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, $v=(v_x^2+v_y^2+v_z^2)^{1/2}$ $dv=?$