2.16: S20 Equilibrium Statistical Mechanics SOLUTIONS
Exercise 20.1.1
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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 20.1.2
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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 20.1.3
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I have not produced a written solution for this yet.
Exercise 20.2.1
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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 20.2.2
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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 20.2.3
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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.