I encountered a question:

"Find the ratio of Kinetic Energy and Total Energy of a star, made out of a monoatomic ideal gas, You may not consider density to be uniform. Kinetic energy is the associated KE of gas particles w.r.t to Temperature. You may assume uniform temperature."

The solution is fine until they use $K_E = {3\over 2}k_BT$ per the monoatomic gas particle. Isn't this $K_E$ associated with Kinetic Gas Theory which neglects all attractive forces. But we have to calculate the self-energy of this star, which certainly arises due to gravitational interactions between these particles and is certainly not neglectable.

Is it a valid assumption that $K_E = {3\over 2}k_BT$?

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    $\begingroup$ Yes, the kinetic energy can reasonably-well be assumed $E_{kin} = \frac{3}{2} kT$. Gravity or its absence has no influence on the kinetic energy. The question asks you to assume a isothermal polytropic star with a finite central density. $\endgroup$ Commented Aug 29, 2021 at 18:11
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    $\begingroup$ I thought that this result is only valid when the gas is under no influence of attractive forces, but here there are attractive forces right? While deriving kinetic energy, we assume the absence of external forces. $\endgroup$
    – Anmoldeep
    Commented Aug 29, 2021 at 20:06

1 Answer 1


I will use some concepts form statistical mechanics, I hope you are familiar with some of the concepts.

Consider a gas of $N$ particles of mass $m$ with Hamiltonian function

$$H(\bar{q}, \bar{p}) = \sum_{i=1}^N {|\vec{p}_i|^2 \over 2 m} + U(\bar{q})$$

Here, $\bar{q} = (\vec{q}_1,\vec{q}_2,... \vec{q}_N)$ are the position of the particles and $\bar{p} = (\vec{p}_1,\vec{p}_2,... \vec{p}_N)$ are their momenta. $U(\bar{q})$ is the potential, it may for example be the gravitational potential:

$$U(\bar{q}) = {1 \over 2}\sum_{i \neq j} G{m^2 \over |\vec{q}_i-\vec{q}_j|}$$

If the temperature $T$ is fixed, it means that we can work in the Canonical Ensemble and the probability that our system is in a particular configuration $(\bar{q}, \bar{p})$ is given by

$$\mathcal{P}(\bar{q}, \bar{p}) = \frac{e^{-\beta H(\bar{q}, \bar{p})}}{\displaystyle \int {d\bar{q}d\bar{p} \over h^{3N}N!} \ e^{-\beta H(\bar{q}, \bar{p})}}$$

where $\beta = {1 \over k_B T}$.

The average kinetic energy of the system is obtained by integrating the kinetic energy over the distribution

$$\left< K_E \right> = \displaystyle \int K_E(\bar{p})\mathcal{P}(\bar{q}, \bar{p}) {d\bar{q}d\bar{p} \over h^{3N}N!} = \displaystyle \int \left(\sum_{i=1}^N {|\vec{p}_i|^2 \over 2 m} \right)\mathcal{P}(\bar{q}, \bar{p}) {d\bar{q}d\bar{p} \over h^{3N}N!}$$

Now, since $K_E$ doesn't depend on $\bar{q}$, we can separate the integral in two parts, one over the momenta and one over the positions.

$$\left< K_E \right> = \frac{\displaystyle \int d\bar{q}d\bar{p} \left(\sum_{i=1}^N {|\vec{p}_i|^2 \over 2 m} \right)e^{-\beta H(\bar{q}, \bar{p})}}{\displaystyle \int d\bar{q}d\bar{p}\ e^{-\beta H(\bar{q}, \bar{p})}} = \frac{\displaystyle \int d\bar{p} \left(\sum_{i=1}^N {|\vec{p}_i|^2 \over 2 m} \right)e^{-\sum_{i=1}^N \beta{|\vec{p}_i|^2 \over 2 m}}}{\displaystyle \int d\bar{p}\ e^{-\sum_{i=1}^N \beta{|\vec{p}_i|^2 \over 2 m}}} \frac{\displaystyle \int d\bar{q}e^{-\beta U(\bar{q})}}{\displaystyle \int d\bar{q}\ e^{-\beta U(\bar{q})}}$$

The integral on $\bar{q}$ simplifies, and therefore the average kinetic energy doesn't depend on the potential. Continuing the calculation, we find

$$\require{cancel}\left< K_E \right> = \frac{\displaystyle \int d\bar{p} \left(\sum_{i=1}^N {|\vec{p}_i|^2 \over 2 m} \right)\prod_{i=1}^N e^{- \beta{|\vec{p}_i|^2 \over 2 m}}} {\left( \displaystyle \int d\vec{p}\ e^{-\beta{|\vec{p}|^2 \over 2 m}}\right)^N} = {\displaystyle \int_{-\infty}^{+\infty} dp {p^2 \over 2 m} e^{- \beta{p^2 \over 2 m}}} \frac{3N \left(\displaystyle \int_{-\infty}^{+\infty} dp\ e^{- \beta{p^2 \over 2 m}}\right)^{\cancel{3N}-1}} {\cancel{\left(\displaystyle \int_{-\infty}^{+\infty} dp\ e^{-\beta{p^2 \over 2 m}}\right)^{3N}}}$$

The gaussian integrals are readily solved and yield

$$\left< K_E \right> = 3N {\sqrt{2\pi m} \over 2}(k_BT)^{{3\over 2}}\frac{1}{\sqrt{2\pi m k_B T}} = {3 \over 2}Nk_BT$$

I know these calculations may look cumbersome, if you have never seen them. Feel free to ask clarifications on passages that I have done too hastily.

  • $\begingroup$ Thanks a lot kind sir, yes the calculations are a bit beyond my level of understanding since I am only in the 12th year of high school, but the calculations satisfy what I needed to ask. God bless you once again.... $\endgroup$
    – Anmoldeep
    Commented Aug 30, 2021 at 6:59

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