Compressibility of radiation
We may give one other example of the kinetic theory of a gas, one which is not used in chemistry so much, but is used in astronomy. We have a large number of photons in a box in which the temperature is very high. (The box is, of course, the gas in a very hot star. The sun is not hot enough; there are still too many atoms, but at still higher temperatures in certain very hot stars, we may neglect the atoms and suppose that the only objects that we have in the box are photons.) Now then, a photon has a certain momentum $\FLPp$. (We always find that we are in terrible trouble when we do kinetic theory: $p$ is the pressure, but $p$ is the momentum; $v$ is the volume, but $v$ is the velocity; $T$ is the temperature, but $T$ is the kinetic energy or the time or the torque; one must keep one’s wits about one!) This $\FLPp$ is momentum, it is a vector. Going through the same analysis as before, it is the $x$-component of the vector $\FLPp$ which generates the “kick,” and twice the $x$-component of the vector $\FLPp$ is the momentum which is given in the kick. Thus $2p_x$ replaces $2mv_x$, and in evaluating the number of collisions, $v_x$ is still $v_x$, so when we get all the way through, we find that the pressure in Eq. (39.4) is, instead, \begin{equation} \label{Eq:I:39:15} P = 2np_xv_x. \end{equation} Then, in the averaging, it becomes $n$ times the average of $p_xv_x$ (the same factor of $2$) and, finally, putting in the other two directions, we find \begin{equation} \label{Eq:I:39:16} PV = N\avg{\FLPp\cdot\FLPv}/3. \end{equation} This checks with the formula (39.9), because the momentum is $m\FLPv$; it is a little more general, that is all. The pressure times the volume is the total number of atoms times $\tfrac{1}{3}(\FLPp\cdot\FLPv)$, averaged. Now, for photons, what is $\FLPp\cdot\FLPv$? The momentum and the velocity are in the same direction, and the velocity is the speed of light, so this is the momentum of each of the objects, times the speed of light. The momentum times the speed of light of every photon is its energy: $E = pc$, so these terms are the energies of each of the photons, and we should, of course, take an average energy, times the number of photons. So we have $\tfrac{1}{3}$ of the energy inside the gas: \begin{equation} \label{Eq:I:39:17} PV = U/3\text{ (photon gas)}. \end{equation} For photons, then, since we have $\tfrac{1}{3}$ in front, $(\gamma - 1)$ in (39.11) is $\tfrac{1}{3}$, or $\gamma = \tfrac{4}{3}$, and we have discovered that radiation in a box obeys the law \begin{equation} \label{Eq:I:39:18} PV^{4/3} = C. \end{equation} So we know the compressibility of radiation! That is what is used in an analysis of the contribution of radiation pressure in a star, that is how we calculate it, and how it changes when we compress it. What wonderful things are already within our power!