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Q & A: Maxwell energy distribution

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Most recent answer: 02/27/2014
Q:
what is wrong with my attempt to gain a distribution function for energy from Maxwell velocity distribution? f(v)dv=f(E)dE f(E)=f(v)/mv now when I substitute v with (2E/m)^0.5 I gain a function with E^-0.5 while it must be E^0.5 please help...
- Anonymous
A:

For other readers: this question is about the distribution of the (translational) kinetic energies of the particles in a classical gas.

Your reasoning is good as far as it goes. You've shown that the probability density function for kinetic energy (which you call f(E)) has a factor of E-0.5 , multiplied by the density function for the corresponding speeds, which you call f(v). Now you need to remember that  f(v) depends on how many different states have speed near v. If you think of the states as distributed uniformly for all different vector velocities, v, in three dimensions, you can see that the ones with speed near v form a circular shell of area v2. That gives a factor of v2 in f(v). Since E is proportional to v2, that gives another factor of E in f(E). Combined with the E-0.5  you calculated, that gives a net factor of E0.5 for the density of states near E. It sounds as if you're having no trouble obtaining the other factor, the exponential cutoff (e-E/kT).

Mike W. 


(published on 02/26/2014)

Follow-Up #1: Maxwell-Boltzmann distribution

Q:
well,when I substitute f(v)=(m/2pi kT)exp(-mv^2/2kT) and in the equation obtained substitue mv^2 with E there is no factor of E gained to change E^-0.5 toE^0.5 and my problem with it continues...please write me the calculations...
- Anonymous
A:

That's because you are using a distribution formula for a single component of the velocity, e.g. vx. However velocity is a vector. That's why you need a 3D picture of the possible velocities. The ones with fixed magnitude (i.e. speed) form a spherical shell around the origin. Its area goes as v2. That's the factor missing from your calculation. 

If you were interested in only motion along one direction and the kinetic energy associated with it, your calculation would be right, including the E-0.5 factor.

Mike W.


(published on 02/27/2014)

Follow-Up #2: converting distribution functions

Q:
thank you for your answer.i will try to make it.but I have got another problem, why is f(v)dv equal to f(E)dE? and what would that mean physically and mathematically???please explain...
- Anonymous
A:

Here for clarity we'd better use careful notation: fv(v) and fE(E) for the two probability density functions,, which do not have the same functional form.

 fv(v)dv tells you what fraction of the particles have speeds between v and v+dv.  fE(E)dE tells you what fraction of the particles have kinetic energies between E and E+dE. These are closely related because E is a monotonic function of v. The number of particles with speed between v and v+dv is just the same as the number with KE between mv2/2 and m(v+dv)2/2. Since dv is very small, that's just the number with KE between mv2/2 and mv2/2+d(mv2/2), i.e. between E and E+dE.

Mike W.


(published on 02/27/2014)

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