- #1

Derivator

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Hi,

according to the equipartition theorem

[itex]<p_k \frac{\partial H}{\partial p_k}> =k_bT[/itex] (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

[itex]<\sum_k^f p_k \frac{\partial H}{\partial p_k}> =\sum_k^f k_bT[/itex]

<=>

[itex]<\sum_k^f p_k \frac{\partial H}{\partial p_k}> = f k_b T[/itex]

now using [itex]f=3N-d[/itex] where N is the number of particles and d the number of 'constraints', one gets:

[itex]<\sum_k^{3N-d} p_k \frac{\partial H}{\partial p_k}> = (3N-d) k_b T[/itex]

using [itex]\frac{\partial H}{\partial p_k}=2\frac{p_k}{2m}[/itex] one gets:

[itex]<\sum_k^{3N-d} \frac{p_k^2}{2 m}> = \frac{3N-d}{2} k_b T[/itex]

however, the common result is:

[itex]<\sum_k^{N} \frac{\vec p_k^2}{2 m}> = \frac{3N-d}{2} k_b T[/itex] (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:

[itex]<\sum_k^{N} \frac{\vec p_k^2}{2 m}>[/itex]

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

[itex]<\sum_k^{3N} \frac{p_k^2}{2 m}> = <\sum_k^{N} \frac{\vec p_k^2}{2 m}>[/itex]

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

[itex]<\sum_k^{3N-d} \frac{p_k^2}{2 m}> = <\sum_k^{N} \frac{\vec p_k^2}{2 m}>[/itex]for an example, please see page 11 of http://www.physics.buffalo.edu/phy411-506/topic3/lec-3-1.pdf (where d=3)

according to the equipartition theorem

[itex]<p_k \frac{\partial H}{\partial p_k}> =k_bT[/itex] (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

[itex]<\sum_k^f p_k \frac{\partial H}{\partial p_k}> =\sum_k^f k_bT[/itex]

<=>

[itex]<\sum_k^f p_k \frac{\partial H}{\partial p_k}> = f k_b T[/itex]

now using [itex]f=3N-d[/itex] where N is the number of particles and d the number of 'constraints', one gets:

[itex]<\sum_k^{3N-d} p_k \frac{\partial H}{\partial p_k}> = (3N-d) k_b T[/itex]

using [itex]\frac{\partial H}{\partial p_k}=2\frac{p_k}{2m}[/itex] one gets:

[itex]<\sum_k^{3N-d} \frac{p_k^2}{2 m}> = \frac{3N-d}{2} k_b T[/itex]

however, the common result is:

[itex]<\sum_k^{N} \frac{\vec p_k^2}{2 m}> = \frac{3N-d}{2} k_b T[/itex] (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:

[itex]<\sum_k^{N} \frac{\vec p_k^2}{2 m}>[/itex]

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

[itex]<\sum_k^{3N} \frac{p_k^2}{2 m}> = <\sum_k^{N} \frac{\vec p_k^2}{2 m}>[/itex]

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

[itex]<\sum_k^{3N-d} \frac{p_k^2}{2 m}> = <\sum_k^{N} \frac{\vec p_k^2}{2 m}>[/itex]for an example, please see page 11 of http://www.physics.buffalo.edu/phy411-506/topic3/lec-3-1.pdf (where d=3)

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