Statistical Mechanics – Average Value of Energy in Statistical Mechanics

density-of-statesenergystatistical mechanics

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)}

Best Answer

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

Related Solutions

Statistical Mechanics – Rigorous Derivation of the Mean Free Path in a Gas

The following is a self answer concerning the rigorous derivation of the relative velocity in a Maxwellian gas. It is here for comparison against other answers.

Assumptions are

Two particle collisions are the most probable ones, and collisions involving more than this, contribute to the mean relative velocity by a negligible amount.
The mean relative velocity is strictly positive.

I think that this derivation differs from the one by @Thorondor because within it, $\mathbf{v_1}^2+\mathbf{v_2}^2\neq\mathbf{v_r}^2$. Which is to say that the dot product of $\mathbf{v_1}$, and $\mathbf{v_2}$ is not assumed to be zero. This forces us to integrate over all possible angles between the velocities, which may account for the ~0.07 difference. It is also why the derivation is significantly longer.

The mean free path of an atom/molecule in a Maxwellian gas, depends upon the average relative velocity of each particle to one another. In order to obtain it, we first find the magnitude of velocity $\mathbf{v}_1$ relative to all other particles moving with $\mathbf{v}_2$. This expression is then averaged for all values of $\mathbf{v}_2$, from zero to infinity, producing a mean relative speed given $\mathbf{v}_1$ exists.

Following modus ponens, the mean is multiplied by the probability that an atom/molecule has velocity $\mathbf{v}_1$, and then averaged for all $\mathbf{v}_1$ from zero to infinity.

Consider two velocities $\mathbf{v}_1$ and $\mathbf{v}_2$ inclined at an angle $\theta$, there relative velocity will be described by

\begin{align} (v_r)_{12}&=\sqrt{v_1^2+v_2^2-2v_1v_2cos(\theta)}. \end{align}

All directions are equally probable for $\mathbf{v}_2$. To calculate the average of $(v_r)_{12}$, we multiply it by the probability that it lies within some solid angle $d\Omega$.

\begin{align} \langle(v_r)_{12}\rangle&=\int_{\Omega}\frac{d\Omega}{4\pi}(v_r)_{12}\\ &=\int_{\Omega}\frac{2\pi sin(\theta)d\theta}{4\pi}(v_r)_{12}\\ &=\frac{1}{2}\int_{0}^{\pi}sin(\theta)d\theta(v_1^2+v_2^2-2v_1v_2cos(\theta))^\frac{1}{2} \end{align}

Consider a change of basis, where $cos(\theta)=x$, so $sin(\theta)d\theta=-dx$. The limits of integration will change to $cos(0)=1$ and $cos(\pi)=-1$.

\begin{align} \therefore~\langle(v_r)_{12}\rangle&=\frac{1}{2}\int_{1}^{-1}dx(v_1^2+v_2^2-2v_1v_2x)^\frac{1}{2}\\ &=\frac{1}{3}\frac{(v_1^2+v_2^2-2v_1v_2x)^\frac{3}{2}}{2v_1v_2}\Biggr|_{1}^{-1}\\ &=\frac{1}{6v_1v_2}\left((v_1^2+v_2^2+2v_1v_2)^\frac{3}{2}-(v_1^2+v_2^2-2v_1v_2)^\frac{3}{2}\right)\\ &=\frac{1}{6v_1v_2}\left((v_1+v_2)^\frac{3}{2}(v_1+v_2)^\frac{3}{2}+(v_1-v_2)^\frac{3}{2}(v_1-v_2)^\frac{3}{2}\right)\\ &=\frac{1}{6v_1v_2}\left((v_1+v_2)^3+|v_1-v_2|^3\right)\\ \end{align}

We take the magnitude of $v_1-v_2$ because $\langle(v_r)_{12}\rangle$ should always be positive, and therefore $\left((v_1-v_2)^2\right)^\frac{3}{2}$ must also be. This splits the average into two parts.

\begin{align} \langle(v_r)_{12}\rangle&=\frac{1}{6v_1v_2}\left((v_1+v_2)^3-(v_1-v_2)^3\right),~when~v_1\geq v_2,\\ &=\frac{1}{6v_1v_2}\left[2v_2^3+6v_1^2v_2\right]\\ &=\frac{3v_1^2+v_2^2}{3v_1}.\\ \langle(v_r)_{12}\rangle&=\frac{1}{6v_1v_2}\left((v_1+v_2)^3+(v_1-v_2)^3\right),~when~v_2>v_1,\\ &=\frac{3v_2^2+v_1^2}{3v_2}. \end{align}

The average relative velocity of an atom/molecule possessing magnitude and direction $\mathbf{v_1}$, with respect to all other particles moving with $\mathbf{v_2}$, lying within speeds of zero to infinity is then

\begin{align} \langle(v_r)_1\rangle&=\int_{0}^{\infty}P(v_2)\langle(v_r)_{12}\rangle dv_2,\\ P(v_2)dv_2&=4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}e^{-\frac{mv_2^2}{2k_BT}}v_2^2dv_2. \end{align}

Note that $\langle(v_r)_1\rangle$ has two distinct forms depending on the difference between speeds. Because of this we break the integral into two parts.

\begin{align} \langle(v_r)_1\rangle=4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}&\left[\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\ &+\left.\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right] \end{align}

To arrive at the relative velocity of any atom with another, we form a product of the above, and the probability a particle has velocity $v_1+dv_1$.

\begin{align} \langle v_r \rangle&=\int_{0}^{\infty}P(v_1)\langle(v_r)_1\rangle dv_1 \end{align}

\begin{align} \langle v_r \rangle=\left(4\pi\left[\frac{m}{2\pi k_BT}\right]^\frac{3}{2}\right)^2\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2&\left[\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\ &+\left.\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right] \end{align}

which may be re-written as

\begin{align} \langle v_r\rangle=\left(\frac{2m}{k_BT}\right)^3&\left[\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\ &+\left.\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right], \end{align}

\begin{align} \langle v_r\rangle&=\left(\frac{2m}{k_BT}\right)^3\left[I_1+I_2\right], \end{align}

where $I_1$ stands for the first integral in the square brackets of the above, and $I_2$ the second one.

\begin{align} I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\\ I_2&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2 \end{align}

We wish to show that these two definite integrals are equivalent. To do so, a substitution, then change in the order of integration following Fubini's theorem is employed.

Consider the piecewise function,

\begin{align} \mathbb{I}_{\{v_2\leq v_1\}}&=\begin{cases} 1&v_1\leq v_2\\ 0&v_1>v_2\\ \end{cases} \end{align}

which is zero outside the limit of the second integral of $I_1$. Using it, we can rewrite equation $I_1$ as

\begin{align} I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{\infty}dv_2\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2.\\ \end{align}

We can combine these two integrals with a substitution $a=v_1$, allowing us to write the following

\begin{align} I_1&=\int_{0}^{\infty}\int_{0}^{\infty}\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{ma^2}{2k_BT}}a^2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2dv_2da.\\ \end{align}

Since the integrals limits are constant, we can apply Fubini's theorem exchange there order.

\begin{align} I_1&=\int_{0}^{\infty}\int_{0}^{\infty}\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{ma^2}{2k_BT}}a^2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2dadv_2\\ &=\int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^2\int_{0}^{\infty}dv_1\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{mv_1^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_1^2\\ &=\int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^2\int_{v_2}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_1^2\\ \end{align}

Then, because the integral is definite, we can interchange $v_1$ with $v_2$, and obtain an equivalent volume.

\begin{align} I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\\ &=I_2 \end{align}

Hence, the average relative velocity can be rewritten as

\begin{align} \langle v_r\rangle&=2\left(\frac{2m}{k_BT}\right)^3I_2. \end{align}

To evaluate $I_2$, first consider the integral

\begin{align} I_0&=\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2. \end{align}

We can break this into parts,

\begin{align} I_0&=\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^3+\frac{v_1^2}{3}\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2, \end{align}

then consider the substitution $\frac{m}{2k_BT}v_2^2=x$, so $dv_2=\frac{k_BT}{mv_2}$, and the lower limit of integration changes to $\frac{mv_1^2}{2k_BT}$.

\begin{align} I_0&=\int_{\frac{mv_1^2}{2k_BT}}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^3+\frac{v_1^2}{3}\int_{\frac{mv_1^2}{2k_BT}}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2\\ &=2\left(\frac{k_BT}{m}\right)^2\int_{\frac{mv_1^2}{2k_BT}}^{\infty}e^{-x}xdx+\frac{v_1^2}{3}\frac{k_BT}{m}\int_{\frac{mv_1^2}{2k_BT}}^{\infty}e^{-x}dx \end{align}

Since

\begin{align} \int_{a}^{b}xe^{-x}dx&=-(x+1)e^{-x}\Biggr|_{a}^{b}, \end{align}

$I_0$ simplifies to

\begin{align} I_0&=2\left(\frac{k_BT}{m}\right)^2\left[\frac{m}{2k_BT}v_1^2+1\right]e^{\frac{-mv_1^2}{2k_BT}}\\ &=e^{-\frac{mv_1^2}{2k_BT}}\left[\frac{4}{3}\left(\frac{k_BT}{m}\right)v_1^2+2\left(\frac{k_BT}{m}\right)^2\right] \end{align}

Substituting this into $I_2$ yields

\begin{align} I_2=&\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\left[\frac{4}{3}\left(\frac{k_BT}{m}\right)v_1^2+2\left(\frac{k_BT}{m}\right)^2\right] \\=&2\left(\frac{k_BT}{m}\right)^2\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\\ &+\frac{4}{3}\frac{k_BT}{m}\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^4\\ \end{align}

$I_2$ is actually a standard integral of the form,

\begin{align} \int_{0}^{\infty}e^{-ax^2}x^n&=\frac{1}{2a^{\frac{n+1}{2}}}\Gamma\left(\frac{n+1}{2}\right), \end{align}

where $\Gamma$ is the gamma function, a generalization of the factorial. $I_2$ now simplifies to

\begin{align} I_2&=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\left[\Gamma\left(\frac{3}{2}\right)+\frac{2}{3}\Gamma\left(\frac{5}{2}\right)\right]\\ &=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\left[\frac{\sqrt{\pi}}{2}+\frac{2}{3}\frac{3\sqrt{\pi}}{4}\right]\\ &=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\sqrt{\pi} \end{align}

Hence, the average relative speed of any Maxwellian gas particle with respect to any other is

\begin{align} \langle v_r\rangle&=2\left[4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}\right]^2\left(\frac{k_BT}{m}\right)^\frac{7}{2}\sqrt{\pi},\\ &=\frac{2\sqrt{\pi}\cdot16\pi^2}{8\pi^3}\left(\frac{k_BT}{m}\right)^{-3}\left(\frac{k_BT}{m}\right)^\frac{7}{2},\\ &=\frac{4}{\sqrt{\pi}}\sqrt{\frac{k_BT}{m}},\\ &=\sqrt{\frac{16}{\pi}\frac{k_BT}{m}}. \end{align}

And the ratio of mean velocity $\langle v_r\rangle$ to mean relative velocity $\langle v\rangle$ is

\begin{align} \frac{\langle v\rangle}{\langle v_r\rangle}&=\frac{\sqrt{\frac{8}{\pi}\frac{k_BT}{m}}}{\sqrt{\frac{16}{\pi}\frac{k_BT}{m}}}\\ &=\frac{1}{\sqrt{2}} \end{align}

And thus the mean free path $\ell$ is

\begin{align} \ell=\frac{1}{n\sigma\sqrt{2}} \end{align}

Energy Statistical Mechanics – Deriving the Relationship: ?E? = kT Where k is the Boltzmann Constant

Probably, part of the explanation passed unnoticed, or it was missing. The formula $$ p(E)=\frac{e^{-\frac{E}{kT}}}{kT} \tag{1} $$ has not the same validity as the general probability of finding the system in a state of energy $E$ in a canonical ensemble, which can be obtained from your formula for $n_i$ dividing by $N$.

However, once one has justified the use of the partition function for a set of non-interacting one-dimensional harmonic oscillators$^*$, formula ($1$) is the correct energy distribution for the $1D$ harmonic oscillator, the energy density of states is a constant (it does not depend on $E$). Therefore, if $g_i=constant$, the probability is proportional to $e^{-\frac{E}{kT}}$, and the normalization factor of the density distribution in energy is $\frac{1}{kT}$. If we evaluate the partition function of a classical harmonic oscillator in $1D$, we find $$ Z=\frac{kT}{\hbar \omega}, \tag{2} $$ differing from the previous normalization by an energy factor $\hbar \omega$, necessary for dimensional reasons (the probability density of states in the phase space is dimensionless, while the probability density of energy has the dimension of inverse energy). In any case, a partition function depending linearly on $T$ is all we need to get the average energy equal to $kT$.

Summarizing, equation $(1)$ is valid only for the $1D$ classical harmonic oscillator. However, the final result $\left< E \right>= kT$, is a special case of the equipartition theorem, stating that the equilibrium average of each quadratic degree of freedom in the Hamiltonian is equal to $\frac{kT}{2}$.

$^*$ I'm afraid I disagree with the justification for using a set of harmonic oscillators the way you reported (electrons on the walls of the black body are modeled as 1-dimensional harmonic oscillators). The main reasons are the following:

In the case of blackbody radiation, the system of interest is the radiation field in the cavity, not the electrons;
It is the electromagnetic field, expressed in terms of normal modes, that is described exactly as a set of non-interacting harmonic oscillators. Any electronic system could be only approximately treated that way.
The so-called ultraviolet catastrophe is due to the infinite number of normal modes required to describe the field in the cavity. The number of electrons, even if macroscopical, is finite.

Best Answer

Related Solutions

Statistical Mechanics – Rigorous Derivation of the Mean Free Path in a Gas

Energy Statistical Mechanics – Deriving the Relationship: ?E? = kT Where k is the Boltzmann Constant

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