Partition Function and BlackBody Radiation

Question

I'll start with a few definitions: $$\beta \equiv \frac{1}{k_bT}$$

Where T is the temperature of a system.

And the partition function:

$$Z \equiv \sum_{j}e^{-\beta \epsilon_j}=\int D(\epsilon)d\epsilon e^{-\beta\epsilon}$$

Where s is an index over all distinguishable microstates of the system (in the case of discrete states) and $\epsilon$ is the energy of each of the microstates. In the continuous case, $D(\epsilon)$ is the density of states.

So, if we consider the partition function of a photon gas of temperature $T$ where the energy of a photon w/ frequency $\omega$ is $\epsilon = \hbar\omega$, the total energy of an assemblage of photons is given by:

$$E = \sum_i s_i \hbar\omega_i$$

Where i is the index over all possible frequencies $\omega_i$ and $s_i$ is the number of photons with frequency $\omega_i$.

We can then see that the partition function for a photon gas can be expressed as:

$$Z = \sum_{k}e^{-\beta E_k} = \sum_{s_1,s_2,...=0}^{\infty} e^{-\beta\hbar(s_1\omega_1 + s_2\omega_2 +...)} = \sum_{s_1,s_2,...=0}^{\infty} e^{-\beta\hbar s_1\omega_1}e^{ -\beta \hbar s_2\omega_2}...$$

The second equivalence is established by noting that each state $\Psi_k$ is uniquely determined by the number of photons in each mode $M_i$ (each of which has a corresponding frequency $\omega_i$).

$$Z = \sum_{s_1=0}^{\infty}\sum_{s_2=0}^{\infty}... (e^{-\beta\hbar s_1\omega_1}e^{ -\beta \hbar s_2\omega_2}...) = \sum_{s_1=0}^{\infty}e^{-\beta\hbar s_1 \omega_1}\sum_{s_2=0}^{\infty}\sum_{s_3=0}^{\infty}... (e^{-\beta\hbar s_2\omega_2}e^{ -\beta \hbar s_3\omega_3}...)$$

By pulling out the $\omega_1$ term. We can continue this line to arrive at:

$$Z = (\sum_{s_1=0}^{\infty}e^{-\beta\hbar s_1\omega_1})*(\sum_{s_2=0}^{\infty}e^{-\beta\hbar s_2\omega_2})*... = \prod_{i=1}Z_{\omega_i}$$

Where $Z_{\omega}$ is the partition function $\sum_{s=0}^{\infty} e^{-\beta\hbar s \omega}$ for a system consisting of photons only of frequency $\omega$.

However, the partition function of thermal blackbody radiation ends up being very different. For some reason, it is given by:

$$Z = \sum_i Z_{\omega_i} = \int D(\omega) d\omega Z_{\omega}$$

Where $Z_{\omega}$ is the same as defined above. This seems to be a contradiction. So I suppose my question is: Which of these solutions is correct?

Answer 1

In some cases we can approximate the sum by an integral appropriate:

$$\sum_j e^{-\beta\epsilon_j} =\int e^{-\beta E}D(\epsilon)d\epsilon=\int e^{-\beta E(\omega)}D(\omega)d\omega$$

In fact that is how we determine the density function. And I think there is no reason we can write $\sum_i Z_{\omega_i}=\int Z_\omega D(\omega)d\omega$ with the same $D(\omega)$ from the egality above.

And also there is no reason to believe that $Z=\sum_i Z_{\omega_i}$, in fact $Z=\prod_i Z_{\omega_i}$ like you have demonstrated. So that $\ln Z=\sum_i \ln Z_{\omega_i}$, therefore:

$$\ln Z=\int (\ln Z_{\omega_i}) D(\omega)d\omega $$

In fact $D(\omega)d\omega$ counts the number of states having frequency between $\omega$ and $\omega+d\omega$, and similar for $D(\epsilon)d\epsilon$. You can imagine that $$Z=\prod_{\omega_i}Z_{\omega_i}^{D(\omega)d\omega}$$

which mean we have approximate all the frequency in $[\omega,\omega+d\omega]$ by $\omega$.

Answer 2

Your answer works for 1D case. In 1D k-space, any "sn", or equivalently "frequency" has only one corresponding standing wave mode (one dot in the 1D K-space). Another answer works for 3D case, where the K-space is 3D, so more than one K-modes in the 3D K-space can have the same frequency. Therefore, when calculate Z function, we have to consider all those modes. In the second equation, the $D(\omega) \text{d}\omega$ is the number of modes in the K-space which share the same frequency of $\omega$. It is obvious that those modes populate inside a spherical shell with a radius of $k = \omega/C$ in the K-space.