# Bose-Einstein Stats and Planck Formula

1. Nov 7, 2013

### modulus

I worked out the Planck Black-Body Radiation Formula using Bose-Einstein Statistics, but I feel there is something conceptual I am missing here.

When Planck derived the formula, he started out with the Boltzmann distribution function, and assumed that there were discrete energy levels, instead of a continuous spread. That's it.

But Bose-Einstein statistics assumes that the particles which fill these energy levels (in this case, photons), are non-distinguishable. Yet when we proceed with that assumption, we end up with the Planck formula (only the density of states expression, which when multiplied by hv, gives the final expression).

So is making energy levels in the Boltzmann Distribution discrete somehow equivalent to assuming non-distinguishable particles? What am I missing here???

2. Nov 7, 2013

### vanhees71

It's not true that Planck started from the Boltzmann statistics but that he invented Bose statistics in a somewhat hidden way.

The modern derivation of Bose statistics of an ideal gas is much simpler. For simplicity let's assume uncharged free Klein-Gordon particles. Consider a finite cubic box of length $L$ and assume periodic boundary conditions. Then the occupation-number basis with respect to the single-particle momentum eigenstates, $|\{N(\vec{p}) \}_{\vec{p}} \rangle$ with $N(\vec{p}) \in \mathbb{N}_0.$. The momenta run over the discrete set $\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3$.

Then the canonical partion sum is given by
$$Z=\mathrm{Tr} \exp(-\beta \hat{H}).$$
Now
$$\hat{H}=\sum_{\vec{p}} E_{\vec{p}} \hat{N}(\vec{p}).$$
Thus the partition sum can be easily evaluated in the occupation-number basis
$$Z=\prod_{\vec{p}} \sum_{N(\vec{p})=0}^{\infty} \exp[-\beta E(\vec{p}) N(\vec{p})].$$
The geometric series is easily summed to
$$Z=\prod_{\vec{p}} \frac{1}{1-\exp[-\beta E(\vec{p})]}.$$
The total energy is given by
$$\mathcal{E}=-\frac{\partial}{\partial \beta} \ln Z=\sum_{\vec{p}} \frac{E(\vec{p})}{\exp[\beta E(\vec{p})]-1}.$$
Finally, in the large volume limit, you can approximate the sum by an integral, using
$$\mathrm{d} \rho=V \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3}.$$
This gives
$$\mathcal{E}=V \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} E(\vec{p}) f_{\mathrm{B}}(\vec{p})$$
with the Bose-distribution function
$$f_{\mathrm{B}}(\vec{p})=\frac{1}{\exp[\beta E(\vec{p})]-1}.$$

3. Nov 7, 2013

### DrDu

Planck was not taking about particles but that the energy of vibrations comes in discrete quanta.
These quanta were already perceived as indistinguishable by Planck.

4. Nov 7, 2013

### Jano L.

Yes, these two approaches lead to the same result. One can either use the Boltzmann distribution for discrete states of distinguishable material oscillators, or reinterpret the result as Bose-Einstein distribution of non-distinguishable elements of energy over these oscillators. Mathematically they lead to the same spectrum. For Planck, the material oscillators were more real than their energy elements - the latter were just a way to do probability calculations. For Einstein and Bose, the elements of energy were more real, especially if the energy element was that of electromagnetic field - quantum of light.

I do not think so. They are two independent things, and both are needed in both approaches. The indistinguishability of energy elements is an assumption about how to calculate entropy of energy distribution, and could be in principle made also for system with continuous energy. Only it is hard to define and calculate entropy for such distributions, so Boltzmann and Planck used discrete approach, which is much easier.

5. Nov 8, 2013

### DrDu

As far as I understand, Planck was basically quantizing harmonic oscillators which he used as a model for the matter interacting with light and using these as a probe to determine the equilibrium distribution function of light. Quantizing the harmonic oscillator gives rise to indistinguishable bosonic quasi-particles, namely the phonons. So my answer to the OP's question would be rather affirmative.

6. Jan 14, 2016

### dhelicon

The BE distribution factor IS the Planck distribution, utilized for the same physical system. That is an inescapable fact. Whether you integrate for energy or count probabilities does not change the physical system or this distribution.

7. Jan 14, 2016

### mpresic

The best reference I have ever seen for a complete treatment is in "From c-numbers to Q-numbers" by Darrigol. Warning this is a real complete treatment that is really involved and time-consuming.