Virial Theorem and Equipartition Theorem

  • Thread starter Derivator
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  • #1
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Hi folks,

in this book (correct page should open, if not: p.199): http://books.google.com/books?id=12...r thermodynamics&pg=PA199#v=onepage&q&f=false

it says (formula (7.168)):

[tex]<\sum_i \vec{r}_i \vec{F}_i > = -p \oint{\vec{r} d\vec{S}}[/tex]

It is explained, why dF=-p dS, but I dont see the connection between the mean value of the sum on the left hand side and the ring integral on the right hand side.

Best,
derivator
 

Answers and Replies

  • #2
Jano L.
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The trick is that in the sum on the left-hand side, you take all the forces, internal and external(due to the wall). It turns out that the internal forces cancel out and you are left with the forces due to that wall. Now averaging can be performed on the surface of the wall, which allow you to reformulate the sum as an integral of the surface forces -pdS.
 
  • #3
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Now averaging can be performed on the surface of the wall, which allow you to reformulate the sum as an integral of the surface forces -pdS.
Hi,

could you please explain a litle bit more detailed, how this averaging is done? That is, why is the given integral an approximation to the left ahnd side average of the sum?

best,
derivator
 
  • #4
Jano L.
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Because the pressure p is an approximate expression for the molecular forces the wall exerts on molecules. I think writing the averaging procedure formally is not so easy, but it is obvious that the result is the integral given above. Try to write it and let us know!
 
  • #5
Jano L.
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Hi Derivator,
I've just came across the virial theorem and realized that what I wrote earlier is not correct. In fact, the internal forces do not cancel out generally. For the ideal gas, however, there are only contact forces during the collisions, so they cancel out. The only contribution to the sum is due to external forces. Here is the derivation:
The sum has nonzero contributions only from the molecules being repulsed by the wall in the instant considered (only these molecules feel external force of the wall). We choose some small patch of area dS and calculate the contribution by the molecules near this patch. The average force on the surface dS is

-pd\mathbf S,

so the contribution is

\sum_i \mathbf r_i \mathbf F_i (i near \Delta S) \approx - \mathbf r p \Delta S


here \mathbf r is the radius vector of the patch. Finally we sum the contributions from all possible patches sovering the surface enclosing the gas:

\sum_i \mathbf r_i \mathbf F_i (whole surface) \approx -\oint p d\mathbf S

Sorry for text formulae, but tex wouldn't work properly.

Jano
 

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