# Problems with the derivation of Planck's radiation law

I have tried to understand the derivation of Planck's radiation law, but there's a lot I don't quite understand. The derivation I've been reading is this one: https://edisciplinas.usp.br/pluginfile.php/48089/course/section/16461/qsp_chapter10-plank.pdf

Here are my questions:

1. In the derivation, you consider standing wave modes given by eq. 10.2 in the article. But these modes can't have well-defined energies (given by $$E_n=nh\nu$$ later in the article), because that energy will always vary according to the square of the field strength, as explained by Maxwell's theory.
2. In the article, they also claim that there are only two independent polarization states of the modes, and they explain this by referring to plane waves of EM-radiation. But I thought we considered standing waves, not plane waves?
3. In deriving the Rayleigh-Jeans law, you use the equipartition theorem to find that the average energy is $$kT$$. But that would require two independent quadratic energy modes. But isn't there only one? Sure, we have electric and magnetic fields, but in a plane wave (which they used earlier in the article, as explained above), these fields aren't independent. In particular, the electric field is proportional to the magnetic field, the constant of proportionality being $$c$$, and they are always 90 degrees on each other and the direction of propagation.
4. Later it's also claimed that these modes can be considered to be harmonic oscillators, giving the same energies as those of particles trapped in a harmonic oscillator potential, whether it's a classical or quantum oscillator. Why can you make such an assumption?

1. The revolutionary insight here is precisely that the prediction from classical electrodynamics - which would suggest that the allowed energies for each mode are continuous - is wrong. If you instead restrict the allowed energies to discrete multiples of $$h\nu$$ (where $$\nu$$ is the frequency of the mode in question and $$h$$ is some constant), then the resulting prediction matches experiment. Einstein suggested that we take this apparently ad-hoc discretization seriously, which was one of the major stepping stones toward quantum mechanics.
2. Standing waves and plane waves are not mutually exclusive phenomena. $$\mathbf E = \sin(kx)\sin(\omega t) \hat y$$ is a plane wave (the wavefronts are planes $$x=const.$$) and also a standing wave.
3. The two quadratic degrees of freedom in the Hamiltonian are not quite $$\mathbf E$$ and $$\mathbf B$$. If you formulate electromagnetism in the Hamiltonian framework, then (choosing the Coulomb gauge as per knzhou's nice answer here) the Hamiltonian can be expressed in terms of $$\mathbf E$$ and the vector potential $$\mathbf A$$ as $$H \sim \frac{1}{2}(E^2+\omega^2A^2)$$. It turns out that $$\mathbf E$$ is the canonically conjugate "momentum" to $$\mathbf A$$, and it is in this sense that the electromagnetic field is analogous to a harmonic oscillator.
4. This is answered by $$(3)$$.

The electromagnetic field Lagrangian density is given by

$$\mathcal L = \frac{B^2}{2\mu_0}-\frac{\epsilon_0E^2}{2}$$ Writing $$\mathbf E = -\nabla \phi - \dot{\mathbf A}$$ and $$\mathbf B = \nabla \times \mathbf A$$, we can express this in terms of the vector potential as

$$\mathcal L = \frac{1}{2\mu_0}(\nabla\times\mathbf A)^2-\frac{\epsilon_0}{2}\left(-\nabla \phi - \dot{\mathbf A}\right)^2$$

Treating $$\mathbf A$$ like our generalized coordinates, we define the canonical momenta $$\boldsymbol \Pi$$ to be $$\boldsymbol \Pi = \frac{\partial \mathcal L}{\partial \dot{\mathbf A}} = \epsilon_0 (-\nabla\phi - \dot{\mathbf A})$$ $$\iff \dot{\mathbf A} = -\frac{1}{\epsilon_0}\boldsymbol\Pi + \nabla \phi$$

Performing the requisite Legendre transformation to move to the Hamiltonian formalism yields the Hamiltonian density:

$$\mathcal H = \boldsymbol \Pi \cdot \dot{\mathbf A} - \mathcal L = \frac{1}{2\epsilon_0}\Pi^2 + \frac{1}{2\mu_0} (\nabla\times \mathbf A)^2 + \boldsymbol{\Pi} \cdot \nabla \phi$$

Once we integrate over space, we'll be able to integrate the last term by parts. If the fields go to zero at infinity, it simply vanishes in the absence of charges, yielding

$$\mathcal H = \frac{1}{2\epsilon_0}\Pi^2+ \frac{1}{2\mu_0}(\nabla \times \mathbf A)^2$$

The last thing is to rewrite $$(\nabla \times \mathbf A)^2$$. This is somewhat tedious in general, but the idea is that if we're looking at a single electromagnetic wave mode in a cavity, then $$\mathbf A$$ will be a plane wave of the form $$\mathbf A_0 e^{i\mathbf k \cdot \mathbf r - \omega t}$$, where $$\omega = ck$$ and $$c$$ is the speed of light. If that's the case, then $$\nabla \times \mathbf A = i\mathbf k \times \mathbf A$$. If we further work in the Coulomb gauge, then $$\nabla \cdot \mathbf A =i \mathbf k\cdot \mathbf A = 0$$; this allows us to write (after some algebra) $$(\nabla \times \mathbf A) = k^2 A^2$$ and so $$\mathcal H = \frac{1}{2\epsilon_0}\Pi^2 + \frac{k^2}{2\mu_0} A^2$$ Looking back, you should identify $$\boldsymbol \Pi = \epsilon_0 \mathbf E$$. Factoring out the $$\epsilon_0$$ and noting that $$\frac{1}{\epsilon_0\mu_0}=c^2$$, we have finally that $$\mathcal H = \epsilon_0 \left(\frac{1}{2}E^2+ \frac{1}{2}\omega^2 A^2\right)$$

• Thank you for the answer. I can't say my Lagrangian/Hamiltonian mech skills are very good, but I'm reading Goldstein's classical mech right now as well, so eventually I'll probably be able to fully appreciate your answer. But regardless, do you have an article I could one day read where they derive the form of the Hamiltonian in terms of the vector potential and electric field, as stated in your answer? – Physics2718 Jan 28 at 7:15
• @Physics2718 I've updated my answer with the derivation, skipping over a bit of algebra. Goldstein (briefly) addresses Hamiltonian field theory in chapter 13, but I don't remember if he derives this particular form. – J. Murray Jan 28 at 12:47