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 $\frac{k_{B}T}{2}$.


2 Answers 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.

  • $\begingroup$ 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 ?" $\endgroup$ Commented Mar 30, 2014 at 17:58
  • $\begingroup$ No problem. You can use whatever ensemble you like to find $p_q$ using lagrange multipliers. $\endgroup$
    – Julien
    Commented Mar 30, 2014 at 18:39
  • $\begingroup$ 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 $\dagger$ $\endgroup$
    – Julien
    Commented Mar 30, 2014 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

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


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