# Maxwell-Boltzmann Distribution

• Heisenberg2001
In summary, the OP is trying to find the contribution of one particle to the partition function. There should be a factor of ##V## and the exponential should have a minus sign.
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}##

Last edited by a moderator:
Heisenberg2001 said:
I am lost at this point of the solution. Can someone assist me with this?
Change to spherical coordinates and integrate the angles out.

vela said:
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.

gurbir_s said:
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.

vela said:
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.

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

vanhees71 said:
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.

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