# How are degrees of freedom and energy related in classical theory?

How are degrees of freedom and energy related in classical theory? How do we come to know that each quadratic degree of freedom classically contributes a factor of $\frac{k_{B}T}{2}$.

The short answer is that $\bar E = -\frac{1}{Z}\frac{\partial Z}{\partial \beta}$ ($\dagger$). $\bar E$ is the average energy of a quadratic degree of freedom in this case and $Z$ is the partition function which sums over all possible values of $E$ associated with your degree of freedom $q$. So it's just how the math works out. There is a short derivation near the bottom of this wikipedia article.

Here is what I consider an intuitive justification of ($\dagger$):

The mean energy of a degree of freedom q is clearly equal to: $\bar E = \Sigma_q p_q E_q$, where $p_q$ is the probability associated with the state $E_q$. (From the principle that entropy is maximized in equilibrium we can derive $p_q$.) The probability (or weight) is: $$p_q=\frac{e^{-\beta E_q}}{\Sigma_q e^{-\beta E_q}}$$ Now if you look carefully at $\bar E$: $$\bar E = \frac{\Sigma_qE_qe^{-\beta E_q}}{\Sigma_q e^{-\beta E_q}}$$

You will see that it is quivalent to the formula $\bar E = -\frac{1}{Z}\frac{\partial Z}{\partial \beta}$ which yields $\frac{1}{2}k_B T$ in 1D for a quadratic degree of freedom. Here is a hint if you would like to do this calculation yourself: you will have to take the continuum limit and use a Gaussian integral to evaluate the partition function.

• Thanks for the help. However I can't understand why the canonical ensemble picture was required in this case. I mean that seems something special ( granted in large N limit all ensembles are same). I mean "is there any other method somewhat general ?" Mar 30 '14 at 17:58
• No problem. You can use whatever ensemble you like to find $p_q$ using lagrange multipliers. Mar 30 '14 at 18:39
• Here is a link to a decent derivation for the grand canonical ensemble xbeams.chem.yale.edu/~batista/vaa/node16.html, this doesn't really address your question though, I was just trying to justify the form of $\dagger$ Mar 30 '14 at 18:44

The argument goes something like this:

Suppose we have a system in contact with a large reservoir at temperature $T$ with which it can exchange energy. The Boltzmann distribution tells us that the probability of finding the system in some small region of phase space $\mathrm{d}V$ is given by

$$\rho(q_i,p_i)\,\mathrm{d}{V}= \frac{1}{Z} \exp(-E(q_i,p_i)/k_B T) \,\mathrm{d}{V} \,,$$

where $q_i$ and $p_i$ are the position and momentum variables that define the system, and which the energy $E$ depends on. $Z$ is a normalisation factor. If you're unfamiliar with this equation, derivations can be found on pages concerning 'Maxwell-Boltzmann statistics' and 'the canonical ensemble'. Additionally, if you were unaware, phase space is just the space of position and momentum variables --- that is, a point in phase space just corresponds to some particular values of the $q_i$ and the $p_i$.

$Z$ can be determined from requiring the total probability to be one:

$$Z = \int_\mathrm{all\ phase\ space} \exp(- E(q_i,p_i)/k_B T) \, \mathrm{d} V \,.$$

Now suppose the energy of the system takes the form

$$E = \alpha z^2 + E' \,,$$

where $\alpha$ is some constant, $z$ is some variable, and $E'$ just corresponds to some other terms that don't depend on $z$. In words, suppose that the energy is separable and quadratic in $z$. The probability that $z$ lies between $z$ and $z+\mathrm{d}z$ is obtained by integrating out all other phase space variables besides $z$ in the expression for $\rho(q_i,p_i)$. If we do this, we will find that these integrals cancel with the big integral $Z$, leaving $$\rho(z)\,\mathrm{d}z = \exp(-\alpha z^2/k_B T)\,\mathrm{d}z / \int_{\mathrm{all\ }z} \exp(-\alpha z^2/k_B T) \,\mathrm{d}z \,.$$

If we want to find the average value of this term (which I will denote with angle brackets), we can use the result from the theory of continuous probability distributions that

$$\langle \alpha z^2 \rangle = \int_{\mathrm{all\ }z} \alpha z^2 \rho(z) \, \mathrm{d}z \,.$$

Evaluating the integrals (which are Gaussian integrals, or related), gives the result

$$\langle \alpha z^2 \rangle = \int_{\mathrm{all\ }z} \alpha z^2 \exp(-\alpha z^2/k_B T)\,\mathrm{d}z / \int_{\mathrm{all\ }z} \exp(-\alpha z^2/k_B T) \,\mathrm{d}z = \frac{1}{2}k_B T \,.$$