Stat. mech. says: temperature ALWAYS infinite; what goes wrong?

[nonequilibrium]

So let's be clear: the result of the following calculation is non-sensical. What I'm trying to figure out is: what step is making it non-sensical.

I'm working in a microcanonical ensemble with energy E. The corresponding phase volume is $|\Omega| = \int \delta(E- \mathcal H(X)) \mathrm d X$ (H is the Hamiltonian of course, X a microstate) and the entropy is $S = k_B \ln |\Omega|$.

Now, (inverse) temperature is $\beta = \frac{\partial S}{\partial E} = \frac{k_B}{|\Omega|} \cdot \frac{\partial}{\partial E} \int \delta(E- \mathcal H(X)) \mathrm d X$.

Assuming I can bring the derivative inside (we always do in physics...), we get

$\beta \propto \int \delta'(E- \mathcal H(X)) \mathrm d X$

But this is zero!

If this last step is not immediately clear, note that $\int f(x) \delta'(g(x)-a) \mathrm d x = - \int f'(x) \delta(g(x)-a) \mathrm d x$ due to partiel integration (explained in e.g. Griffiths' Introduction to Electrodynamics)

EDIT: an easier way to see the delta-derivative integral is zero: the derivative of an even function (delta function!) is odd.

 

[Whovian]

I can't tell for sure since I don't know much about this branch of physics, but it seems it might be related to the "any warm body emits infinite radiation" problem, which was solved by QM?

 

[nonequilibrium]

No the result itself is wrong, I just need to find out where in the computation it went wrong.

 

[fzero]

The measure $dX$ over the microstates is not independent of the energy. Not only is the term that you write down not zero in general, but there will be an additional term where the energy derivative acts on the measure. You can see how it works for a gas of free particles if you look over the computations at http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_6/node2.html.

 

[nonequilibrium]

I'm pretty sure dX cannot depend on E... It seems like that statement doesn't make a lot of sense. What do you even mean by it?

But can you help me out in this specific example:

Take a 1D harmonic oscillator with a 2D phase space. The Hamiltonian is $\mathcal H(X) = x^2 + p^2$. Hence $|\Omega| = \iint \delta(x^2+p^2-E) \mathrm d x \mathrm d p$. We can calculate it explicitly:

$|\Omega| = 2 \pi \int \delta(r^2-E) \cdot r \cdot \mathrm dr = \pi \int \delta(r^2- E) \mathrm d( r^2) = \pi$

No E-dependence...

 

[fzero]

I'm pretty sure dX cannot depend on E... It seems like that statement doesn't make a lot of sense. What do you even mean by it?

But can you help me out in this specific example:

Take a 1D harmonic oscillator with a 2D phase space. The Hamiltonian is $\mathcal H(X) = x^2 + p^2$. Hence $|\Omega| = \iint \delta(x^2+p^2-E) \mathrm d x \mathrm d p$. We can calculate it explicitly:

$|\Omega| = 2 \pi \int \delta(r^2-E) \cdot r \cdot \mathrm dr = \pi \int \delta(r^2- E) \mathrm d( r^2) = \pi$

No E-dependence...

You are right, in the classical case, the measure is not explicitly a function of the energy. However, we can see the problem, since a single harmonic oscillator is not an ensemble. Consider $N$ of them, subject to the condition

$$\sum_{i=1}^N \left( p_i^2 + x_i^2 \right) = 2 E.$$

Then the phase space is a $2N$-dimensional sphere with volume

$$\Omega = \frac{\pi^N}{N!} \int \delta(R^2-2E) R^{2N-1} dR \sim \frac{(2\pi)^N}{N!}E^{N-1}.$$

Note that the entire concept of temperature relies on the notion of having a large number of microscopic systems in near thermal equilibrium. We cannot define the temperature of a single particle.

 

[nonequilibrium]

Yes, I agree the physics only makes sense for an ensemble, I simply hadn't expected that whether or not it would depend on E (at all) would depend on dimensionality, but I see I was wrong! My example was misleading since $E^{1-1} = 1$. Thank you for clearing that up!

And I also believe I understand what you mean about my original claim that that certain integral was zero. It seems it doesn't need to be: do you agree that if we have an n-dimensional phase space then

$\int \delta'(E- \mathcal H(X)) \mathrm d X = - \int \frac{\partial}{\partial H} \delta(E-H) \cdot H^{(n-1)/2} \cdot \mathrm d H \mathrm d^{n-1} (\cdots)$

where I've done a change in coordinates and one of the new coordinates is H which is equal in value to the Hamiltonian in a certain point (i.e. the new coordinate system labels each energy shell; the other coordinates aren't written down explicitly as they are not important for what follows) Partial integration:

$= \int \delta(E-H) \cdot \frac{n-1}{2} H^{(n-3)/2} \cdot \mathrm d H \mathrm d^{n-1} (\cdots) = \frac{n-1}{2E} \int \delta(E-\mathcal H(X)) \mathrm d X$

Hmmm... wait this should give zero for n = 2 (as we just cleared up)

Also this calculation is probably only valid for special kinds of Hamiltonians? Namely ones where there are only quadratic terms in phase space variables?

 

[fzero]

If the Hamiltonian was not quadratic, the constant energy surface will not be a sphere. For a more complicated Hamiltonian, it's might be easier to compute the phase space volume $V(E)$ for all states satisfying $\mathcal{H}\leq E$. Then you can compute the volume of the piece with energies between $E$ and $E+dE$ from $\Omega(E) = dV(E)/dE$. This bypasses whatever funny dependence the argument of the delta function has, but the total volume integral might not be much easier to do.

 

[nonequilibrium]

But say I'm not specifically interested in calculating the volume, but in doing a calculation as shown in my previous post. Are there then constraints on my Hamiltonian? Note that that calculation is equivalent with $\beta = \frac{n-1}{2E}$ with n the dimension of phase space (although I feel I must've made a minor mistake somewhere since it doesn't give the right answer for my harmonic oscillator example)

Thank you for your time by the way.

 

[fzero]

But say I'm not specifically interested in calculating the volume, but in doing a calculation as shown in my previous post.

Are there then constraints on my Hamiltonian? Note that that calculation is equivalent with $\beta = \frac{n-1}{2E}$ with n the dimension of phase space (although I feel I must've made a minor mistake somewhere since it doesn't give the right answer for my harmonic oscillator example).

No, there aren't additional constraints on the Hamiltonian. Let me first say that in the microcanonical ensemble the temperature is always given in terms of the amount of energy per particle, proportional to $E/N$ in the large $N$ limit. This is just because the energy is the input and there is no external system available to exchange energy with, which would have allowed us to define a temperature thermodynamically.

Now, as you show, it is possible to derive this result by choosing a new coordinate $H$, equal to the classical Hamiltonian. The Jacobian of the transformation is a function of $H$. In the large $n$ (your phase space dimension) limit, on general grounds, we expect that the Jacobian reduces to some monomial $H^{\alpha n}$, for some constant $\alpha$ that depends on the precise Hamiltonian. I think the way to argue the scaling with $n$ is just to note that the vector $\partial H/\partial p_i$ is rank $n$, while $\partial H/\partial x_i$ is rank $n$ (or zero for a free particle). So each phase space coordinate contributes a factor to the leading term.

As we saw above, for the harmonic oscillator, we found $H^{n-1}$, where it wasn't necessary to use the large $n$ limit at that point in the computation. For a free particle, we'll find $H^{(n-1)/2}$, which is the same factor that you used. For the reasons just given, I don't think that this is a general result, which should explain your suspicion of a mistake. The mistake was in assuming a particular precise form of the Jacobian.