# Wavefunctions and the twin paradox

My previous post was awfully long, so I must assume many of my readers may have started to read it, but… Well… Gave up halfway or even sooner. 🙂 I added a footnote, though, which is interesting to reflect upon. Also, I know many of my readers aren’t interested in the math—even if they understand one cannot really appreciate quantum theory without the math. But… Yes. I may have left some readers behind. Let me, therefore, pick up the most interesting bit of all of the stories in my last posts in as easy a language as I can find.

We have that weird 360/720° symmetry in quantum physics or—to be precise—we have it for elementary matter-particles (think of electrons, for example). In order to, hopefully, help you understand what it’s all about, I had to explain the often-confused but substantially different concepts of a reference frame and a representational base (or representation tout court). I won’t repeat that explanation, but think of the following.

If we just rotate the reference frame over 360°, we’re just using the same reference frame and so we see the same thing: some object which we, vaguely, describe by some ei·θ function. Think of some spinning object. In its own reference frame, it will just spin around some center or, in ours, it will spin while moving along some axis in its own reference frame or, seen from ours, as moving in some direction while it’s spinning—as illustrated below.

To be precise, I should say that we describe it by some Fourier sum of such functions. Now, if its spin direction is… Well… In the other direction, then we’ll describe it by by some ei·θ function (again, you should read: a Fourier sum of such functions). Now, the weird thing is is the following: if we rotate the object itself, over the same 360°, we get a different object: our ei·θ and ei·θ function (again: think of a Fourier sum, so that’s a wave packet, really) becomes a −e±i·θ thing. We get a minus sign in front of it. So what happened here? What’s the difference, really?

Well… I don’t know. It’s very deep. Think of you and me as two electrons who are watching each other. If I do nothing, and you keep watching me while turning around me, for a full 360° (so that’s a rotation of your reference frame over 360°), then you’ll end up where you were when you started and, importantly, you’ll see the same thing: me. 🙂 I mean… You’ll see exactly the same thing: if I was an e+i·θ wave packet, I am still an an e+i·θ wave packet now. Or if I was an ei·θ wave packet, then I am still an an ei·θ wave packet now. Easy. Logical. Obvious, right?

But so now we try something different: turn around, over a full 360° turn, and you stay where you are and watch me while I am turning around. What happens? Classically, nothing should happen but… Well… This is the weird world of quantum mechanics: when I am back where I was—looking at you again, so to speak—then… Well… I am not quite the same any more. Or… Well… Perhaps I am but you see me differently. If I was e+i·θ wave packet, then I’ve become a −e+i·θ wave packet now.

Not hugely different but… Well… That minus sign matters, right? Or If I was wave packet built up from elementary a·ei·θ waves, then I’ve become a −ei·θ wave packet now. What happened?

It makes me think of the twin paradox in special relativity. We know it’s a paradox—so that’s an apparent contradiction only: we know which twin stayed on Earth and which one traveled because of the gravitational forces on the traveling twin. The one who stays on Earth does not experience any acceleration or deceleration. Is it the same here? I mean… The one who’s turning around must experience some force.

Can we relate this to the twin paradox? Maybe. Note that a minus sign in front of the e−±i·θ functions amounts a minus sign in front of both the sine and cosine components. So… Well… The negative of a sine and cosine is the sine and cosine but with a phase shift of 180°: −cosθ = cos(θ ± π) and −sinθ = sin(θ ± π). Now, adding or subtracting a common phase factor to/from the argument of the wavefunction amounts to changing the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360° and 720° symmetries are, effectively, related. 🙂

Post scriptumGoogle honors Max Born’s 135th birthday today. 🙂 I think that’s a great coincidence in light of the stuff I’ve been writing about lately (possible interpretations of the wavefunction). 🙂

# Field energy and field momentum

This post goes to the heart of the E = mc2, equation. It’s kinda funny, because Feynman just compresses all of it in a sub-section of his Lectures. However, as far as I am concerned, I feel it’s a very crucial section. Pivotal, I’d say, which would fit with its place in all of the 115 Lectures that make up the three volumes, which is sort of mid-way, which is where we are here. So let’s get go for it. 🙂

Let’s first recall what we wrote about the Poynting vector S, which we calculate from the magnetic and electric field vectors E and B by taking their cross-product:

This vector represents the energy flow, per unit area and per unit time, in electrodynamical situations. If E and/or are zero (which is the case in electrostatics, for example, because we don’t have magnetic fields in electrostatics), then S is zero too, so there is no energy flow then. That makes sense, because we have no moving charges, so where would the energy go to?

I also made it clear we should think of S as something physical, by comparing it to the heat flow vector h, which we presented when discussing vector analysis and vector operators. The heat flow out of a surface element da is the area times the component of perpendicular to da, so that’s (hn)·da = hn·da. Likewise, we can write (Sn)·da = Sn·da. The units of S and h are also the same: joule per second and per square meter or, using the definition of the watt (1 W = 1 J/s), in watt per square meter. In fact, if you google a bit, you’ll find that both h and S are referred to as a flux density:

1. The heat flow vector h is the heat flux density vector, from which we get the heat flux through an area through the (hn)·da = hn·da product.
2. The energy flow is the energy flux density vector, from which we get the energy flux through the (Sn)·da = Sn·da product.

So that should be enough as an introduction to what I want to talk about here. Let’s first look at the energy conservation principle once again.

#### Local energy conservation

In a way, you can look at my previous post as being all about the equation below, which we referred to as the ‘local’ energy conservation law:

Of course, it is not the complete energy conservation law. The local energy is not only in the field. We’ve got matter as well, and so that’s what I want to discuss here: we want to look at the energy in the field as well as the energy that’s in the matter. Indeed, field energy is conserved, and then it isn’t: if the field is doing work on matter, or matter is doing work on the field, then… Well… Energy goes from one to the other, i.e. from the field to the matter or from the matter to the field. So we need to include matter in our analysis, which we didn’t do in our last post. Feynman gives the following simple example: we’re in a dark room, and suddenly someone turns on the light switch. So now the room is full of field energy—and, yes, I just mean it’s not dark anymore. :-). So that means some matter out there must have radiated its energy out and, in the process, it must have lost the equivalent mass of that energy. So, yes, we had matter losing energy and, hence, losing mass.

Now, we know that energy and momentum are related. Respecting and incorporating relativity theory, we’ve got two equivalent formulas for it:

1. E− p2c2 = m02c4
2. pc = E·(v/c) ⇔ p = v·E/c= m·v

The E = mc2 and m = ·m0·(1−v2/c2)−1/2 formulas connect both expressions. So we can look at it in either of two ways. We could use the energy conservation law, but Feynman prefers the conservation of momentum approach, so let’s see where he takes us. If the field has some energy (and, hence, some equivalent mass) per unit volume, and if there’s some flow, so if there’s some velocity (which there is: that’s what our previous post was all about), then it will have a certain momentum per unit volume. [Remember: momentum is mass times velocity.] That momentum will have a direction, so it’s a vector, just like p = mv. We’ll write it as g, so we define g as:

g is the momentum of the field per unit volume.

What units would we express it in? We’ve got a bit of choice here. For example, because we’re relating everything to energy here, we may want to convert our kilogram into eV/cor J/cunits, using the mass-energy equivalence relation E = mc2. Hmm… Let’s first keep the kg as a measure of inertia though. So we write: [g] = [m]·[v]/m= (kg·m/s)/m3. Hmm… That doesn’t show it’s energy, so let’s replace the kg with a unit that’s got newton and meter in it, cf. the F = ma law. So we write: [g] = (kg·m/s)/m= (kg/s)/m= [(N·s2/m)/s]/m= N·s/m3. Well… OK. The newton·second is the unit of momentum indeed, and we can re-write it including the joule (1 J = 1 N·m), so then we get [g] = (J·s/m4), so what’s that? Well… Nothing much. However, I do note it happens to be the dimension of S/c2, so that’s [S/c2] = [J/(s·m2)]·(s2/m2) = (J·s/m4). 🙂 Let’s continue the discussion.

Now, momentum is conserved, and each component of it is conserved. So let’s look at the x-direction. We should have something like:

If you look at this carefully, you’ll probably say: “OK. I understood the thing with the dark room and light switch. Mass got converted into field energy, but what’s that second term of the left?”

Good. Smart. Right remark. Perfect. […] Let me try to answer the question. While all of the quantities above are expressed per unit volume, we’re actually looking at the same infinitesimal volume element here, so the example of the light switch is actually an example of a ‘momentum outflow’, so it’s actually an example of that second term of the left-hand side of the equation above kicking in! 🙂

Indeed, the first term just sort of reiterates the mass-energy equivalence: the energy that’s in the matter can become field energy, so to speak, in our infinitesimal volume element itself, and vice versa. But if it doesn’t, then it should get out and, hence, become ‘momentum outflow’. Does that make sense? No?

Hmm… What to say? You’ll need to look at that equation a couple of times more, I guess. But I need to move on, unfortunately. [Don’t get put off when I say things like this: I am basically talking to myself, so it means I’ll need to re-visit this myself. :-/]

Let’s look at all of the three terms:

1. The left-hand side (i.e. the time rate-of-change of the momentum of matter) is easy. It’s just the force on it, which we know is equal to Fq(E+v×B). Do we know that? OK… I’ll admit it. Sometimes it’s easy to forget where we are in an analysis like this, but so we’re looking at the electromagnetic force here. 🙂 As we’re talking infinitesimals here and, therefore, charge density rather than discrete charges, we should re-write this as the force per unit volume which is ρE+j×B. [This is an interesting formula which I didn’t use before, so you should double-check it. :-)]
2. The first term on the right-hand side should be equally obvious, or… Well… Perhaps somewhat less so. But with all my rambling on the Uncertainty Principle and/or the wave-particle duality, it should make sense. If we scrap the second term on the right-hand side, we basically have an equation that is equivalent to the E = mc2 equation. No? Sorry. Just look at it, again and again. You’ll end up understanding it. 🙂
3. So it’s that second term on the right-hand side. What the hell does that say? Well… I could say: it’s the local energy or momentum conservation law. If the energy or momentum doesn’t stay in, it has to go out. 🙂 But that’s not very satisfactory as an answer, of course. However, please just go along with this ‘temporary’ answer for a while.

So what is that second term on the right-hand side? As we wrote it, it’s an x-component – or, let’s put it differently, it is or was part of the x-component of the momentum density – but, frankly, we should probably allow it to go out in any direction really, as the only constraint on the left-hand side is a per second rate of change of something. Hence, Feynman suggest to equate it to something like this:

What a, b and c? The components of some vector? Not sure. We’re stuck. This piece really requires very advanced math. In fact, as far as I know, this is the only time where Feynman says: “Sorry. This is too advanced. I’ll just give you the equation. Sorry.” So that’s what he does. He explains the philosophy of the argument, which is the following:

1. On the left-hand side, we’ve got the time rate-of-change of momentum, so that obeys the F = dp/dt = d(mv)/dt law, with the force Fper unit volume, being equal to F(unit volume) = ρE+j×B.
2. On the right-hand side, we’ve got something that can be written as:

So we’d need to find a way to ρE+j×B in terms of and B only – eliminating ρ and j by using Maxwell’s equations or whatever other trick  – and then juggle terms and make substitutions to get it into a form that looks like the formula above, i.e. the right-hand side of that equation. But so Feynman doesn’t show us how it’s being done. He just mentions some theorem in physics, which says that the energy that’s flowing through a unit area per unit time divided by c2 – so that’s E/cper unit area and per unit time – must be equal to the momentum per unit volume in the space, so we write:

g = S/c2

He illustrates the general theorem that’s used to get the equation above by giving two examples:

OK. Two good examples. However, it’s still frustrating to not see how we get the g = S/c2 in the specific context of the electromagnetic force, so let’s do a dimensional analysis at least. In my previous post, I showed that the dimension of S must be J/(m2·s), so [S/c2] = [J/(m2·s)]/(m2/s2) = [N·m/(m2·s)]·(s2/m2) = [N·s/m3]. Now, we know that the unit of mass is 1 kg = N/(m/s2). That’s just the force law: a force of 1 newton will give a mass of 1 kg an acceleration of 1 m/s per second, so 1 N = 1 kg·(m/s2). So the [N·s/m3] dimension is equal to [kg·(m/s2)·s/m3] = [(kg·(m/s)/m3] = [(kg·(m/s)]/m3, which is the dimension of momentum (p = mv) per unit volume, indeed. So, yes, the dimensional analysis works out, and it’s also in line with the p = v·E/c2 = m·v equation, but… Oh… We did a dimensional analysis already, where we also showed that [g] = [S/c2] = (J·s/m4). Well… In any case… It’s a bit frustrating to not see the detail here, but let us note the the Grand Result once again:

The Poynting vector S gives us the energy flow as well as the momentum density= S/c2.

But what does it all mean, really? Let’s go through Einstein’s illustration of the principle. That will help us a lot. Before we do, however, I’d like to note something. I’ve always wondered a bit about that dichotomy between energy and momentum. Energy is force times distance: 1 joule is 1 newton × 1 meter indeed (1 J = 1 N·m). Momentum is force times time, as we can express it in N·s. Planck’s constant combines all three in the dimension of action, which is force times distance times time: ≈ 6.6×10−34 N·m·s, indeed. I like that unity. In this regard, you should, perhaps, quickly review that post in which I explain that is the energy per cycle, i.e. per wavelength or per period, of a photon, regardless of its wavelength. So it’s really something very fundamental.

We’ve got something similar here: energy and momentum coming together, and being shown as one aspect of the same thing: some oscillation. Indeed, just see what happens with the dimensions when we ‘distribute’ the 1/cfactor on the right-hand side over the two sides, so we write: c·= S/c and work out the dimensions:

1. [c·g = (m/s)·(N·s)/m= N/m= J/m3.
2. [S/c] = (s/m)·(N·m)/(s·m2) = N/m= J/m3.

Isn’t that nice? Both sides of the equation now have a dimension like ‘the force per unit area’, or ‘the energy per unit volume’. To get that, we just re-scaled g and S, by c and 1/c respectively. As far as I am concerned, this shows an underlying unity we probably tend to mask with our ‘related but different’ energy and momentum concepts. It’s like E and B: I just love it we can write them together in our Poynting formula = ε0c2E×B. In fact, let me show something else here, which you should think about. You know that c= 1/(ε0μ0), so we can write also as SE×B0. That’s nice, but what’s nice too is the following:

1. S/c = c·= ε0cE×B = E×B/μ0c
2. S/g = c= 1/(ε0μ0)

So, once again, Feynman may feel the Poynting vector is sort of counter-intuitive when analyzing specific situations but, as far as I am concerned, I feel the Poyning vector makes things actually easier to understand. Instead of two E and B vectors, and two concepts to deal with ‘energy’ (i.e. energy and momentum), we’re sort of unifying things here. In that regard – i.e in regard of feeling we’re talking the same thing really – I’d really highlight the S/g = c2 = 1/(ε0μ0) equation. Indeed, the universal constant acts just like the fine-structure constant here: it links everything to everything. 🙂

And, yes, it’s also about time we introduce the so-called principle of least action to explain things, because action, as a concept, combines force, distance and time indeed, so it’s a bit more promising than just energy, of just momentum. Having said that, you’ll see in the next section that it’s sometimes quite useful to have the choice between one formula or the other. But… Well… Enough talk. Let’s look at Einstein’s car.

#### Einstein’s car

Einstein’s car is a wonderful device: it rolls without any friction and it moves with a little flashlight. That’s all it needs. It’s pictured below. 🙂 So the situation is the following: the flashlight shoots some light out from one side, which is then stopped at the opposite end of the car. When the light is emitted, there must be some recoil. In fact, we know it’s going to be equal to 1/c times the energy because all we need to do is apply the pc = E·(v/c) formula for v = c, so we know that p = E/c. Of course, this momentum now needs to move Einstein’s car. It’s frictionless, so it should work, but still… The car has some mass M, and so that will determine its recoil velocity: v = p/M. We just apply the general p = mv formula here, and v is not equal to c here, of course! Of course, then the light hits the opposite end of the car and delivers the same momentum, so that stops the car again. However, it did move over some distance x = vt. So we could flash our light again and get to wherever we want to get. [Never mind the infinite accelerations involved!] So… Well… Great! Yes, but Einstein didn’t like this car when he first saw it. In fact, he still doesn’t like it, because he knows it won’t take you very far. 🙂

The problem is that we seem to be moving the center of gravity of this car by fooling around on the inside only. Einstein doesn’t like that. He thinks it’s impossible. And he’s right of course. The thing is: the center of gravity did not change. What happened here is that we’ve got some blob of energy, and so that blob has some equivalent mass (which we’ll denote by U/c2), and so that equivalent mass moved all the way from one side to the other, i.e. over the length of the car, which we denote by L. In fact, it’s stuff like this that inspired the whole theory of the field energy and field momentum, and how it interacts with matter.

What happens here is like switching the light on in the dark room: we’ve got matter doing work on the field, and so matter loses mass, and the field gains it, through its momentum and/or energy. To calculate how much, we could integrate S/c or c·over the volume of our blob, and we’d get something in joule indeed, but there’s a simpler way here. The momentum conservation says that the momentum of our car and the momentum of our blob must be equal, so if T is the time that was needed for our blob to go to the other side – and so that’s, of course, also the time during which our car was rolling – then M·v = M·x/T must be equal to (U/c2= (U/c2)·L/T. The 1/T factor on both sides cancel, so we write: M·x = (U/c2)·L. Now, what is x? Yes. In case you were wondering, that’s what we’re looking for here. 🙂 Here it is:

x = vT = vL/c = (p/M)·(L/c) = [U/c)/M]·(L/c) = (U/c2)·(L/M)

So what’s next? Well… Now we need to show that the center-of-mass actually did not move with this ‘transfer’ of the blob. I’ll leave the math to you here: it should all work out. And you can also think through the obvious questions:

1. Where is the energy and, hence, the mass of our blob after it stops the car? Hint: think about excited atoms and imagine they might radiate some light back. 🙂
2. As the car did move a little bit, we should be able to move it further and further away from its center of gravity, until the center of gravity is no longer in the car. Hint: think about batteries and energy levels going down while shooting light out. It just won’t happen. 🙂

Now, what about a blob of light going from the top to the bottom of the car? Well… That involves the conservation of angular momentum: we’ll have more mass on the bottom, but on a shorter lever-arm, so angular momentum is being conserved. It’s a very good question though, and it led Einstein to combine the center-of-gravity theorem with the angular momentum conservation theorem to explain stuff like this.

It’s all fascinating, and one can think of a great many paradoxes that, at first, seem to contradict the Grand Principles we used here, which means that they would contradict all that we have learned so far. However, a careful analysis of those paradox reveals that they are paradoxes indeed: propositions which sound true but are, in the end, self-contradictory. In fact, when explaining electromagnetism over his various Lectures, Feynman tasks his readers with a rather formidable paradox when discussing the laws of induction, he solves it here, ten chapters later, after describing what we described above. You can busy yourself with it but… Well… I guess you’ve got something better to do. If so, just take away the key lesson: there’s momentum in the field, and it’s also possible to build up angular momentum in a magnetic field and, if you switch it off, the angular momentum will be given back, somehow, as it’s stored energy.

That’s also why the seemingly irrelevant circulation of S we discussed in my previous post, where we had a charge next to an ordinary magnet, and where we found that there was energy circulating around, is not so queer. The energy is there, in the circulating field, and it’s real. As real as can be. 🙂

# Another post for my kids: introducing (special) relativity

In my previous post, I talked about energy, and I tried to keep it simple – but also accurate. However, to be completely accurate, one must, of course, introduce relativity at some point. So how does that work? What’s ‘relativistic’ energy? Well… Let me try to convey a few ideas here.

The first thing to note is that the energy conservation law still holds: special theory or not, the sum of the kinetic and potential energies in a (closed) system is always equal to some constant C. What constant? That doesn’t matter: Nature does not care about our zero point and, hence, we can add or subtract any (other) constant to the equation K.E. + P.E. = T + U = C.

That being said, in my previous post, I pointed out that the constant depends on the reference point for the potential energy term U: we will usually take infinity as the reference point (for a force that attracts) and associate it with zero potential (U = 0). We then get a function U(x) like the one below: for gravitational energy we have U(x) = –GMm/x, and for electrical charges, we have U(x) = q1q2/4πε0x. The mathematical shape is exactly the same but, in the case of the electromagnetic forces, you have to remember that likes repel, and opposites attract, so we don’t need the minus sign: the sign of the charges takes care of it.

Minus sign? In case you wonder why we need that minus sign for the potential energy function, well… I explained that in my previous post and so I’ll be brief on that here: potential energy is measured by doing work against the force. That’s why. So we have an infinite sum (i.e. an integral) over some trajectory or path looking like this: U = – ∫F·ds.

For kinetic energy, we don’t need any minus sign: as an object picks up speed, it’s the force itself that is doing the work as its potential energy is converted into kinetic energy, so the change in kinetic energy will equal the change in potential energy, but with opposite sign: as the object loses potential energy, it gains kinetic energy. Hence, we write ΔT = –ΔU = ∫F·ds..

That’s all kids stuff obviously. Let’s go beyond this and ask some questions. First, why can we add or subtract any constant to the potential energy but not to the kinetic energy? The answer is… Well… We actually can add or subtract a ‘constant’ to the kinetic energy as well. Now you will shake your head: Huh? Didn’t we have that T = mv2/2 formula for kinetic energy? So how and why could one add or subtract some number to that?

Well… That’s where relativity comes into play. The velocity v depends on your reference frame. If another observer would move with and/or alongside the object, at the same speed, that observer would observe a velocity equal to zero and, hence, its kinetic energy – as that observer would measure it – would also be zero. You will object to that, saying that a change of reference frame does not change the force, and you’re right: the force will cause the object to accelerate or decelerate indeed, and if the observer is not subject to the same force, then he’ll see the object accelerate or decelerate indeed, regardless of his reference frame is a moving or inertial frame. Hence, both the inertial as well as the moving observer will see an increase (or decrease) in its kinetic energy and, therefore, both will conclude that its potential energy decreases (or increases) accordingly. In short, it’s the change in energy that matters, both for the potential as well as for the kinetic energy. The reference point itself, i.e. the point from where we start counting so to say, does not: that’s relative. [This also shows in the derivation for kinetic energy which I’ll do below.]

That brings us to the second question. We all learned in high school that mass and energy are related through Einstein’s mass-energy relation, E = mc2, which establishes an equivalence between the two: the mass of an object that’s picking up speed increases, and so we need to look at both speed and mass as a function of time. Indeed, remember Newton’s Law: force is the time rate of change of momentum: F = d(mv)/dt. When the speed is low (i.e. non-relativistic), then we can just treat m as a constant and write that  F = mdv/dt = ma (the mass times the acceleration). Treating m as a constant also allows us to derive the classical (Newtonian) formula for kinetic energy:

So if we assume that the velocity of the object at point O is equal to zero (so vo = 0), then ΔT will be equal to T and we get what we were looking for: the kinetic energy at point P will be equal to T = mv2/2.

Now, you may wonder why we can’t do that same derivation for a non-constant mass? The answer to that question is simple: taking the m factor out of the integral can only be done if we assume it is a constant. If not, then we should leave it inside. It’s similar to taking a derivative. If m would not be constant, then we would have to apply the product rule to calculate d(mv)/dt, so we’d write d(mv)/dt = (dm/dt)v + m(dv/dt). So we have two terms here and it’s only when m is constant that we can reduce it to d(mv)/dt = m(dv/dt).

So we have our classical kinetic energy function. However, when the velocity gets really high – i.e. if it’s like the same order of magnitude as the velocity of light – then we cannot assume that mass is constant. Indeed, the same high-school course in physics that taught you that E = mc2 equation will probably also have taught you that an object can never go faster than light, regardless of the reference frame. Hence, as the object goes faster and faster, it will pick up more momentum, but its rate of acceleration should (and will) go down in such way that the object can never actually reach the speed of light. Indeed, if Newton’s Law is to remain valid, we need to correct it such a way that m is no longer constant: m itself will increase as a function of its velocity and, hence, as a function of time. You’ll remember the formula for that:

This is often written as m = γm0, with m0 denoting the mass of the object at rest (in your reference frame that is) and γ = (1 – v2/c2)–1/2 the so-called Lorentz factor. The Lorentz factor is named after a Dutch physicist who introduced it near the end of the 19th century in order to explain why the speed of light is always c, regardless of the frame of reference (moving or not), or – in other words – why the speed of light is not relative. Indeed, while you’ll remember that there is no such thing as an absolute velocity according to the (special) theory of relativity, the velocity of light actually is absolute ! That means you will always see light traveling at speed c regardless of your reference frame. To put it simply, you can never catch up with light and, if you would be traveling away from some star in a spaceship with a velocity of 200,000 km per second, and a light beam from that star would pass you, you’d measure the speed of that light beam to be equal to 300,000 km/s, not 100,000 km/s. So is an absolute speed that acts as an absolute speed limit regardless of your reference frame. [Note that we’re talking only about reference frames moving at a uniform speed: when acceleration comes into play, then we need to refer to the general theory of relativity and that’s a somewhat different ball game.]

The graph below shows how γ varies as a function of v. As you can see, the mass increase only becomes significant at speeds of like 100,000 km per second indeed. Indeed, for v = 0.3c, the Lorentz factor is 1.048, so the increase is about 5% only. For v = 0.5c, it’s still limited to an increase of some 15%. But then it goes up rapidly: for v = 0.9c, the mass is more than twice the rest mass: m ≈ 2.3m0; for v = 0.99c, the mass increase is 600%: m ≈ 7m0; and so on. For v = 0.999c – so when the speed of the object differs from c only by 1 part in 1,000 – the mass of the object will be more than twenty-two times the rest mass (m ≈ 22.4m0).

You probably know that we can actually reach such speeds and, hence, verify Einstein’s correction of Newton’s Law in particle accelerators: the electrons in an electron beam in a particle accelerator get usually pretty close to c and have a mass that’s like 2000 times their rest mass. How do we know that? Because the magnetic field needed to deflect them is like 2000 times as great as their (theoretical) rest mass. So how fast do they go? For their mass to be 2000 times m0, 1 – v2/c2 must be equal to 1/4,000,000. Hence, their velocity v differs from c only by one part in 8,000,000. You’ll have to admit that’s very close.

Other effects of relativistic speeds

So we mentioned the thing that’s best known about Einstein’s (special) theory of relativity: the mass of an object, as measured by the inertial observer, increases with its speed. Now, you may or may not be familiar with two other things that come out of relativity theory as well:

1. The first is length contraction: objects are measured to be shortened in the direction of motion with respect to the (inertial) observer. The formula to be used incorporates the reciprocal of the Lorentz factor: L = (1/γ)L0. For example, a meter stick in a space ship moving at a velocity v = 0.6c will appear to be only 80 cm to the external/inertial observer seeing it whizz past… That is if he can see anything at all of course: he’d have to take like a photo-finish picture as it zooms past ! 🙂
2. The second is time dilation, which is also rather well known – just like the mass increase effect – because of the so-called twin paradox: time will appear to be slower in that space ship and, hence, if you send one of two twins away on a space journey, traveling at such relativistic speed, he will come back younger than his brother. The formula here is a bit more complicated, but that’s only because we’re used to measure time in seconds. If we would take a more natural unit, i.e. the time it takes light to travel a distance of 1 m, then the formula will look the same as our mass formula: t = γt0 and, hence, one ‘second’ in the space ship will be measured as 1.25 ‘seconds’ by the external observer. Hence, the moving clock will appear to run slower – to the external (inertial) observer that is.

Again, the reality of this can be demonstrated. You’ll remember that we introduced the muon in previous posts: muons resemble electrons in the sense that they have the same charge, but their mass is more than 200 times the mass of an electron. As compared to other unstable particles, their average lifetime is quite long: 2.2 microseconds. Still, that would not be enough to travel more than 600 meters or so – even at the speed of light (2.2 μs × 300,000 km/s = 660 m). But so we do detect muons in detectors down here that come all the way down from the stratosphere, where they are created when cosmic rays hit the Earth’s atmosphere some 10 kilometers up. So how do they get here if they decay so fast? Well, those that actually end up in those detectors, do indeed travel very close to the speed of light and, hence, while from their own point of view they live only like two millionths of a second, they live considerably longer from our point of view.

Relativistic energy: E = mc2

Let’s go back to our main story line: relativistic energy. We wrote above that it’s the change of energy that matters really. So let’s look at that.

You may or may not remember that the concept of work in physics is closely related to the concept of power. In fact, you may actually remember that power, in physics at least, is defined as the work done per second. Indeed, we defined work as the (dot) product of the force and the distance. Now, when we’re talking a differential distance only (i.e. an infinitesimally small change only), then we can write dT = F·ds, but when we’re talking something larger, then we have to do that integral: ΔT = ∫F·ds. However, we’re interested in the time rate of change of T here, and so that’s the time derivative dT/dt which, as you easily verify, will be equal to dT/dt = (F·ds)/dt = F·(ds/dt) = F·and so we can use that differential formula and we don’t need the integral. Now, that (dot) product of the force and the velocity vectors is what’s referred to as the power. [Note that only the component of the force in the direction of motion contributes to the work done and, hence, to the power.]

OK. What am I getting at? Well… I just want to show an interesting derivation: if we assume, with Einstein, that mass and energy are equivalent and, hence, that the total energy of a body always equals E = mc2, then we can actually derive Einstein’s mass formula from that. How? Well… If the time rate of change of the energy of an object is equal to the power expended by the forces acting on it, then we can write:

dE/dt = d(mc2)/dt = F·v

Now, we cannot take the mass out of those brackets after the differential operator (d) because the mass is not a constant in this case (relativistic speeds) and, hence, dm/dt ≠ 0. However, we can take out c2 (that’s an absolute constant, remember?) and we can also substitute F using Newton’s Law (F = d(mv)/dt), again taking care to leave m between the brackets, not outside. So then we get:

d(mc2)/dt = c2dm/dt = [d(mv)/dt]·v = d(mv)/dt

In case you wonder why we can replace the vectors (bold face) v and d(mv) by their magnitudes (or lengths) v and d(mv): v and mv have the same direction and, hence, the angle θ between them is zero, and so v·v =v││v│cosθ =v2. Likewise, d(mv) and v also have the same direction and so we can just replace the dot product by the product of the magnitudes of those two vectors.

Now, let’s not forget the objective: we need to solve this equation for m and, hopefully, we’ll find Einstein’s mass formula, which we need to correct Newton’s Law. How do we do that? We’ll first multiply both sides by 2m. Why? Because we can then apply another mathematical trick, as shown below:

c2(2m)·dm/dt = 2md(mv)/dt ⇔ d(m2c2)/dt = d(m2v2)/dt

However, if the derivatives of two quantities are equal, then the quantities themselves can only differ by a constant, say C. So we integrate both sides and get:

m2c= m2v+ C

Be patient: we’re almost there. The above equation must be true for all velocities v and, hence, we can choose the special case where v = 0 and call this mass m0, and then substitute, so we get m0c= m00+ C = C. Now we put this particular value for C back in the more general equation above and we get:

mc= mv+ m0c⇔ m = mv2/c2 +m⇔ m(1 – v2/c2) = m⇔ m = m0/(1 – v2/c2)–1/2

So there we are: we have just shown that we get the relativistic mass formula (it’s on the right-hand side above) if we assume that Einstein’s mass-energy equivalence relation holds.

Now, you may wonder why that’s significant. Well… If you’re disappointed, then, at the very least, you’ll have to admit that it’s nice to show how everything is related to everything in this theory: from E = mc2, we get m0/(1 – v2/c2)–1/2. I think that’s kinda neat!

In addition, let us analyze that mass-energy relation in another way. It actually allows us to re-define kinetic energy as the excess of a particle over its rest mass energy, or – it’s the same expression really – or the difference between its total energy and its rest energy.

How does that work? Well… When we’re looking at high-speed or high-energy particles, we will write the kinetic energy as:

K.E. = mc– m0c= (m – m0)c= γm0c– m0c= m0c2(γ – 1).

Now, we can expand that Lorentz factor γ = (1 – v2/c2)–1/2 into a binomial series (the binomial series is an infinite Taylor series, so it’s not to be confused with the (finite) binomial expansion: just check it online if you’re in doubt). If we do that, we we can write γ as an infinite sum of the following terms:

γ = 1 + (1/2)v2/c+ (3/8)v4/c+ (5/16)v6/c+ …

Now, when we plug this back into our (relativistic) kinetic energy equation, we can scrap a few things (just do it) to get where I wanted to get:

K.E. = (1/2)m0v+ (3/8)m0v4/c+ (5/16)m0v6/c+ …

Again, you’ll wonder: so what? Well… See how the non-relativistic formula for kinetic energy (K.E. = m0v2/2) appears here as the first term of this series and, hence, how the formula above shows that our ‘Newtonian’ formula is just an approximation. Of course, at low speeds, the second, third etcetera terms represent close to nothing and, hence, then our Newtonian ‘approximation is obviously pretty good of course !

OK… But… Now you’ll say: that’s fine, but how did Einstein get inspired to write E = mc2 in the first place? Well, truth be told, the relativistic mass formula was derived first (i.e. before Einstein wrote his E = mc2 equation), out of a derivation involving the momentum conservation law and the formulas we must use to convert the space-time coordinates from one reference frame to another when looking at phenomena (i.e. the so-called Lorentz transformations). And it was only afterwards that Einstein noted that, when expanding the relativistic mass formula, that the increase in mass of a body appeared to be equal to the increase in kinetic energy divided by c2 (Δm = Δ(K.E.)/c2). Now, that, in turn, inspired him to also assign an equivalent energy to the rest mass of that body: E0 = m0c2. […] At least that’s how Feynman tells the story in his 1965 Lectures… But so we’ve actually been doing it the other way around here!

Hmm… You will probably find all of this rather strange, and you may also wonder what happened to our potential energy. Indeed, that concept sort of ‘disappeared’ in this story: from the story above, it’s clear that kinetic energy has an equivalent mass, but what about potential energy?

That’s a very interesting question but, unfortunately, I can only give a rather rudimentary answer to that. Let’s suppose that we have two masses M and m. According to the potential energy formula above, the potential energy U between these two masses will then be equal to U = –GMm/r. Now, that energy is not interpreted as energy of either M or m, but as energy that is part of the (M, m) system, which includes the system’s gravitational field. So that energy is considered to be stored in that gravitational field. If the two masses would sit right on top of each other, then there would be no potential energy in the (M, m) system and, hence, the system as a whole would have less energy. In contrast, when we separate them further apart, then we increase the energy of the system as a whole, and so the system’s gravitational field then increases. So, yes, the potential energy does impact the (equivalent) mass of the system, but not the individual masses M and m. Does that make sense?

For me , it does, but I guess you’re a bit tired by now and, hence, I think I should wrap up here. In my next (and probably last) post on relativity, I’ll present those Lorentz transformations that allow us to ‘translate’ the space and time coordinates from one reference frame to another, and in that post I’ll also present the other derivation of Einstein’s relativistic mass formula, which is actually based on those transformations. In fact, I realize I should have probably started with that (as mentioned above, that’s how Feynman does it in his Lectures) but, then, for some reason, I find the presentation above more interesting, and so that’s why I am telling the story starting from another angle. I hope you don’t mind. In any case, it should be the same, because everything is related to everything in physics – just like in math. That’s why it’s important to have a good teacher. 🙂