[MUSIC] They're familiar with the idea that hot objects emit radiation. For example, a light bulb. Now, in the hot wire filament, and electron originally in an excited state, drops down to a lower energy state. And the energy difference is simply given out as a photon. We're also familiar with the absorption of radiation by surfaces. For example, clothes in the summer absorbs photons from the sun and heat up. Black clothes, absorb more radiation than lighter ones. This means, of course, that light colored clothes reflect a large fraction of the light falling on them. Let's try to analyze the statistical thermodynamics of photons. Now, let's consider a box of photons. The very first observation is that the number of photons in the gas is not fixed. That is, the photons are absorbed and re-emitted by the enclosure walls. Now, what about the interacting nature? What do you think, are the photons interacting, or non-interacting? The photons of course, non-interacting. It is by absorption and re-emission that equilibrium is maintained in the photon gas. Now, let's consider a closed isothermal ensemble that is the canonical ensemble. What is the thermodynamic potential, associated with this ensemble? Well, this is of course the Helmholtz free energy, F. Now for the photon gas, the Helmholtz free energy is constant in equilibrium that is at constant temperature in volume. However, n, the number of photon changes. From this analysis, it immediately follows that the change in the Helmholtz free energy as you change the number of particles is zero. Simply put the chemical potential is zero for a photon gas. Now, what kind of a particle is a photon? Is it a fermion or is it a boson? It turns out that the photon gas is a boson with the characteristic chemical potential mu equals zero. We can write down the canonical partition function in the following way. Now from this, we can find the average occupation number using the expression. The energy of the photon is simply given as the planks constant h, times the frequency new. The distribution of energy density as a function of the frequency mu in a photon gas, holds a special place in the development of quantum mechanics. The energy density can be written as a product of three parts, the energy of a photon of frequency multiplied by the average number of photons of that energy, multiplied by the density of states g of. How do we evaluate the density of states g of? We can evaluate this analogous to the discussions of the model for a phonon. This yields that the density of states scales as the square of the frequency. From this analysis, we get the famous Planck's radiation law. We can also recover Rayleigh-Jeans laws as limit of Planck's law. Now, let's consider the case for long wavelengths. This leads to the fact that the energy density as a function of wavelength scales as the inverse 4th power. This is, of course, known as the Wien's law that's valid at long wavelengths. Now, let's consider short wavelengths. That is, h c over lambda far, far greater than one. This now yields a relation that the energy density scales as the inverse fifth power of wavelength, weighted by an exponential whose exponent depends inversely on the wavelength. This is the law valid at short wavelengths. We can readily obtain from Planck's radiation law, the Stefan-Boltzmann law. We simply need to integrate the energy density over the entire frequency range. Now, using available substitution of x, given as beta h u was the integral. Now, from this, we get the famous Stefan Boldsman law, that is T to the power of 4. But more importantly, we've now found a precise evaluation of the Stefan's constant, sigma. This equation tells us that a precise measurement of the Stefan's constant could then be used for example, to determine Planck's constant. In fact, Plank actually determined h as the constant needed, in order for his radiation law to fit the observed spectral distribution law. This gave him a value h of 6.55 times to the minus 27 [INAUDIBLE]. Which compares very well with the present precise value of 6.625 times 10 to the -27. Having derived all of these expressions, we can evaluate all the thermodynamic quantities for black body radiation through the partition function, Zed. Now the Helmholtz free energy is simply given as minus kBT logarithm of the partition function. Now, Z here is the partition function of the photon gas. Remember, that the logarithm of a product can be written as the summation of the logarithms. This allows us to write the partition function as the sum over the logarithm of the individual bosons tying the occupation factors. We can carry art a usual conversion of the summation to an integral, note that in this expression, epsilon s is the energy of the single photon state. And the factor of 2 arises from the two polarizations available to each photon. Now, this can be integrated in a variety of ways. Perhaps the most direct way to integrate is over the phase space, and write down the energy as the momentum times the speed of light. Introducing the dimensionless variable x given as beta p c, the Helmholtz free energy can be written as. The dimensionless integral can be evaluated to be pi over 15. This gives that the Helmholtz free energy scales as the fourth power of temperature. From the Helmholtz free energy, we can evaluate the entropy as the derivative of the Helmholtz free energy with respect to temperature holding the volume constant. This shows that the entropy scales as the third power of the temperature. We know the entropy and the Helmholtz field energy. From this, the internal energy can be evaluated very easily. But what about the pressure of the photon gas? While going back to our definitions, pressure is simply defined as negative of the derivative of the hand hold free energy, with respect to volume. This shows that the pressure scales as the 4th power of the temperature. Finally, what about the Gibbs free energy? We can use thermodynamic relations to find out that this is zero. Now, there's a more intuitive way in which we could have argued that the Gibbs free energy for our photon gas is zero. This is because the Gibbs free energy G, is merely given as the chemical potential times the number of particles, and as we had shown, the chemical potential for a photon gas is zero, and hence the Gibbs free energy is zero. To summarize, we've obtained the spectral distribution from the Bose occupation in much the same way as we obtained the Maxwell distribution for a classical gas. The only other required ingredient was the density of states. And this allowed us to determine all the thermodynamic quantities associated with the photon gas. And from this, we derived the famous Planck's radiation law.
