Why is there a $1/2$ in the definition of energy per degree of freedom $E=(1/2)kT$? I was looking for an authoritative definition of Boltzmann's Constant.  That led me to this discussion on NIST's site: Kelvin: Thermodynamic Temperature

Thus, internal energy and temperature are different, though directly related. The SI unit of energy is the joule. A “derived” SI unit, the joule itself is defined in terms of three SI base units — the kilogram, the meter and the second. But thermodynamic temperature is expressed in kelvins. There needs to be a way to connect the two.


The bridge between those two realms is the Boltzmann constant ($k_B$, or often just $k$), which relates the kinetic energy content ($E$) of matter to its temperature ($T$): $E = k_{B}T$. For the simplest collection of particles such as atoms, the average kinetic energy is $(1/2) m v^2$ distributed over the three degrees of freedom, where $m$ is the mass and $v$ is the velocity, so the total translational energy is $(3/2) k_{B}T.$

I can't think of any good reason to have the factor of $1/2$ in this primitive formula.  Can someone explain where this came from?  I'm guessing it's an artifact of history.
 A: It's not actually a definition, but rather a result called the equipartition theorem. If the Hamiltonian of a thermodynamical system in contact with a heat reservoir at temperature $T$ depends on some degree of freedom $x$, then the equipartition theorem says that
$$\left< x \frac{\partial H}{\partial x}\right>= k_B T$$
where $\langle \cdot \rangle$ denotes the ensemble average and $H$ is the Hamiltonian of the system.
The Hamiltonian of an ideal gas of $N$ in 3 dimensions is $H = \sum_{i=1}^N \frac{\mathbf p_i^2}{2m}$ where $\mathbf p_i =(p_{ix},p_{iy},p_{iz})$.  Accordingly via the equipartition theorem,
$$\left<p_{ix}\frac{\partial H}{\partial p_{ix}}\right> = \langle p_{ix}^2/m\rangle = k_bT \implies \langle p_{ix}^2\rangle = \langle p_{iy}^2\rangle = \langle p_{iz}^2\rangle = m k_b T$$
The average kinetic energy of the $i^{th}$ particle is then
$$\left< \frac{p_{ix}^2+p_{iy}^2+p_{iz}^2}{2m}\right> = \frac{3}{2} k_B T$$
A: I recall this argument from Giancoli, so grain of salt, but good for intuition.
Suppose you have gas in a box. It has length $L$ in the $x$ direction and speed $v_x$. Consider a single molecule that has a perfectly elastic collision with the faces of the box at $x=0$ and $x=L$. For a molecule just bouncing off at $x=0$ a full return trip will take time $2L/v_x$. During that trip it will have a change in momentum of $-2mv_x$, so $\Delta p/\Delta t=\frac{mv_x^2}{L}$ is an average force. Divide by the cross sectional area you get a pressure $P$ and the product of length and area is volume $V$, so  $PV=mv_x^2$. Each particle has K.E. $\frac{1}{2}mv^2$ where $v^2=v_x^2+v_y^2+v_z^2$. If we assume the same speed in each direction then $v_x^2=(1/3)v^2$.
Combined we have $PV=\frac{2}{3}N\left<\frac{1}{2}mv^2\right>=NkT$ by the ideal gas law. If we have $E=N\left<\frac{1}{2}mv^2\right>$, then $\frac{E}{N}=\left<\frac{1}{2}mv^2\right>=\frac{3}{2}kT$.
A: Here is an elementary approach ...
The ideal gas equation can be expressed in terms of the number, $N$ of molecules (rather then the number of moles) as
$$pV=NkT$$
in which $k$ is the Boltzmann's constant.
But the kinetic theory gives us
$$pV=\tfrac 13 Nm\overline {c^2}.$$
Therefore we have
$$\tfrac 13 m\overline {c^2}=kT\ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ \tfrac 12 m\overline {c^2}=\tfrac 32 kT$$
You see now how the factor of $\tfrac 12$ enters. It is ultimately because we are expressing a relationship between temperature and kinetic energy, and there is a factor of $\tfrac 12$ in the kinetic energy formula. According to the classical result of equipartition of energy we can assign a mean energy of $\tfrac 12 kT$ to each degree of freedom: the three degrees of translational freedom, plus any degrees of rotational and vibrational freedom.
A: After I posted, I realized I had just gone through the derivation of the result.  All that was needed was to connect the following three equations.
$$\left\langle \mathcal{KE}\right\rangle =\frac{3}{2}kT$$
$$P=\frac{2}{3}n\left\langle \mathcal{KE}\right\rangle $$
$$PV=NkT$$
My notes are written in Mathematica, so it's difficult to post them as MathJax.  I hope these screen-scrapes aren't too offensive.  The notes were written for my own review, so they are a bit hand-wavy.




