# 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.

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, $$\left = \langle p_{ix}^2/m\rangle = k_bT \implies \langle p_{ix}^2\rangle = \langle p_{iy}^2\rangle = \langle p_{iz}^2\rangle = m k_b T$$

The average kinetic energy of the $$i^{th}$$ particle is then $$\left< \frac{p_{ix}^2+p_{iy}^2+p_{iz}^2}{2m}\right> = \frac{3}{2} k_B T$$

• After I posted I realized I had just gone through a derivation of the result. It comes from the connection between the ideal gas law $PV=nkT$ and kinetic theory. Jan 20 at 19:31

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 $$\frac{E}{N}=\left<\frac{1}{2}mv^2\right>=\frac{3}{2}kT$$.

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.

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.