Statistical Mechanics in an even potential

I dug this up from a Midterm examination I had on Statistical Mechanics a few years ago.

Suppose we have a system of particles confined to a potential given by

$U(x) = \alpha x^{2n}$

for integer n and $\alpha$ a real valued constant of appropriate units. If the system contains N particles at temperature T, we would like to calculate the heat capacity and the entropy.

Let's start by calculating the partition function. By definition,

$Z = \Sigma e^{-\beta E}$

Where $\beta = \frac{1}{k_b T}$ as usual, and we are summing over all energy states E. Since our potential function is continuous, the energy states take the form:

$Z = \int e^{-\beta ax^{2n} dx$

Now let $x = u (\beta a)^{\frac{-1}{2n}}$. Then $dx = du (\beta a)^{\frac{-1}{2n}$ and $x^{2n} = u^{2n} (\beta a)}$. This gives us

$Z = \int e^{-u^{2n}}(\beta a)^{\frac{-1}{2n}} du$

This integral looks pretty hard, particularly because we don't know anything specific about the value of n. If n is one, we can come up with an answer. For anything else, we're in some trouble. But what we do know is that, whatever the answer is, it is some number. So let's just call that number $\lambda$ and not worry about it. Thus

$Z = \lambda (\beta a)^{\frac{-1}{2n}}$

The heat capacity is given by

$C_v = \frac{1}{k_b T^2}\frac{\partial^2 \ln Z}{\partial \beta^2}$

This is very convenient, because the factors a and $\lambda$ will drop out in this step:

$C_v = \frac{1}{k_b T^2}\frac{\partial^2}{\partial \beta^2} \big( \ln \lambda + \frac{-1}{2n}\ln a + \frac{-1}{2n}\ln \beta \big)$

$C_v = \frac{1}{k_b T^2}\frac{1}{2n}\frac{1}{\beta^2} = \frac{k_b}{2n}$

The entropy is

$S = \frac{\partial}{\partial T}(k_b T \ln Z)$

$S = \frac{k_b}{2n}\big(\ln\beta + 1 \big)$

I skipped a few steps on the last line, but it's just calculating some derivatives. The limiting case here is $n = 1$, which is just a harmonic oscillator. You can see that as the potential gets more anharmonic, we decrease the entropy and the heat capacity. This should not be that surprising--these anharmonic oscillators require far more energy for a displacement, and hence the number of available states is confined for increasing n.

Wormhole Geometry

I dug this one up from a General Relativity class I took a few years back. I'm a little rusty on the subject, so I can't guarantee that this will be that sensible to follow. Problem courtesy of Werner Israel.

A "wormhole" is a staple of science fiction writing, but it actually describes a particular structure in general relativity that has, if not real world applications, at least academic ones. The wormhole is described by the following metric:

$ds^2 = dr^2 + (r^2 + a^2)d\Omega^2 -dt^2$

Where s a line element in this space, r is a radial coordinate, $\Omega$ is an element representing both $\theta \& \phi$, and t is time. a is a constant. Note that a normal spherical coordinate metric has $a = 0$, of course. What does this thing look like? Well, at any r, we sweep out a sphere of radius $\sqrt{r^2 + a^2}$ in constant time. Or, if you prefer, at constant t and $\theta$, it sweeps out circles of radius $\sqrt{r^2 + a^2}$. I've drawn a picture of what that looks like below

Don't worry too much about what the axis mean. What I'm really drawing is the shape of the coordinate system that I've described, for any $\theta$. Most importantly, you can see that the coordinate system allows for non-trivial solutions at $r = 0$. By comparison, the spherical coordinate system, drawn in this manner, is a cone, and vanishes exactly at $r = 0$. This is why the former is describes as a wormhole: it has no singular points, and thus may in principle allow passage from negative to positive values of r.

We can calculate the energy density and radial pressure for such an object to exist. The energy density required to maintain the wormhole is given by the zero component of the stress-energy tensor, and radial pressure by the first component:

$\rho = T_0^0$

$P_r = T_1^1$

We would like to convert this to a spherically symmetric metric, because metrics of the following form can be computed relatively easily.

$ds^2 = e^\alpha(r)dr^2 + r^2 d\Omega^2 - e^\gamma(r)dt^2$

To do so, let

$r^2 = u^2 - a^2$

$dr^2 = \frac{u^2 du^2}{u^2-a^2}$

So that

$ds^2 =\frac{u^2 du^2}{u^2-a^2} + u^2 d\Omega^2 - dt^2$

Then the functions

$e^\alpha = \frac{u^2 du^2}{u^2-a^2}$

$e^\gamma = 1$

The zero component is then given by

$8\pi u^2 T_0^0 = \frac{d}{du}\big(u e^{-\alpha} - u\big)$

This equation can be derived by calculating $R_{00}$ of the curvature tensor. I won't show that here, as it is a bit of work. It is fairly standard. This simplifies very nicely.

$8\pi u^2 T_0^0 = \frac{d}{du}\big(\frac{-a^2}{u}\big)$

Thus

$T_0^0 = \rho = -\frac{a^2}{8\pi u^4}$

The energy density is negative. Objects with negative energy density might be considered gravitationally repulsive--or have a negative mass. That is, our wormhole geometry would only exist in a dark energy or inflationary regime.

Likewise, we can calculate the radial pressure from a similar expression.

$8\pi u^2 T_1^1 = e^{-\alpha}(1 + u\gamma^\prime) - 1$

$\Gamma = 0$ for our case, so we end up with an identical expression for the pressure:

$P_r = \frac{-a^2}{8\pi u^4}$

I think it is basically coincidental that these two terms work out to be equal. They aren't actually equal in magnitude, of course, since I'm using units in which $c = G = 1$. The pressure is negative, which again, suggests that this object is inflationary--namely that there is an outward pressure that will drive objects away from the centre of the wormhole.

Now, as to the question of whether wormholes exist, well, I can't really say. We live in an expanding universe, and such a universe would have a negative pressure driving it. But beyond that, it's hard to say much. We haven't found the existence of any localized objects that display these sorts of properties. It's also unclear that you would ever be able to "pass through" the wormhole in any meaningful sense--as you approach $r=0$, the outward pressure tends to infinity, so anything approaching the wormhole would be blown out of it. As such, wormholes in the science fiction sense will probably mostly remain as just that--fiction--but maybe, possibly, somewhere in the universe, there might be an object with similar properties to this.