Velocities for a modified maxwell distribution (an interesting problem)

[HappyEuler2]

Homework Statement

Consider a momentum distribution such that:

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

where $$\alpha$$ is the angle between $$\overline{p}$$ and $$\alpha$$

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

Homework Equations

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

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.

 

[mark726]

Hello,

Thank you for your detailed post. I am also a scientist and here are my thoughts on your questions:

a) The expectation value for U can be calculated by taking the integral of the momentum distribution multiplied by the velocity components, as you have done in your attempt at a solution. However, instead of using spherical coordinates, I would recommend using Cartesian coordinates, as it will simplify the integration. The expectation value for U_x can be calculated by integrating over the x-component of the momentum distribution multiplied by the velocity in the x-direction. The same can be done for U_y and U_z. Since the cosine term is an odd function, it will integrate to zero, as you have correctly noted. The final result for the expectation value of U will be the sum of the three components, as you have also correctly noted.

b) To calculate the number of particles passing through a unit area at any given time in the positive x-direction, we can use the same approach as in part a). We will integrate the momentum distribution multiplied by the velocity in the x-direction over a unit area in the y-z plane. However, since the integration is only over the positive x-direction, we can set the lower bound of the integration to zero and the upper bound to infinity. This will give us the total number of particles passing through the unit area in the positive x-direction. To calculate the number of particles passing through the unit area in the negative x-direction, we can simply take the negative of the result obtained from the integration.

I hope this helps. Let me know if you have any further questions or if you need any clarification. Good luck with your work!