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Velocities for a modified maxwell distribution (an interesting problem)

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a momentum distribution such that:

    f([tex]\overline{p}[/tex]) = (2*[tex]\pi[/tex]*k*T*m)^(3/2)*exp(-p^2/(2*m*k*T))*(1+[tex]\epsilon[/tex]*cos([tex]\alpha[/tex])

    where [tex]\alpha[/tex] is the angle between [tex]\overline{p}[/tex] and [tex]\alpha[/tex]

    a) what is the expectation value associated with U = <V_x> + <V_y> + <V_z>
    b) How many particles are passing though a unit area at any given time in the positive x-direction? The negative direction? Is this related to U_x

    2. Relevant equations

    The standard maxwell distribution is the first term in the momentum distribution given


    3. The attempt at a solution
    So to start this isn't too terrible, at least at first: I start by noting that the cosine term is really just a dot product of the p-vector and the x-unit vector divided by their magnitudes and note the exp(-P^2/(2*m*k*T)) * (2*Pi*m*k*T)^(3/2) is the standard maxwell distribution and expand out into a sum of two terms:

    f(P) = f(P)_standard distribution + (P_x)/sqrt((P_x)^2+(P_y)^2+(P_z)^2)*exp(-P^2/(2*m*k*T)*(2*Pi*m*k*T)^(3/2)

    Now since I am looking only for one velocity component at a time I swap to the velocity representation. HOWEVER, I can not simply swap the last term in my sum into three separate terms for each velocity component because of the sqrt term associated with the exponential.

    So before I continue on I deal with the first term in my sum, the standard f(v). Well since it is the standard form, the value of <V_i> is well known to be zero (due to the distribution being Gaussian with a mean of zero), so I can just drop it away and concentrate the the last term in the sum.

    Since I can't get away with separating out the components I rewrite the equation in terms of spherical coordinates:

    (the extra part:) f(v) = (m/(2*Pi*k*T))^(3/2)*v^2*(cos(theta)^2*sin(phi)^2)/v *exp(-m*v^2/(2*k*T))

    I should note it may not be clear where the P_x went: I switch to the velocity distribution, and then note that in the spherical representation v_x = v*cos(theta)*sin(phi)

    Ok so now I multiply by the volume element dV in the modified spherical coordinates to get

    <V_x> = integral(integral(integral(v^3*cos(theta)^2*sin(phi)^3*exp(-m*v^2/(2*k*T)*(m/(2*pi*k*T)^(3/2)dtheta (from 0 to 2*Pi) dphi (from 0 to Pi) dV (from 0 to infinity))

    which converges (I made a mistake in my earlier calculation that I just caught while typing this up, so now I am not sure how to make this integral work correctly, but I do know that it is finite, I think its a gamma function but I'll need to check again).

    Anyway, following this same line of thought, if I now go ahead and subsitute in the values for V_y or V_z in spherical and multiply by the distribution and integrate I will find these terms are zero due to the orthogonality of the two spherical harmonics that are multiplied by the spherical harmonic for the expansion of the V_x term. (I.E for those that dislike the spherical harmonic argument I have at least one odd function under an integral after these multiplications that when integrated over a period vanish to zero, thus eliminating the rest of the integral).

    So that more or less solves a).

    What I am stuck on is solving b).

    If I assume that I can take the vector distribution I have generated, I just need to integrate over the x-direction multiplied by a fixed area set by the y-z conditions. However, this raises a problem, since the last term is unable to be uncoupled the integration must still be over x,y, and z components; only this time I can't use the spherical argument to integrate because the y,z components can only be integrated between 0 and 1 not 0 to infinity such as the x direction. I thought about reworking the bounds in spherical coordinates to take this into account, but I feel this is a fool's pursuit. Any ideas?

    Urgent help is needed on this one.
     
  2. jcsd
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