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