# How do you Find the Variance of Temperature in a Maxwell Boltzmann Distribution?

In Understanding Molecular Simulation, the following Maxwell Boltzmann distribution for momentum is given:

$$\mathcal{P}(p) = \left(\frac{\beta}{2\pi m}\right)^{3/2}\text{exp}\left[\frac{-\beta p^2}{2m}\right]$$

From this the relative variance of the kinetic energy can be calculated as:

$$\frac{\sigma_{p^2}^2}{\left< \ p^2\right>^2} \equiv \frac{\left< \ p^4\right> - \left< \ p^2\right>^2}{\left< \ p^2\right>^2} = \frac{2}{3}$$

This is used to motivate the fact that the instantaneous temperature of a system in the canonical ensemble will fluctuate also, and its variance can be given as:

$$\begin{split}\frac{\sigma_{T_k}^2}{\left< \ T_k \right>^2_{NVT}} & \equiv \frac{\left< \ T_k^2\right>_{NVT} - \left< \ T_k\right>_{NVT}^2}{\left< \ T_k\right>_{NVT}^2} \\ & = \frac{N\left< \ p^4\right> + N(N-1)\left< \ p^2\right>\left< \ p^2\right>- N^2\left< \ p^2\right>^2}{N^2\left< \ p^2\right>^2} \\ & = \frac{1}{N} \frac{\left< \ p^4\right> - \left< \ p^2\right>^2}{\left< \ p^2\right>^2} = \frac{2}{3N}\end{split}$$

This seems to indicate that:

$$\left< \ T_k^2\right>_{NVT} = N\left< \ p^4\right> + N(N-1)\left< \ p^2\right>\left< \ p^2\right>$$

The instantaneous temperature is:

$$k_B T = T_k = \sum^N_i{\frac{p_i^2}{2mN_f}}$$

Where $$N_f$$ represents the degrees of freedom and is given as $$N_f = 3N-3$$. Given this expression, $$T_k^2$$ is the following:

$$\begin{split}T_k^2 & = \left(\sum_i^N{\frac{p_i^2}{2mN_f}}\right)^2 \\ & = \sum_i^N\sum_j^N{\frac{p_i^2p_j^2}{4m^2N_f^2}} \\ & = \frac{1}{4m^2N_f^2}\left[(p_1^4+p_2^4+... +p_N^4) + (p_1^2p_2^2 + p_1^2p_3^2 + ... + p_N^2p_{N-1}^2)\right]\end{split}$$

When taking the average does this actually simplify down to what is given above? If it does why is it okay to simply say each particles momentum can be represented by the average? If this is incorrect, what method should be used to reach the desired equation for $$\left< \ T_k^2 \right>$$?

• Does this help at all: physics.stackexchange.com/a/218643/59023 Sep 27 '21 at 22:27
• No sorry, at first glance I'm not sure it does. Which part is supposed to help here? Sep 28 '21 at 15:27
• The general premise of calculating moments of a distribution function, which is for what you are looking. Sep 28 '21 at 15:51
• Okay, so how would one go about this? I understand how this makes sense for some variable quantity like $p$ or $p^2$ alone, but how would you do it for a quantity like instantaneous temperature which has numerous fixed values of momentum? It seems to me like the step your advocating is skipped because the way to get the average instantaneous temperature is take the average of every particles momentum and put that into the equation? But is that accurate? Sep 28 '21 at 17:00
• The temperature, whether kinetic or thermodynamic, does not directly result from velocity moment integration of the distribution function. One must assume some sort of equation of state to relate the pressure to the temperature (often this is done "behind the scenes" so it is assumed without explicitly telling you, the reader, what happened). Sep 28 '21 at 19:54

Dimensional analysis tells us $$T_k$$ (which I assume is short for $$k_BT$$, with dimensions of energy) should have the dimensions of $$p^4/m^2$$ rather than $$p^4$$, so your first method clearly didn't work. (Your fallacy was thinking if $$a/b=c/d$$ then $$a=c$$, as opposed to $$a=c/(2m)^2$$.) What it actually shows (assuming you got the rest of it right) is $$\langle T_k^2\rangle=\frac{3N+2}{3N}\langle T_k\rangle^2=\frac{3N+2}{6Nm}\langle p^2\rangle$$.
• Yes I believe it is short for that, I also missed off a key part in my equation, that is the instantaneous temperature should be written: $$T_k = \sum^N_i{\frac{p_i^2}{2m N_f}}$$ Where $N_f$ is the number of degrees of freedom and can be written $3N-3$. Perhaps that has some bearing on the above? I'm not sure what you're saying in the next part of your answer, we can see that $\left< \ T_k\right>^2$ is in the denominator and it is given as $N^2\left< \ p^2\right>^2$, so surely the remaining terms are due to $\left< \ T_k^2\right>$? Sep 28 '21 at 8:37