Boltzmann distribution - error?

  • #1
509
6

Answers and Replies

  • #2
Rap
814
9
Why do you expect proportionality to E^-1/2 ? What would be your full expression for the Maxwell distribution and why?
 
  • #3
509
6
I expect that p^2/(2m)=E, so p=(2Em)^1/2 so
dp/dE = proportional to E^-1/2
Maybe px, py, pz, demand a different formula?
 
  • #4
649
2
I suspect the following, although i haven't really analyzed it in detail:

The distribution f_p used in that calculation is the distribution over the magnitude of p, not the distribution over the three components px, py, pz. So there is an extra factor p^2 due to the integration measure p^2 dp dphi dtheta, and then you integrates over phi/theta. The extra factor p^2 causes an extra factor E , that turns your E^-1/2 into the E^1/2 in the formula.
 
  • #5
509
6
I suspect the following, although i haven't really analyzed it in detail:

The distribution f_p used in that calculation is the distribution over the magnitude of p, not the distribution over the three components px, py, pz. So there is an extra factor p^2 due to the integration measure p^2 dp dphi dtheta, and then you integrates over phi/theta. The extra factor p^2 causes an extra factor E , that turns your E^-1/2 into the E^1/2 in the formula.
Yes, now I see that it is so. Above in my link is calculation with distrubution of speed v. If I use dn/dv and dv/dE, and f(v) the above is clear.

It was also unclear to me, that distribution of potential energy of gas in vertical tube has not factor E^1/2. Because this factor is a consequence of three dimensions.
Is it OK comparision, or it should be something better for a distribution without additional factor E^1/2?
 

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