Maxwellian Distribution of Velocities

[cahill8]

Homework Statement

A group of stars in a Maxwellian distribution have a one-dimensional velocity dispersion \sigma. The number of objects within an element d^3v is

dN=F(v) d^3 v=\beta\hspace{2pt}Exp[-v^2/2 \sigma^2]d^3v where \beta is a constant

Find that the mean speed \bar{v}=\sqrt{8/\pi} \text{ }\sigma and \bar{v^2}=3\sigma^2

Homework Equations

I think I need to use these:

\bar{v}=<v>=\int^\infty_0 v F(v) dv
\bar{v^2}=<v^2>=\int^\infty_0 v^2 F(v) dv

The Attempt at a Solution

I'm trying to find the right approach, neither of the above integrals yield the correct answer. Here's what I tried:

F(v) can be found in the equation for dN
F(v)=\beta Exp[-v^2/2\sigma^2]

\bar{v}=\beta \int^\infty_0 v Exp[-v^2/2\sigma^2] dv
\bar{v}=\beta [4 \sigma^4]
which can equal \sqrt{8/\pi} if \beta=\sqrt{2}/2\sigma^3

However then \bar{v^2}=\beta \int^\infty_0 v^2 Exp[-v^2/2\sigma^2] dv
\bar{v^2}=\beta [16 \sigma^6]

Putting in \beta=\sqrt{2}/2\sigma^3:
\bar{v^2}=\sqrt{2}\hspace{3pt} 8 \sigma^3

while the given answer above was \bar{v^2}=3\sigma^2

Where have I gone wrong? Thanks

 

[ideasrule]

Homework Statement

Find that the mean speed \bar{v}=\sqrt{8/\pi} \text{ }\sigma and \bar{v^2}=3\sigma^2

Are you sure that's the answer? It is for the 3-dimensional case, but we're only talking about a 1-dimensional distribution.

\bar{v}=\beta \int^\infty_0 v Exp[-v^2/2\sigma^2] dv
\bar{v}=\beta [4 \sigma^4]

You made a mistake somewhere in the integration.

However then \bar{v^2}=\beta \int^\infty_0 v^2 Exp[-v^2/2\sigma^2] dv
\bar{v^2}=\beta [16 \sigma^6]

This is also not correct. How did you integrate these functions?

 

[cahill8]

Thanks for the reply. That is the answer given (from the textbook galaxy dynamics, problem 4.18)

Your right those answers were wrong, I calculated them using mathematica when writing this topic but inputted the equation wrong.

The first can be done by making a u substitution u=v^2/2\sigma^2 which leads to an answer of \bar{v}=\beta \sigma^2. This can be correct if \beta=\dfrac{\sqrt{8/\pi}}{\sigma}

In order to check this, I calculate \beta \int^\infty_0 v^2 exp[-v^2/2\sigma^2] dv in mathematica, this gives \bar{v^2}=\beta \sqrt{\pi/2} \hspace{2pt}\sigma^3 and after substituting the above \beta:

\bar{v^2}=\sqrt{\dfrac{8\pi}{2\pi}}\sigma^2=2 \sigma^2 While close, the correct answer is 3\sigma^2.

Can you see where I went wrong?

 

[ideasrule]

I got the exact same answer, so as long as the question is one-dimensional, I think we're both right and the answer key is wrong.

The reason the answer is different for 3 dimensions is that dN has an extra factor of v^2, to account for the fact that the number of combinations of one-dimensional velocities that give a total speed of "v" increases as v^2. However, no such factor exists for the one-dimensional case.

 

[cahill8]

Well another part of the question says the mean of one component of velocity, \bar{v_x^2} = \sigma^2

could I simply continue and say \bar{v_x^2}=\beta\int^\infty_0 v_x^2\hspace{2pt} exp\left[-(\sqrt{v_x^2+v_y^2+v_z^2})^2/2\sigma^2\right] dv= \beta\int^\infty_0 v_x^2\hspace{2pt} exp\left[-(v_x^2+v_y^2+v_z^2)/2\sigma^2\right] dv
Not so sure about dv though, v=(v_x^2+v_y^2+v_z^2)^{1/2} dv=?