Make use of Gaussian symmetry?

[AStaunton]

The question is:

for a gas of weakly interacting molecules show that <v_x>=0

where <v_x> is the average velocity in the x direction.

the probability of a molecule having a velocity v is given by:

$$p(v_{X})=\sqrt{\frac{m}{2\pi kT}}e^{-\frac{mv_{x}^{2}}{2kT}}$$

The above is a Gaussian curve with max amplitude sqrt(m/2(pi)kT) at v_x=0 and it decays symmetrically either way. and of course due to the symmetry ie. the Gaussian is an even function the positive velocities will cancel the negative velocites and thus the average v_x will be 0...

This is the only way I can show that <v_x>=0. can anyone suggest a more formal/"proofy" way to show this?

 

[JaWiB]

Definition of expectation value: $$<v_x>=\int_{-\infty}^{\infty}v_x{\rho}(v_x)dv_x$$

At a glance, I don't think it's a difficult integral to calculate

 

[JaWiB]

In fact, you probably don't even need to do any calculation since v_x is an odd function and rho(v_x) is an even function, you should get an odd function when you multiply them.

 

[AStaunton]

I notice that you're integrating from -infinity to infinity...

The integral I have seen to give <v> goes from 0 to infinity, alll else is the same as what you have shown, also I assume you meant dv_x instead of dx?

Why do you integrate from -infinity instead of 0?

 

[AStaunton]

not to fire too many questions at ya, but can you explain how you know v_x is an odd function?

 

[JaWiB]

The particles can travel in the negative direction. If you don't allow that, then all of them have a positive velocity and of course the average will not be zero. And yes I meant dv_x; I tried to change it but maybe it didn't go through quickly enough...

not to fire too many questions at ya, but can you explain how you know v_x is an odd function?

Let f(v_x) = v_x. Then f(-v_x) = -v_x = -f(v_x). Since f(-v_x) = -f(v_x) it must be odd (unless I'm remembering the definition of an odd function wrongly)

 

[AStaunton]

just to clarify - what I said that the way to calculate probability was to go from 0 to infinity as in:

$$\int_{0}^{\infty}vf(v)dv$$

where:

$$f(v)=4\pi(\frac{m}{2\pi kT})^{\frac{3}{2}}v^{2}e^{-mv^{2}/2kT}$$

The difference is that here we're talking about speed where as with the Gaussian plot, that was with velocity which is why you integrate from -infty to infty? and of course speed can't be negative