How is the Maxwell-Boltzmann Distribution Derived?

[Heisenberg2001]

Homework Statement

Consider a monoatomic gas, whose momentum relation is $\vec{p}=m\vec{v}$, where $\vec{v}$ is the three-dimensional velocity of the gas particles.


1. Show that the classical partition function can be written as $$z=\frac{4\pi m^3}{h^3}\int dv v^2e^{\beta \frac{mv^2}{2}}.$$

2. Let $f(v)=Av^2e^{\beta \frac{mv^2}{2}}.$ Find $A$ such that $\int_{0}^{\infty}dv f(v)=1.$

Relevant Equations

$\vec{p}=m\vec{v}\, , \,z=\frac{4\pi m^3}{h^3}\int dv v^2e^{\beta \frac{mv^2}{2}}$

$f(v)=Av^2e^{\beta \frac{mv^2}{2}}\, , \,\int_{0}^{\infty}dv f(v)=1$

1.

$\vec{p}=m\vec{v}$

$H=\frac{\vec{p}^2}{2m}+V=\frac{1}{2}m\vec{v}^2$

$z=\frac{1}{(2\pi \hbar)^3}\int d^3\vec{q}d^3\vec{p}e^{-\beta H(\vec{p},\vec{q})}$

$z=\frac{Vm^3}{(2\pi \hbar)^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}$

$z=\frac{Vm^3}{(2\pi \frac{h}{2\pi})^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}$

$z=\frac{Vm^3}{h^3}\int d^3 \vec{v}e^{-\beta \frac{mv^2}{2}}$

I am lost at this point of the solution. Can someone assist me with this?

2.

$\int_{0}^{\infty}dv f(v)=1$

$\int_{0}^{\infty}Av^2e^{\beta \frac{mv^2}{2}}dv=1$

$A\int_{0}^{\infty}v^2e^{\beta \frac{mv^2}{2}}dv=1$

$A\frac{1}{4\left(\frac{\beta m}{2} \right)}\sqrt{\frac{\pi}{\frac{\beta m}{2}}}=1$

$A=\sqrt{\frac{4\beta ^3m^3}{2\pi}}$

$A=4\pi \sqrt {\left(\frac{\beta m}{2\pi} \right)^3}$

$A=4\pi \left( \frac{m}{2\pi k_BT} \right)^\frac{3}{2}$

 

[fresh_42]

I edited your post. Please see: https://www.physicsforums.com/help/latexhelp/

 

[vela]

I am lost at this point of the solution. Can someone assist me with this?

Change to spherical coordinates and integrate the angles out.

 

[gurbir_s]

Change to spherical coordinates and integrate the angles out.

There would still be a minus sign left in the exponential and a volume term V outside.

 

[vela]

There would still be a minus sign left in the exponential and a volume term V outside.

It appears $z$ is the contribution from just one particle to the partition function, and there should be a factor of $V$. Also, the exponential should have a minus sign. It could simply be typos by the OP.

 

[gurbir_s]

It appears $z$ is the contribution from just one particle to the partition function, and there should be a factor of $V$. Also, the exponential should have a minus sign. It could simply be typos by the OP.

Yes, $z$ should be the contribution from one particle only as the integral is over a 6d phase space.

 

[vanhees71]

Also note that a - is missing in the exponential ;-).

 

[Heisenberg2001]

Also note that a - is missing in the exponential ;-).

How do I edit the equations that I have typed? I can't see an edit button.

 

[jim mcnamara]

After a short period of time, posts cannot be edited. Paste your latex into a fresh new post, then edit it to make your changes. Or add comments....