I want to compute the average kinetic energy of a particle at a certain temperature T given by the Hamiltonian:
$$
H = \sum_{i=1}^{N}\frac{\mathbf{p}_i^{2}}{2m}+V(\mathbf{r}_{1},…\mathbf{r}_{N}),
$$
and show that the average kinetic energy is:
$$
\langle K_{i}\rangle=\frac {3}{2}k_{b}T
$$
To start with I would compute the single particle partition function for a certain configuration of all the other particles.
$$
Q_{1}=\int exp\left ( -\frac{\mathbf{p}_{i}^{2}/2m+V(\mathbf{r}_{1},…\mathbf{r}_{N})}{k_{b}T} \right )d^3pd^{3N}q=\int exp\left ( -\frac{\mathbf{p}_{i}^{2}}{2mk_{b}T} \right )d^3p*\int exp \left( -\frac{V(\mathbf{r}_{1},…\mathbf{r}_{N})}{k_{b}T} \right )d^{3N}q
$$
The integral over the potential part takes some value P depending on the considered configuration. This will factor out in the end anyway. The kinetic part is just a Gaussian integral and hence the single particle partition function is
$$
Q1=P*(2m\pi k_{b}T)^{3/2}
$$
Now from the definition of the canonical partition function I can write the average kinetic energy as a derivative
$$
\langle K_{i}\rangle = -\frac{k_{b}T}{2m}\frac{\partial}{\partial\frac{1}{2m}}log
\left ( P\int exp\left ( -\frac{\mathbf{p}_{i}^{2}}{2mk_{b}T} \right )d^3p \right )=\frac {P\int\mathbf{p}_{i}^{2}exp\left ( -\frac{\mathbf{p}_{i}^{2}}{2mk_{b}T} \right )d^3p}{Q_{1}}
$$.
This factor out the potential part of the partition function. And hence I would just have to apply the derivative to the term
$$
Q^{'}_{1}=(2m\pi k_{b}T)^{3/2}
$$
and hence:
$$
\langle K_{i}\rangle = -\frac{k_{b}T}{2m}\frac{\partial}{\partial\frac{1}{2m}}log(Q^{'}_{1})
$$
My problem is now I dont't really know how to compute this derivative since
2m occurs reciprocal in the derivative but non reciprocal in the term to differentiate. I am also not 100% sure if this approach works. Does anybody know how to compute this derivative? Or if I am completely wrong what would be a nice way to do this computation?
Best Answer
Your derivation is basically fine, and the last step is straightforward. Define $\mu = \frac{1}{2m}$, and then $$ Q_1' \propto \mu^{-3/2} \quad\Rightarrow\quad \ln Q_1' = -\frac{3}{2} \ln\mu +C $$ where $C$ is a constant, and so $$ \frac{\partial}{\partial\mu} \ln Q_1' = -\frac{3}{2} \frac{1}{\mu} = -3m $$ which leads to $\langle K_i\rangle = \frac{3}{2}k_BT$.
This is actually a version of one of the standard ways of deriving the equipartition of energy, or rather the "$\frac{1}{2}k_BT$ per quadratic degree of freedom" formula. If the energy can be written $E=\kappa x^2 + E_0$, where $x$ is one coordinate or one component of momentum, $\kappa$ is a constant, and $E_0$ is independent of $x$, then the classical partition function can always be factorized into an integral over $x$ and "the rest". When we write down the expression for $\langle x^2\rangle$, the integrals over the other coordinates will appear in the numerator and the denominator, and will cancel, as you have noted. We will just be left with a ratio of integrals over $x$. We can treat each of the three Cartesian components of momentum of each particle separately.
In that case we are just using the mathematical result $$ \langle x^2\rangle = \frac{\int_{-\infty}^{\infty} dx \, x^2 \exp(-\alpha x^2)}{\int_{-\infty}^{\infty} dx \,\exp(-\alpha x^2)} = \frac{1}{2\alpha} $$ where, here, $\alpha=\kappa/k_BT$. This leads to $\langle \kappa x^2\rangle = \frac{1}{2}k_BT$ as expected. The above ratio of integrals can be tackled by parameter differentiation (as mentioned, for instance in Mathematical Methods of Physics by J Mathews and RL Walker, 2nd edition, p61):
Your method is equivalent to this.