Velocity distribution of molecules in a gas - integral help

[kettle0fish]

Homework Statement

Hullo

I'm working through a textbook, and I've come across this expression which I don't follow at all.. the section is headed 'Velocity distribution of moledcules in a gas'. I follow it up until this integral, but I've not got a clue how they do it. I've had a couple of stabs at it myself, but failing that I googled a similar integral and found something saying that it was only possible under certain limits (something to do with a gaussian integral?) Can anyone help, please?!

([mean]u^2) ̅=(∫(u^2)exp(-(mu^2)/2kT)du)/(∫exp(-(mu^2)/2kT)du)=(1/2 (2kT/m)^(3/2))/(2kT/m)^(1/2) =kT/m.

Homework Equations

u is one component of the velocity of a molecule in the gas, k is the Boltzmann constant, and T is the thermodynamic temperature.

I'm a year before university in the UK, so if the solution is some crazy maths I haven't got a hope of understanding then just say so, and I'll have to grin and bear it and just take it on authority..

Thanks!

Kettle0fish

P.S. this isn't homework! I'm on a gap year, I don't have to do homework anymore! but the silly forum is making me post it here...

The Attempt at a Solution

had a stab at the denominator, intergrated by parts but ended up with something involving the numerator..

 

[vela]

Are you familiar with doing integrals over the xy plane using polar coordinates?

 

[kettle0fish]

I'm not, no.. but it sounds like something I could follow (I've done a bit of integrating functions like y = f(x,z) and quite a bit of polar co-ordinates) :)

 

[vela]

OK, there's a trick to evaluating the basic Gaussian integral. Consider the integrals

I = \int_{-\infty}^{\infty} e^{-\alpha x^2}\,dx=\int_{-\infty}^{\infty} e^{-\alpha y^2}\,dy

You can write

I^2 = \int_{-\infty}^{\infty} e^{-\alpha x^2}\,dx \int_{-\infty}^{\infty} e^{-\alpha y^2}\,dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\alpha(x^2+y^2)}\,dx\,dy

You can think of the last integral as one over the entire xy plane. Now if you switch to polar coordinates, you get

I^2 = \int_0^\infty \int_0^{2\pi} e^{-\alpha r^2} r\,d\theta\,dr

which you can easily evaluate. The factor of r in the integrand comes from changing variables from rectangular to polar coordinates. If you're not familiar with that, you may just want to take it on faith for now that dx dy = r dr d\theta.

To evaluate the numerator, you can use the fact that

u^2 e^{-\alpha u^2} = -\frac{\partial}{\partial \alpha} e^{-\alpha u^2}

The idea here is two interchange the order of integration and differentiation:

\int u^2 e^{-\alpha u^2} du = -\int \frac{\partial}{\partial \alpha} e^{-\alpha u^2} du = -\frac{\partial}{\partial \alpha} \int e^{-\alpha u^2} du

If that seems too weird, you can just integrate by parts. (Actually, that might be easier now that I think about it.)

 

[kettle0fish]

That all looks as if it follows, I'll give it a proper look-over later

Thanks very much! It's always so much better to be able to know why :)

Kettle0fish