# Maxwell distribution of relative velocities

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1. Nov 20, 2015

### Violina

I want to calculate the distribution function for the relative velocity v_r = v_1 - v_2 of two particles. Each particle velocity follows the Maxwell distribution for velocities. They have the same mass and temperature.
Can I just multiplicate the Maxwell distributions of the two particles (and normalize it)? Can someone help me?
Thank you!

2. Nov 20, 2015

### Buzz Bloom

Hi Violina:

If I understand what you are trying to do, you want the distribution of the relative speed between two particles randomly chosen from those in a gas in equilibrium at a given temperature. You asked:
I think the problem is more complicated than that. Each particle has a velocity vector with a magnitude (speed) defined by the distribution function for the particular temperature, and an random direction orientation equally distributed over all possible angles, that is, the angular distribution between a reference point on a unit spherical surface, and a random point on that surface. Then you use the law of cosines:
to get the length of the difference vector. Then you have to integrate over four variables: 1 and 2 are the product of two Maxwell distributions for the two particles, and 3 and 4 are the longitude and latitude over a spherical surface to get the angle distribution between the two vectors.

Good luck.

Regards,
Buzz

3. Nov 23, 2015

### Violina

Thank you Buzz.
I think I don't need the cosine term because v_1 and v_2 are vectors and v_r = v_1-v_2 is also a vector. But I think I can't just set in v_r in the Maxwell distribution for velocities?
And I don't know which mass I should set in. Maybe something like: m*|(v_1-v_2)| to get something like a relative mass?

4. Nov 23, 2015

### Buzz Bloom

Hi Violina:

For some reason today I am having trouble using the quote tool, so I will quote what I want to discuss using copy and paste.
I think I don't need the cosine term because v_1 and v_2 are vectors​
I think we are on different channels. You need to calculate the joint distribution of two 3D random vectors. In spherical coordinates. Each of the two distributions is a Maxwell distribution of a radial length times a uniform distribution over a unit sphere.
And I don't know which mass I should set in.​
The problem you described in post #1 was about velocities, not momenta. I am not sure how you want to define a momentum problem. If you want the momentum of one particle relative to the position of a second particle, you use the mass of the first particle.

Regards,
Buzz

Last edited: Nov 23, 2015
5. Nov 23, 2015

### Violina

Thank you very much.
Here is the original exercise. Maybe then you can better understand, whats my problem:

Consider a gas in which the velocities of the particles are isotropically distributed and follow the
Maxwell distribution

f(vi) = 4(pi) *(m/(2 pi k T))^3/2 (v_i)^2 * exp (- mv_i^2/(2kT))

For two random particles i = 1 and i = 2, the (vectors of the) velocity of the centre of mass and the
relative velocity are defined as

V =(v_1 + v_2) /2
v_r = v_1 - v_2

Calculate the distribution F(vr) of the relative velocity in analogy to the Maxwell distribution. Compare
F(vr) and f(v). HINT: Express v1 and v2 in terms of V and vr.

6. Nov 23, 2015

### Buzz Bloom

Hi Violina:

I am having some difficulty reading the equation in the quote and matching it with the equation below from Wikipedia.

Here the vector vi is represented in Cartesian coordinates. Note that you have two elements in your equation that are absent from the equation just above.
(1) 4π to the left of the first "*"
(2) vi2 to the left of the second "*"

(1) I am guessing that your 4π is because your equation gives the distribution just for the length of the vector vi while ignoring the various angles of the vector. Remember that 4π is the surface area of a unit sphere, would result from ingratiating the angle distribution over all possible angles.
(2) I have no idea where your vi2 comes from.

The suggestion
is good. The result you want is the distribution of vr. V can be ignored because it is irrelevant to the distribution of vr.

Regards,
Buzz

7. Nov 23, 2015

### Staff: Mentor

You should have looked higher up on that page, as Violina's equation is the first one given.

(1) and (2) are because the equation we are considering the Maxwell-Boltzmann distribution of speed. The 4π is indeed due to the angular distribution, while the v2 is due to the density of states: the higher the speed, the more possibilities you have to decompose that speed into different components.

8. Nov 23, 2015

### Buzz Bloom

Hi DrClaud:

I appreciate your pointing out to me where Violina's equation came from. I apologize for my carelessness in not noticing this on the Wikipedia page.

I also thank your for explanation about vi2, although I am still somewhat confused. I get that the form with vi2 is part of the difference, along with the 4π, between a distribution of speeds and a distribution of velocities. I am guessing that an additional integration of the vector form in spherical coordinates, with one of the two particles at the origin, would result in the vi2, but I don't understand what that integral would look like from your "the higher the speed, the more possibilities you have to decompose that speed into different components". Can you give a bit more detail about this explanation?

Regards,
Buzz

Last edited: Nov 23, 2015
9. Nov 23, 2015

### Staff: Mentor

My answer was maybe too complicated for nothing. It appears the same way when going from Cartesian to spherical coordinates:
$$\mathrm{d}^3\mathbf{r} \Rightarrow r^2 \sin \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\phi$$

10. Nov 23, 2015

### Buzz Bloom

Hi DrClaude:

I think I get it now. The r2 becomes is the vi2 because r is the speed of the velocity vector.

Thank you very much for your response to my question.

Regards,
Buzz

11. Nov 23, 2015

### Buzz Bloom

Hi Violina:

I want to suggest again you that use the cosine law, and calculate the distribution of the angle. It comes from choosing a particular orientation angle for v1, like the North pole and allowing the other orientation angle to be any point on a unit sphere. Integrate over the longitude angle, φ, in polar coordinates (see the RHS of post #9 with r2 dr=1) to get the distribution of the angle between v1, the North pole, and v2, which is α = π/2 - θ. Thus
cos α.= sin θ.​

I have had an insight following the digestion of DrClaud's posts, together with something I read in the Wikipedia article:
It can also be seen that the Maxwell–Boltzmann velocity distribution for the vector velocity
[vx, vy, vz] is the product of the distributions for each of the three directions:​
Also, each of the three coordinate distributions, f(x), f(y) and f(z) are all the same Gaussian distribution. This means that my suggestion to use the Cosine Law and polar coordinates is a bad idea. Think about combining the factored form of the distribution for three coordinates, x, y, z, for two vectors. Think about the combined distribution for each coordinate separately.

Hope this helps.

Regards,
Buzz

Last edited: Nov 23, 2015