This is an extract from my textbook Physics for Degree Students B.Sc Second Year By C L Arora. [Image Source]

"For every independent quadratic term appearing in the energy expression of the system, the average energy corresponding to temperature $T$ of the system is $\frac{1}{2}kT$"

Again let us consider a monoatomic gas with $N$ molecules forming a system. We need $f$ position coordinates and $f$ momentum coordinates to specify the system uniquely. The total energy can be expressed as


Now imagine that the energy expression is so arranged that there is a term $E_i(p_i)$ which depends on the coordinate $p_i$ alone and this term does not appear in any other energy term.

Then we may write $E=E_i(p_i)+E'(q_1,q_2,...q_f,p_1,p_2,..,p_{i-1},p_{i+1},...,p_f)$ where $E'$ is not a function of $p_i$.

The mean value of $E_i$ can now be obtained as

$$\bar{E_i}=\frac{\sum E_ie^{-\beta E}}{\sum e^{-\beta E}}$$ where $\beta=\frac{1}{kT}$ and the summation extends to all microstates which pertains to total energy $E$.

Here, to find $E_i$ why do they use $\sum E_i P(E)$ and not $\sum E_i P(E_i)$? Why are they finding expectation value of $E_i$ using probability of the system having total energy $E$ ?

I'm not really sure how they got and what they mean by $$\bar{E_i}=\frac{\sum E_ie^{-\beta E}}{\sum e^{-\beta E}}$$

Shouldn't it be

$$\bar{E_i}=\frac{\sum E_ie^{-\beta E_i}}{\sum e^{-\beta E_i}}$$

instead where $P(E_i)$ i.e. probability of occurrence (for a molecule) of energy $E_i$ is $\frac{e^{-\beta E_i}}{\sum e^{-\beta E_i}}$ ?


2 Answers 2


I think that all those doubts share the same root. The probability density of the system states depends on the system energy ($E$).

The author is splitting the energy in two additive terms, just for convenience. The occurrence of a state do not depends on such arbitrary partition.

  • why do they use $\sum E_i P(E)$ and not $\sum E_i P(E_i)$?

Just because of the stated above.

  • I'm not really sure how they got and what they mean by $$\bar{E_i}=\frac{\sum E_ie^{-\beta E}}{\sum e^{-\beta E}}$$

The probability of a state (leaving aside the fact that the variables are continuous) with energy $E$ is then $P(E)=\frac{e^{-\beta E}}{\sum e^{-\beta E}}$, so the expected value of $E_i$ (represented by $\bar E_i$) must be (for the considered distribution)

$$\bar{E_i}=\frac{\sum E_ie^{-\beta E}}{\sum e^{-\beta E}}$$

where the sums goes for all the system states.

Edit in response to comments

Time and space are considered continuous in both, classical and quantum treatments. Thinking in terms of the former, that means (for your case) that you can find the gas molecules in almost any place (except where it is forbidden because of very high repulsive potentials (molecules too close for example)). That is, they are not constrained to a grid of discrete space points. Also, if time is continuous, by definition the linear momentum are continuous variables.

As said before, the summations goes for all system states. Each system state is specified by a set of coordinates. They are the position and momentum values. If coordinates are discrete variable, theoretically you can look the set of possible values of them ($q_{1,k},q_{2,k},\dots,q_{f,k},p_{1,k},p_{2,k},\dots,p_{f,k}$ for $k=1,2,\dots$) one by one and count them (even if they are infinite), like state 1, state 2, etc. and sum over each state ($\sum$). But if the variables are continuous each coordinate can take any real value (or a subset of $\mathcal R$ where it is still continuous), they are not countable and you cannot simply sum over all states but integrate ($\int$). Please, check this section: https://en.wikipedia.org/wiki/Riemann_sum#Connection_with_integration I do not know the deeper mathematical reasons of the above and I think that they are far from being trivial.

  • $\begingroup$ Thanks. I got that part. Could you please explain why $p_n$ and $q_n$ (where $n =\{1,2,3,...,f\}$) can be considered to be continuous over here? $\endgroup$
    – user139621
    Oct 6, 2017 at 10:52
  • $\begingroup$ Please, can you rephrase? I am not sure if I understand what you are asking for in your comment. $\endgroup$ Oct 6, 2017 at 10:56
  • $\begingroup$ I mean why can $p$ and $q$ be considered as continuously variable and converted to integral form from summation form? (As in here) $\endgroup$
    – user139621
    Oct 6, 2017 at 10:58
  • $\begingroup$ I edited my answer in response to your comment $\endgroup$ Oct 6, 2017 at 12:16

In general if you want the expected value of variable $X$ you'd have to calculate

$$ \overline{X} = \frac{\sum Xe^{-\beta E}}{\sum e^{-\beta E}} $$

In your case just take $X = E_i$

  • $\begingroup$ How is $E_i$ a function of $E$ ? i.e. how is $E_i=E_i(E)$ ? $\endgroup$
    – user139621
    Oct 6, 2017 at 7:18
  • $\begingroup$ @Blue Sorry, that's poor notation, what I mean is the sum is over all states with different energy $\endgroup$
    – caverac
    Oct 6, 2017 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.