# Trying to understand my prof's derivation of Planck's Law

• I

## Main Question or Discussion Point

Not many people understood his proof in class, and the textbook's proof wasn't very clear so we went by with other derivations online. Then he filled half the midterm with his method, so I'm trying to understand how he did things.

Looking back it seems very similar to the proofs we found online but different in a lot of places.

The goal is to come to the spectral energy density form of
$$u(f)df=\frac{8\pi h f^3}{c^3}\frac{1}{e^\frac{hf}{k_B T}-1}df$$
The closest proof to his method was this: qsp_chapter10-plank.pdf which divides the law into two parts: the number volume density $\frac{8\pi f^2 df}{c^3}$ and the average energy per state $\frac{hf}{e^\frac{hf}{k_B T}-1}$. I'll refer to this proof as the PDF proof.

The professor divided Planck's Law into three parts instead: number volume density, energy per state $hf$ and state occupation $\frac{1}{e^\frac{hf}{k_B T}-1}$.

Then he derived the number volume density. The PDF proof gives that since $n$ in all directions is positive we're only concerned with only the positive octant of the reciprocal space, resulting in a factor of 1/8.

The professor does not have that factor. Instead, he starts the proof with $k=\frac{n\pi}{L}$ from the wave in a box scenario, then all of a sudden switches to $k=\frac{2\pi n}{L}$. The form then looks like:
$$N = 2\big( \frac{L}{2\pi}\big)^3\int_0^{4\pi}\int_0^{k_{max}}k^2 dk d\Omega$$
where $\frac{L}{2\pi}$ is, in his words, the density per dimension and the integral is the volume of the k-space. He specifically put them down separately, raising the power of $\frac{L}{2\pi}$ for each dimension instead of going through (or rather, skipping steps for) the volume integral like the PDF proof.

So from all of this I can't understand:
• Does state occupation refer to the average state that the photon occupies? How does this state occupation translate from a photon in a box to unbound free space?
• How did he go from $k=\frac{n\pi}{L}$ to $k=\frac{2\pi n}{L}$ and eliminate the 1/8 factor?
• He says $\frac{L}{2\pi}$ is density per dimension, and that it's cubed because there's three dimensions. Am I right in interpreting this as the maximum allowed length per state? Or is this a volume of some sort instead of density?
• He separated physical dimensions $\frac{L}{2\pi}$ from the volume integral. Then he manipulated the integral in another problem as if it referred to physical space instead of reciprocal space. Taken by itself, what does the k-space integral $\int_0^{4\pi}\int_0^{k_{max}}k^2 dk d\Omega$ mean?

Related Classical Physics News on Phys.org
Homework Helper
Gold Member
I like to use periodic boundary that Reif introduces in his Statistical and Thermal Physics textbook: $e^{i k_x x }=e^{i k_x(x+L_x)}$ to get $k_x L_x=n_x (2 \pi)$, and similarly for $y$ and $z$. $\\$ This makes $\Delta n=(\Delta n_x)(\Delta n_y)(\Delta n_z)=\frac{V \, \Delta^3 k }{(2 \pi)^3}=\frac{V \, \Delta^3 p}{h^3}$. $\\$ Since you are on the subject of the Planck blackbody function, you might find this post of interest, since, in practice, the more important use of the Planck function is to determine the power radiated per unit area from a blackbody surface or from the aperture of a blackbody cavity. https://www.physicsforums.com/threads/radiance-and-energy-density-of-a-black-body.956343/#post-6063569

Last edited:
I like to use periodic boundary that Reif introduces in his Statistical and Thermal Physics textbook: $e^{i k_x x }=e^{i k_x(x+L_x)}$ to get $k_x L_x=n_x (2 \pi)$, and similarly for $y$ and $z$. $\\$ This makes $\Delta n=(\Delta n_x)(\Delta n_y)(\Delta n_z)=\frac{V \, \Delta^3 k }{(2 \pi)^3}=\frac{V \, \Delta^3 p}{h^3}$. $\\$ Since you are on the subject of the Planck blackbody function, you might find this post of interest, since, in practice, the more important use of the Planck function is to determine the power radiated per unit area from a blackbody surface or from the aperture of a blackbody cavity. https://www.physicsforums.com/threads/radiance-and-energy-density-of-a-black-body.956343/#post-6063569
Thanks! I took a look at Reif, and he specifically explains the differences between his approach and mine. It's making much more sense.

The prof. and Reif both do not take into account reflections at the boundaries, which restricts allowed values of $x$ to multiples of the whole wavelength and not half-wavelength. In exchange, this also allows for negative $k$-values which I think corresponds to inverted phase of the wave equation.

On the other hand, by making $k=\frac{nx}{L}$ I'm accounting for reflections and that reflection will invert phase, therefore only allowing positive $k$-values and introducing the one-eighth factor. As we know both approaches give the identical result.

Reif justified ignoring reflection by saying since $\lambda$ is on the scale of 1Å and $\lambda \gg L$, the boundary effects are negligible. I don't quite like this justification since $L$ can arbitrarily be made so this condition is not met, but nonetheless it works.

As for the rest of the equation (deviating from Reif's notation),
$$N=d^3 n = dn_x dn_y dn_z = \rho (k) d^3 k$$
where $\rho(k)$ is the density of states (number density per $k$-state) and $d^3 k$ is the allowed volume element of $k$-space.
Using the relationship $k=\frac{n\pi}{L}$ we find that the density of states is:
$$\rho(k)=\big( \frac{L}{\pi}\big) ^3 = \frac{V}{\pi^3}$$ for $k=\frac{n\pi}{L}$, and the volume element is:
$$d^3 k = \frac{1}{8}\int_0^{4\pi} d\Omega k^2 dk$$
The product of these two counts the actual number of states (or particles).

Combining the occupation factor, energy per photon and the polarization factor of 2, Planck's law for spectral energy density can be expressed as:
$$u(k)dk=2\frac{1}{V}\rho(k) \bar{E}\phi d^3 k$$
where $\bar{E}=\frac{h}{\nu}=\frac{hc}{\lambda}$ is the photon energy and $\phi = \frac{1}{e^{\frac{hf}{k_B T}}-1} = \frac{1}{e^{\frac{hc}{k_B \lambda T}}-1}$ is the occupation factor.

As for going from spectral energy density to spectral radiance, finding the c/4 factor was a homework problem itself. Your post does it in physical dimensions, but having this division of Planck's law in $k$-space makes the problem much easier to solve (this bit was what I was referring to when I said prof. was treating reciprocal space like physical space).

Between the spectral energy density and spectral radiance forms of Planck's law:
We know that the number of photons in a cavity of unit volume 1 (spectral energy density) is
$$N_u=\rho (k) d^3 k = \frac{V}{\pi ^3}\frac{1}{8}\int_0^{4\pi} d\Omega k^2 dk \\ N_u=\frac{1}{8\pi^3}k^2 dk\int_0^{2\pi}\int_0^\pi \sin(\theta) d\theta d\phi$$
For spectral radiance, we consider the upper hemisphere of the $k$-space sphere with the aperture at its apex. I think of this like a projection of outbound rays onto the ray from the center directly through the apex. The cavity can be approximated as a prism or cylinder with volume 1, top area $A$ and length $c∆t$:
$$N_B=\rho (k)d^3 k = \frac{V}{\pi ^3}\frac{1}{8}\int_0^{4\pi} Ac∆t \cos (\theta) d\Omega k^2 dk \\ N_B=Ac∆t\frac{1}{\pi ^3}\frac{1}{8}k^2 dk\int_0^{2\pi}\int_0^{\frac{\pi}{2}} \cos (\theta) \sin (\theta) d\theta d\phi$$
where the limits for the polar angle $\theta$ is from $0$ to $\frac{\pi}{2}$ because it's the upper hemisphere, and the cosine is the "projection."
Calculating the spectral radiance integral gives $N_B=\frac{Ac∆t}{4} N_u$, and plugging this into the above form for Planck's law you can find the c/4 factor.

Last edited: