script and x as a superscript and so we know that’s exponentiation. But so we’re not so economical for the logarithmic function. Not at all. In fact, we use three symbols for the logarithmic function: (1) ‘log’ (which is quite verbose as a symbol in itself, because it consists of three letters), (2) 2 and (3) x. That’s not economical at all! Indeed, why don’t we just write y = 2x or something? So that’s a subscript in front, instead of a superscript behind. It would work. It’s just a matter of getting used to it, i.e. it’s just a convention in other words.

Of course, I am joking a bit here but you get my point: in essence, the logarithmic function should not come across as being more ‘difficult’ or less ‘natural’ than the exponential function: exponentiation involves two numbers – a base and an exponent – and, hence, it’s logical that we have two inverse operations, rather than one. [You’ll say that a sum or a product involves (at least) two terms or two factors as well, so why don’t they have two inverse operations? Well… Addition and multiplication are commutative operations: a+b = b+a, and a·b = b·a. Exponentiation isn’t: a≠ na. That’s why. Check it: 2×3 = 3×2, but 23 = 8 ≠ 3= 9.]

Now, apart from us ‘liking’ exponential functions more than logarithmic functions because of the non-relevant fact that we learned about log functions only much later in our life, we will usually also have a strong preference for one or the other base for an exponential. The most preferred base is, obviously, ten (10). We use that base in so-called scientific notations for numbers. For example: the elementary charge (i.e. the charge of an electron) is approximately –1.6×10−19 coulombs. […] Oh… We have a minus sign in the exponent here (–19). So what’s that? Sorry. I forgot to mention that. But it’s easy: a–n = (an)–1 = 1/an.

Our most preferred base is 10 because we have a decimal system, and we have a decimal system because we have ten fingers. Indeed, the Maya used a base-20 system because they used their toes to count as well (so they counted in twenties instead of tens), and it also seems that some tribes had octal (base-8) systems because they used the spaces between their fingers, rather than the fingers themselves. And, of course, we all know that computers use a base-2 system because… Well… Because they’re computers. In any case, 10 is called the common base, because… Well… Because it’s common.

However, by now you know that, in physics and mathematics, we prefer that strange numberas a base. However, remember it’s not that strange: it’s just a number like π. Why do we call it ‘natural’? Because of that nice property: the derivative of the exponential function ecomes back to itself: d(ex)/dt = ex. That’s not the case for 10x. In fact, taking the derivative of 10is pretty easy too: we just need to put a coefficient in front. To be specific, we need to put the logarithm (to base e) of the base of our exponential function (i.e. 10) in front: d(10x)/dt = 10xln(10). [Ln(10) is yet another notation that has been introduced, it seems, to confuse young kids and ensure they hate logarithms: ln(10) is just loge(10) or, if I would have had my way in terms of conventions (which would ensure an ‘economic’ use of symbols), we could also write ln(10) = e10. :-)]

Stop! I am going way too fast here. We first need to define what irrational powers are! Indeed, from all that I’ve written so far, you can imagine what am/n is (am/n  = am)1/n, but what if m is not an integer? What if m equals the square root of 2, for example? In other words, what is 10or ex  or 2or whatever for irrational exponents?

We all sort of ‘know’ what irrationals are: it involves limits, infinitesimals, fractions of fractions, Dedekind cuts. Whatever, even if you don’t understand a word of what I am writing here, you do – intuitively: irrationals can be approximated by fractions of fractions. The grand idea is that we divide some number by 2, and then we divide by 2 once again (so we divide by 4), and then once again (so we take 1/8), and again (1/16), and so on and so on. These are Dedekind cuts. Of course, dividing by two is a pretty random way of cutting things up. Why don’t we divide by three, or by four, for example? Well… It’s the same as with those other ‘natural’ numbers: we have to start somewhere and so this  ‘binary’ way of cutting things up is probably the most ‘natural’. 🙂 [Have you noticed how many ‘natural’ numbers we’ve mentioned already: 10, e, π, 2… And one (1) itself of course. :-)]

So we’ll use something like Dedekind cuts for irrational powers as well. We’ll define them as a sort of limit (in fact, that’s exactly what they are) and so we have to find some approximation (or convergence) process that allows us to do so.

We’ll start with base 10 here because, as mentioned above, base 10 comes across as more ‘natural’ (or ‘common’) to us non-mathematicians than the so-called ‘natural’ base e. However, I should note that the base doesn’t matter much because it’s quite easy to switch from one base to another. Indeed, we can always write a= (bk)= bks = bt with a = band t = k·s (as for k, k is obviously equal to logb(a). From this simple formula, you can see that changing base amounts to changing the horizontal scale: we replace s by t = k·s. That’s it. So don’t worry about our choice of base. 🙂

Complex exponentials from an algebraic point of view: well… Not the introduction 🙂

Ouf! So much stuff! But so here we go. We take base 10 and see how such an approximation of an irrational power of 10 (10x) looks like. Of course, we can write any irrational number x as some (positive or negative) integer plus an endless series of decimals after the zero (e.g. e = 2 + 0.7182818284590452… etc). So let’s just focus on numbers between 0 and 1 as for now (so we’ll take the integer out of the total, so to speak). In fact, before we start, I’ll cheat and show you the result, just to make sure you can follow the argument a bit.

Yes. That’s how 10x looks like, but so we don’t know that yet because we don’t know what irrational powers are, and so we can’t make a graph like that–yet. We only know very general things right now, such as:

• 100 = 1 and 101 = 10 etcetera.
• Most importantly, we know that 10m/n  = (10m)1/n = (101/n)for integer m and n.

In fact, we’ll use the second fact to calculate 10x for x = 1/2, 1/4, 1/8, 1/16, and so on and so on. We’ll go all the way down to where x becomes a fraction very close to zero: that’s the table below. Note that the x values in the table are rational fractions 1/2, 1/4, 1/8 etcetera indeed, so x is not an irrational exponent: x is a real number but rational, so x can be expressed either as a fraction of two integers m and n (m = 1 and n = 1, 4, 8, 16, 32 and so on here), or as a decimal number with a finite number of decimals behind the decimal point (0.5, 0.25, 0.125, 0.0625 etcetera).

The third column gives the value 10x for these fractions x = 1/2, 1/4, 1/8 etcetera. How do we get these? Hmm… It’s true. I am jumping over another hurdle here. The key assumption behind the table is that we know how to take the square root of a number, so that we can calculate 101/2, to quite some precision indeed, as 101/2 = 3.162278 (and there’s more decimals but we’re not too interested in them right now), and then that we can take the square root of that value (3.162278). That’s quite an assumption indeed.

However, if we don’t want this post to become a book in itself, then I must assume we can do that. In fact, I’ve done it with a calculator here but, before there were calculators, this kind of calculations could and had to be done with a table of logarithms. That’s because of a very convenient property of logarithms: logc(ab) =logc(a) + logc(b). However, as said, I should be writing a post here only, not a book. [Already now, this post beats the record in terms of length and verbosity…] So I’ll just ask you to accept that – at this stage – we know how to calculate the square root of something and, therefore, to accept that we can take the square root not only of 10 but of any number really, including 3.162278, and then the root of that number, and then the root of that result, and so and so on. So that gives us the values in the third column of the table above: they’re successive square roots. [Please do double-check! It will help you to understand what I am writing about here.]

So… Back to the main story. What we are doing in the table above is to take the square root in succession, so that’s (101/2)1/2 = 101/4, and then again: (101/4)1/2 = 101/8 , and then again: (101/8)1/2 = 101/16 , so we get 101/2, 101/4, 101/8, 101/16, 101/32 and so on and so on. All the way down. Well… Not all the way down. In fact, in the table above, we stop after ten iterations already, so that’s when x = 1/1024. [Note that 1/1024 is 2 to the power minus 10: 2–10 = 1/210   = 1/1024. I am just throwing that in here because that little ‘fact’ will come in handy later.]

Why do we stop after ten iterations? Well… Actually, there’s no real good reason to stop at exactly ten iterations. We could have 15 iterations: then x would be 1/215 = 1/32768. Or 20 (x = 1/1048576). Or 39 (x = 1/too many digits to write down). Whatever. However, we start to notice something interesting that actually allows us to stop. We note that 10 to the power x (10x) tends to one as x becomes very small.

Now you’re laughing. Well… Surely ! That’s what we’d expect, isn’t it? 10= 1. Is that the grand conclusion?

No.

The question is how small should x be? That’s where the fourth column of the table above comes in. We’re calculating a number there that converges to some value quite near to 2.3 as x goes to zero and – importantly – it converges rather quickly. In fact, if you’d do the calculations yourself, you’d see that it converges to 2.302585 after a while. [With Excel or some similar application, you can do 20 or more iterations in no time, and so that’s what you’ll find.]

Of course, we can keep going and continue adding zillions of decimals to this number but we don’t want to do that: 2.302585 is fine. We don’t need any more decimals. Why? Well… We’re going to use this number to approximate 10near x = 0: it turns out that we can get a real good approximation of 10x near x = 0 using that 2.302585 factor, so we can write that

10≈ 1 + 2.302585x

That approximation is the last column in the table above. In order to show you how good it is as an ‘approximation’, I’ve plotted the actual values for 10x (blue markers) and the approximated values for 10x (black markers) using that 1 + 2.302585x formula. You can see it’s a pretty good match indeed if x is small. And ‘small’ here is not that small: a ratio like x = 1/8 (i.e. x = 0.125) is good enough already! In fact, the graph below shows that 1/16 = 0.0625 is almost perfect! So we don’t need to ‘go down’ too far: ten iterations is plenty!

I’ve probably ‘lost’ you by now. What are we doing here really? How did we get that linear approximation formula, and why do we need it? Well… See the last column: we calculate (10x–1)/x, so that’s the difference between 10and 1 divided by the (fractional) exponent x and we see, indeed, that that number converges to a value very near to 2.302585. Why? Well… What we are actually doing is calculating the gradient of 10x, i.e. the slope of the tangent line to the (non-linear) 10x curve. That’s what’s shown in the graph below.

Working backwards, we can then re-write (10x–1)/x ≈ 2.302585 as 10≈ 1 + 2.302585x indeed.

So what we’ve got here is quite standard: we know we can approximate a non-linear curve with a linear curve, using the gradient near the point that we’re observing (and so that’s near the point x = 0 in this case) and so that‘s what we’re doing here.

Of course, you should remember that we cannot actually plot a smooth curve like that, for the moment that is, because we can only calculate 10x for rational real numbers. However, it’s easy to generalize and just ‘fill the gaps’ so to speak, and so that’s how irrational powers are defined really.

Hmm… So what’s the next step? Well… The next step is not to continue and continue and continue and continue etcetera to show that the smooth curve above is, indeed, the graph of 10x. No. The next step is to use that linear approximation to algebraically calculate the value of 10is, so that’s a power of 10 with a complex exponent.

HUH!?

Yes. That’s the gem I found in Feynman’s 1965 Lectures. [Well… One of the gems, I should say. There are many. :-)]

It’s quite interesting. In his little chapter on ‘algebra’ (Lectures, I-22), Feynman just assumes that this ‘law’ that 10= 1 + 2.302585x is not only ‘correct’ for small real fractions x but also for very small complex fractions, and then he just reverses the procedure above to calculate 10ifor larger values of x. Let’s see how that goes.

However, let’s first switch the variable from x to s, because we’re talking complex numbers now. Indeed, I can’t use the symbol x as I used it above anymore because x is now the real part of some complex number 10is. In addition, I should note that Feynman introduces this delta (Δ). The idea behind is to make things somewhat easier to read by relating s to an integer: Δ = 1024s, so Δ = 1, 2, 4, 8,… 1024 for s = 1/1024, 1/512, 1/256 etcetera (see the second column in the table below). I am not entirely sure why he does that: Feynman must think fractions are harder to ‘read’. [Frankly, the introduction of this Δ makes Feynman’s exposé somewhat harder to ‘read’ IMHO – but that’s just a matter of taste, I guess.] Of course, the approximation then becomes

10= 1 + 2.302585·Δ/1024 = 1 + 0.0022486Δ.

The table below is the one that Feynman uses. The important thing is that you understand the first line in this table: 10i/1024 = 1 + 0.00225i·Δ1 + 0.00225i·1 = 1 + 0.00225i. And then we go to the second line: 10i/512 = 10i/1024·10i/1024 = 102i/1024 = 10i/512, so we’re doing the reverse thing here: we don’t take square roots but we square what we’ve found already. So we multiply 1 + 0.00225i with itself and get (1+0.00225i)(1+0.00225i) =  1 + 2·0.00225i + 0.002252i2 = 1 – 0.000005 + 0.45i ≈ 0.999995 + 0.45i ≈ 1 + 0.0045i.

Let’s go to the third line now. In fact, what we’re doing here is working our way back up, i.e. all the way from s = 1/1024 to s = 1. And that’s where the ‘magic’ of i (i.e. the fact that i2 = –1) is starting to show: (0.999995+0.0045i)2 =  0.99999 + 2·0.999995·0.0045i + 0.00452i= 0.99997 + 0.009i. So the real part of 10iis changing as well – it is decreasing in fact! Why is that? Because of the term with the ifactor! [I write 0.99997 instead of 0.99996 because I round up here, while Feynman consistently rounds down.]

So now the game is clear: we take larger and larger fractions s (i/512, i/256, i/128, etcetera), and calculate 10iby squaring the previous result. After ten iterations, we get the grand result for s = i/1 = i:

10is = –0.66928 + 0.74332i (more or less that is)

Note the minus sign in front of the real part, and look at the intermediate values for x and y too. Isn’t that remarkable?

OK. Waw ! But… So what? What’s next?

Well… To graph 10is, we should not just keep squaring things because that amounts to doubling the exponent again and again and so that means the argument is just making larger and larger jumps along the positive real axis really (see that graph that I made above: the distance between the successive values of x gets larger and larger, and so that’s a bad recipe for a smooth graph).

So what can we do? Well… We should just take a sufficiently small power, i/8 for example, and multiply that with 1, 2, 3 etcetera so we get something more ‘regular’. That’s what’s done in the table below and what’s represented in the graph underneath (to get the scale of the horizontal axis, note that s = p/8).

Hey! Look at that! There we are! That’s the graph we were looking for: it shows a (complex) exponential (10is) as a periodic (complex-valued) function with the real part behaving like a cosine function and the imaginary part behaving like as a sine function.

Note the upper and lower bounds: +1 and –1. Indeed, it doesn’t seem to matter whether we use 10 or as a base: the x and y part oscillate between −1 and +1. So, whatever the base, we’ll see the same pattern: the base only changes the scale of the horizontal axis (i.e. s). However, that being said, because of this scale factor, I do need to say like a cosine/sine function when discussing that graph above. So I cannot say they are a cosine and a sine function. Feynman calls these functions algebraic sine and cosine functions.

But – remember! – we can always switch base through a clever substitution so 10is = eit and recalculate stuff to whatever number of decimals behind the decimal point we’d want. So let’s do that: let’s switch to base e. WOW! What happens?

We then [Finally! you’ll say!] get values that – Surprise ! Surprise ! – correspond to the real cosine and sine function. That then, in turn, allows us to just substitute the ‘algebraic’ cosine and sine function for the ‘real’ cosine in an equation that – Yes! – is Euler’s formula itself:

ei= cos(t) + isin(t)

So that’s it. End of story.

[…]

You’ll say: So what? Well… Not sure what to say. I think this is rather remarkable. This is not the formal mathematical proof of Euler’s formula (at least not of the kind that you’ll find in a textbook or on Wikipedia). No, we are just calculating the values x and y of ei= x + iy using an approximation process used to calculate real powers and then, well… Just some bold assumption involving infinitesimals really.

I think this is amazing stuff (even if I’ll downplay that statement a bit in my post scriptum). I really don’t understand these things the way I would like to understand them. I guess I just haven’t got the right kind of brain for these things. 😦 Indeed, just think about it: when we have the real exponential ex, then we’ve got that typical ‘rocket’ graph (i.e. the blue one in the graph below): just something blasting away indeed. But when we put in the exponent (eix), then we get two components oscillating up and down like the cosine and sine function. Well… Not only like the cosine and sine function: the green and red line– i.e. the real and imaginary part of eix!– actually are the cosine and sine function!

Do you understand this in an intuitive way? Yes? You do? Waw ! Please write me and tell me how. I don’t. 😦

Oh well… The good thing about it is… Well… At least complex numbers will always stay ‘magical’ to me. 🙂

Post scriptum: When I write, above, that I don’t understand this in an intuitive way, I don’t mean to say it’s not logical. In fact, it is. It has to be, of course, because we’re talking math here! 🙂

The logic is pretty clear indeed. We have an exponential function here (y = 10x) and we’re evaluating that function in the neighborhood of x = 0 (we do it on the positive side only but we could, of course, do the same analysis on the other side as well). So then we use that very general mathematical procedure of calculating approximate values for the (non-linear) 10x curve using the gradient. So we plug in some differential value for x (in differential terms, we’d write Δx – but so the delta symbol here has nothing to do with Feynman’s Δ above) and, of course, we find Δy = 2.302585·Δx. So we add that to 1 (the value of 10at point x = 0) and, then, we go through these iterations, not using that linear equation any more, but the very fundamental property of an exponential function that 102x = (10x)2. So we start with an approximate value, but then the value we plug into these iterative calculations is the square of the previous value. So, to calculate the next points, we do not use an approximation method any more, but we just square the first result, and then the second and so on and so on, and that’s just calculation, not approximation.

[In fact, you may still wonder and think that it’s quite remarkable that the points we calculate using this process are so accurate, but that’s due to the rapid convergence of that value we found for the gradient. Well… Yes and no. Here I must admit that Feynman (and I) cheated a bit because we used a rather precise value for the gradient: 2.302585, so that’s six significant digits after the decimal point. Now, that value is actually calculated based on twenty (rather than 10) iterations when ‘going down’. But that little factoid is not embarrassing because it doesn’t change much: the argument itself is sound. Very sound.]

OK… That’s easy enough to understand. The thing that is not easy to understand – intuitively that is – is that we can just insert some complex differential Δs into that Δy = 2.302585·Δx equation. Isn’t it ‘weird’, indeed, that we can just use a complex fraction s = i/1024 to calculate our first point, instead of a real fraction x = 1/1024? It is. That’s the only thing really. Indeed, once we’ve done that, it’s plain sailing again: we just square the result to get the next result, and then we square that again, and so on and so on. However, that being said, the difference is that the ‘magic’ of i comes into play indeed. When squaring, we do not get a 4a2 result but an (a+bi)= a– b2 + 2abi. So it’s that minus sign and the i that give an entirely different ‘dynamic’ to how the function evolves from there (i.e. different as compared to working with a real base only). It’s all quite remarkable really because we start off with a really tiny value b here: 0.00225 to be precise, so that’s (less than) 1/445 ! [Of course, the real part a, at the point from where we start doing these iterations, is one.]

But so that first step is ‘weird’ indeed. Why is it no problem whatsoever to insert the complex fraction s = i/1024 into 1 + 2.302585o·s, instead of the real fraction 1/1024, and then afterwards, to square these complex numbers that we’re getting, instead of real numbers?

It just doesn’t feel right, does it? I must admit that, at first, I felt that Feynman was doing something ‘illegal’ too. But, obviously, he’s not. It’s plain mathematical logic. We have two functions here: one is linear (y = 1 + 2.302585·x), and the other is quadratic (y = x2) and so what’s happening really is that, at the point x = 0, we change the function. We substitute not x for ix really but y = 10for y = 10ix. So we still have an independent real variable x but, instead of a real-valued y = 10function, we now have a complex-valued y = 10ifunction.

However, the ‘output’ of that function, of course, is a complex y, not a real y. In our case, because we’re plotting a function really–to be precise, we’re calculating the exponential function y = 10through all these iterations–we get a complex-valued function of the shape that, by now, we know so well.

So it is ‘discontinuous’ in a way, and so I can’t say all that much about it. Look at the graph below where, once again, we have the real exponential function ex and then the two components of the complex exponential eix. This time, I’ve plotted them on both sides of the zero point because they’re continuous on both sides indeed. Imagine we’re walking along this blue ex curve from some negative x to zero. We’re familiar with the path. It has, for instance, that property we exploited above: as we doubled the ‘input’ (so from x we went to 2x), the ‘output’ went up not as the double but as the square of the original value: e2x = (ex)2. And then we also know that, around the point x = o, we can approximate it with a linear function. In fact, in this case, the linear approximation is super-simple: y = 1 + x. Indeed, the gradient for ex at point x = 0 is equal to 1! So, yes, we know and understand that blue curve. But then we arrive at point x = 0 and we decide something radical: we change the function!

Yes. That’s what we’re really doing in that very lengthy story above: ei is a complex-valued function of the real variable x. That’s something different. However, we continue to say that the approximation y = 1 + x must also be valid for complex x and y. So we say that ei= 1 + ix. Is that wrong? No. Not at all. Functional forms are functional forms and gradients are gradients: d(eix)/dx = ieix, and ieix at x = 0 is equal to ie0 = i! Hence, ei= 1 + ix is a perfectly legitimate linear approximation. And then it’s just the same thing again: we use that iteration mechanism to calculate successive squares of complex numbers because, for complex exponentials as well, we have e2(ix) = (eix)2.

So. The ‘magic’ is a lot of ‘confusion’ really. The point to note is that we do have a different function here: eiand e‘look’ similar–it’s just that i, right?but, in fact, when we replace x by ix in the exponent of e, that’s quite a radical change. We can use the same linear approximation at x = ix = 0 but then it’s over. Our blue graph stops: we’re no longer walking along it. I can’t even say it bifurcates, so to say, into the red and the green one, because it doesn’t. We’re talking apples and oranges indeed, and so the comparison is quickly done: they’re different. Full stop.

Is there any geometrical relationship between all these curves? Well… Yes and no. I can see one, at the very start: the gradient of our ex function at x = 0 is equal to unity (i.e. 1), and so that’s the same gradient as the gradient of the imaginary part of our new eifunction (the gradient of the real part is zero, before it becomes negative). But that’s just… I mean… That just comes out of Euler’s formula: e= cos(0) + isin(0). Honestly, it’s no use to try to be smart here and think about stuff like that. We’re no longer walking on the blue curve. We’re looking at a new function: a complex-valued function eix (instead of a real-valued function ex) of a real variable (x). That’s it. Just don’t try to relate the two too much: you switched functions. Full stop. It’s like changing trains! 🙂

So… What’s the conclusion? Well… I’d say: “Complex numbers can be analyzed as extensions of real numbers, so to say, but – frankly – they are different.

[…]

I’ll probably never understand complex numbers in the way I would like to understand them–that is like I understand that one plus one is two. However, this rather lengthy forage in the complex forest has helped me somewhat. I hope it helped you too.

# Differential equations revisited: the math behind oscillators

When wrapping up my previous post, I said that I might be tempted to write something about how to solve these differential equations. The math behind them is pretty essential indeed. So let’s revisit the oscillator from a formal-mathematical point of view.

Modeling the problem

The simplest equation we used was the one for a hypothetical ‘ideal’ oscillator without friction and without any external driving force. The equation for a mechanical oscillator (i.e. a mass on a spring) is md2x/dt2 = –kx. The k in this equation is a factor of proportionality: the force pulling back is assumed to be proportional to the amount of stretch, and the minus sign is there because the force is pulling back indeed. As for the equation itself, it’s just Newton’s Law: the mass times the acceleration equals the force: ma = F.

You’ll remember we preferred to write this as d2x/dt2 = –(k/m)x = –ω02x with ω0= k/m. You’ll also remember that ωis an angular frequency, which we referred to as the natural frequency of the oscillator (because it determines the natural motion of the spring indeed). We also gave the general solution to the differential equation: x(t) = x0cos(ω0t + Δ). That solution basically states that, if we just let go of that spring, it will oscillate with frequency ω0 and some (maximum) amplitude x0, the value of which depends on the initial conditions. As for the Δ term, that’s just a phase shift depending on where x is when we start counting time: if x would happen to pass through the equilibrium point at time t = 0, then Δ would be π/2. So Δ allows us to shift the beginning of time, so to speak.

In my previous posts, I just presented that general equation as a fait accompli, noting that a cosine (or sine) function does indeed have that ‘nice’ property of come back to itself with a minus sign in front after taking the derivative two times: d2[cos(ω0t)]/dt2 = –ω02cos(ω0t). We could also write x(t) as a sine function because the sine and cosine function are basically the same except for a phase shift: x0cos(ω0t + Δ) = x0sin(ω0t + Δ + π/2).

Now, the point to note is that the sine or cosine function actually has two properties that are ‘nice’ (read ‘essential’ in the context of this discussion):

1. Sinusoidal functions are periodic functions and so that’s why they represent an oscillation–because that’s something periodic too!
2. Sinusoidal functions come back to themselves when we derive them two times and so that’s why it effectively solves our second-order differential equation.

However, in my previous post, I also mentioned in passing that sinusoidal functions share that second property with exponential functions: d2et/dt= d[det/dt]/dt = det/dt = et. So, if it we would not have had that minus sign in our differential equation, our solution would have been some exponential function, instead of a sine or a cosine function. So what’s going on here?

Solving differential equations using exponentials

Let’s scrap that minus sign and assume our problem would indeed be to solve the d2x/dt2 = ω02x equation. So we know we should use some exponential function, but we have that coefficient ω02. Well… That’s actually easy to deal with: we know that, when deriving an exponential function, we should bring the exponent down as a coefficient: d[eω0t]/dt = ω0eω0t. If we do it two times, we get d2[eω0t]/dt2 = ω02eω0t, so we can immediately see that eω0is a solution indeed.

But it’s not the only one: e–ω0t is a solution too: d2[e–ω0t]/dt2 = (–ω0)(–ω0)e–ω0t = ω02e–ω0t. So e–ω0solves the equation too. It is easy to see why: ω02 has two square roots–one positive, and one negative.

But we have more: in fact, every linear combination c1eω0+ c2e–ω0is also a solution to that second-order differential equation. Just check it by writing it all out: you’ll find that d2[c1eω0+ c2e–ω0t]/dt2 = ω02[c1eω0+c2e–ω0t] and so, yes, we have a whole family of functions here, that are all solutions to our differential equation.

Now, you may or may not remember that we had the same thing with first-order differential equations: we would find a whole family of functions, but only one would be the actual solution or the ‘real‘ solution I should say. So what’s the real solution here?

Well… That depends on the initial conditions: we need to know the value of x at time t= 0 (or some other point t = t1). And that’s not enough: we have two coefficients (cand c2), and, therefore, we need one more initial condition (it takes two equations to solve for two variables). That could be another value for x at some other point in time (e.g. t2) but, when solving problems like this, you’ll usually get the other ‘initial condition’ expressed in terms of the first derivative, so that’s in terms of dx/dt = v. For example, it is not illogical to assume that the initial velocity v0 would be zero. Indeed, we can imagine we pull or push the spring and then let it go. In fact, that’s what we’ve been assuming here all along in our example! Assuming that v0 = 0 is equivalent to writing that

d[c1eω0+ c2e–ω0t]/dt = 0 for t = 0

⇒ ω0c1 – ω0c2 = 0 (e= 1) ⇔  c1 = c2

Now we need the other initial condition. Let’s assume the initial value of x is equal to x0 = 2 (it’s just an example: we could take any value, including negative values). Then we get:

c1eω0+ c2e–ω0t = 2 for t = 0 ⇔ c1 + c= 2 (again, note that e= 1)

Combining the two gives us the grand result that c1 = c= 1 and, hence, the ‘real’ or actual solution is x = eω0e–ω0t. The graph below plots that function for ω= 1 and ω= 0.5 respectively. We could take other values for ω0 but, whatever the value, we’ll always get an exponential function like the ones below. It basically graphs what we expect to happen: the mass just accelerates away from its equilibrium point. Indeed, the differential equation is just a description of an accelerating object. Indeed, the e–ω0t term quickly goes to zero, and then it’s the eω0term that rockets that object sky-high – literally. [Note that the acceleration is actually not constant: the force is equal to kx and, hence, the force (and, therefore, the acceleration) actually increases as the mass goes further and further away from its equilibrium point. Also note that if the initial position would have been minus 2, i.e. x= –2, then the object would accelerate away in the other direction, i.e. downwards. Just check it to make sure you understand the equations.]

The point to note is our general solution. More formally, and more generally, we get it as follows:

• If we have a linear second-order differential equation ax” + bx’ + cx = 0 (because of the zero on the right-hand side, we call such equation homogeneous, so it’s quite a mouthful: a linear and homogeneous DE of the second order), then we can find an exponential function ert that will be a solution for it.
• If such function is a solution, then plugging in it yields ar2ert + brert + cert = 0 or (ar2 + br + c)ert = 0.
• Now, we can read that as a condition, and the condition amounts to ar2 + br + c = 0. So that’s a quadratic equation we need to solve for r to find two specific solutions r1 and r2, which, in turn, will then yield our general solution:

x(t) = c1er1+ c2er2t

Note that the general solution is based on the principle of superposition: any linear combination of two specific solutions will be a solution as well. I am mentioning this here because we’ll use that principle more than once.

Complex roots

The steps as described above implicitly assume that the quadratic equation above (i.e. ar2ert + brert + cert = 0), which is better known as the characteristic equation, does yield two real and distinct roots r1 and r2. In fact, it amounts to assuming that that exponential ert is a real-valued exponential function. We know how to find these real roots from our high school math classes: r = (–b ± [b– 4ac]1/2)/2a. However, what happens if the discrimant b– 4ac is negative?

If the disciminant is negative, we will still have two roots, but they will be complex roots. In fact, we can write these two complex roots as r = α ± βi, with i the imaginary unit. Hence, the two complex roots are each other’s complex conjugate and our er1and er2t can be written as:

er1= e(α+βi)t and er2e(α–βi)t

Also, the general solution based on these two particular solutions will be c1e(α+βi)t + c2e(α–βi)t.

[You may wonder why complex roots have to be complex conjugates from each other. Indeed, that’s not so obvious from the raw r = (–b ± [b– 4ac]1/2)/2a formula. But you can re-write it as r = –b/2a ± [b– 4ac]1/2)/2a and, if b– 4ac is negative, as r = –b/2a ± [(−b2+4ac)1/2/2a]. So that gives you the α and β and shows that the two roots are, in effect, each other’s complex conjugate.]

We should briefly pause here to think about what we are doing here really: if we allow r to be complex, then what we’re doing really is allow a complex-valued function (to be precise: we’re talking the complex exponential functions e(λ±μi)t, or any linear combination of the two) of a real variable (the time variable t) to be part of our ‘solution set’ as well.

Now, we’ve analyzed complex exponential functions before–long time ago: you can check out some of my posts last year (November 2013). In fact, we analyzed even more complex – in fact, I should say more complicated rather than more complex here: complex numbers don’t need to be complicated! 🙂 – because we were talking complex-valued functions of complex variables there! That’s not the case here: the argument t (i.e. the input into our function) is real, not complex, but the output – or the function itself – is complex-valued. Now, any complex exponential e(α+βi)t can be written as eαteiβt, and so that’s easy enough to understand:

1. The first factor (i.e. eαt) is just a real-valued exponential function and so we should be familiar with that. Depending on the value of α (negative or positive: see the graph below), it’s a factor that will create an envelope for our function. Indeed, when α is negative, the damping will cause the oscillation to stop after a while. When α is positive, we’ll have a solution resembling the second graph below: we have an amplitude that’s getting bigger and bigger, despite the friction factor (that’s obviously possible only because we keep reinforcing the movement, so we’re not switching off the force in that case). When α is equal to zero, then eαt is equal to unity and so the amplitude will not change as the spring goes up and down over time: we have no friction in that case.

2. The second factor (i.e. eiβt) is our periodic function. Indeed, eiβt is the same as eiθ and so just remember Euler’s formula to see what it is really:

eiθ = cos(θ) + isin(θ)

The two graphs below represent the idea: as the phase θ = ωt + Δ (the angular frequency or velocity times the time is equal to the phase, plus or minus some phase shift) goes round and round and round (i.e. increases with time), the two components of eiθ, i.e. the real and imaginary part eiθ, oscillate between –1 and 1 because they are both sinusoidal functions (cosine and sine respectively). Now, we could amplify the amplitude by putting another (real) factor in front (a magnitude different than 1) and write reiθ = r·cos(θ) + r·sin(θ) but that wouldn’t change the nature of this thing.

But so how does all of this relate to that other ‘general’ solution which we’ve found for our oscillator, i.e. the one we got without considering these complex-valued exponential functions as solutions. Indeed, what’s the relation between that x = x0cos(ω0t + Δ) equation and that rather frightening c1e(α+βi)t + c2e(α–βi)t equation? Perhaps we should look at x = x0cos(ω0t + Δ) as the real part of that monster? Yes and no. More no than yes actually. Actually… No. We are not going to have some complex exponential and then forget about the imaginary part. What we will do, though, is to find that general solution – i.e. a family of complex-valued functions – but then we’ll only consider those functions for which the imaginary part is zero, so that’s the subset of real-valued functions only.

I guess this must sound like Chinese. Let’s go step by step.

Using complex roots to find real-valued functions

If we re-write d2x/dt2 = –ω02x in the more general ax” + bx’ + cx = 0 form, then we get x” + ω02x = 0 and so the discriminant b– 4ac is equal to –4ω02, and so that’s a negative number. So we need to go for these complex roots. However, before solving this, let’s first restate what we’re actually doing. We have a differential equation that, ultimately, depends on a real variable (the time variable t), but so now we allow complex-valued functions er1e(α+βi)t and er2e(α–βi)t as solutions. To be precise: these are complex-valued functions x of the real variable t.

That being said, it’s fine to note that real numbers are a subset of the complex numbers and so we can just shrug our shoulders and say all that we’re doing is switch to complex-valued functions because we got stuck with that negative determinant and so we had to allow for complex roots. However, in the end, we do want a real-valued solution x(t). So our x(t) = c1e(α+βi)t + c2e(α–βi)t has to be a real-valued function, not a complex-valued function.

That means that we have to take a subset of the family of functions that we’ve found. In other words, the imaginary part of  c1e(α+βi)t + c2e(α–βi)t has to be zero. How can it be zero? Well… It basically means that c1e(α+βi)t and c2e(α–βi)t have to be complex conjugates.

OK… But how do we do that? We need to find a way to write that c1e(α+βi)t + c2e(α–βi)t sum in a more manageable ζ + η form. We can do that by using Euler’s formula once again to re-write those two complex exponentials as follows:

• e(α+βi)t = eαteiβt = eαt[cos(βt) + isin(βt)]
• e(α–βi)t = eαte–iβt = eαt[cos(–βt) + isin(–βt)] = eαt[cos(βt) – isin(βt)]

Note that, for the e(α–βi)t expression, we’ve used the fact that cos(–θ) = cos(θ) and that sin(–θ) = –sin(θ). Also note that α and β are real numbers, so they do not have an imaginary part–unlike cand c2, which may or may not have an imaginary part (i.e. they could be pure real numbers, but they could be complex as well).

We can then re-write that c1e(α+βi)t + c2e(α–βi)t sum as:

c1e(α+βi)t + c2e(α–βi)t = c1eαt[cos(βt) + isin(βt)] + c2eαt[cos(βt) – isin(βt)]

= (c1 + c2)eαtcos(βt) + (c1 – c2)ieαtsin(βt)

So what? Well, we want that imaginary part in our solution to disappear and so it’s easy to see that the imaginary part will indeed disappear if c1 – c2 = 0, i.e. if c1 = c= c. So we have a fairly general real-valued solution x(t) = 2c·eαtcos(βt) here, with c some real number. [Note that c has to be some real number because, if we would assume that cand c(and, therefore, c) would be equal complex numbers, then the c1 – c2 factor would also disappear, but then we would have a complex c1 + c2 sum in front of the eαtcos(βt) factor, so that would defeat the purpose of finding real-valued function as a solution because (c1 + c2)eαtcos(βt) would still be complex! […] Are you still with me? :-)]

So, OK, we’ve got the solution and so that should be it, isn’t it? Well… No. Wait. Not yet. Because these coefficients  c1 and c2 may be complex, there’s another solution as well. Look at that formula above. Let us suppose that c1 would be equal to some (real) number c divided by i (so c= c/i), and that cwould be its opposite, so c= –c(i.e. minus c1). Then we would have two complex numbers consisting of an imaginary part only: c= c/i and c= –c= –c/i, and they would be each other’s complex conjugate. Indeed, note that 1/i = i–1= –i and so we can write c= –c·and c= c·i. Then we’d get the following for that c1e(α+βi)t + c2e(α–βi)t sum:

(c1 + c2)eαtcos(βt) + (c1 – c2)ieαtsin(βt)

= (c/i – c/i)eαtcos(βt) + (c/i + c/i)ieαtsin(βt) = 2c·eαtsin(βt)

So, while cand c2 are complex, our grand result is a real-valued function once again or – to be precise – another family of real-valued functions (that’s because c can take on any value).

Are we done? Yes. There are no other possibilities. So now we just need to remember to apply the principle of superposition: any (real) linear combination of 2c·eαtcos(μt) and 2c·eαtsin(μt) will also be a (real-valued) solution, so the general (real-valued) solution for our problem is:

x(t) = a·2c·eαtcos(βt) + b·2c·eαtsin(βt) = Aeαtcos(βt) + Beαtsin(βt)

eαt[Acos(βt) + Bsin(βt)]

So what do we have here? Well, the first factor is, once again, an ‘envelope’ function: depending on the value of α, (i) negative, (ii) positive or (iii) zero, we have an oscillation that (i) damps out, (ii) goes out of control, or (iii) keeps oscillating in the same steady way forever.

The second part is equivalent to our ‘general’ x(t) = x0cos(ω0t + Δ) solution. Indeed, that x(t) = x0cos(ω0t + Δ) solution is somewhat less ‘general’ than the one above because it does not have the eαt factor. However, x(t) = x0cos(ω0t + Δ) solution is equivalent to the Acos(βt) + Bsin(βt) factor. How’s that? We can show how they are related by using the trigonometric formula for adding angles: cos(α + β) = cos(α)cos(β) – sin(α)sin(β). Indeed, we can write:

x0cos(ω0t + Δ) = x0cos(Δ)cos(ω0t) – x0sin(Δ)sin(ω0t) = Acos(βt) + Bsin(βt)

with A = x0cos(Δ), B = – x0sin(Δ) and, finally, μ = ω0

Are you convinced now? If not… Well… Nothing much I can do, I feel. In that case, I can only encourage you to do a full ‘work-out’ by reading the excellent overview of all possible situations in Paul’s Online MathNotes (tutorial.math.lamar.edu/Classes/DE/Vibrations.aspx).

Feynman’s treatment of second-order differential equations

Feynman takes a somewhat different approach in his Lectures. He solves them in a much more general way. At first, I thought his treatment was too confusing and, hence, I would not have mentioned it. However, I like the logic behind, even if his approach is somewhat more messy in terms of notations and all that. Let’s first look at the differential equation once again. Let’s take a system with a friction factor that’s proportional to the speed: Ff = –c·dx/dt. [See my previous post for some comments on that assumption: the assumption is, generally speaking, too much of a simplification but it makes for a ‘nice’ linear equation and so that’s why physicists present it that way.] To ease the math, c is usually written as c = mγ. Hence, γ = c/m is the friction per unit of mass. That makes sense, I’d think. In addition, we need to remember that ω02 = k/m, so k = mω02. Our differential equation then becomes m·d2x/dt2 = –γm·dx/dt – kx (mass times acceleration is the sum of the forces) or m·d2x/dt2 + γm·dx/dt + mω02·x = 0. Dividing the mass factor away gives us an even simpler form:

d2x/dt2 + γdx/dt + ω02x = 0

You’ll remember this differential equation from the previous post: we used it to calculate the (stored) energy and the Q of a mechanical oscillator. However, we didn’t show you how. You now understand why: the stuff above is not easy–the length of the arguments involved is why I am devoting an entire post to it!

Now, instead of assuming some exponential ert as a solution, real- or complex-valued, Feynman assumes a much more general complex-valued function as solution: he substitutes x for x = Aeiαt, with A a complex number as well so we can write A as A = A0eiΔ. That more general assumption allows for the inclusion of a phase shift straight from the start. Indeed, we can write x as x = A0eiΔeiαt = = A0ei(αt+Δ). Does that look complicated? It probably does, because we also have to remember that α is a complex number! So we’ve got a very general complex-valued exponential function indeed here!

However, let’s not get ahead of ourselves and follow Feynman. So he plugs in that complex-valued x = Aeiαt and we get:

(–α+ iγα + ω02)Aeiαt = 0

So far, so good. The logic now is more or less the same as the logic we developed above. We’ve got two factors here: (1) a quadratic equation –αiγα + ω02 (with one complex coefficient iγ) and (2) a complex exponential function Aeiαt. The second factor (Aeiαt) cannot be zero, because that’s x and we assume our oscillator is not standing still. So it’s the first factor (i.e. the quadratic equation in α with a complex coefficient iγ) which has to be zero. So we solve for the roots α and find

α = –iγ/(–2) ± [(–(iγ)2–4ω02)1/2/(-2)] = iγ/2 ± [(γ2–4ω02)1/2/(-2)]

= iγ/2 ± (ω0– γ2/4)1/2 iγ/2 ± ωγ

[We get this by bringing i and –2 inside of the square root expression. It’s not very straightforward but you should be able to figure it out.]

So that’s an interesting expression: the imaginary part of α is iγ/2 and its real part is (ω0– γ2/4)1/2, which we denoted as ωγ in the expression above. [Note that we assume there’s no problem with the square root expression: γ2/4 should be smaller than ω02 so ωγ is supposed to be some real positive number.] And so we’ve got the two solutions xand x2:

x= Aei(iγ/2 + ωγ)t =  Ae–γt/2+iωγ= Ae–γt/2eiωγ

x= Bei(iγ/2 – ωγ)t =  Be–γt/2–iωγ= Be–γt/2e–iωγt

Note, once again, that A and B can be any (complex) number and that, because of the principle of superposition, any linear combination of these two solutions will also be a solution. So the general solution is

x = Ae–γt/2eiωγ+ Be–γt/2e–iωγ= e–γt/2(Aeiωγ+ Be–iωγt)

Now, we recognize the shape of this: a (real-valued) envelope function e–γt/2 and then a linear combination of two exponentials. But so we want something real-valued in the end so, once again, we need to impose the condition that Aeiωγand Be–iωγare complex conjugates of each other. Now, we can see that eiωγand e–iωγare complex conjugates but what does this say about A and B? Well… The complex conjugate of a product is the product of the complex conjugates of the factors involved: (z1z2)* = (z1*)(z1*). That implies that B has to be the complex conjugate of A: B = A*. So the final (real-valued) solution becomes:

x = e–γt/2(Aeiωγ+ A*e–iωγt)

Now, I’ll leave it to you to prove that the second factor in the product above (Aeiωγ+ A*e–iωγt) is a real-valued function of the real variable t. It should be the same as x0cos(Δ)cos(ω0t) – x0sin(Δ)sin(ω0t), and that gives you a graph like the one below. However, I can readily imagine that, by now, you’re just thinking: Oh well… Whatever! 🙂

So the difference between Feynman’s approach and the one I presented above (which is the one you’ll find in most textbooks) is the assumption in terms of the specific solution: instead of substituting x for ert, with allowing r to take on complex values, Feynman substitutes x for Aeiαt, and allows both A and α  to take on complex values. It makes the calculations more complicated but, when everything is said and done, I think Feynman’s approach is more consistent because more encompassing. However, that’s subject to taste, and I gather, from comments on the Web, that many people think that this chapter in Feynman’s Lectures is not his best. So… Well… I’ll leave it to you to make the final judgment.

Note: The one critique that is relevant, in regard to Feynman’s treatment of the matter, is that he devotes quite a bit of time and space to explain how these oscillatory or periodic displacements can be viewed as being the real part of a complex exponential. Indeed, cos(ωt) is the real part of eiωt. But so that’s something different than (1) expanding the realm of possible solutions to a second-order differential equation from real-valued functions to complex-valued functions in order to (2) then, once we’ve found the general solution, consider only real-valued functions once again as ‘allowable’ solutions to that equation. I think that’s the gist of the matter really. It took me a while to fully ‘get’ this. I hope this post helps you to understand it somewhat quicker than I did. 🙂

Conclusion

I guess the only thing that I should do now is to work some examples. However, I’ll refer you Paul’s Online Math Notes for that once again (see the reference above). Indeed, it is about time I end my rather lengthy exposé (three posts on the same topic!) on oscillators and resonance. I hope you enjoyed it, although I can readily imagine that it’s hard to appreciate the math involved.

It is not easy indeed: I actually struggled with it, despite the fact that I think I understand complex analysis somewhat. However, the good thing is that, once we’re through it, we can really solve a lot of problems. As Feynman notes: “Linear (differential) equations are so important that perhaps fifty percent of the time we are solving linear equations in physics and engineering.” So, bearing in that mind, we should move on to the next.

# The electric oscillator

My previous post was too short to do justice to the topic (resonance phenomena). That’s why I’ll approach the topic using the relatively easy example of an electric oscillator. In addition, in this post I’ll also talk about the Q of an oscillator and the concept of a transient.

My father knew everything about electric motors, for example. Single-phase, split-phase, three-phase; synchronous or asynchronous; with two, four, six or eight poles; wound rotors or squirrel-cage rotors; centrifugal switches, capacitors… Electric motors (and engines in general) had no secrets for him. While I would understand the basic principle of the electric motor (he actually helped me build a little one – just using copper wire, a horsehoe magnet and a huge nail, and a piece of iron – to demonstrate in school), I had difficulty with the huge number of wires coming out of these things. [We had plenty of motors, because my father would bring old washing machines home to get the parts out.] Part of the problem was that he would never take the time to explain me how the capacitor that one needs to start a single-phase motor actually works.

Now I know, because I looked it up: single-phase electric (induction) motors have an auxiliary winding because they do not need have a starting torque. The magnetic field does not rotate: it just pulsates between 0 and 180 degrees and, hence, the rotor doesn’t know in which direction to go and, hence, if there’s no fuse to protect it, the wiring will start burning. [To explain why the wiring does not get (too) hot when it’s rotating is another story–which I won’t tell here because it involves changing electric and magnetic fields and so that’s a bit more complicated.] So now I have a bit more of an inkling of why there’s so many wires coming out of a simple (single-phase) electric motor:

• We have wires coming from the rotor (or, to be precise, from the carbon brushes). [Not always though: a lot of those old electric motors had so-called squirrel-cage rotors, instead of wound rotors.]
• We have wires going to the ‘run’ or ‘main’ winding in the stator (i.e. the stationary part of the motor).
• We have wires going to the ‘start’ or ‘auxiliary’ winding. In fact, with single phase, the ‘run’ and ‘start’ winding will share one common ‘end’ and so there will be three wires only: usually black, brown and blue in Europe and, to make things complicated, the same wires will usually be red, yellow and black in the US. 🙂
• We have wires coming from the capacitor and, most probably, also from some fuse somewhere, and then there’s a centrifugal switch to switch the auxiliary winding off once the motor is running, so that’s one or two more wires.
• And then we also need to control the speed of the motor and so that implies even more wires and little control boxes.

Phew! Things become complicated here. The primitive way to change the speed of single-phase motor is to change the number of poles. How? Well… We can separate the windings and/or by placing taps in-between. In short, more wires. A motor with two poles only will run at 3000 rpm when supplies with 50 Hz power, but we can also have 4, 6, 8 and more poles. More poles, means slower velocity. For example, if we switch to 10 poles, then the motor will run at 600 rpm (yes, 10/2 = 3000/600 = 5, so it’s the same factor). However, changing the number of poles while the motor is running is rather impractical so, in practice, speed control is done through a device referred to as a variable frequency drive (VFD). But so my father would just cut the wires and we’d end up with a motor running at one speed only–not very handy because these things spin incredibly fast–and with too many wires.

I have to admire him for making sense of all those wires. He would do so by measuring the resistance off all the circuits. So he’d just pick two wires and measure the resistance from one end to the other. For example, the main winding has less resistance–typically less than 5 Ω (Ohm)–than the auxiliary winding–typically 10 to 20 Ω (Ohm). Why? The wiring used to run the motor will typically be thicker and, hence, offer less resistance. With a bit of knowledge like that, he’d figure out the wiring in no time, while I would just sit and stare and wonder how he did it.

In any case, let me explain here what I would have liked my father to explain to me, and that’s the components of an electric circuit, and how an electric oscillator works–more or less at least.

The electric oscillator

In an electric circuit, we can have passive and active elements. An example of an active element would be a generator. That’s not passive. So what’s passive?

First, we have a resistor. A resistor is any piece of some substance through we have some current flowing and which offers resistance to that flow of electric current. What does that mean? The resistance (denoted by the symbol R) will determine the flow of current (I) through the circuit as a function of the potential difference (i.e. the voltage) V across. In fact, the resistance is defined as the factor of proportionality between V and I. So that’s Ohm’s Law really:

V = RI = R(dq/dt)

As for the current (I) being equal to I = dq/dt, that’s the definition of electric current itself: a current transports electric charge through a wire, so we can measure the current at any point in the electric current as the time-rate of change dq/dt. Current is Coulomb per second, i.e. in amperes. One ampere amounts to 6.241×1018 unit charges (electrons), i.e. one Coulomb passing through the wire per second, so 1 A = 1 C/s.

As for voltage, we’ve encountered that in previous posts. It’s a difference in potential indeed. Potential is that scalar number Φ which we associated with the potential energy U of a particle with charge q: Φ = U/q. So it’s like the potential energy of the unit charge, and we calculated by using the electric field vector to calculate the amount of work we needed to do to bring a unit charge to some point r: Φ(r) = –∫ds (the minus sign is there because we’re doing work against the electromagnetic force). We’ve actually calculated the difference in potential, or the voltage (difference) for something that’s called a capacitor: two parallel plates with a lack of electrons on one, and too many on the other (see below). As a result, there’s a strong electric field between both, and a difference in potential, and we’ve calculated the voltage as V = ΔΦ = σd0 = qd0A, with the d the plate separation (distance between the two plates), σ the (surface) charge per unit area, ε0 the electric constant and A the area of the plates. So it’s like a battery… For now at least–I’ll correct this statement later.

If we connect the two plates with a wire, i.e. a conductor, then we’ll have a current. Increasing the resistance of the circuit, by putting a resistor in, for example, will reduce the current and, hence, save the battery life somewhat. Of course, the resistor could be something that actually does work for us, a lamp, for example, or an electric motor.

Let me now correct that statement about a capacitor being like a battery. That statement is true and not true–but I must immediately add that it’s much more not true than true. 🙂 A battery is an active circuit element: it generates a voltage because of a chemical reaction that drives electrons through the circuit, and it will continue to provide power until all the reagents have been used up and the chemical reaction stops. In contrast, a capacitor is not active. There is a voltage only because charge has been stored on it (or, to be precise, because charges have been separated on it). Hence, when you connect the capacitor to a passive circuit,  the current will only flow until all of the charge has been drained. So there’s no active element. Also, unlike a battery, the voltage on a capacitor is variable: it’s proportional to the amount of charge stored on it.

OK. So we’ve got a resistor, a capacitor and a voltage source, e.g. a battery but, because we want to look at resonance phenomena, we’ll not have a battery but a voltage source that drives the circuit with a nice sine wave oscillation. Why a sine wave? Well… First, it makes the mathematical analysis easier (we’ll have second-order differential equations again and so d2cosx/dt2 = –cosx and so that’s nice). Second, the AC current that comes into our houses is a nice sine wave indeed. So let’s put it all together now, including our AC generator (instead of a battery). The circuit can then be represented as follows:

In this circuit, the charge q on the capacitor is analogous to the displacement x of the mass on that oscillating spring we analyzed in the previous post. Likewise:

• I = dq/dt is analogous to the velocity v = dx/dt
• The resistance R is analogous to the resistive coefficient γ
• From our formula V = ΔΦ = σd0, it is easy to see that V is proportional to the charge q: V = q/C, with 1/C the factor of proportionality, aka as the capacitance of the capacitor. In other words, 1/C is analogous to the spring constant k.

But we’re missing something: what’s the analogy to the mass or intertia factor in this circuit? Well… There’s one passive element in this circuit which we haven’t explained as yet: the self-inductance L. The phenomenon of self-inductance is the following: a changing electric current in a coil builds up a changing magnetic field, and that induces a current (and, hence, a voltage) that’s opposite to the primary current (and, hence, an opposite voltage). So it resists the change in current and, as such, it’s analogous to mass indeed. The illustration below explains how it works. I’ve also inserted a diagram showing how transformers work, because that’s based on the same principle of changing currents inducing changing magnetic fields that, in turn, generate another current. What’s going on in transformers is referred to as mutual inductance and note, indeed, that it doesn’t work with DC (i.e. steady) current.

Now, I know that’s not all that easy to understand, but I should limit myself here to just giving the formula: the induced voltage is such a coil is proportional to the time-rate of change of the current I = dq/dt. So we have a second-order derivative here:

V = LdI/dt = L(d2q/dt2)

So now we’re finally ready to put it all together. In that ‘basic electric circuit’ above, we’ve got the three passive circuit elements – resistor, capacitor and self-inductance – connected in series, and so then we apply a sine wave voltage to the whole circuit. Of course, all the voltages – i.e. over the resistor, over the capacitor, and over the self-inductance – must add up to the total voltage we apply to the circuit (which we’ll denote by V(t), as it’s a changing voltage), taking into account their sign. We have: VR = RI = R(dq/dt); VC = q/C; and VL = L(dI/dt) = L(d2q/dt2). Hence, we get:

L(d2q/dt2) + R(dq/dt) + q/C = V(t)

This is, once again, a differential equation of the second-order, and its mathematical form is the same as that equation for the oscillating spring (with a driving force and damping). [I repeat the equation below (in the section on the Q and the energy of an oscillator, so you don’t need to scroll too far.] So the solution is going to be the same and we’re going to have resonance if the angular frequency ω of our sine wave (i.e. the AC voltage generated by our generator) is close or equal to some kind of natural frequency characterizing the circuit. So what is that natural frequency ω0? Well… Just like ω0was equal to k/m for our mechanical oscillator, we here get the grand result that ω0= 1/LC, and our friction parameter γ corresponds to R/L.

The Q and the energy of an oscillator

There’s another point I did not develop in my previous post, and that was the energy of an oscillator. To explain that, we’ll take the example of our mechanical spring once again. The equation for that one was:

m(d2x/dt2) + γm(dx/dt) + mω02x = F(t)

Now, from my posts on energy concepts, you’ll remember that a force does work, and that the work done is the product of the force and the displacement (i.e. the distance over which the force is doing work). Work done is energy, potential or kinetic (one gets converted into the other). In addition, you may or may not remember that the work done per second gives us the power, so the concept of power relates energy to time, rather than distance.

For infinitesimal quantities (i.e. using differentials), we can write that the differential work done in a time dt is equal to F·dx. The power that’s expended by the force is then F·dx/dt, so that turns out to be the product of the force and the velocity (dx/dt = v): P = F·v. Now, if we substitute F for that differential equation above, and re-arrange the terms a bit, we get a fairly monstrously looking equation:

P = F·(dx/dt) = m[(d2x/dt2)(dx/dt) + ω02x(dx/dt)] + γm(dx/dt)2

Now it turns out that we can write the first two terms on the left on this monstrous equation as d/dt[m(dx/dt)2/2 + mω02x2/2]. So we have a time derivative here of a sum of two terms we recognize: the first is the kinetic energy (mv2/2) and the second (mω02x2/2) is the potential energy of the spring. [I would need to show that to you but I hope you believe me here.] Both of them taken together are the energy that’s stored in the oscillation, i.e. the stored energy. Now, in the long run, this driving force will not add any more energy to this quantity (the spring will oscillate back and forth, but so we’ll have stable motion and that’s it really). In other words, this derivative must be zero.

But so that driving force continues to do work and so the power must go somewhere. Where? It must all go to that other term: γm(dx/dt)2. What is that term? Well… It’s the energy that gets lost in friction: these are so-called resistive losses, and they usually get dissipated through heating. Hence, what happens is that most of the power of an external force is first used to build up the oscillation, thereby storing energy in the oscillator, but, once that’s done, the system only needs a little bit of energy to compensate for the heating (resistive) losses. Now the interesting thing is to calculate how much energy an oscillator can store. We can calculate that as follows:

• The energy carried by a physical wave is proportional to the square of its amplitude: E ∝ A2. Now, if it is a sinusoidal wave, we’ll need to take the average of the square of a sine or cosine function. Because sin2x and cos2x are the same functions really except for a phase difference of π/2, we can see that the average value for both functions should be 0.5 = 1/2. Hence, for any function Acosx, we can see that the average value of that square amplitude will be A2/2.
• From your statistics classes, you may also remember that the mean of a product of a variable and some constant (e.g. γm(dx/dt)2) will be equal to the product of that constant and the mean of the variable. So we can write 〈γm(dx/dt)2〉 =  γm〈(dx/dt)2〉. Now, taking into account that the solution x for the differential equation is a cosine function x = x0cos(ωt+Δ), its derivative will also be a sinusoidal function but with ω in the amplitude as well. To make a long story short, 〈(dx/dt)2〉 is equal to ω2x02/2, and so we can write 〈γm(dx/dt)2〉 = γmω2x02/2.
• So the expression above gives the energy being absorbed by the oscillator on a permanent basis, and we’ll denote that by 〈P〉 = γmω2x02/2. How much energy is stored?
• Now that we’ve calculated 〈(dx/dt)2〉, we can calculate that too now. We’ll denote it by 〈E〉, and so 〈E〉 = 〈m(dx/dt)2/2 + mω02x2/2〉 = (1/2)m〈(dx/dt)2 + (1/2)mω02〈x2〉 = m(ω2 + ω02)x02/2. So what? Well… From the previous chapter, we know that x0 becomes very large if ω is near to ω0 (that’s what’s resonance is all about) and, hence, the stored energy will be quite large in that case. So the point is that we can get a large stored energy from a relatively small force, which is what you’d expect.

Now, the last thing I need to explain is the Q of an oscillator. The Q of an oscillator compares the stored energy with the amount of work that is done per cycle, multiplied by 2π for some historical reason I don’t understand to well:

Q = 2π·〈E〉/[〈P〉·2π/ω] = (ω2 + ω02)/2γω

Note that 2π/ω is the period, i.e. the time T0 that is needed to go through one cycle of the oscillation. As mentioned above, I am not sure about that 2π factor but it doesn’t matter too much: it’s just a constant and so we could divide by 2π and the result would not be substantially different: the Q is a relative number obviously, used to compare the efficiency of various oscillators when it comes to storing energy. Indeed, Q stands for quality: higher Q indicates a lower rate of energy loss relative to the stored energy of the resonator. So it implies that you do not need a lot of power to keep the oscillation going and, if the external driving force stops, that the oscillations will die out much more slowly. For example, a pendulum on a high-quality bearing, oscillating in air, will have a high Q, while a pendulum immersed in oil will have a low one.

But let me go back to the electric oscillator: we substitute m for L, R for mγ, and 1/C for mω02, and then we can see that, for ω = ω02 (so we calculate the Q at resonance), we find that Q = Lω/R, with ω the resonance frequency. Again, a circuit with high Q means that the circuit can store a very large amount of energy as compared to the work done per cycle of the voltage driving the oscillation.

An application of the Q: transients

Throughout this and my previous posts, I’ve managed to skirt around a more rigorous (i.e. mathematical) treatment of the subject-matter by not actually solving these second-order differential equations. So I won’t suddenly change tack and try to do that now. So this will, once again, be a rather intuitive approach. If you’d want a formal treatment, let me refer you to Paul’s Online Math Notes and, more in particular, the chapter on second-order DEs, which he wraps up with an overview of all differential equations you could possibly encounter when analyzing mechanical springs. But so here we go for the informal approach.

Above, we noted that the Q of a system is the ratio of (1) the stored energy (E) and (2) the work done per cycle, multiplied by 2π. Now, if we’d suddenly switch off the force, then no more work is being done, but the system will lose energy. Let’s suppose we have a system – an oscillating mechanical spring – for which we have a Q equal to 1000·2π, so we have Q/2π = 1000. So that means that the work done per cycle – when that driving force is still one – is one thousandth of its total energy. Hence, it’s not unreasonable to suggest that such system would also lose one thousandth of its total energy per cycle if we would just switch off the force and let go of it. Writing that assumption in terms of differential changes yields the following simple (first-order) differential equation:

dE/dt = –ωE/Q

Huh? Yes. Just think about it. A differential dE is associated with a differential dt. Now, the number of radians that the phase will go through during the infinitesimally short dt time interval is –ωdt, so the change in energy must be equal to dE = –ωdt·(E/Q) (the minus sign is there because we’re talking an energy loss obviously). So that gives us the equation above.

But what about ω? Well… If we just let that oscillator do what we would expect it to do, then it is not unreasonable to assume it would oscillate at its natural frequency. Hence, ω is likely to equal ω0. Combining these two assumptions (i.e. that differential equation above and the ω = ωassumption) gives us the following formula for E:

E = E0e–tω/Q = E0e–γt

[Note that γ is the same friction coefficient: Q = (ω2 + ω02)/2γω and, hence, if ω = ω0, then we get ω0/Q = γ indeed.]

Now, the energy goes as the square of the amplitude of the oscillation (i.e. the displacement x), so we would expect to find the square root of that e–γt in the solution for x, so that’s a e–γt/2 factor. If we’d formally solve it, we’d find the following solution for x indeed:

x = A0e–γt/2cos(ω0t + Δ)

The diagram below shows the envelope curve e–γt/2 as well as the x = e–γt/2cos(ω0t) curve (A0 and Δ depend on the initial conditions obviously). So that’s what’s called a transient: a solution of the differential equation when there is no force present.

Now, I could bombard you with even more equations, more concepts (like the concept of impedance indeed), but I won’t do that here. I hope this post managed to get the most important ideas across and, hence, I’ll conclude this mini-series (i.e. two successive posts) on resonance. As for my next post, I may be tempted to treat the topic of second-order differential equations more formally, that is from a purely mathematical perspective. But let’s see. 🙂

Post scriptum:

The idea of applying only a little bit of power to build up a large amounts of stored energy may or may not trigger some thoughts on how a photo flash works and, in fact, you’re right. A photo flash uses both a transformer (to step up voltage) as well as an oscillator circuit to store up energy. You can find the details on the Web. See, for example, http://electronics.howstuffworks.com/camera-flash3.htm 🙂

# Resonance phenomena

One of the most common behaviors of physical systems is the phenomenon of resonance: a body (not only a tuning fork but any body really, such as a body of water, such as the ocean for example) or a system (e.g. an electric circuit) will have a so-called natural frequency, and an external driving force will cause it to oscillate. How it will behave, then, can be modeled using a simple differential equation, and the so-called resonance curve will usually look the same, regardless of what we are looking at. Besides the standard example of an electric circuit consisting of (i) a capacitor, (ii) a resistor and (iii) an inductor, Feynman also gives the following non-standard examples:

1. When the Earth’s atmosphere was disturbed as a result of the Krakatoa volcano explosion in 1883, it resonated at its own natural frequency, and its period was measured to be 10 hours and 20 minutes.

[In case you wonder how one can measure that, an explosion such as that one creates all kinds of waves, but the so-called infrasonic waves are the one we  are talking about here. They circled the globe at least seven times, shattering windows hundreds of miles away. They did not only shatter windows in a radius , but they were also recorded worldwide. That’s how they could be measured a second, third, etc time. How? There was no wind or so, but the infrasonic waves (i.e. ‘sounds’ beneath the lowest limits of human hearing (about 16 or 17 Hz), down to 0.001 Hz) of such oscillation cause minute changes in the atmospheric pressure which can be measured by microbarometers. So the ‘ringing’ of the atmosphere was measurable indeed. A nice article on infrasound waves is journal.borderlands.com/1997/infrasound. Of course, the surface of the Earth was ‘ringing’ as well, and such seismic shocks then produce tsunami waves, which can also be analyzed in terms of natural frequencies.]

2. Crystals can be made to oscillate in response to a changing external electric field, and this crystal resonance phenomenon is used in quartz clocks: the quartz crystal resonator in a basic quartz wristwatch is usually in the shape of a very small tuner fork. Literally: there’s a tiny tuning fork in your wristwatch, made of quartz, that has been laser-trimmed to vibrate at exactly 32,768 Hz, i.e. 215 cycles per second.

3. Some quantum-mechanical phenomena can be analyzed in terms of resonance as well, but then it’s the energy of the interfering particles that assumes the role of the frequency of the external driving force when analyzing the response of the system. Feynman gives the example of gamma radiation from lithium as a function of the energy of protons bombarding the lithium nuclei to provoke the reaction. Indeed, when graphing the intensity of the gamma radiation emitted as a function of the energy, one also gets a resonance curve, as shown below. [Don’t you just love the fact it’s so old? A Physical Review article of 1948! There’s older stuff as well, because this journal actually started in 1893.]

However, let us analyze the phenomenon first in its most classical appearance: an oscillating spring.

Basics

We’ve seen the equation for an oscillating spring before. From a math point of view, it’s a differential equation (because one of the terms is a derivative of the dependent variable x) of the second order (because the derivative involved is of the second order):

m(d2x/dt2) = –kx

What’s written here is simply Newton’s Law: the force is –kx (the minus sign is there because the force is directed opposite to the displacement from the equilibrium position), and the force has to equal the oscillating mass on the spring times its acceleration: F = ma.

Now, this can be written as d2x/dt2 = –(k/m)x = –ω02x with ω0= k/m. This ωsymbol uses the Greek omega once again, which we used for the angular velocity of a rotating body. While we do not have anything that’s rotating here, ωis still an angular velocity or, to be more precise, it’s an angular frequency. Indeed, the solution to the differential equation above is

x = x0cos(ω0t + Δ)

The xfactor is the maximum amplitude and that’s, quite simply, determined by how far we pulled or pushed the spring when we started the motion. Now, ω0t + Δ = θ is referred to as the phase of the motion, and it’s easy to see that ωis an angular frequency indeed, because ωequals the time derivative dθ/dt. Hence, ωis the phase change, measured in radians, per second, and that’s the definition of angular frequency or angular velocity. Finally, we have Δ. That’s just a phase shift, and it basically depends on our t = 0 point.

Something on the math

I’ll do a separate post on the math that’s associated with this (second-order differential equations) but, in this case, we can solve the equation in a simple and intuitive way. Look at it: d2x/dt2 = –ω02x. It’s obvious that x has to be a function that comes back to itself after two derivations, but with a minus sign in front, and then we also have that coefficient –ω02. Hmm… What can we think of? An exponential function comes back to itself, and if there’s a coefficient in the exponent, then it will end up as a coefficient in front too: d(eat)/dt = aeat and, hence, d2(eat)/dt2 = a2eat. Waw ! That’s close. In fact, that’s the same equation as the one above, except for the minus sign.

In fact, if you’d quickly look at Paul’s Online Math Notes, you’ll see that we can indeed get the general solution for such second-order differential equation (to be precise: it’s a so-called linear and homogeneous second-order DE with constant coefficients) using that remarkable property of exponentials indeed. However, because of the minus sign, our solution for the equation above will involve complex exponentials, and so we’ll get a general function in a complex variable. However, we’ll then impose that our solution has to be real only and, hence, we’ll take a subset of our more general solution. However, don’t worry about that here now. There’s an easier way.

Apart from the exponential function, there are two other functions that come back to themselves after two derivatives: the sine and cosine functions. Indeed, d2cos(t)/dt2 = –cos(t) and d2sin(t)/dt2 = –sin(t). In fact, the sine and cosine function are obviously the same except for a phase shift equal π/2: cos(t) = sin(t + π/2), so we can choose either. Let’s work with the cosine as for now (we can always convert it to a sine function using that cos(t) = sin(t + π/2) identity). The nice thing about the cosine (and sine) function is that we do get that minus sign when deriving it two times, and we also get that coefficient in front. Indeed: d2cos(ω0t)/dt2 = –ω02cos(ω0t). In short, cos(ω0t) is the right function. The only thing we need to add is that xand Δ, i.e. the amplitude and some phase shift but, as mentioned above, it is easy to understand these will depend on the initial conditions (i.e. the value of x at point t = 0 and the initial pull or push on the spring). In short, x = x0cos(ω0t + Δ) is the complete general solution of the  simple (differential) equation we started with (i.e. m(d2x/dt2) = –kx).

Introducing a driving force

Now, most real-life oscillating systems will be driven by an external force, permanently or just for a short while, and they will also lose some of their energy in a so-called dissipative process: friction or, in an electric circuit, electrical resistance will cause the oscillation to slowly lose amplitude, thereby damping it.

Let’s look at the friction coefficient first. The friction will often be proportional to the speed with which the object moves. Indeed, in the case of a mass on a spring, the drag (i.e. the force that acts on a body as it travels through air or a fluid) is dependent on a lot of things: first and foremost, there’s the fluid itself (e.g. a thick liquid will create more drag than water), and then there’s also the size, shape and velocity of the object. I am following the treatment you’ll find in most textbooks here and so that includes an assumption that the resistance force is proportional to the velocity: Ff = –cv = –c(dx/dt). Furthermore, the constant of proportionality c will usually be written as a product of the mass and some other coefficient γ, so we have Ff = –cv = –mγ(dx/dt). That makes sense because we can look at γ = c/m as the friction per unit of mass.

That being said, the simplification as a whole (i.e. the assumption of proportionality with speed) is rather strange in light of the fact that drag forces are actually proportional to the square of the velocity. If you look it up, you’ll find a formula resembling FD = ρCDAv2/2, with ρ the fluid density, CD the drag coefficient of drag (determined by the shape of the object and a so-called Reynolds number, which is determined from experiments), and A the cross-section area. It’s also rather strange to relate drag to mass by writing c as c = mγ because drag has nothing to do with mass. What about dry friction? So that would be kinetic friction between two surfaces, like when the mass is sliding on a surface? Well… In that case, mass would play a role but velocity wouldn’t, because kinetic friction is independent of the sliding velocity.

So why do physicists use this simplification? One reason is that it works for electric circuits: the equivalent of the velocity in electrical resonance is the current I = dq/dt, so that’s the time derivative of the charge on the capacitor. Now, I is proportional to the voltage difference V, and the proportionality coefficient is the resistance R, so we have V = RI = R(dq/dt). So, in short, the resistance curve we’re actually going to derive below is one for electric circuits. The other reason is that this assumption makes it easier to solve the differential equation that’s involved: it makes for a linear differential equation indeed. In fact, that’s the main reason. After all, professors are professors and so they have to give their students stuff that’s not too difficult to solve. In any case, let’s not be bothered too much and so we’ll just go along with it.

Modeling the driving force is easy: we’ll just assume it’s a sinusoidal force with angular frequency ω (and ω is, obviously, more likely than not somewhat different than the natural frequency ω0). If F is sinusoidal force, we can write it as F = F0cos(ωt + Δ). [So we also assume there is some phase shift Δ.] So now we can write the full equation for our oscillating spring as:

m(d2x/dt2) + γm(dx/dt) + kx = F ⇔ (d2x/dt2)+ γ(dx/dt) + ω02x = F

How do  we solve something like that for x? Well, it’s a differential equation once again. In fact, it’s, once again, a linear differential equation with constant coefficients, and so there’s a general solution method for that. As I mentioned above, that general solution method will involve exponentials and, in general, complex exponentials. I won’t walk you through that. Indeed, I’ll just write the solution because this is not an exercise in solving differential equations. I just want you to understand the solution:

x = ρF0cos(ωt + Δ + θ)

ρ in this equation has nothing to do with some density or so. It’s a factor which depends on m, ω and ω0, in a fairly complicated way in fact:

As we can see from the equation above, the (maximum) amplitude of the oscillation is equal to ρF0. So we have the magnitude of the force F here multiplied by ρ. Hence, ρ is a magnification factor which, multiplied with F0, gives us the ‘amount’ of oscillation.

As for the θ in the equation above, we’re using this Greek letter (theta) not to refer to the phase, as we usually do, because the phase here is the whole ωt + Δ + θ expression, not just theta! The theta (θ) here is a phase shift as compared to the original force phase ωt + Δ, and θ also depends on ω and ω0. Again, I won’t show how we derived this solution but just accept it as for now:

These three equations, taken together, should allow you to understand what’s going on really. We’ve got an oscillation x = ρF0cos(ωt + Δ + θ), so that’s an equation with this amplification or magnification factor ρ and some phase shift θ. Both depend on the difference between ωand ω, and the two graphs below show how exactly.

The first graph shows the resonance phenomenon and, hence, it’s what’s referred to as the resonance curve: if the difference between ωand ω is small, we get an enormous amplification effect. It would actually go to infinity if it weren’t for the frictional force (but, of course, if the frictional force was not there, the spring would just break as the oscillation builds up and the swings get bigger and bigger).

The second graph shows the phase shift θ. It is interesting to note that the lag θ is equal –π/2 when ω0 is equal to ω, but I’ll let you figure out why this makes sense. [It’s got something to do with that cos(t) = sin(t + π/2) identity, so it’s nothing ‘deep’ really.]

I guess I should, perhaps, also write something about the energy that gets stored in an oscillator like this because, in that resonance curve above, we actually have ρ squared on the vertical axis, and that’s because energy is proportional to the square of the amplitude: E ∝ A2. I should also explain a concept that’s closely related to energy: the so-called Q of an oscillator. It’s an interesting topic, if only because it helps us to understand why, for instance, the waves of the sea are such tremendous stores of energy! Furthermore, I should also write something about transients, i.e. oscillations that dampen because the driving force was turned off so to say. However, I’ll leave that for you to look it up if you’re interested in this topic. Here, I just wanted to present the essentials.

[…] Hey ! I managed to keep this post quite short for a change. Isn’t that good? 🙂

# Understanding gyroscopes

You know a gyroscope: it’s a spinning wheel or disk mounted in a frame that itself is free to alter in direction, so the axis of rotation is not affected as the mounting tilts or moves about. Therefore, gyroscopes are used to provide stability or maintain a reference direction in navigation systems. Understanding a gyroscope itself is simple enough: it only involves a good understanding of the so-called moment of inertia. Indeed, in the previous post, we introduced a lot of concepts related to rotational motion, notably the concepts of torque and angular momentum but, because that post was getting too long, I did not talk about the moment of inertia and gyroscopes. Let me do that now. However, I should warn you: you will not be able to understand this post if you haven’t read or didn’t understand the previous post. So, if you can’t follow, please go back: it’s probably because you didn’t get the other post.

The moment of inertia and angular momentum are related but not quite the same. Let’s first recapitulate angular momentum. Angular momentum is the equivalent of linear momentum for rotational motion:

1. If we want to change the linear motion of an object, as measured by its momentum p = mv, we’ll need to apply a force. Changing the linear motion means changing either (a) the speed (v), i.e. the magnitude of the velocity vector v, (b) the direction, or (c) both. This is expressed in Newton’s Law, F = m(dv/dt), and so we note that the mass is just a factor of proportionality measuring the inertia to change.
2. The same goes for angular momentum (denoted by L): if we want to change it, we’ll need to apply a force, or a torque as it’s referred to when talking rotational motion, and such torque can change either (a) L’s magnitude (L), (b) L’s direction or (c) both.

Just like linear momentum, angular momentum is also a product of two factors: the first factor is the angular velocity ω, and the second factor is the moment of inertia. The moment of inertia is denoted by I so we write L = Iω. But what is I? If we’re analyzing a rigid body (which is what we usually do), then it will be calculated as follows:

This is easy enough to understand: the inertia for turning will depend not just on the masses of all of the particles that make up the object, but also on their distance from the axis of rotation–and note that we need to square these distances. The L = Iω formula, combined with the formula for I above, explains why a spinning skater doing a ‘scratch spin’ speeds up tremendously when drawing in his or her arms and legs. Indeed, the total angular momentum has to remain the same, but I becomes much smaller as a result of that r2 factor in the formula. Hence, if I becomes smaller, then ω has to go up significantly in order to conserve angular momentum.

Finally, we note that angular momentum and linear momentum can be easily related through the following equation:

That’s all kids stuff. To understand gyroscopes, we’ll have to go beyond that and do some vector analysis. In the previous post, we explained that rotational motion is usually analyzed in terms of torques than forces, and we detailed the relations between force and torque. More in particular, we introduced a torque vector τ with the following components:

τ = (τyz, τzx, τxy) = (τx, τy, τz) with

τx = τyz = yFz – zFy

τy = τzx = zFx – xFz

τz = τxy = xFy – yFx.

We also noted that this torque vector could be written as a cross product of a radius vector and the force: τ = F. Finally, we also pointed out the relation between the x-, y- and z-components of the torque vector and the plane of rotation:

(1) τx = τyz is rotational motion about the x-axis (i.e. motion in the yz-plane)

(2) τy = τzx is rotational motion about the y-axis (i.e. motion in the zx plane)

(3) τz = τxy is rotational motion about the z-axis (i.e. motion in the xy-plane)

The angular momentum vector L will have the same direction as the torque vector, but it’s the cross product of the radius vector and the momentum vector: L = p. For clarity, I reproduce the animation I used in my previous post once again.

How do we get that cross vector product for L? We noted that τ (i.e. the Greek tau) = dL/dt. So we need to take the time derivative of all three components of L. What are the components of L? They look very similar to those of τ:

L = (Lyz, Lzx, Lxy) = (Lx, Ly, Lz) with

Lx = Lyz = ypz – zpy

Ly = Lzx = zpx – xpz

Lz = Lxy = xpy – ypx.

Now, just check the time derivatives of Lx, Ly, and Lz and you’ll find the components of the torque vector τ. Together with the formulas above, that should be sufficient to convince you that L is, indeed, a vector cross product of r and p: L = p.

Again, if you feel this is too difficult, please read or re-read my previous post. But if you do understand everything, then you are ready for a much more difficult analysis, and that’s an explanation of why a spinning top does not fall as it rotates about.

In order to understand that explanation, we’ll first analyze the situation below. It resembles the experiment with the swivel chair that’s often described on ‘easy physics’ websites: the man below holds a spinning wheel with its axis horizontal, and then turns this axis into the vertical. As a result, the man starts to turn himself in the opposite direction.

Let’s now look at the forces and torques involved. These are shown below.

This looks very complicated–you’ll say! You’re right: it is quite complicated–but not impossible to understand. First note the vectors involved in the starting position: we have an angular momentum vector L0 and an angular velocity vector ω0. These are both axial vectors, as I explained in my previous post: their direction is perpendicular to the plane of motion, i.e. they are arrows along the axis of rotation. This is in line with what we wrote above: if an object is rotating in the zx-plane (which is the case here), then the angular momentum vector will have a y-component only, and so it will be directed along the y-axis. Which side? That’s determined by the right-hand screw rule. [Again, please do read my previous post for more details if you’d need them.]

So now we have explained L0 and ω0. What about all the other vectors? First note that there would be no torque if the man would not try to turn the axis. In that case, the angular momentum would just remain what it is, i.e. dL/dt = 0, and there would be no torque. Indeed, remember that τ = dL/dt, just like F = dp/dt, so dL/dt = 0, then τ = 0. But so the man is turning the axis of rotation and, hence, τ = dL/dt ≠ 0. What’s changing here is not the magnitude of the angular momentum but its direction. As usual, the analysis is in terms of differentials.

As the man turns the spinning wheel, the directional change of the angular momentum is defined by the angle Δθ, and we get a new angular momentum vector L1. The difference between L1 and L0 is given by the vector ΔL. This ΔL vector is a tiny vector in the L0L1 plane and, because we’re looking at a differential displacement only, we can say that, for all practical purposes, this ΔL is orthogonal to L0 (as we move from L0 to L1, we’re actually moving along an arc and, hence, ΔL is a tangential vector). Therefore, simple trigonometry allows us to say that its magnitude ΔL will be equal to L0Δθ. [We should actually write sin(Δθ) but, because we’re talking differentials and measuring angles in radians (so the value reflects arc lengths), we can equate sin(Δθ) with Δθ).]

Now, the torque vector τ has the same direction as the ΔL vector (that’s obvious from their definitions), but what is its magnitude? That’s an easy question to answer: τ = ΔL/Δt = L0Δθ/Δt = L0 (Δθ/Δt). Now, this result induces us to define another axial vector which we’ll denote using the same Greek letter omega, but written as a capital letter instead of in lowercase: Ω. The direction of Ω is determined by using that right-hand screw rule which we’ve always been using, and Ω‘s magnitude is equal to Ω = Δθ/Δt. So, in short, Ω is an angular velocity vector just like ω: its magnitude is the speed with which the man is turning the axis of rotation of the spinning wheel, and its direction is determined using the same rules. If we do that, we get the rather remarkable result that we can write the torque vector τ as the cross product of Ω and L0:

τ = Ω×L0

Now, this is not an obvious result, so you should check it yourself. When doing that, you’ll note that the two vectors are orthogonal and so we have τ = Ω×L0 = Ω×L0 =|Ω||L0|sin(π/2)n = ΩL0n with n the normal unit vector given, once again, by the right-hand screw rule. [Note how the order of the two factors in a cross product matters: b = –a.]

You’re probably tired of this already, and so you’ll say: so what?

Well… We have a torque. A torque is produced by forces, and a torque vector along the z-axis is associated with rotation about the z-axis, i.e. rotation in the xy-plane. Such rotation is caused by the forces F and –F that produce the torque, as shown in the illustration. [Again, their direction is determined by the right-hand screw rule – but I’ll stop repeating that from now on.] But… Wait a minute. First, the direction is wrong, isn’t it? The man turns the other way in reality. And, second, where do these forces come from? Well… The man produces them, and the direction of the forces is not wrong: as the man applies these forces, with his hands, as he holds the spinning wheel and turns it into the vertical direction, equal and opposite forces act on him (cf. the action-reaction principle), and so he starts to turn in the opposite direction.

So there we are: we have explained this complex situation fully in terms of torques and forces now. So that’s good. [If you don’t believe the thing about those forces, just get one of your wheels out of your mountainbike, let it spin, and try to change the plane in which it is spinning: you’ll see you’ll need a bit of force. Not much, but enough, and it’s exactly the kind of force that the man in the illustration is experiencing.]

Now, what if we would not be holding the spinning wheel? What if we would let it pivot, for example? Well… It would just pivot, as shown below.

But… Why doesn’t it fall? Hah! There we are! Now we are finally ready for the analysis we really want to do, i.e. explaining why these spinning tops (or gyros as they’re referred to in physics) don’t fall.

Such spinning top is shown in the illustration below. It’s similar to the spinning wheel: there’s a rotational axis, and we have the force of gravity trying to change the direction of that axis, so it’s like the man turning that spinning wheel indeed, but so now it’s gravity exerting the force that’s needed to change the angular momentum. Let’s associate the vertical direction with the z-axis, and the horizontal place with the xy-axis, and let’s go step-by-step:

1. The gravitational force wants to pull that spinning top down. So the ΔL vector points downward this time, not upward. Hence, the torque vector will point downward too. But so it’s a torque pointing along the z-axis.
2. Such torque along the z-axis is associated with a rotation in the xy-plane, so that’s why the spinning top will slowly revolve about the z-axis, parallel to the xy-plane. This process is referred to as precession, and so there’s a precession torque and a precession angular velocity.

So that explains precession and so that’s all there is to it. Now you’ll complain, and rightly so: what I write above, does not explain why the spinning top does not actually fall. I only explained that precession movement. So what’s going on? That spinning top should fall as it precesses, shouldn’t it?

It actually does fall. The point to note, however, is that the precession movement itself changes the direction of the angular momentum vector as well. So we have a new ΔL vector pointing sideways, i.e. a vector in the horizontal plane–so not along the z axis. Hence, we should have a torque in the horizontal plane, and so that implies that we should have two equal and opposite forces acting along the z-axis.

In fact, the right-hand screw rule gives us the direction of those forces: if these forces were effectively applied to the spinning top, it would fall even faster! However, the point to note is that there are no such forces. Indeed, it is not like the man with the spinning wheel: no one (or nothing) is pushing or applying the forces that should produce the torque associated with this change in angular momentum. Hence, because these forces are absent, the spinning top begins to ‘fall’ in the opposite direction of the lacking force, thereby counteracting the gravitational force in such a way that the spinning top just spins about the z-axis without actually falling.

Now, this is, most probably, very difficult to understand in the way you would like to understand it, so just let it sink in and think about it for a while. In this regard, and to help the understanding, it’s probably worth noting that the actual process of reaching equilibrium is somewhat messy. It is illustrated below: if we hold a spinning gyro for a while and then, suddenly, we let it fall (yes, just let it go), it will actually fall. However, as it’s falling, it also starts turning and then, because it starts turning, it also starts ‘falling’ upwards, as explained in that story of the ‘missing force’ above. Initially, the upward movement will overshoot the equilibrium position, thereby slowing the gyro’s speed in the horizontal plane. And so then, because its horizontal speed becomes smaller, it stops ‘falling upward’, and so that means it’s falling down again. But then it starts turning again, and so on and so on. I hope you grasp this–more or less at least. Note that frictional effects will cause the up-and-down movement to dampen out, and so we get a so-called cycloidal motion dampening down to the steady motion we associate with spinning tops and gyros.

That, then, is the ‘miracle’ of a spinning top explained. Is it less of a ‘miracle’ now that we have explained it in terms of torques and missing forces? That’s an appreciation which each of us has to make for him- or herself. I actually find it all even more wonderful now that I can explain it more or less using the kind of math I used above–but then you may have a different opinion.

In any case, let us – to wrap it all up – ask some simple questions about some other spinning objects. What about the Earth for example? It has an axis of rotation too, and it revolves around the Sun. Is there anything like precession going on?

The first answer is: no, not really. The axis of rotation of the Earth changes little with respect to the stars. Indeed, why would it change? Changing it would require a torque, and where would the required force for such torque come from? The Earth is not like a gyro on a pivot being pulled down by some force we cannot see. The Sun attracts the Earth as a whole indeed. It does not change its axis of rotation. That’s why we have a fairly regular day and night cycle.

The more precise answer is: yes, there actually is a very slow axial precession. The whole precessional cycle takes approximately 26,000 years, and it causes the position of stars – as perceived by us, earthlings, that is – to slowly change. Over this cycle, the Earth’s north axial pole moves from where it is now, in a circle with an angular radius of about 23.5 degrees, as illustrated below.

What is this precession caused by? There must be some torque. There is. The Earth is not perfectly spherical: it bulges outward at the equator, and the gravitational tidal forces of the Moon and Sun apply some torque here, attempting to pull the equatorial bulge into the plane of the ecliptic, but instead causing it to precess. So it’s a quite subtle motion, but it’s there, and it’s got also something to do with the gravitational force. However, it’s got nothing to do with the way gravitation makes a spinning top do what it does. [The most amazing thing about this, in my opinion, is that, despite the fact that the precessional movement is so tiny, the Greeks had already discovered it: indeed, the Greek astronomer and mathematician Hipparchus of Nicaea gave a pretty precise figure for this so-called ‘precession of the equinoxes’ in 127 BC.]

What about electrons? Are they like gyros rotating around some pivot? Here the answer is very simple and very straightforward: No, not at all! First, there are no pivots in an atom. Second, the current understanding of an electron – i.e. the quantum-mechanical understanding of a electron – is not compatible with the classical notion of spin. Let me just copy an explanation from Georgia State University’s HyperPhyics website. It basically says it all:

“Experimental evidence like the hydrogen fine structure and the Stern-Gerlach experiment suggest that an electron has an intrinsic angular momentum, independent of its orbital angular momentum. These experiments suggest just two possible states for this angular momentum, and following the pattern of quantized angular momentum, this requires an angular momentum quantum number of 1/2. With this evidence, we say that the electron has spin 1/2. An angular momentum and a magnetic moment could indeed arise from a spinning sphere of charge, but this classical picture cannot fit the size or quantized nature of the electron spin. The property called electron spin must be considered to be a quantum concept without detailed classical analogy.

So… I guess this should conclude my exposé on rotational motion. I am not sure what I am going to write about next, but I’ll see. 🙂

Post scriptum:

The above treatment is largely based on Feynman’s Lectures.(Vol. I, Chapter 18, 19 and 20). The subject could also be discussed using the concept of a force couple, aka pure moment. A force couple is a system of forces with a resultant moment but no resultant force. Hence, it causes rotation without translation or, more generally, without any acceleration of the centre of mass. In such analysis, we can say that gravity produces a force couple on the spinning top. The two forces of this couple are equal and opposite, and they pull at opposite ends. However, because one end of the top is fixed (friction forces keep the tip fixed to the ground), the force at the other end makes the top go about the vertical axis.

The situation we have is that gravity causes such force couple to appear, just like the man tilting the spinning wheel causes such force couple to appear. Now, the analysis above shows that the direction of the new force is perpendicular to the plane in which the axis of rotation changes, or wants to change in the case of our spinning top. So gravity wants to pull the top down and causes it to move sideways. This horizontal movement will, in turn, create another force couple. The direction of the resultant force, at the free end of the axis of rotation of the top, will, once again, be vertical, but it will oppose the gravity force. So, in a very simplified explanation of things, we could say:

1. Gravity pulls the top downwards, and causes a force that will make the top move sideways. So the new force, which causes the precession movement, is orthogonal to the gravitation force, i.e. it’s a horizontal force.
2. That horizontal force will, in turn, cause another force to appear. That force will also be orthogonal to the horizontal force. As we made two 90 degrees turns, so to say, i.e. 180 degrees in total, it means that this third force will be opposite to the gravitational force.
3. In equilibrium, we have three forces: gravity, the force causing the precession and, finally, a force neutralizing gravity as the spinning top precesses about the vertical axis.

This approach allows for a treatment that is somewhat more intuitive than Feynman’s concept of the ‘missing force.’

# Spinning: the essentials

When introducing mirror symmetry (P-symmetry) in one of my older posts (time reversal and CPT-symmetry), I also introduced the concept of axial and polar vectors in physics. Axial vectors have to do with rotations, or spinning objects. Because spin – i.e. turning motion – is such an important concept in physics, I’d suggest we re-visit the topic here.

Of course, I should be clear from the outset that the discussion below is entirely classical. Indeed, as Wikipedia puts it: “The intrinsic spin of elementary particles (such as electrons) is quantum-mechanical phenomenon that does not have a counterpart in classical mechanics, despite the term spin being reminiscent of classical phenomena such as a planet spinning on its axis.” Nevertheless, if we don’t understand what spin is in the classical world – i.e. our world for all practical purposes – then we won’t get even near to appreciating what it might be in the quantum-mechanical world. Besides, it’s just plain fun: I am sure you have played, as a kid of as an adult even, with one of those magical spinning tops or toy gyroscopes and so you probably wonder how it really works in physics. So that’s what this post is all about.

The essential concept is the concept of torque. For rotations in space (i.e. rotational motion), the torque is what the force is for linear motion:

• It’s the torque (τ) that makes an object spin faster or slower, just like the force would accelerate or decelerate that very same object when it would be moving along some curve (as opposed to spinning around some axis).
• There’s also a similar ‘law of Newton’ for torque: you’ll remember that the force equals the time rate-of-change of a vector quantity referred to as (linear) momentum: F = dp/dt = d(mv)/dt = ma (the mass times the acceleration). Likewise, we have a vector quantity that is referred to as angular momentum (L), and we can write: τ (i.e. the Greek tau) = dL/dt.
• Finally, instead of linear velocity, we’ll have an angular velocity ω (omega), which is the time rate-of-change of the angle θ defining how far the object has gone around (as opposed to the distance in linear dynamics, describing how far the object has gone along). So we have ω = dθ/dt. This is actually easy to visualize because we know that θ is the length of the corresponding arc on the unit circle. Hence, the equivalence with the linear distance traveled is easily ascertained.

There are numerous other equivalences. For example, we also have an angular acceleration: α = dω/dt = d2θ/dt2; and we should also note that, just like the force, the torque is doing work – in its conventional definition as used in physics – as it turns an object:

ΔW = τ·Δθ

However, we also need to point out the differences. The animation below does that very well, as it relates the ‘new’ concepts – i.e. torque and angular momentum – to the ‘old’ concepts – i.e. force and linear momentum.

So what do we have here? We have vector quantities once again, denoted by symbols in bold-face. However, τ, L and ω are special vectors: axial vectors indeed, as opposed to the polar vectors F, p and v. Axial vectors are directed along the axis of spin – so that is, strangely enough, at right angles to the direction of spin, or perpendicular to the ‘plane of the twist’ as Feynman calls it – and the direction of the axial vector is determined by the direction of spin through one of two conventions: the ‘right-hand screw rule’ or the ‘left-hand screw rule’. Physicists have settled on the former.

If you feel very confused now (I did when I first looked at it), just step back and go through the full argument as I develop it here. It helps to think of torque (also known, for some obscure reason, as the moment of the force) as a twist on an object or a plane indeed: the torque’s magnitude is equal to the tangential component of the force, i.e. F·sin(Δθ), times the distance between the object and the axis of rotation (we’ll denote this distance by r). This quantity is also equal to the product of the magnitude of the force itself and the length of the so-called lever arm, i.e. the perpendicular distance from the axis to the line of action of the force (this lever arm length is denoted by r0). So we can write τ as:

1. The product of the tangential component of the force times the distance r: τ = r·Ft = r·F·sin(Δθ)
2. The product of the length of the lever arm times the force: τ = r0·F
3. The torque is the work done per unit of distance traveled: τ = ΔW/Δθ or τ = dW/dθ in the limit.

So… These are actually only the basics, which you should remember from your high-school physics course. If not, have another look at it. We now need to go from scalar quantities to vector quantities to understand that animation above. Torque is not a vector like force or velocity, not a priori at least. However, we can associate torque with a vector of a special type, an axial vector. Feynman calls vectors such as force or (linear) velocity ‘honest’ or ‘real’ vectors. The mathematically correct term for such ‘honest’ or ‘real’ vectors is polar vector. Hence, axial vectors are not ‘honest’ or ‘real’ in some sense: we derive them from the polar vectors. They are, in effect, a so-called cross product of two ‘honest’ vectors. Here we need to explain the difference between a dot and a cross product between two vectors once again:

(1) A dot product, which we denoted by a little dot (·), yields a scalar quantity: b = |a||b|cosα = a·b·cosα with α the angle between the two vectors a and b. Note that the dot product of two orthogonal vectors is equal to zero, so take care:  τ = r·Ft = r·F·sin(Δθ) is not a dot product of two vectors. It’s a simple product of two scalar quantities: we only use the dot as a mark of separation, which may be quite confusing. In fact, some authors use ∗ for a product of scalars to avoid confusion: that’s not a bad idea, but it’s not a convention as yet. Omitting the dot when multiplying scalars (as I do when I write |a||b|cosα) is also possible, but it makes it a bit difficult to read formulas I find. Also note, once again, how important the difference between bold-face and normal type is in formulas like this: it distinguishes vectors from scalars – and these are two very different things indeed.

(2) A cross product, which we denote by using a cross (×), yields another vector: τ = r×F =|r|·|F|·sinα·n = r·F·sinα·n with n the normal unit vector given by the right-hand rule. Note how a cross product involves a sine, not a cosine – as opposed to a dot product. Hence, if r and F are orthogonal vectors (which is not unlikely), then this sine term will be equal to 1. If the two vectors are not perpendicular to each other, then the sine function will assure that we use the tangential component of the force.

But, again, how do we go from torque as a scalar quantity (τ = r·Ft) to the vector τ = r×F? Well… Let’s suppose, first, that, in our (inertial) frame of reference, we have some object spinning around the z-axis only. In other words, it spins in the xy-plane only. So we have a torque around (or about) the z-axis, i.e. in the xy-plane. The work that will be done by this torque can be written as:

ΔW = FxΔx + FyΔy = (xFy – yFx)Δθ

Huh? Yes. This results from a simple two-dimensional analysis of what’s going on in the xy-plane: the force has an x- and a y-component, and the distance traveled in the x- and y-direction is Δx = –yΔθ and Δy = xΔθ respectively. I won’t go into the details of this (you can easily find these elsewhere) but just note the minus sign for Δx and the way the x and y get switched in the expressions.

So the torque in the xy-plane is given by τxy = ΔW/Δθ = xFy – yFx. Likewise, if the object would be spinning about the x-axis – or, what amounts to the same, in the yz-plane – we’d get τyz = yFz – zFy. Finally, for some object spinning about the y-axis (i.e. in the zx-plane – and please note I write zx, not xz, so as to be consistent as we switch the order of the x, y and z coordinates in the formulas), then we’d get τzx = zFx – xFz. Now we can appreciate the fact that a torque in some other plane, at some angle with our Cartesian planes, would be some combination of these three torques, so we’d write:

(1)    τxy = xFy – yFx

(2)    τyz = yFz – zFy and

(3)    τzx = zFx – xFz.

Another observer with his Cartesian x’, y’ and z’ axes in some other direction (we’re not talking some observer moving away from us but, quite simply, a reference frame that’s being rotated itself around some axis not necessarily coinciding with any of the x-, y- z- or x’-, y’- and z’-axes mentioned above) would find other values as he calculates these torques, but the formulas would look the same:

(1’) τx’y’ = x’Fy’ – y’Fx’

(2’) τy’z’ = y’Fz’ – z’Fy’ and

(3’) τz’x’ = z’Fx’ – x’Fz’.

Now, of course, there must be some ‘nice’ relationship that expresses the τx’y’, τy’z’ and τz’x’ values in terms of τxy, τyz, just like there was some ‘nice’ relationship between the x’, y’ and z’ components of a vector in one coordination system (the x’, y’ and z’ coordinate system) and the x, y, z components of that same vector in the x, y and z coordinate system. Now, I won’t go into the details but that ‘nice’ relationship is, in fact, given by transformation expressions involving a rotation matrix. I won’t write that one down here, because it looks pretty formidable, but just google ‘axis-angle representation of a rotation’ and you’ll get all the details you want.

The point to note is that, in both sets of equations above, we have an x-, y- and z-component of some mathematical vector that transform just like a ‘real’ vector. Now, if it behaves like a vector, we’ll just call it a vector, and that’s how, in essence, we define torque, angular momentum (and angular velocity too) as axial vectors. We should note how it works exactly though:

(1) τxy and τx’y’ will transform like the z-component of a vector (note that we were talking rotational motion about the z-axis when introducing this quantity);

(2) τyz and τy’z’ will transform like the x-component of a vector (note that we were talking rotational motion about the x-axis when introducing this quantity);

(3) τzx and τz’x’ will transform like the y-component of a vector (note that we were talking rotation motion when introducing this quantity). So we have

τ = (τyz, τzx, τxy) = (τx, τy, τz) with

τx = τyz = yFz – zFy

τy = τzx = zFx – xFz

τz = τxy = xFy – yFx.

[This may look very difficult to remember but just look at the order: all we do is respect the clockwise order x, y, z, x, y, z, x, etc. when jotting down the x, y and z subscripts.]

Now we are, finally, well equipped to once again look at that vector representation of rotation. I reproduce it once again below so you don’t have to scroll back to that animation:

We have rotation in the zx-plane here (i.e. rotation about the y-axis) driven by an oscillating force F, and so, yes, we can see that the torque vector oscillates along the y-axis only: its x- and z-components are zero. We also have L here, the angular momentum. That’s a vector quantity as well. We can write it as

L = (Lyz, Lzx, Lxy) = (Lx, Ly, Lz) with

Lx = Lyz = ypz – zpy (i.e. the angular momentum about the x-axis)

Ly = Lzx = zpx – xpz (i.e. the angular momentum about the y-axis)

Lz = Lxy = xpy – ypx (i.e. the angular momentum about the z-axis),

And we note, once again, that only the y-component is non-zero in this case, because the rotation is about the y-axis.

We should now remember the rules for a cross product. Above, we wrote that τ = r´F =|r|×|F|×sina×n = = r×F×sina×n with n the normal unit vector given by the right-hand rule. However, a vector product can also be written in terms of its components: c = a´b if and only

cx = aybz – azby,

cy = azbx – axbz, and

cz = axby – aybx.

Again, if this looks difficult, remember the trick above: respect the clockwise order when jotting down the x, y and z subscripts. I’ll leave it to you to work out r´F and r´p in terms of components but, when you write it all out, you’ll see it corresponds to the formulas above. In addition, I will also leave it to you to show that the velocity of some particle in a rotating body can be given by a similar vector product: v = ω´r, with ω being defined as another axial vector (aka pseudovector) pointing along the direction of the axis of rotation, i.e. not in the direction of motion. [Is that strange? No. As it’s rotational motion, there is no ‘direction of motion’ really: the object, or any particle in that object, goes round and round and round indeed and, hence, defining some normal vector using the right-hand rule to denote angular velocity makes a lot of sense.]

I could continue to write and write and write, but I need to stop here. Indeed, I actually wanted to tell you how gyroscopes work, but I notice that this introduction has already taken several pages. Hence, I’ll leave the gyroscope for a separate post. So, be warned, you’ll need to read and understand this one before reading my next one.