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# S19 Equilibrium Statistical Mechanics SOLUTIONS

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Exercise 19.1.1: From the definition of $$f$$, derive the above expression for the number density $$n({\vec x})$$.

Solution: Since $$f/h^3$$ is the density of particles in phase space, the number in some small volume V, small enough such that $$f$$ does not change much throughout the volume, is given by N = V\int d^3p f/h^3. So the number density is $$n = \int d^3p f/h^3.$$ If every internal degree of freedom has the same value of $$f$$ and if we want to count all the particles, regardless of internal state, then we have $$n = g \int d^3p f/h^3.$$

Exercise 19.1.2: From the definition of $$f$$, derive the above expression for the energy density $$\epsilon({\vec x})$$.

Solution: Same as above except instead of calculating the total number in some small volume, we want the total energy. Therefore we just insert E(p) into the integral, to add up the energy from each region of momentum space, instead of just the number of particles from each region of momentum space.

Exercise 19.1.3:  (Optional) From the definition of $$f$$, derive the above expression for the pressure $$P({\vec x})$$.  This one is significantly harder. You need to recall that pressure is force per unit area. The force on a wall from particles hitting it in time interval $$\Delta t$$ is equal to the sum of the changes in each particle's momentum as it bounces off the wall, divided by $$\Delta t$$. First show that for a wall perpendicular to the $$x$$ axis this force is given by $$F =\frac{1}{2} \int d^3p f A (v_x \Delta t) (2p_x)/\Delta t$$ so that $$P = \int d^3p f v_x p_x$$. Then use the fact that $$v_x = p_xc^2/E$$ (see footnote2) and that $$\int d^3p f p_x^2 = 1/3\int d^3p f p^2$$ as long as $$f$$ only depends on $$p^2 = p_x^2+p_y^2 + p_z^2$$ to get $$P = \int d^3p f p^2c^2/(3E)$$. If there are $$g$$ internal degrees of freedom, that's that many more particles doing exactly the same thing ($$f$$ tells us the distribution for each internal degree of freedom) so we get the desired result including the factor of $$g$$.

Solution: I have not produced a written solution for this yet.

A gas of photons ($$g=2$$) in kinetic equilibrium with $$\mu = 0$$ has a contribution to its number density from particles with magntiude of momentum between $$p$$ and $$p+dp$$ equal to

$n(p)dp = \frac{8\pi }{h^3} \frac{p^2dp}{\exp(pc/(k_B T)) -1}$.

Note that by $$n(p)$$ we mean the function of the magnitude of momentum that when integrated over $$p$$ gives number density $$n = \int_0^\infty dp n(p)$$.

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Exercise 19.2.1: Derive the above equation for $$n(p) dp$$. Start from $$n = \frac{g}{h^3}\int d^3p f$$. Because the energy of each particle (and therefore the whole integrand) only depends on the magnitude of the momentum, $$p = \sqrt{p_x^2+p_y^2+p_z^2}$$, switch from Cartesian to spherical coordinates and integrate over the angular variables. That is, replace $$d^3p = dp_xdp_ydp_z$$ with $$p^2dp d(\cos\theta_p)d\phi_p$$ and integrate over the angular variables $$\theta_p$$ and $$\phi_p$$. Remember that $$E(p) = pc$$ for photons.

Solution: The integral over angular variables results in $$n = \frac{g}{h^3} \int_0^\infty 4\pi p^2 dp f$$. Now the integral is adding up numbers of particles from momentum space shells of momentum space volume $$4\pi p^2 dp$$. We need only substitute in the appropriate expression for $$f$$, set $$E=pc, \mu = 0$$ and $$g=2$$ to get the answer.

Exercise 19.2.2: If you sampled one photon out of the distribution, there is a probability that it will have energy between $$p$$ and $$p+dp$$. For what value of $$p$$ does this probability peak?

Solution: To find the peak of the distribution set $$dn(p)/dp = 0$$ and solve for $$p$$. One ends up with a transcendental equation to solve. Setting $$x = pc/kT$$ makes it possible to write it down fairly compactly as $$(2-x)e^x = 2$$. One can narrow in on the solution numerically with a calculator -- especially a graphing one if you just plot the left-hand side and choose the value of $$x$$ that gives 2. I used a calculator and in a few tries had $$x=1.6$$ which is pretty close as $$(2-1.6)e^{1.6} = 1.98.$$ So the most probable $$p$$ is $$p = 1.6 k_B T/c$$.

Exercise 19.2.3: How does that most probable $$p$$ depend on temperature?  Notice that this is qualitatively consistent with what you expect for temperature.

Solution: The most probable $$p$$ corresponds to an energy that is $$pc = 1.6 k_B T$$. This corresponds to what we expect since we see that $$\simeq k_B T$$ is a typical particle kinetic energy.