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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?

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    $\begingroup$ I'm confused by the first part of the question. "independent of the potential energy part" isn't this how kinetic energy is defined? Perhaps you want to show that the average kinetic energy of the particles doesn't vary if the potential energy term varies? What do you mean by 'independent'? $\endgroup$ – Al Nejati Sep 26 '18 at 20:42
  • $\begingroup$ @ AlNejati In the end one should obtain $\langle K_{i}\rangle=\frac {3}{2}k_{b}T$. But yes you are right the average kinetic energy should not vary if the potential energy varies. This is what I meant $\endgroup$ – zodiac Sep 27 '18 at 4:55
  • $\begingroup$ @zodiac $\langle K \rangle = \frac{3}{2}k_{B}T$ is for mono-atomic ideal gas. Your system consists of interaction with potential $V$. So the result should be different from what you expected. $\endgroup$ – K_inverse Sep 27 '18 at 5:35
  • $\begingroup$ Actually @K_inverse the result is true whether or not there is a potential energy term $V$. It applies whenever the energy includes some quadratic terms in coordinates or momenta which do not appear elsewhere in the energy. For any classical mechanical system, this is the case for the translational kinetic energy terms. See my answer for more details. $\endgroup$ – user197851 Sep 27 '18 at 5:49
  • $\begingroup$ @LonelyProf, oh yes, the equipartition theorem. I almost forgot. Thanks $\endgroup$ – K_inverse Sep 27 '18 at 12:15
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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):

  1. Know that $\int_{-\infty}^{\infty} dx \,\exp(-\alpha x^2) = \sqrt{\pi/\alpha}$
  2. Differentiate both sides with respect to $\alpha$.

Your method is equivalent to this.

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