Re-visiting the speed of light, Planck’s constant, and the fine-structure constant

Note: I have published a paper that is very coherent and fully explains what the fine-structure constant actually is. There is nothing magical about it. It’s not some God-given number. It’s a scaling constant – and then some more. But not God-given. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

A brother of mine sent me a link to an article he liked. Now, because we share some interest in physics and math and other stuff, I looked at it and…

Well… I was disappointed. Despite the impressive credentials of its author – a retired physics professor – it was very poorly written. It made me realize how much badly written stuff is around, and I am glad I am no longer wasting my time on it. However, I do owe my brother some explanation of (a) why I think it was bad, and of (b) what, in my humble opinion, he should be wasting his time on. 🙂 So what it is all about?

The article talks about physicists deriving the speed of light from “the electromagnetic properties of the quantum vacuum.” Now, it’s the term ‘quantum‘, in ‘quantum vacuum’, that made me read the article.

Indeed, deriving the theoretical speed of light in empty space from the properties of the classical vacuum – aka empty space – is a piece of cake: it was done by Maxwell himself as he was figuring out his equations back in the 1850s (see my post on Maxwell’s equations and the speed of light). And then he compared it to the measured value, and he saw it was right on the mark. Therefore, saying that the speed of light is a property of the vacuum, or of empty space, is like a tautology: we may just as well put it the other way around, and say that it’s the speed of light that defines the (properties of the) vacuum!

Indeed, as I’ll explain in a moment: the speed of light determines both the electric as well as the magnetic constants μand ε0, which are the (magnetic) permeability and the (electric) permittivity of the vacuum respectively. Both constants depend on the units we are working with (i.e. the units for electric charge, for distance, for time and for force – or for inertia, if you want, because force is defined in terms of overcoming inertia), but so they are just proportionality coefficients in Maxwell’s equations. So once we decide what units to use in Maxwell’s equations, then μand ε0 are just proportionality coefficients which we get from c. So they are not separate constants really – I mean, they are not separate from c – and all of the ‘properties’ of the vacuum, including these constants, are in Maxwell’s equations.

In fact, when Maxwell compared the theoretical value of c with its presumed actual value, he didn’t compare c‘s theoretical value with the speed of light as measured by astronomers (like that 17th century Ole Roemer, to which our professor refers: he had a first go at it by suggesting some specific value for it based on his observations of the timing of the eclipses of one of Jupiter’s moons), but with c‘s value as calculated from the experimental values of μand ε0! So he knew very well what he was looking at. In fact, to drive home the point, it may also be useful to note that the Michelson-Morley experiment – which accurately measured the speed of light – was done some thirty years later. So Maxwell had already left this world by then—very much in peace, because he had solved the mystery all 19th century physicists wanted to solve through his great unification: his set of equations covers it all, indeed: electricity, magnetism, light, and even relativity!

I think the article my brother liked so much does a very lousy job in pointing all of that out, but that’s not why I wouldn’t recommend it. It got my attention because I wondered why one would try to derive the speed of light from the properties of the quantum vacuum. In fact, to be precise, I hoped the article would tell me what the quantum vacuum actually is. Indeed, as far as I know, there’s only one vacuum—one ’empty space’: empty is empty, isn’t it? 🙂 So I wondered: do we have a ‘quantum’ vacuum? And, if so, what is it, really?

Now, that is where the article is really disappointing, I think. The professor drops a few names (like the Max Planck Institute, the University of Paris-Sud, etcetera), and then, promisingly, mentions ‘fleeting excitations of the quantum vacuum’ and ‘virtual pairs of particles’, but then he basically stops talking about quantum physics. Instead, he wanders off to share some philosophical thoughts on the fundamental physical constants. What makes it all worse is that even those thoughts on the ‘essential’ constants are quite off the mark.

So… This post is just a ‘quick and dirty’ thing for my brother which, I hope, will be somewhat more thought-provoking than that article. More importantly, I hope that my thoughts will encourage him to try to grind through better stuff.

On Maxwell’s equations and the properties of empty space

Let me first say something about the speed of light indeed. Maxwell’s four equations may look fairly simple, but that’s only until one starts unpacking all those differential vector equations, and it’s only when going through all of their consequences that one starts appreciating their deep mathematical structure. Let me quickly copy how another blogger jotted them down: 🙂

god-said-maxwell-equation

As I showed in my above-mentioned post, the speed of light (i.e. the speed with which an electromagnetic pulse or wave travels through space) is just one of the many consequences of the mathematical structure of Maxwell’s set of equations. As such, the speed of light is a direct consequence of the ‘condition’, or the properties, of the vacuum indeed, as Maxwell suggested when he wrote that “we can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena”.

Of course, while Maxwell still suggests light needs some ‘medium’ here – so that’s a reference to the infamous aether theory – we now know that’s because he was a 19th century scientist, and so we’ve done away with the aether concept (because it’s a redundant hypothesis), and so now we also know there’s absolutely no reason whatsoever to try to “avoid the inference.” 🙂 It’s all OK, indeed: light is some kind of “transverse undulation” of… Well… Of what?

We analyze light as traveling fields, represented by two vectors, E and B, whose direction and magnitude varies both in space as well as in time. E and B are field vectors, and represent the electric and magnetic field respectively. An equivalent formulation – more or less, that is (see my post on the Liénard-Wiechert potentials) – for Maxwell’s equations when only one (moving) charge is involved is:

E

B

This re-formulation, which is Feynman’s preferred formula for electromagnetic radiation, is interesting in a number of ways. It clearly shows that, while we analyze the electric and magnetic field as separate mathematical entities, they’re one and the same phenomenon really, as evidenced by the B = –er×E/c equation, which tells us the magnetic field from a single moving charge is always normal (i.e. perpendicular) to the electric field vector, and also that B‘s magnitude is 1/times the magnitude of E, so |B| = B = |E|/c = E/c. In short, B is fully determined by E, or vice versa: if we have one of the two fields, we have the other, so they’re ‘one and the same thing’ really—not in a mathematical sense, but in a real sense.

Also note that E and B‘s magnitude is just the same if we’re using natural units, so if we equate c with 1. Finally, as I pointed out in my post on the relativity of electromagnetic fields, if we would switch from one reference frame to another, we’ll have a different mix of E and B, but that different mix obviously describes the same physical reality. More in particular, if we’d be moving with the charges, the magnetic field sort of disappears to re-appear as an electric field. So the Lorentz force F = Felectric + Fmagnetic = qE + qv×B is one force really, and its ‘electric’ and ‘magnetic’ component appear the way they appear in our reference frame only. In some other reference frame, we’d have the same force, but its components would look different, even if they, obviously, would and should add up to the same. [Well… Yes and no… You know there’s relativistic corrections to be made to the forces to, but that’s a minor point, really. The force surely doesn’t disappear!]

All of this reinforces what you know already: electricity and magnetism are part and parcel of one and the same phenomenon, the electromagnetic force field, and Maxwell’s equations are the most elegant way of ‘cutting it up’. Why elegant? Well… Click the Occam tab. 🙂

Now, after having praised Maxwell once more, I must say that Feynman’s equations above have another advantage. In Maxwell’s equations, we see two constants, the electric and magnetic constant (denoted by μand ε0 respectively), and Maxwell’s equations imply that the product of the electric and magnetic constant is the reciprocal of c2: μ0·ε= 1/c2. So here we see εand only, so no μ0, so that makes it even more obvious that the magnetic and electric constant are related one to another through c.

[…] Let me digress briefly: why do we have c2 in μ0·ε= 1/c2, instead of just c? That’s related to the relativistic nature of the magnetic force: think about that B = E/c relation. Or, better still, think about the Lorentz equation F = Felectric + Fmagnetic = qE + qv×B = q[E + (v/c)×(E×er)]: the 1/c factor is there because the magnetic force involves some velocity, and any velocity is always relative—and here I don’t mean relative to the frame of reference but relative to the (absolute) speed of light! Indeed, it’s the v/c ratio (usually denoted by β = v/c) that enters all relativistic formulas. So the left-hand side of the μ0·ε= 1/c2 equation is best written as (1/c)·(1/c), with one of the two 1/c factors accounting for the fact that the ‘magnetic’ force is a relativistic effect of the ‘electric’ force, really, and the other 1/c factor giving us the proper relationship between the magnetic and the electric constant. To drive home the point, I invite you to think about the following:

  • μ0 is expressed in (V·s)/(A·m), while εis expressed in (A·s)/(V·m), so the dimension in which the μ0·εproduct is expressed is [(V·s)/(A·m)]·[(A·s)/(V·m)] = s2/m2, so that’s the dimension of 1/c2.
  • Now, this dimensional analysis makes it clear that we can sort of distribute 1/c2 over the two constants. All it takes is re-defining the fundamental units we use to calculate stuff, i.e. the units for electric charge, for distance, for time and for force – or for inertia, as explained above. But so we could, if we wanted, equate both μ0 as well as εwith 1/c.
  • Now, if we would then equate c with 1, we’d have μ0 = ε= c = 1. We’d have to define our units for electric charge, for distance, for time and for force accordingly, but it could be done, and then we could re-write Maxwell’s set of equations using these ‘natural’ units.

In any case, the nitty-gritty here is less important: the point is that μand εare also related through the speed of light and, hence, they are ‘properties’ of the vacuum as well. [I may add that this is quite obvious if you look at their definition, but we’re approaching the matter from another angle here.]

In any case, we’re done with this. On to the next!

On quantum oscillations, Planck’s constant, and Planck units 

The second thought I want to develop is about the mentioned quantum oscillation. What is it? Or what could it be? An electromagnetic wave is caused by a moving electric charge. What kind of movement? Whatever: the charge could move up or down, or it could just spin around some axis—whatever, really. For example, if it spins around some axis, it will have a magnetic moment and, hence, the field is essentially magnetic, but then, again, E and B are related and so it doesn’t really matter if the first cause is magnetic or electric: that’s just our way of looking at the world: in another reference frame, one that’s moving with the charges, the field would essential be electric. So the motion can be anything: linear, rotational, or non-linear in some irregular way. It doesn’t matter: any motion can always be analyzed as the sum of a number of ‘ideal’ motions. So let’s assume we have some elementary charge in space, and it moves and so it emits some electromagnetic radiation.

So now we need to think about that oscillation. The key question is: how small can it be? Indeed, in one of my previous posts, I tried to explain some of the thinking behind the idea of the ‘Great Desert’, as physicists call it. The whole idea is based on our thinking about the limit: what is the smallest wavelength that still makes sense? So let’s pick up that conversation once again.

The Great Desert lies between the 1032 and 1043 Hz scale. 1032 Hz corresponds to a photon energy of Eγ = h·f = (4×10−15 eV·s)·(1032 Hz) = 4×1017 eV = 400,000 tera-electronvolt (1 TeV = 1012 eV). I use the γ (gamma) subscript in my Eγ symbol for two reasons: (1) to make it clear that I am not talking the electric field E here but energy, and (2) to make it clear we are talking ultra-high-energy gamma-rays here.

In fact, γ-rays of this frequency and energy are theoretical only. Ultra-high-energy gamma-rays are defined as rays with photon energies higher than 100 TeV, which is the upper limit for very-high-energy gamma-rays, which have been observed as part of the radiation emitted by so-called gamma-ray bursts (GRBs): flashes associated with extremely energetic explosions in distant galaxies. Wikipedia refers to them as the ‘brightest’ electromagnetic events know to occur in the Universe. These rays are not to be confused with cosmic rays, which consist of high-energy protons and atomic nuclei stripped of their electron shells. Cosmic rays aren’t rays really and, because they consist of particles with a considerable rest mass, their energy is even higher. The so-called Oh-My-God particle, for example, which is the most energetic particle ever detected, had an energy of 3×1020 eV, i.e. 300 million TeV. But it’s not a photon: its energy is largely kinetic energy, with the rest mass m0 counting for a lot in the m in the E = m·c2 formula. To be precise: the mentioned particle was thought to be an iron nucleus, and it packed the equivalent energy of a baseball traveling at 100 km/h! 

But let me refer you to another source for a good discussion on these high-energy particles, so I can get get back to the energy of electromagnetic radiation. When I talked about the Great Desert in that post, I did so using the Planck-Einstein relation (E = h·f), which embodies the idea of the photon being valid always and everywhere and, importantly, at every scale. I also discussed the Great Desert using real-life light being emitted by real-life atomic oscillators. Hence, I may have given the (wrong) impression that the idea of a photon as a ‘wave train’ is inextricably linked with these real-life atomic oscillators, i.e. to electrons going from one energy level to the next in some atom. Let’s explore these assumptions somewhat more.

Let’s start with the second point. Electromagnetic radiation is emitted by any accelerating electric charge, so the atomic oscillator model is an assumption that should not be essential. And it isn’t. For example, whatever is left of the nucleus after alpha or beta decay (i.e. a nuclear decay process resulting in the emission of an α- or β-particle) it likely to be in an excited state, and likely to emit a gamma-ray for about 10−12 seconds, so that’s a burst that’s about 10,000 times shorter than the 10–8 seconds it takes for the energy of a radiating atom to die out. [As for the calculation of that 10–8 sec decay time – so that’s like 10 nanoseconds – I’ve talked about this before but it’s probably better to refer you to the source, i.e. one of Feynman’s Lectures.]

However, what we’re interested in is not the energy of the photon, but the energy of one cycle. In other words, we’re not thinking of the photon as some wave train here, but what we’re thinking about is the energy that’s packed into a space corresponding to one wavelength. What can we say about that?

As you know, that energy will depend both on the amplitude of the electromagnetic wave as well as its frequency. To be precise, the energy is (1) proportional to the square of the amplitude, and (2) proportional to the frequency. Let’s look at the first proportionality relation. It can be written in a number of ways, but one way of doing it is stating the following: if we know the electric field, then the amount of energy that passes per square meter per second through a surface that is normal to the direction in which the radiation is going (which we’ll denote by S – the s from surface – in the formula below), must be proportional to the average of the square of the field. So we write S ∝ 〈E2〉, and so we should think about the constant of proportionality now. Now, let’s not get into the nitty-gritty, and so I’ll just refer to Feynman for the derivation of the formula below:

S = ε0c·〈E2

So the constant of proportionality is ε0c. [Note that, in light of what we wrote above, we can also write this as S = (1/μ0·c)·〈(c·B)2〉 = (c0)·〈B2〉, so that underlines once again that we’re talking one electromagnetic phenomenon only really.] So that’s a nice and rather intuitive result in light of all of the other formulas we’ve been jotting down. However, it is a ‘wave’ perspective. The ‘photon’ perspective assumes that, somehow, the amplitude is given and, therefore, the Planck-Einstein relation only captures the frequency variable: Eγ = h·f.

Indeed, ‘more energy’ in the ‘wave’ perspective basically means ‘more photons’, but photons are photons: they have a definite frequency and a definite energy, and both are given by that Planck-Einstein relation. So let’s look at that relation by doing a bit of dimensional analysis:

  • Energy is measured in electronvolt or, using SI units, joule: 1 eV ≈ 1.6×10−19 J. Energy is force times distance: 1 joule = 1 newton·meter, which means that a larger force over a shorter distance yields the same energy as a smaller force over a longer distance. The oscillations we’re talking about here involve very tiny distances obviously. But the principle is the same: we’re talking some moving charge q, and the power – which is the time rate of change of the energy – that goes in or out at any point of time is equal to dW/dt = F·v, with W the work that’s being done by the charge as it emits radiation.
  • I would also like to add that, as you know, forces are related to the inertia of things. Newton’s Law basically defines a force as that what causes a mass to accelerate: F = m·a = m·(dv/dt) = d(m·v)/dt = dp/dt, with p the momentum of the object that’s involved. When charges are involved, we’ve got the same thing: a potential difference will cause some current to change, and one of the equivalents of Newton’s Law F = m·a = m·(dv/dt) in electromagnetism is V = L·(dI/dt). [I am just saying this so you get a better ‘feel’ for what’s going on.]
  • Planck’s constant is measured in electronvolt·seconds (eV·s) or in, using SI units, in joule·seconds (J·s), so its dimension is that of (physical) action, which is energy times time: [energy]·[time]. Again, a lot of energy during a short time yields the same energy as less energy over a longer time. [Again, I am just saying this so you get a better ‘feel’ for these dimensions.]
  • The frequency f is the number of cycles per time unit, so that’s expressed per second, i.e. in herz (Hz) = 1/second = s−1.

So… Well… It all makes sense: [x joule] = [6.626×10−34 joule]·[1 second]×[f cycles]/[1 second]. But let’s try to deepen our understanding even more: what’s the Planck-Einstein relation really about?

To answer that question, let’s think some more about the wave function. As you know, it’s customary to express the frequency as an angular frequency ω, as used in the wave function A(x, t) = A0·sin(kx − ωt). The angular frequency is the frequency expressed in radians per second. That’s because we need an angle in our wave function, and so we need to relate x and t to some angle. The way to think about this is as follows: one cycle takes a time T (i.e. the period of the wave) which is equal to T = 1/f. Yes: one second divided by the number of cycles per second gives you the time that’s needed for one cycle. One cycle is also equivalent to our argument ωt going around the full circle (i.e. 2π), so we write:  ω·T = 2π and, therefore:

ω = 2π/T = 2π·f

Now we’re ready to play with the Planck-Einstein relation. We know it gives us the energy of one photon really, but what if we re-write our equation Eγ = h·f as Eγ/f = h? The dimensions in this equation are:

[x joule]·[1 second]/[cyles] = [6.626×10−34 joule]·[1 second]

⇔ = 6.626×10−34 joule per cycle

So that means that the energy per cycle is equal to 6.626×10−34 joule, i.e. the value of Planck’s constant.

Let me rephrase truly amazing result, so you appreciate it—perhaps: regardless of the frequency of the light (or our electromagnetic wave, in general) involved, the energy per cycle, i.e. per wavelength or per period, is always equal to 6.626×10−34 joule or, using the electronvolt as the unit, 4.135667662×10−15 eV. So, in case you wondered, that is the true meaning of Planck’s constant!

Now, if we have the frequency f, we also have the wavelength λ, because the velocity of the wave is the frequency times the wavelength: = λ·f and, therefore, λ = c/f. So if we increase the frequency, the wavelength becomes smaller and smaller, and so we’re packing the same amount of energy – admittedly, 4.135667662×10−15 eV is a very tiny amount of energy – into a space that becomes smaller and smaller. Well… What’s tiny, and what’s small? All is relative, of course. 🙂 So that’s where the Planck scale comes in. If we pack that amount of energy into some tiny little space of the Planck dimension, i.e. a ‘length’ of 1.6162×10−35 m, then it becomes a tiny black hole, and it’s hard to think about how that would work.

[…] Let me make a small digression here. I said it’s hard to think about black holes but, of course, it’s not because it’s ‘hard’ that we shouldn’t try it. So let me just mention a few basic facts. For starters, black holes do emit radiation! So they swallow stuff, but they also spit stuff out. More in particular, there is the so-called Hawking radiation, as Roger Penrose and Stephen Hawking discovered.

Let me quickly make a few remarks on that: Hawking radiation is basically a form of blackbody radiation, so all frequencies are there, as shown below: the distribution of the various frequencies depends on the temperature of the black body, i.e. the black hole in this case. [The black curve is the curve that Lord Rayleigh and Sir James Jeans derived in the late 19th century, using classical theory only, so that’s the one that does not correspond to experimental fact, and which led Max Planck to become the ‘reluctant’ father of quantum mechanics. In any case, that’s history and so I shouldn’t dwell on this.]

600px-Black_body

The interesting thing about blackbody radiation, including Hawking radiation, is that it reduces energy and, hence, the equivalent mass of our blackbody. So Hawking radiation reduces the mass and energy of black holes and is therefore also known as black hole evaporation. So black holes that lose more mass than they gain through other means are expected to shrink and ultimately vanish. Therefore, there’s all kind of theories that say why micro black holes, like that Planck scale black hole we’re thinking of right now, should be much larger net emitters of radiation than large black holes and, hence, whey they should shrink and dissipate faster.

Hmm… Interesting… What do we do with all of this information? Well… Let’s think about it as we continue our trek on this long journey to reality over the next year or, more probably, years (plural). 🙂

The key lesson here is that space and time are intimately related because of the idea of movement, i.e. the idea of something having some velocity, and that it’s not so easy to separate the dimensions of time and distance in any hard and fast way. As energy scales become larger and, therefore, our natural time and distance units become smaller and smaller, it’s the energy concept that comes to the fore. It sort of ‘swallows’ all other dimensions, and it does lead to limiting situations which are hard to imagine. Of course, that just underscores the underlying unity of Nature, and the mysteries involved.

So… To relate all of this back to the story that our professor is trying to tell, it’s a simple story really. He’s talking about two fundamental constants basically, c and h, pointing out that c is a property of empty space, and h is related to something doing something. Well… OK. That’s really nothing new, and surely not ground-breaking research. 🙂

Now, let me finish my thoughts on all of the above by making one more remark. If you’ve read a thing or two about this – which you surely have – you’ll probably say: this is not how people usually explain it. That’s true, they don’t. Anything I’ve seen about this just associates the 1043 Hz scale with the 1028 eV energy scale, using the same Planck-Einstein relation. For example, the Wikipedia article on micro black holes writes that “the minimum energy of a microscopic black hole is 1019 GeV [i.e. 1028 eV], which would have to be condensed into a region on the order of the Planck length.” So that’s wrong. I want to emphasize this point because I’ve been led astray by it for years. It’s not the total photon energy, but the energy per cycle that counts. Having said that, it is correct, however, and easy to verify, that the 1043 Hz scale corresponds to a wavelength of the Planck scale: λ = c/= (3×10m/s)/(1043 s−1) = 3×10−35 m. The confusion between the photon energy and the energy per wavelength arises because of the idea of a photon: it travels at the speed of light and, hence, because of the relativistic length contraction effect, it is said to be point-like, to have no dimension whatsoever. So that’s why we think of packing all of its energy in some infinitesimally small place. But you shouldn’t think like that. The photon is dimensionless in our reference frame: in its own ‘world’, it is spread out, so it is a wave train. And it’s in its ‘own world’ that the contradictions start… 🙂

OK. Done!

My third and final point is about what our professor writes on the fundamental physical constants, and more in particular on what he writes on the fine-structure constant. In fact, I could just refer you to my own post on it, but that’s probably a bit too easy for me and a bit difficult for you 🙂 so let me summarize that post and tell you what you need to know about it.

The fine-structure constant

The fine-structure constant α is a dimensionless constant which also illustrates the underlying unity of Nature, but in a way that’s much more fascinating than the two or three things the professor mentions. Indeed, it’s quite incredible how this number (α = 0.00729735…, but you’ll usually see it written as its reciprocal, which is a number that’s close to 137.036…) links charge with the relative speeds, radii, and the mass of fundamental particles and, therefore, how this number also these concepts with each other. And, yes, the fact that it is, effectively, dimensionless, unlike h or c, makes it even more special. Let me quickly sum up what the very same number α all stands for:

(1) α is the square of the electron charge expressed in Planck units: α = eP2.

(2) α is the square root of the ratio of (a) the classical electron radius and (b) the Bohr radius: α = √(re /r). You’ll see this more often written as re = α2r. Also note that this is an equation that does not depend on the units, in contrast to equation 1 (above), and 4 and 5 (below), which require you to switch to Planck units. It’s the square of a ratio and, hence, the units don’t matter. They fall away.

(3) α is the (relative) speed of an electron: α = v/c. [The relative speed is the speed as measured against the speed of light. Note that the ‘natural’ unit of speed in the Planck system of units is equal to c. Indeed, if you divide one Planck length by one Planck time unit, you get (1.616×10−35 m)/(5.391×10−44 s) = m/s. However, this is another equation, just like (2), that does not depend on the units: we can express v and c in whatever unit we want, as long we’re consistent and express both in the same units.]

(4) α is also equal to the product of (a) the electron mass (which I’ll simply write as me here) and (b) the classical electron radius re (if both are expressed in Planck units): α = me·re. Now think that’s, perhaps, the most amazing of all of the expressions for α. [If you don’t think that’s amazing, I’d really suggest you stop trying to study physics. :-)]

Also note that, from (2) and (4), we find that:

(5) The electron mass (in Planck units) is equal me = α/r= α/α2r = 1/αr. So that gives us an expression, using α once again, for the electron mass as a function of the Bohr radius r expressed in Planck units.

Finally, we can also substitute (1) in (5) to get:

(6) The electron mass (in Planck units) is equal to me = α/r = eP2/re. Using the Bohr radius, we get me = 1/αr = 1/eP2r.

So… As you can see, this fine-structure constant really links all of the fundamental properties of the electron: its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, its mass (and, hence, its energy),…

So… Why is what it is?

Well… We all marvel at this, but what can we say about it, really? I struggle how to interpret this, just as much – or probably much more 🙂 – as the professor who wrote the article I don’t like (because it’s so imprecise, and that’s what made me write all what I am writing here).

Having said that, it’s obvious that it points to a unity beyond these numbers and constants that I am only beginning to appreciate for what it is: deep, mysterious, and very beautiful. But so I don’t think that professor does a good job at showing how deep, mysterious and beautiful it all is. But then that’s up to you, my brother and you, my imaginary reader, to judge, of course. 🙂

[…] I forgot to mention what I mean with ‘Planck units’. Well… Once again, I should refer you to one of my other posts. But, yes, that’s too easy for me and a bit difficult for you. 🙂 So let me just note we get those Planck units by equating not less than five fundamental physical constants to 1, notably (1) the speed of light, (2) Planck’s (reduced) constant, (3) Boltzmann’s constant, (4) Coulomb’s constant and (5) Newton’s constant (i.e. the gravitational constant). Hence, we have a set of five equations here (ħ = kB = ke = G = 1), and so we can solve that to get the five Planck units, i.e. the Planck length unit, the Planck time unit, the Planck mass unit, the Planck energy unit, the Planck charge unit and, finally (oft forgotten), the Planck temperature unit. Of course, you should note that all mass and energy units are directly related because of the mass-energy equivalence relation E = mc2, which simplifies to E = m if c is equated to 1. [I could also say something about the relation between temperature and (kinetic) energy, but I won’t, as it would only further confuse you.]

OK. Done! 🙂

Addendum: How to think about space and time?

If you read the argument on the Planck scale and constant carefully, then you’ll note that it does not depend on the idea of an indivisible photon. However, it does depend on that Planck-Einstein relation being valid always and everywhere. Now, the Planck-Einstein relation is, in its essence, a fairly basic result from classical electromagnetic theory: it incorporates quantum theory – remember: it’s the equation that allowed Planck to solve the black-body radiation problem, and so it’s why they call Planck the (reluctant) ‘Father of Quantum Theory’ – but it’s not quantum theory.

So the obvious question is: can we make this reflection somewhat more general, so we can think of the electromagnetic force as an example only. In other words: can we apply the thoughts above to any force and any movement really?

The truth is: I haven’t advanced enough in my little study to give the equations for the other forces. Of course, we could think of gravity, and I developed some thoughts on how gravity waves might look like, but nothing specific really. And then we have the shorter-range nuclear forces, of course: the strong force, and the weak force. The laws involved are very different. The strong force involves color charges, and the way distances work is entirely different. So it would surely be some different analysis. However, the results should be the same. Let me offer some thoughts though:

  • We know that the relative strength of the nuclear force is much larger, because it pulls like charges (protons) together, despite the strong electromagnetic force that wants to push them apart! So the mentioned problem of trying to ‘pack’ some oscillation in some tiny little space should be worse with the strong force. And the strong force is there, obviously, at tiny little distances!
  • Even gravity should become important, because if we’ve got a lot of energy packed into some tiny space, its equivalent mass will ensure the gravitational forces also become important. In fact, that’s what the whole argument was all about!
  • There’s also all this talk about the fundamental forces becoming one at the Planck scale. I must, again, admit my knowledge is not advanced enough to explain how that would be possible, but I must assume that, if physicists are making such statements, the argument must be fairly robust.

So… Whatever charge or whatever force we are talking about, we’ll be thinking of waves or oscillations—or simply movement, but it’s always a movement in a force field, and so there’s power and energy involved (energy is force times distance, and power is the time rate of change of energy). So, yes, we should expect the same issues in regard to scale. And so that’s what’s captured by h.

As we’re talking the smallest things possible, I should also mention that there are also other inconsistencies in the electromagnetic theory, which should (also) have their parallel for other forces. For example, the idea of a point charge is mathematically inconsistent, as I show in my post on fields and charges. Charge, any charge really, must occupy some space. It cannot all be squeezed into one dimensionless point. So the reasoning behind the Planck time and distance scale is surely valid.

In short, the whole argument about the Planck scale and those limits is very valid. However, does it imply our thinking about the Planck scale is actually relevant? I mean: it’s not because we can imagine how things might look like  – they may look like those tiny little black holes, for example – that these things actually exist. GUT or string theorists obviously think they are thinking about something real. But, frankly, Feynman had a point when he said what he said about string theory, shortly before his untimely death in 1988: “I don’t like that they’re not calculating anything. I don’t like that they don’t check their ideas. I don’t like that for anything that disagrees with an experiment, they cook up an explanation—a fix-up to say, ‘Well, it still might be true.'”

It’s true that the so-called Standard Model does not look very nice. It’s not like Maxwell’s equations. It’s complicated. It’s got various ‘sectors’: the electroweak sector, the QCD sector, the Higgs sector,… So ‘it looks like it’s got too much going on’, as a friend of mine said when he looked at a new design for mountainbike suspension. 🙂 But, unlike mountainbike designs, there’s no real alternative for the Standard Model. So perhaps we should just accept it is what it is and, hence, in a way, accept Nature as we can see it. So perhaps we should just continue to focus on what’s here, before we reach the Great Desert, rather than wasting time on trying to figure out how things might look like on the other side, especially because we’ll never be able to test our theories about ‘the other side.’

On the other hand, we can see where the Great Desert sort of starts (somewhere near the 1032 Hz scale), and so it’s only natural to think it should also stop somewhere. In fact, we know where it stops: it stops at the 1043 Hz scale, because everything beyond that doesn’t make sense. The question is: is there actually there? Like fundamental strings or whatever you want to call it. Perhaps we should just stop where the Great Desert begins. And what’s the Great Desert anyway? Perhaps it’s a desert indeed, and so then there is absolutely nothing there. 🙂

Hmm… There’s not all that much one can say about it. However, when looking at the history of physics, there’s one thing that’s really striking. Most of what physicists can think of, in the sense that it made physical sense, turned out to exist. Think of anti-matter, for instance. Paul Dirac thought it might exist, that it made sense to exist, and so everyone started looking for it, and Carl Anderson found in a few years later (in 1932). In fact, it had been observed before, but people just didn’t pay attention, so they didn’t want to see it, in a way. […] OK. I am exaggerating a bit, but you know what I mean. The 1930s are full of examples like that. There was a burst of scientific creativity, as the formalism of quantum physics was being developed, and the experimental confirmations of the theory just followed suit.

In the field of astronomy, or astrophysics I should say, it was the same with black holes. No one could really imagine the existence of black holes until the 1960s or so: they were thought of a mathematical curiosity only, a logical possibility. However, the circumstantial evidence now is quite large and so… Well… It seems a lot of what we can think of actually has some existence somewhere. 🙂

So… Who knows? […] I surely don’t. And so I need to get back to the grind and work my way through the rest of Feynman’s Lectures and the related math. However, this was a nice digression, and so I am grateful to my brother he initiated it. 🙂

Planck’s constant (II)

My previous post was tough. Tough for you–if you’ve read it. But tough for me too. 🙂

The blackbody radiation problem is complicated but, when everything is said and done, what the analysis says is that the the ‘equipartition theorem’ in the kinetic theory of gases ‘theorem (or the ‘theorem concerning the average energy of the center-of-mass motion’, as Feynman terms it), is not correct. That equipartition theorem basically states that, in thermal equilibrium, energy is shared equally among all of its various forms. For example, the average kinetic energy per degree of freedom in the translation motion of a molecule should equal that of its rotational motions. That equipartition theorem is also quite precise: it also states that the mean energy, for each atom or molecule, for each degree of freedom, is kT/2. Hence, that’s the (average) energy the 19th century scientists also assigned to the atomic oscillators in a gas.

However, the discrepancy between the theoretical and empirical result of their work shows that adding atomic oscillators–as radiators and absorbers of light–to the system (a box of gas that’s being heated) is not just a matter of adding additional ‘degree of freedom’ to the system. It can’t be analyzed in ‘classical’ terms: the actual spectrum of blackbody radiation shows that these atomic oscillators do not absorb, on average, an amount of energy equal to kT/2. Hence, they are not just another ‘independent direction of motion’.

So what are they then? Well… Who knows? I don’t. But, as I didn’t quite go through the full story in my previous post, the least I can do is to try to do that here. It should be worth the effort. In Feynman’s words: “This was the first quantum-mechanical formula ever known, or discussed, and it was the beautiful culmination of decades of puzzlement.” And then it does not involve complex numbers or wave functions, so that’s another reason why looking at the detail is kind of nice. 🙂

Discrete energy levels and the nature of h

To solve the blackbody radiation problem, Planck assumed that the permitted energy levels of the atomic harmonic oscillator were equally spaced, at ‘distances’ ħωapart from each other. That’s what’s illustrated below.

Equally space energy levels

Now, I don’t want to make too many digressions from the main story, but this En = nħω0 formula obviously deserves some attention. First note it immediately shows why the dimension of ħ is expressed in joule-seconds (J·s), or electronvolt-seconds (J·s): we’re multiplying it with a frequency indeed, so that’s something expressed per second (hence, its dimension is s–1) in order to get a measure of energy: joules or, because of the atomic scale, electronvolts. [The eV is just a (much) smaller measure than the joule, but it amounts to the same: 1 eV ≈ 1.6×10−19 J.]

One thing to note is that the equal spacing consists of distances equal to ħω0, not of ħ. Hence, while h, or ħ (ħ is the constant to be used when the frequency is expressed in radians per second, rather than oscillations per second, so ħ = h/2π) is now being referred to as the quantum of action (das elementare Wirkungsquantum in German), Planck referred to it as as a Hilfsgrösse only (that’s why he chose the h as a symbol, it seems), so that’s an auxiliary constant only: the actual quantum of action is, of course, ΔE, i.e. the difference between the various energy levels, which is the product of ħ and ω(or of h and ν0 if we express frequency in oscillations per second, rather than in angular frequency). Hence, Planck (and later Einstein) did not assume that an atomic oscillator emits or absorbs packets of energy as tiny as ħ or h, but packets of energy as big as ħωor, what amounts to the same (ħω = (h/2π)(2πν) = hν), hν0. Just to give an example, the frequency of sodium light (ν) is 500×1012 Hz, and so its energy is E = hν. That’s not a lot–about 2 eV only– but it still packs 500×1012 ‘quanta of action’ !

Another thing is that ω (or ν) is a continuous variable: hence, the assumption of equally spaced energy levels does not imply that energy itself is a discrete variable: light can have any frequency and, hence, we can also imagine photons with any energy level: the only thing we’re saying is that the energy of a photon of a specific color (i.e. a specific frequency ν) will be a multiple of hν.

Probability assumptions

The second key assumption of Planck as he worked towards a solution of the blackbody radiation problem was that the probability (P) of occupying a level of energy E is P(EαeE/kT. OK… Why not? But what is this assumption really? You’ll think of some ‘bell curve’, of course. But… No. That wouldn’t make sense. Remember that the energy has to be positive. The general shape of this P(E) curve is shown below.

graph

The highest probability density is near E = 0, and then it goes down as E gets larger, with kT determining the slope of the curve (just take the derivative). In short, this assumption basically states that higher energy levels are not so likely, and that very high energy levels are very unlikely. Indeed, this formula implies that the relative chance, i.e. the probability of being in state E1 relative to the chance of being in state E0, is P1/Pe−(E1–E0)k= e−ΔE/kT. Now, Pis n1/N and Pis n0/N and, hence, we find that nmust be equal to n0e−ΔE/kT. What this means is that the atomic oscillator is less likely to be in a higher energy state than in a lower one.

That makes sense, doesn’t it? I mean… I don’t want to criticize those 19th century scientists but… What were they thinking? Did they really imagine that infinite energy levels were as likely as… Well… More down-to-earth energy levels? I mean… A mechanical spring will break when you overload it. Hence, I’d think it’s pretty obvious those atomic oscillators cannot be loaded with just about anything, can they? Garbage in, garbage out:  of course, that theoretical spectrum of blackbody radiation didn’t make sense!

Let me copy Feynman now, as the rest of the story is pretty straightforward:

Now, we have a lot of oscillators here, and each is a vibrator of frequency w0. Some of these vibrators will be in the bottom quantum state, some will be in the next one, and so forth. What we would like to know is the average energy of all these oscillators. To find out, let us calculate the total energy of all the oscillators and divide by the number of oscillators. That will be the average energy per oscillator in thermal equilibrium, and will also be the energy that is in equilibrium with the blackbody radiation and that should go in the equation for the intensity of the radiation as a function of the frequency, instead of kT. [See my previous post: that equation is I(ω) = (ω2kt)/(π2c2).]

Thus we let N0 be the number of oscillators that are in the ground state (the lowest energy state); N1 the number of oscillators in the state E1; N2 the number that are in state E2; and so on. According to the hypothesis (which we have not proved) that in quantum mechanics the law that replaced the probability eP.E./kT or eK.E./kT in classical mechanics is that the probability goes down as eΔE/kT, where ΔE is the excess energy, we shall assume that the number N1 that are in the first state will be the number N0 that are in the ground state, times e−ħω/kT. Similarly, N2, the number of oscillators in the second state, is N=N0e−2ħω/kT. To simplify the algebra, let us call e−ħω/k= x. Then we simply have N1 = N0x, N2 = N0x2, …, N= N0xn.

The total energy of all the oscillators must first be worked out. If an oscillator is in the ground state, there is no energy. If it is in the first state, the energy is ħω, and there are N1 of them. So N1ħω, or ħωN0x is how much energy we get from those. Those that are in the second state have 2ħω, and there are N2 of them, so N22ħω=2ħωN0x2 is how much energy we get, and so on. Then we add it all together to get Etot = N0ħω(0+x+2x2+3x3+…).

And now, how many oscillators are there? Of course, N0 is the number that are in the ground state, N1 in the first state, and so on, and we add them together: Ntot = N0(1+x+x2+x3+…). Thus the average energy is

formula

Now the two sums which appear here we shall leave for the reader to play with and have some fun with. When we are all finished summing and substituting for x in the sum, we should get—if we make no mistakes in the sum—
energy

Feynman concludes as follows: “This, then, was the first quantum-mechanical formula ever known, or ever discussed, and it was the beautiful culmination of decades of puzzlement. Maxwell knew that there was something wrong, and the problem was, what was right? Here is the quantitative answer of what is right instead of kT. This expression should, of course, approach kT as ω → 0 or as → .”

It does, of course. And so Planck’s analysis does result in a theoretical I(ω) curve that matches the observed I(ω) curve as a function of both temperature (T) and frequency (ω). But so what it is, then? What’s the equation describing the dotted curves? It’s given below:

formula blackbody

I’ll just quote Feynman once again to explain the shape of those dotted curves: “We see that for a large ω, even though we have ωin the numerator, there is an e raised to a tremendous power in the denominator, so the curve comes down again and does not “blow up”—we do not get ultraviolet light and x-rays where we do not expect them!”

Is the analysis necessarily discrete?

One question I can’t answer, because I just am not strong enough in math, is the question or whether or not there would be any other way to derive the actual blackbody spectrum. I mean… This analysis obviously makes sense and, hence, provides a theory that’s consistent and in accordance with experiment. However, the question whether or not it would be possible to develop another theory, without having recourse to the assumption that energy levels in atomic oscillators are discrete and equally spaced with the ‘distance’ between equal to hν0, is not easy to answer. I surely can’t, as I am just a novice, but I can imagine smarter people than me have thought about this question. The answer must be negative, because I don’t know of any other theory: quantum mechanics obviously prevailed. Still… I’d be interested to see the alternatives that must have been considered.

Post scriptum: The “playing with the sums” is a bit confusing. The key to the formula above is the substitution of (0+x+2x2+3x3+…)/(1+x+x2+x3+…) by 1/[(1/x)–1)] = 1/[eħω/kT–1]. Now, the denominator 1+x+x2+x3+… is the Maclaurin series for 1/(1–x). So we have:

(0+x+2x2+3x3+…)/(1+x+x2+x3+…) = (0+x+2x2+3x3+…)(1–x)

x+2x2+3x3… –x22x3–3x4… = x+x2+x3+x4

= –1+(1+x+x2+x3…) = –1 + 1/(1–x) = –(1–x)+1/(1–x) = x/(1–x).

Note the tricky bit: if x = e−ħω/kT, then eħω/kis x−1 = 1/x, and so we have (1/x)–1 in the denominator of that (mean) energy formula, not 1/(x–1). Now 1/[(1/x)–1)] = 1/[(1–x)/x] = x/(1–x), indeed, and so the formula comes out alright.

Planck’s constant (I)

If you made it here, it means you’re totally fed up with all of the easy stories on quantum mechanics: diffraction, double-slit experiments, imaginary gamma-ray microscopes,… You’ve had it! You now know what quantum mechanics is all about, and you’ve realized all these thought experiments never answer the tough question: where did Planck find that constant (h) which pops up everywhere? And how did he find that Planck relation which seems to underpin all and everything in quantum mechanics?

If you don’t know, that’s because you’ve skipped the blackbody radiation story. So let me give it to you here. What’s blackbody radiation?

Thermal equilibrium of radiation

That’s what the blackbody radiation problem is about: thermal equilibrium of radiation.

Huh? 

Yes. Imagine a box with gas inside. You’ll often see it’s described as a furnace, because we heat the box. Hence, the box, and everything inside, acquires a certain temperature, which we then assume to be constant. The gas inside will absorb energy and start emitting radiation, because the gas atoms or molecules are atomic oscillators. Hence, we have electrons getting excited and then jumping up and down from higher to lower energy levels, and then again and again and again, thereby emitting photons with a certain energy and, hence, light of a certain frequency. To put it simply: we’ll find light with various frequencies in the box and, in thermal equilibrium, we should have some distribution of the intensity of the light according to the frequency: what kind of radiation do we find in the furnace? Well… Let’s find out.

The assumption is that the box walls send light back, or that the box has mirror walls. So we assume that all the radiation keeps running around in the box. Now that implies that the atomic oscillators not only radiate energy, but also receive energy, because they’re constantly being illuminated by radiation that comes straight back at them. If the temperature of the box is kept constant, we arrive at a situation which is referred to as thermal equilibrium. In Feynman’s words: “After a while there is a great deal of light rushing around in the box, and although the oscillator is radiating some, the light comes back and returns some of the energy that was radiated.”

OK. That’s easy enough to understand. However, the actual analysis of this equilibrium situation is what gave rise to the ‘problem’ of blackbody radiation in the 19th century which, as you know, led Planck and Einstein to develop a quantum-mechanical view of things. It turned out that the classical analysis predicted a distribution of the intensity of light that didn’t make sense, and no matter how you looked at it, it just didn’t come out right. Theory and experiment did not agree. Now, that is something very serious in science, as you know, because it means your theory isn’t right. In this case, it was disastrous, because it meant the whole of classical theory wasn’t right.

To be frank, the analysis is not all that easy. It involves all that I’ve learned so far: the math behind oscillators and interference, statistics, the so-called kinetic theory of gases and what have you. I’ll try to summarize the story but you’ll see it requires quite an introduction.

Kinetic energy and temperature

The kinetic theory of gases is part of what’s referred to as statistical mechanics: we look at a gas as a large number of inter-colliding atoms and we describe what happens in terms of the collisions between them. As Feynman puts it: “Fundamentally, we assert that the gross properties of matter should be explainable in terms of the motion of its parts.” Now, we can do a lot of intellectual gymnastics, analyzing one gas in one box, two gases in one box, two gases in one box with a piston between them, two gases in two boxes with a hole in the wall between them, and so on and so on, but that would only distract us here. The rather remarkable conclusion of such exercises, which you’ll surely remember from your high school days, is that:

  1. Equal volumes of different gases, at the same pressure and temperature, will have the same number of molecules.
  2. In such view of things, temperature is actually nothing but the mean kinetic energy of those molecules (or atoms if it’s a monatomic gas).

So we can actually measure temperature in terms of the kinetic energy of the molecules of the gas, which, as you know, equals mv2/2, with m the mass and v the velocity of the gas molecules. Hence, we’re tempted to define some absolute measure of temperature T and simply write:

T = 〈mv2/2〉

The 〈 and 〉 brackets denote the mean here. To be precise, we’re talking the root mean square here, aka as the quadratic mean, because we want to average some magnitude of a varying quantity. Of course, the mass of different gases will be different – and so we have 〈m1v12/2〉 for gas 1 and 〈m2v22/2〉 for gas 2 – but that doesn’t matter: we can, actually, imagine measuring temperature in joule, the unit of energy, including kinetic energy. Indeed, the units come out alright: 1 joule = 1 kg·(m2/s2). For historical reasons, however, T is measured in different units: degrees Kelvin, centigrades (i.e. degrees Celsius) or, in the US, in Fahrenheit. Now, we can easily go from one measure to the other as you know and, hence, here I should probably just jot down the so-called ideal gas law–because we need that law for the subsequent analysis of blackbody radiation–and get on with it:

PV = NkT

However, now that we’re here, let me give you an inkling of how we derive that law. A classical (Newtonian) analysis of the collisions (you can find the detail in Feynman’s Lectures, I-39-2) will yield the following equation: P = (2/3)n〈mv2/2〉, with n the number of atoms or molecules per unit volume. So the pressure of a gas (which, as you know, is the force (of a gas on a piston, for example) per unit area: P = F/A) is also equal to the mean kinetic energy of the gas molecules multiplied by (2/3)n. If we multiply that equation by V, we get PV = N(2/3)〈mv2/2〉. However, we know that equal volumes of volumes of different gases, at the same pressure and temperature, will have the same number of molecules, so we have PV = N(2/3)〈m1v12/2〉 = N(2/3)〈m2v22/2〉, which we write as PV = NkT with kT = (2/3)〈m1v12/2〉 = (2/3)〈m2v22/2〉.

In other words, that factor of proportionality k is the one we have to use to convert the temperature as measured by 〈mv2/2〉 (i.e. the mean kinetic energy expressed in joules) to T (i.e. the temperature expressed in the measure we’re used to, and that’s degrees Kelvin–or Celsius or Fahrenheit, but let’s stick to Kelvin, because that’s what’s used in physics). Vice versa, we have 〈mv2/2〉 = (3/2)kT. Now, that constant of proportionality k is equal to k 1.38×10–23 joule per Kelvin (J/K). So if T is (absolute) temperature, expressed in Kelvin (K), our definition says that the mean molecular kinetic energy is (3/2)kT.

That k factor is a physical constant referred to as the Boltzmann constant. If it’s one of these constants, you may wonder why we don’t integrate that 3/2 factor in it? Well… That’s just how it is, I guess. In any case, it’s rather convenient because we’ll have 2/3 factors in other equations and so these will cancel out with that 3/2 term. However, I am digressing way too much here. I should get back to the main story line. However, before I do that, I need to expand on one more thing, and that’s a small lecture on how things look like when we also allow for internal motion, i.e. the rotational and vibratory motions of the atoms within the gas molecule. Let me first re-write that PV = NkT equation as

PV = NkT = N(2/3)〈m1v12/2〉 = (2/3)U = 2U/3

For monatomic gas, that U would only be the kinetic energy of the atoms, and so we can write it as U = (2/3)NkT. Hence, we have the grand result that the kinetic energy, for each atom, is equal to (3/2)kT, on average that is.

What about non-monatomic gas? Well… For complex molecules, we’d also have energy going into the rotational and vibratory motion of the atoms within the molecule, separate from what is usually referred to as the center-of-mass (CM) motion of the molecules themselves. Now, I’ll again refer you to Feynman for the detail of the analysis, but it turns out that, if we’d have, for example, a diatomic molecule, consisting of an A and B atom, the internal rotational and vibratory motion would, indeed, also absorb energy, and we’d have a total energy equal to (3/2)kT + (3/2)kT = 2×(3/2)kT = 3kT. Now, that amount (3kT) can be split over (i) the energy related to the CM motion, which must still be equal to (3/2)kT, and (ii) the average kinetic energy of the internal motions of the diatomic molecule excluding the bodily motion of the CM. Hence, the latter part must be equal to 3kT – (3/2)kT = (3/2)kT. So, for the diatomic molecule, the total energy happens to consist of two equal parts.

Now, there is a more general theorem here, for which I have to introduce the notion of the degrees of freedom of a system. Each atom can rotate or vibrate or oscillate or whatever in three independent directions–namely the three spatial coordinates x, y and z. These spatial dimensions are referred to as the degrees of freedom of the atom (in the kinetic theory of gases, that is), and if we have two atoms, we have 2×3 = 6 degrees of freedom. More in general, the number of degrees of freedom of a molecule composed of r atoms is equal to 3rNow, it can be shown that the total energy of an r-atom molecule, including all internal energy as well as the CM motion, will be 3r×kT/2 = 3rkT/2 joules. Hence, for every independent direction of motion that there is, the average kinetic energy for that direction will be kT/2. [Note that ‘independent direction of motion’ is used, somewhat confusingly, as a synonym for degree of freedom, so we don’t have three but six ‘independent directions of motion’ for the diatomic molecule. I just wanted to note that because I do think it causes confusion when reading a textbook like Feynman’s.] Now, that total amount of energy, i.e.  3r(kT/2), will be split as follows according to the “theorem concerning the average energy of the CM motion”, as Feynman terms it:

  1. The kinetic energy for the CM motion of each molecule is, and will always be, (3/2)kT.
  2. The remainder, i.e. r(3/2)kT – (3/2)kT = (3/2)(r–1)kt, is internal vibrational and rotational kinetic energy, i.e. the sum of all vibratory and rotational kinetic energy but excluding the energy of the CM motion of the molecule.

Phew! That’s quite something. And we’re not quite there yet.

The analysis for photon gas

Photon gas? What’s that? Well… Imagine our box is the gas in a very hot star, hotter than the sun. As Feynman writes it: “The sun is not hot enough; there are still too many atoms, but at still higher temperatures in certain very hot stars, we may neglect the atoms and suppose that the only objects that we have in the box are photons.” Well… Let’s just go along with it. We know that photons have no mass but they do have some very tiny momentum, which we related to the magnetic field vector, as opposed to the electric field. It’s tiny indeed. Most of the energy of light goes into the electric field. However, we noted that we can write p as p = E/c, with c the speed of light (3×108). Now, we had that = (2/3)n〈mv2/2〉 formula for gas, and we know that the momentum p is defined as p = mv. So we can substitute mvby (mv)v = pv. So we get = (2/3)n〈pv/2〉 = (1/3)n〈pv〉.

Now, the energy of photons is not quite the same as the kinetic energy of an atom or an molecule, i.e. mv2/2. In fact, we know that, for photons, the speed v is equal to c, and pc = E. Hence, multiplying by the volume V, we get

PV = U/3

So that’s a formula that’s very similar to the one we had for gas, for which we wrote: PV = NkT = 2U/3. The only thing is that we don’t have a factor 2 in the equation but so that’s because of the different energy concepts involved. Indeed, the concept of the energy of a photon (E = pc) is different than the concept of kinetic energy. But so the result is very nice: we have a similar formula for the compressibility of gas and radiation. In fact, both PV = 2U/3 and PV = U/3 will usually be written, more generally, as:

PV = (γ – 1)U 

Hence, this γ would be γ = 5/3 ≈ 1.667 for gas and 4/3 ≈ 1.333 for photon gas. Now, I’ll skip the detail (it involves a differential analysis) but it can be shown that this general formula, PV = (γ – 1)U, implies that PVγ (i.e. the pressure times the volume raised to the power γ) must equal some constant, so we write:

PVγ = C

So far so good. Back to our problem: blackbody radiation. What you should take away from this introduction is the following:

  1. Temperature is a measure of the average kinetic energy of the atoms or molecules in a gas. More specifically, it’s related to the mean kinetic energy of the CM motion of the atoms or molecules, which is equal to (3/2)kT, with k the Boltzmann constant and T the temperature expressed in Kelvin (i.e. the absolute temperature).
  2. If gas atoms or molecules have additional ‘degrees of freedom’, aka ‘independent directions of motion’, then each of these will absorb additional energy, namely kT/2.

Energy and radiation

The atoms in the box are atomic oscillators, and we’ve analyzed them before. What the analysis above added was that average kinetic energy of the atoms going around is (3/2)kT and that, if we’re talking molecules consisting of r atoms, we have a formula for their internal kinetic energy as well. However, as an oscillator, they also have energy separate from that kinetic energy we’ve been talking about alrady. How much? That’s a tricky analysis. Let me first remind you of the following:

  1. Oscillators have a natural frequency, usually denoted by the (angular) frequency ω0.
  2. The sum of the potential and kinetic energy stored in an oscillator is a constant, unless there’s some damping constant. In that case, the oscillation dies out. Here, you’ll remember the concept of the Q of an oscillator. If there’s some damping constant, the oscillation will die out and the relevant formula is 1/Q = (dW/dt)/(ω0W) = γ0, with γ the damping constant (not to be confused with the γ we used in that PVγ = C formula).

Now, for gases, we said that, for every independent direction of motion there is, the average kinetic energy for that direction will be kT/2. I admit it’s a bit of a stretch of the imagination but so that’s how the blackbody radiation analysis starts really: our atomic oscillators will have an average kinetic energy equal to kT/2 and, hence, their total energy (kinetic and potential) should be twice that amount, according to the second remark I made above. So that’s kT. We’ll note the total energy as W below, so we can write:

W = kT

Just to make sure we know what we’re talking about (one would forget, wouldn’t one?), kT is the product of the Boltzmann constant (1.38×10–23 J/K) and the temperature of the gas (so note that the product is expressed in joule indeed). Hence, that product is the average energy of our atomic oscillators in the gas in our furnace.

Now, I am not going to repeat all of the detail we presented on atomic oscillators (I’ll refer you, once again, to Feynman) but you may or may not remember that atomic oscillators do have a Q indeed and, hence, some damping constant γ. So we can use and re-write that formula above as

dW/dt = (1/Q)(ω0W) = (ω0W)(γ/ω0) = γW, which implies γ = (dW/dt)/W

What’s γ? Well, we’ve calculated the Q of an atomic oscillator already: Q = 3λ/4πr0. Now, λ = 2πc/ω(we just convert the wavelength into (angular) frequency using λν = c) and γ = ω0/Q, so we get γ = 4πr0ω0/[3(2πc/ω0)] = (2/3)r0ω02/c. Now, plugging that result back into the equation above, we get

dW/dt = γW = (2/3)(r0ω02kT)/c

Just in case you’d have difficulty following – I admit I did 🙂 – dW/dt is the average rate of radiation of light of (or near) frequency ω02. I’ll let Feynman take over here:

Next we ask how much light must be shining on the oscillator. It must be enough that the energy absorbed from the light (and thereupon scattered) is just exactly this much. In other words, the emitted light is accounted for as scattered light from the light that is shining on the oscillator in the cavity. So we must now calculate how much light is scattered from the oscillator if there is a certain amount—unknown—of radiation incident on it. Let I(ω)dω be the amount of light energy there is at the frequency ω, within a certain range dω (because there is no light at exactly a certain frequency; it is spread all over the spectrum). So I(ω) is a certain spectral distribution which we are now going to find—it is the color of a furnace at temperature T that we see when we open the door and look in the hole. Now how much light is absorbed? We worked out the amount of radiation absorbed from a given incident light beam, and we calculated it in terms of a cross section. It is just as though we said that all of the light that falls on a certain cross section is absorbed. So the total amount that is re-radiated (scattered) is the incident intensity I(ω)dω multiplied by the cross section σ.

OK. That makes sense. I’ll not copy the rest of his story though, because this is a post in a blog, not a textbook. What we need to find is that I(ω). So I’ll refer you to Feynman for the details (these ‘details’ involve fairly complicated calculations, which are less important than the basic assumptions behind the model, which I presented above) and just write down the result:

blackbody radiation formula

This formula is Rayleigh’s law. [And, yes, it’s the same Rayleigh – Lord Rayleigh, I should say respectfully – as the one who invented that criterion I introduced in my previous post, but so this law and that criterion have nothing to do with each other.] This ‘law’ gives the intensity, or the distribution, of light in a furnace. Feynman says it’s referred to as blackbody radiation because “the hole in the furnace that we look at is black when the temperature is zero.” […] OK. Whatever. What we call it doesn’t matter. The point is that this function tells us that the intensity goes as the square of the frequency, which means that if we have a box at any temperature at all, and if we look at it, the X- and gamma rays will be burning out eyes out ! The graph below shows both the theoretical curve for two temperatures (Tand 2T0), as derived above (see the solid lines), and then the actual curves for those two temperatures (see the dotted lines).

Blackbody radation graph

This is the so-called UV catastrophe: according to classical physics, an ideal black body at thermal equilibrium should emit radiation with infinite power. In reality, of course, it doesn’t: Rayleigh’s law is false. Utterly false. And so that’s where Planck came to the rescue, and he did so by assuming radiation is being emitted and/or absorbed in finite quanta: multiples of h, in fact.

Indeed, Planck studied the actual curve and fitted it with another function. That function assumed the average energy of a harmonic oscillator was not just proportional with the temperature (T), but that it was also a function of the (natural) frequency of the oscillators. By fiddling around, he found a simple derivation for it which involved a very peculiar assumption. That assumption was that the harmonic oscillator can take up energies only ħω at the time, as shown below.

Equally space energy levels

Hence, the assumption is that the harmonic oscillators cannot take whatever (continous) energy level. No. The allowable energy levels of the harmonic oscillators are equally spaced: E= nħω. Now, the actual derivation is at least as complex as the derivation of Rayleigh’s law, so I won’t do it here. Let me just give you the key assumptions:

  1. The gas consists of a large number of atomic oscillators, each with their own natural frequency ω0.
  2. The permitted energy levels of these harmonic oscillator are equally spaced and ħωapart.
  3. The probability of occupying a level of energy E is P(Eαe–E/kT.

All the rest is tedious calculation, including the calculation of the parameters of the model, which include ħ (and, hence, h, because h = 2πħ) and are found by matching the theoretical curves to the actual curves as measured in experiments. I’ll just mention one result, and that’s the average energy of these oscillators:

energy

As you can see, the average energy does not only depend on the temperature T, but also on their (natural) frequency. So… Now you know where h comes from. As I relied so heavily on Feynman’s presentation here, I’ll include the link. As Feynman puts it: “This, then, was the first quantum-mechanical formula ever known, or ever discussed, and it was the beautiful culmination of decades of puzzlement. Maxwell knew that there was something wrong, and the problem was, what was right? Here is the quantitative answer of what is right instead of kT.”

So there you go. Now you know. 🙂 Oh… And in case you’d wonder: why the h? Well… Not sure. It’s said the h stands for Hilfsgrösse, so that’s some constant which was just supposed to help him out with the calculation. At that time, Planck did not suspect it would turn out to be one of the most fundamental physical constants. 🙂

Post scriptum: I went quite far in my presentation of the basics of the kinetic theory of gases. You may wonder now. I didn’t use that theoretical PVγ = C relation, did I? And why all the fuss about photon gas? Well… That was just to introduce that PVγ = C relation, so I could note, here, in this post scriptum, that it has a similar problem. The γ exponent is referred to as the specific heat ratio of a gas, and it can be calculated theoretically as well, as we did–well… Sort of, because we skipped the actual derivation. However, their theoretical value also differs substantially from actually measured values, and the problem is the same: one should not assume that a continuous value for 〈E〉. Agreement between theory and experiment can only be reached when the same assumptions as those of Planck are used: discrete energy levels, multiples of ħ and ω: E= nħω. Also, the specific functional form which Planck used to resolve the blackbody radiation problem is also to be used here. For more details, I’ll refer to Feynman too. I can’t say this is easy to digest, but then who said it would be easy? 🙂

The point to note is that the blackbody radiation problem wasn’t the only problem in the 19th century. As Feynman puts it: “One often hears it said that physicists at the latter part of the nineteenth century thought they knew all the significant physical laws and that all they had to do was to calculate more decimal places. Someone may have said that once, and others copied it. But a thorough reading of the literature of the time shows they were all worrying about something.” They were, and so Planck came up with something new. And then Einstein took it to the next level and then… Well… The rest is history. 🙂