Boltzmann Distribution: Mean Energy & Mean Square Energy

[Berko]

Homework Statement

Compute the mean energy and mean square energy of a free particle using the Boltzmann distribution.

Homework Equations

The Attempt at a Solution

I calculated the mean energy to be 3/2 kT by computing the mean square velocity directly from the Boltmann distribution function and then multiplying it by 1/2 m and then again multiplying by 3 to get the answer in three dimensions.

However, I cannot figure out how to compute the mean square energy as E^2 is proportional v^4 and computing it in one dimension does not allow me to simply multiply by 3 to get the answer in 3 dimensions.

 

[Berko]

I just tried <E^2> = <(Ex+Ey+Ez)^2> = 9<Ex^2> since <Ex>=<Ey>=<Ez> and each of the three components are independent of the others.

My result was <E^2> = 27/8 k^2 T^2.

Does this make sense?

 

[dextercioby]

No. Why 9 and not 3 ?

 

[Berko]

No. Why 9 and not 3 ?

Cross terms. (x+y+z)^2 = x^2+y^2+z^2 +2xz +2xy +2yz

 

[dextercioby]

Hmm, you're right. Your answer should be fine then.

 

[Berko]

Well, I just recalculated.

<E^2> = <(x+y+z)^2> = <x^2+y^2+z^2+2xy+2xz+2yz> = 3<x^2> + 6<x><x> = 3<x^2> + 6<x>^2

where I used x for Ex and so on.

So, now my answer for <E^2> is 21/8 k^2 T^2