5
$\begingroup$

I haven't taken any classes in Statistical Mechanics, but in studying Structure of Matter I found some ideas I'm not very familiar with, related with the average value of energy ($E$). Given a $p(E)$ probability density, the average energy is:

\begin{equation} \langle E\rangle=\frac{\int\limits_0^\infty Ep(E)dE}{\int\limits_0^\infty p(E)dE} \tag{1} \end{equation}

Now, in two different cases the average energy is calculated using either the Boltzmann distribution (the average energy per normal mode, when deriving Rayleigh Jeans) \begin{equation} p(E)\propto e^{-\frac{E}{kT}} \end{equation} or the Maxwell-Boltzmann(*) distribution (when deriving $\beta=\frac{1}{kT}$.). \begin{equation} p(E)\propto\sqrt{E}e^{-\frac{E}{kT}} \end{equation} The first case yields:

\begin{equation} E=kT \tag{2} \end{equation}

While the second case yields:

\begin{equation} E=\frac{3}{2}kT \tag{3} \end{equation}

That is familiar from kinetic theory.

I suppose I am calculating average energy in two different situations.

Can you provide some physical (mathematically is quite clear, the distributions being different) insight on why this results are different in $(2)$ and $(3)$?

(*) Actually it is not directly stated. It "builds" average energy starting from
\begin{equation} \langle E\rangle=\frac{\int\limits_0^\infty Edn}{\int\limits_0^\infty dn} \end{equation} and then $dn$ is expressed in term of $dE$ and there is this $\sqrt{E}$ factor, so I assumed the book was implicitly using Maxwell Boltzmann. (Brehm, Introdution to the structure of matter. Chapter 2, section 3, example 2)

$\endgroup$
2
  • $\begingroup$ Possible duplicate here $\endgroup$
    – joseph h
    Oct 13 at 20:06
  • 4
    $\begingroup$ Does this answer your question? $E=kT$ or $\frac32kT$? $\endgroup$
    – Dave
    Oct 13 at 21:02
2
$\begingroup$

For the first case the probability of a certain energy starts high and reduces quickly (red curve), so there is a higher probability of low energy than high energy, for this case.

For the second case (blue), the probability of low energy is low, but for higher energies the probability is greater than for the red curve.

enter image description here

This causes the average energy to be higher for the second case.

$\endgroup$
3
  • 2
    $\begingroup$ If I have understood the OP's question correctly, this answer does not address it at all. They are wondering why they sometimes see textbooks use a plain Boltzmann factor and sometimes see them use an extra $\sqrt{E}$ in the probability. The question states that it's quite clear that the two will give you different results, and the question is more about the interpretation of the two equations. $\endgroup$
    – sasquires
    Oct 14 at 5:08
  • $\begingroup$ If the OP sees your comment, maybe they'll clarify the question $\endgroup$ Oct 14 at 9:50
  • $\begingroup$ I confirm that sasquires is right. I am asking why for each situation we use the respective distribution and some insight about the physical meaning. $\endgroup$
    – Feynman_00
    Oct 15 at 4:37
2
$\begingroup$

Can you provide some physical (mathematically is quite clear, the distributions being different) insight on why this results are different in (2) and (3)?

In the $kT$ case there are two degrees of freedom (e.g., single point particle in two dimensions, e.g., a simple harmonic oscillators in one dimension, etc.). In the $3kT/2$ case there are three degrees of freedom (e.g., single particle in three dimensions, etc.).

Each "degree of freedom" contributes $kT/2$ to the average energy.

For example, in the free particle case: $E(p)=\frac{p^2}{2m}$. And $dE \propto pdp$

So, in two dimensions: $$ E \propto \int dp p e^{-E(p)/kT} \propto \int dE e^{-E/kT}\;. $$

And, in three dimensions: $$ E \propto \int dp p^2 e^{-E(p)/kT} \propto \int dE\sqrt{E} e^{-E/kT}\;. $$

$\endgroup$
4
  • 1
    $\begingroup$ In case anyone is wondering, the reason why there is a $p$ in 2D and a $p^2$ in 3D here is because of the density of states, as I explained in my answer. The number of states with a particular energy $E$ of a free particle in 2D is given by $\frac{p_x^2 + p_y^2}{2m} = E$, which defines a circle in the space of $(p_x, p_y)$ that has a circumference which is proportional to $p = \sqrt{p_x^2 + p_y^2}$. In 3D, you also have $p_z$ and you are finding the surface area of a sphere, which is proportional to $p^2$. $\endgroup$
    – sasquires
    Oct 14 at 5:03
  • $\begingroup$ As far as I understand, degrees of freedom in Statistical Mechanics is the dimension of the phase space (1D oscillator has 2 DoF, while in classical mechanics it has 1 DoF, namely the dimension of the configuration space). As for the answer, the comparison between the 2d and the 3d case shows that each distribution emerges naturally in the description of a free particle, Boltzmann in the former, Maxwell-Boltzmann in the latter. Can you confirm that this is correct? $\endgroup$
    – Feynman_00
    Oct 15 at 4:49
  • 1
    $\begingroup$ No, that example I gave was incorrect. 1D oscillator has 2 DoF. Sorry. I corrected it. $\endgroup$
    – hft
    Oct 15 at 7:15
  • $\begingroup$ Two quadradic terms in the Hamiltonian means two DoF. In both classical and quantum case the Hamiltonian for SHO looks like $p^2/2m + kx^2/2$. $\endgroup$
    – hft
    Oct 15 at 7:17
1
$\begingroup$

Briefly, neither expression is correct in general. There are two basic issues:

  1. Boltzmann factors are not probabilities but are ratios between probabilities.
  2. Boltzmann factors are proportional to the probability that a particular state with energy $E$ is occupied.

Because of this, both integrals need to be taken over the set of all states (or density of states or phase space) rather than over $E$ itself. There is no universal relationship between $E$ and $T$ because of the importance of the density of states.

There are some technicalities depending on which ensemble you’re talking about, but they aren’t worth going into here.

As an aside, entropy is a more fundamental quantity than temperature, and it is also very closely tied to the set of available states. This explains why statistical mechanics is so rich; most of it has to do with the relationship between energy and entropy, which are both closely related to the states available to the system.

$\endgroup$
1
$\begingroup$

TL;DR: Density-of-states

Boltzmann distribution gives a probability of a microstate $n$ in terms of its energy $E_n$: $$p_n=Z^{-1}e^{-\frac{ E_n}{k_BT}}.$$ In case of Maxwell-Boltzmann distribution the states are continuous and labeled by their velocities, so that we have $$ p(v_x,v_y,v_z)=Z^{-1} e^{-\frac{ E(v_x,v_y,v_z)}{k_BT}} $$ The average of any function of energy, $f(E)$, is the given by $$ \langle f(E)\rangle = Z^{-1}\int_{-\infty}^{+\infty} dv_x\int_{-\infty}^{+\infty}dv_y\int_{-\infty}^{+\infty} dv_z f[E(v_x,v_y,v_z)] e^{-\frac{ E(v_x,v_y,v_z)}{k_BT}}=Z^{-1}\int_0^{+\infty}d\epsilon\rho(\epsilon)f(\epsilon)e^{-\frac{\epsilon}{k_BT}}, $$ where the density-of-states is $$ \rho(\epsilon)=\int_{-\infty}^{+\infty} dv_x\int_{-\infty}^{+\infty}dv_y\int_{-\infty}^{+\infty} dv_z\delta[\epsilon-E(v_x,v_y,v_z)]. $$ In case of $E=\frac{m(v_x^2+v_y^2+v_z^2)}{2}$ we obtain $$ \rho(\epsilon)=4\pi\sqrt{\frac{m}{2\epsilon}}. $$ If we now want to calculate the average energy, we set $f(\epsilon)=\epsilon$ in the equations above and obtain $$ \langle E\rangle \propto \int_0^{+\infty}d\epsilon\sqrt{\epsilon}e^{-\frac{\epsilon}{k_BT}}. $$

Remark
Depending on the textbook the density-of-states may be introduced differently (i.e., without using delta-function) and coefficients may differ. The idea however remains the same: replacing the summation over states by integration over energy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.