# Equiparition theorem, kinetic energy, temperatur

1. Jul 11, 2011

### Derivator

Hi,

according to the equipartition theorem

$<p_k \frac{\partial H}{\partial p_k}> =k_bT$ (where H depends on f generalized coodinates p and q and f is the number of degrees of freedom)

the temperature for a system of particles should be given by the result of the following derivation

$<\sum_k^f p_k \frac{\partial H}{\partial p_k}> =\sum_k^f k_bT$
<=>
$<\sum_k^f p_k \frac{\partial H}{\partial p_k}> = f k_b T$

now using $f=3N-d$ where N is the number of particles and d the number of 'constraints', one gets:
$<\sum_k^{3N-d} p_k \frac{\partial H}{\partial p_k}> = (3N-d) k_b T$

using $\frac{\partial H}{\partial p_k}=2\frac{p_k}{2m}$ one gets:
$<\sum_k^{3N-d} \frac{p_k^2}{2 m}> = \frac{3N-d}{2} k_b T$

however, the common result is:
$<\sum_k^{N} \frac{\vec p_k^2}{2 m}> = \frac{3N-d}{2} k_b T$ (note, that p_k is a vector)

That is, what people seem to do is, they don't write the kinetic energy on the left hand side as a sum of the components p_k, but as a sum over the momentum vectors:
$<\sum_k^{N} \frac{\vec p_k^2}{2 m}>$

I don't see, how they manage the upper bound 3N-d of the sum. If the upper bound was only 3N, then the last equation for the kinetic energy would be obvious to me. That is, I would understand the following
$<\sum_k^{3N} \frac{p_k^2}{2 m}> = <\sum_k^{N} \frac{\vec p_k^2}{2 m}>$
where the sum on the left hand side is going over the momentum components of all particles and the sum on the right hand side is going over the momentum-vectors of all particles.

but I don' t see, how people manage to get
$<\sum_k^{3N-d} \frac{p_k^2}{2 m}> = <\sum_k^{N} \frac{\vec p_k^2}{2 m}>$

for an example, please see page 11 of http://www.physics.buffalo.edu/phy411-506/topic3/lec-3-1.pdf [Broken] (where d=3)

Last edited by a moderator: May 5, 2017
2. Jul 12, 2011

### Bill_K

Because, in the thermodynamic limit, N is of order 1023 and d is only 3.

3. Jul 13, 2011

### Derivator

i see, only an approximation...

thank you.