- #1

- 541

- 26

http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution#cite_note-0

In formula 7 we obtain proportionaly to E^1/2 and exp(-E).

But I expect proportionaly to E^-1/2 and exp(-E).

Where I am wrong?

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- Thread starter exponent137
- Start date

- #1

- 541

- 26

http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution#cite_note-0

In formula 7 we obtain proportionaly to E^1/2 and exp(-E).

But I expect proportionaly to E^-1/2 and exp(-E).

Where I am wrong?

- #2

- 814

- 9

- #3

- 541

- 26

dp/dE = proportional to E^-1/2

Maybe px, py, pz, demand a different formula?

- #4

- 649

- 2

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

- 541

- 26

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.

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.

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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