# Derivation of mean kinetic energy

I read from a book that average kinetic energy is equal to $3kT/2$ where $k$ is Boltzmann's constant and $T$ is the kelvin temperature. I don't know how the formula was derived. Any help to gain insights into the derivation of this formula would be helpful.

• How familiar are you with statistical mechanics? – getafix Oct 19 '16 at 12:07

## A simple validation

The result you quoted is the average translational kinetic energy for an ideal gas.

First, let's sketch out a rough derivation for the average kinetic energy of a particles of an ideal gas using nothing more than high school physics. The gas molecules just undergo translations, and don't rotate/vibrate. Also, they don't interact with each other.

Imagine a rectangular box which contains an ideal gas; the gas molecules have a speed $v_x$ in the $x$-direction. Since the box has a uniform cross-section of area $A$, a volume element spanned by a gas molecule in the box is $V'= |v_x\Delta t|A$. On average half the molecules move to the right, and half the molecules move to the left, since there is no preferred direction of motion. Thus, the number of molecules colliding with the container walls is

no. of molecules $= \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t|A$

The momentum transferred per molecule is $2mv_x$, so the total momentum transferred is $\Delta P = \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t| 2mv_x$

Since force is rate of change of momentum, $F = \lim_{\Delta t \to 0} \frac{\Delta P}{\Delta t} = \frac{nMAv_x^2}{V}$

Pressure is force per unit area, so the average pressure is $P = \frac{nM \langle v_x^2\rangle}{V}$

Now, note that the total velocity is $v^2 = v_x^2+v_y^2+v_z^2$, and since there is no preferred direction of motion, one can assume all three components of velocity to be equal. $v_{rms}^2 = 3 \langle v_x^2\rangle$

Thus, $P = \frac{nMv_{rms}^2}{3V}$ also from the ideal gas equation $P = \frac{NkT}{V}$ we get $\frac{mv_{rms}^2}{3} = kT$

$E_{\text{kinetic}} = \frac{mv_{rms}^2}{2} = \frac{3kT}{2}$ Which is the result you wanted.

## Using the Maxwell-Boltzmann Distribution

The same result for an translational kinetic energy can be arrived at using the Maxwell-Boltzmann distribution.

$$f(v) = 4 \pi \left( \frac{m}{2 \pi kT}\right)^{3/2} v^2 e^{\frac{-mv^2}{2kT}}$$

The average kinetic energy is $$\langle E_k \rangle= \langle \frac{mv^2}{2} \rangle = \int_{0}^{\infty}\frac{mv^2}{2} f(v)\mathrm{d}v = \frac{3kT}{2}$$

You can evaluate the integral as an exercise. I have omitted the details for the sake of brevity. It is relatively simple, you can either do it by hand, or look it up in a table or enter it into Mathematica Now, the general statement of the theorem (using words and not math) us

For a system at equilibrium the average value of each quadratic cobtribution to energy is $\frac{kT}{2}$

An example of a quadratic contribution would be kinetic energy ($\frac{p^2}{2m}$) or the harmonic potential ($\frac{kx^2}{2}$)

## A simple statistical mechanics approach

This theorem is strictly valid at "high" temperatures; we require that seperation between energy levels is small, and a lot of states are occupied so that they maybe treated as a continuum. This works well for translational and rotational modes of motion, but is not as reliable for vibrational/electronic modes. Anyway, note that the analysis given below is under the framework of these assumptions.

note: $\beta = \frac{1}{kT}$ For a one dimensional container the translational partition function derived using particle in the box energy levels is given below

$$q^T = \frac{X}{\Lambda} \qquad \text{where} \ \Lambda = \frac{h \beta^{1/2}}{(2 \pi m)^{1/2}}$$

$$\langle E^T \rangle = -\frac{1}{q^T} \left( \frac{\partial q^T}{\partial \beta}\right)_V$$

again, you can evaluate this derivative as an exercise. (Hint: $\Lambda$ is constant multiplied by $\beta^{1/2}$). The result is given below $$\langle E^T \rangle = \frac{kT}{2}$$

This is for translation in one direction (say X). The same process can be repeated for Y and Z, and thus for three dimensions

$$\langle E^T \rangle = \frac{3kT}{2}$$

Now let us examine rotations. Once again, I point out that this analysis is valid in the range $T > \theta^R = \frac{hcB}{k}$

Here, B is the rotational constant.

In this range, an approximate expression for the partition function of a heteronuclear diatmic moeclue is $$q^R = \frac{1}{\beta hcB}$$

$$\langle E^R \rangle = -\frac{1}{q^R} \left( \frac{\partial q^R}{\partial \beta}\right)_V = \frac{1}{\beta} = kT$$

Note: I already pointed out that this result is valid at "high" temperatures, however, I did not give you a sense of scale. For hydrogen molecule $\theta^R = 87.6 \ K$ for chlorine molecule it is $0.351 K$. Similarly, the approximations made while deriving the translational functional are valid at room temperatures and for everyday macroscopic containers whose dimensions are much greater than the thermal wavelength ($\Lambda$) of the molecules.

A similar analysis can be performed for vibrational modes, however the temperature range over which it is valid is much greater ($> 805 K$ for chlorine and $> 6332 K$ for hydrogen). These conditions aren't within the realm of everyday experience.

When using classical statistical mechanics to show that the average energy $\langle E \rangle = kT/2$, in one dimension, some assumptions about the energy must first be made.

We try to be quite general and assume that the energy E is a function of some coordinates $q_k$ and corresponding momenta $p_k$ and q and p range from 1 to n. Two assumptions are now made
(a) that energy can be split additively into two terms $$E=\epsilon (p_i) + E^{’}(q,p)$$ where the first term depends only on some particular $p_i$ and the second terms do not depend on $p_i$ but contains all the rest.

It may seem rather artificial to split the energy this way but as an example consider that $\epsilon (p_i)$ represents the kinetic energy of a molecule $p^2/(2m)$. This is independent of the internal energy of the electrons (chemical bond strength depending on position q and momentum p ) and of internal vibrational normal modes. These in turn are orthogonal to one another and as we normally invoke the Born-Oppenheimer approximation these are independent of rotational and electronic energies.
It happens that precise details do not matter for terms in $E^{’}$ will soon cancel out. Assuming that the energy is additive means also that exponentials can be made into products which greatly simplifies the calculation.

The second assumption (b) is that the first energy term $\epsilon (p_i)$ is quadratic in $p_i$, or $$\epsilon (p_i) = bp_i^2$$ where b is some constant. From now on we label $\epsilon (p_i)$ as $\epsilon$ and $p_i$ as p
The average energy is found in the standard way of finding an average of a quantity, and here we additionally assume that the energy follows a canonical distribution and behaves classically as integrals rather than sums are used. In quantum systems if $E \ll kT$ then this approach is also valid. The following equation is horribly complicated but it soon simplifies $$\langle \epsilon \rangle = \frac{\int \int \epsilon e^{-\beta E} dq_n,dp_n }{ \int \int e^{-\beta E} dq_n,dp_n }$$ where $\beta = (kT)^{-1}$ and k is Boltzmann’s constant and T temperature and the integrals span $\pm \infty$. To evaluate this we now use condition (a) and also separate out terms in the exponential at the same time, $$\langle \epsilon \rangle =\frac{ \int \epsilon e^{-\beta \epsilon} dp \int \int e^{-\beta E^{’}}dq_n,dp_n }{ \int e^{-\beta \epsilon} dp \int \int e^{-\beta E^{’}}dq_n,dp_n }$$ all the second set of integrals spans all the $n-1$ terms remaining in $E^{’}$ and consequently these integrals cancel out as they do not depend on $\epsilon$, leaving $$\langle \epsilon\rangle = \frac{ \int \epsilon e^{-\beta \epsilon} dp }{ \int e^{-\beta \epsilon} dp }$$ This is the standard equation for evaluating a quantity, $\epsilon$ in this case, with a distribution given by $\exp(-\beta \epsilon)$. Evaluation can be achieved by noticing that the numerator is the derivative wrt $\beta$ of the numerator, and that $dy/y = d\ln(y)$ which gives $$\langle \epsilon\rangle=-\frac{ \frac{d}{d\beta}(\int e^{-\beta \epsilon} dp ) }{ \int e^{-\beta \epsilon} dp }$$ $$\langle \epsilon\rangle=- \frac{d}{d\beta}\ln(\int e^{-\beta \epsilon} dp )$$ which leaves one integral to be evaluated. At this point it is necessary now to use the second assumption and substitute $\epsilon =bp^2$ into the integral. The final equation contains a standard integral which can be looked up and differentiated and the result is $$\langle \epsilon\rangle= -\frac{d}{d\beta} \ln(\int_{-\infty}^{+\infty} e^{-\beta bp^2}dp) = 1/(2\beta)$$ or $$\langle \epsilon\rangle =kT/2$$ This is the ‘equipartition theorem’ and applies when energy has a quadratic term, thus it applies to kinetic energy, harmonic potential energy, rotational motion etc. In three dimensions the value is $3kT/2$ because there are three orthogonal directions for kinetic energy, and for example with the (1D) harmonic oscillator, the contribution is $kT/2$ from kinetic energy and additionally $kT/2$ from potential energy as this varies quadratically in bond displacement. To be valid, as mentioned above, any quanta must be much smaller than kT or this calculation breaks down. At room temperature $kT \approx 210~ \pu{cm^{-1}}$ so this indicates that vibrational quanta are generally not going to satisfy the criteria, but rotational motion generally does as these quanta are usually only a few $\pu {cm^{-1}}$.