Maxwell distribution in thermal equilibrium

[Firben]

Homework Statement

In thermal equilibrium, the particle in a gas are distributed in velocity space according to the Maxwell distribution

f(v) = A*exp(-mv^2/(2KT))

What is the average velocity ? What is the most probable velocity ?

Homework Equations

<v> = ∫∫∫vf(v)d³v = (0 to infinty) ∫∫∫v_xf(v)d³vx^{^} + ∫∫∫v_yf(v)d³vy^{^} + ∫∫∫v_zf(v)d³vz^{^}

The Attempt at a Solution

I started with the v_y coordinate

[0 to inf] ∫∫∫v_yAexp(-mv^2/(2KT))dy = ∫[0 to 2pi]∫[0 to pi][0 to inf]v_yA*exp(-mv²/(2KT))*v²_y sin(θ)dθdvdφ = 4piA∫v_y³exp(-mv^2/(2KT)) dv = Here is the problem, if i evaluate the integral i got a non-zero value. I know the 3 integrals should be zero since they are mult. of odd times even. And the most probable velocity is just the derivative of the f(v) function by resp. coordinate right ?

 

[TSny]

If you are going to use spherical coordinates, then you will need to express v_y in terms of v, θ, and φ before you integrate over θ and φ.

Or, you can use Cartesian coordinates and write the volume element as dv_xdv_ydv_z. Then you would integrate over the variables v_x, v_y, and v_z.

 

[Firben]

Yes, i forgot to include that in my latex code, but i have done that. If i integrate with cartesian coordinates i got:
[0 to inf]∫∫∫Ae^{-m/(2KT)*(V_x2+v_y2+v_z2}dv_xdv_ydv_z =[0 to inf] ∫Ae^{-m/(2KT)*v_x2} +--- = non-zero value which is wrong. What goes wrong ?

 

[TSny]

If you are trying to get the average y-component of velocity, then you need to include a factor of v_y in the integrand.

Also, what are the correct limits in the integral for v_x, v_y, and v_z? (Note that the components of velocity can be negative.)

 

[Firben]

Did you mean the integral = 4piA∫vy3exp(-mv^2/(2KT)) dv or [0 to inf] ∫Ae-m/(2KT)*vx2 (x component) ?

 

[Firben]

If i do the following integral

v^- = [inf to -inf]∫vƒ_m^{^}d³v = and put in spherical coordinates and simplify i got 4π(πv_th²)^-3/2*v_th⁴ [inf to -inf] ∫y³ e^-y2dy = [Part. int.] =...= 2*(2KT/mπ)^1/2

It should be zero

 

[TSny]

The Maxwell probability distribution is $Ce^{-m(v_x^2+v_y^2+v_z^2)/(2kT)}$ where $C$ is a constant.

This means that the average value of any function $f(v_x, v_y, v_z)$ is given by $$C\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(v_x, v_y, v_z)e^{-m(v_x^2+v_y^2+v_z^2)/(2kT)}dv_xdv_ydv_z$$

So, to find the average value of $v_x$, you would let $f(v_x, v_y, v_z) = v_x$. Try this and see what you get.

It's important not to confuse the probability distribution for the variables $v_x, v_y, v_z$ (as given above) with the probability distribution for the single variable $v$ (the speed of the particle). The distribution function for $v$ is $Bv^2e^{-mv^2/(2kT)}$, where $B$ is a constant.

 

[Firben]

If i integrate with resp. cartesian coordinates i got it to be equal to C*((2πKT)/m)^3/2*(KT/m). Since the Gaussian integral is equal to (2πKT/m)^1/2. Is something missing here ?

 

[TSny]

Suppose you are trying to find the average value of $v_x$. What does the integrand look like when you are integrating over all $v_x$? Is the integrand an even function or an odd function of $v_x$?

 

[Firben]

If i integrate over v_x: [-inf to inf]C∫∫[-KTv_x/m*exp(-m(vx^2+vy^2+vz^2)/(2KT))](-inf to inf] + KT/m [-inf to inf]∫exp(-m((vx^2+vy^2+vz^2)/(2KT)) dvy dvz. The integral [-inf to inf] Is odd right ?. But i don't know if vx is even or odd

 

[TSny]

$$<v_x> = C\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}v_xe^{-m(v_x^2+v_y^2+v_z^2)/(2kT)}dv_xdv_ydv_z$$

$$<v_x> = C\int_{-\infty}^{\infty}v_xe^{-mv_x^2/(2kT)}dv_x\int_{-\infty}^{\infty}e^{-mv_y^2/(2kT)}dv_y\int_{-\infty}^{\infty}e^{-mv_z^2/(2kT)}dv_z$$

What do you get when you evaluate the first integral on the right?

 

[Firben]

This was the integral i just evaluated above which i got to be equal to C*((2πKT)/m)^3/2*(KT/m). One with part. integration(and gaussian) and the last two integrals is the gaussian integral.

 

[TSny]

Did you integrate the first integral from -∞ to +∞? The integrand is $$v_xe^{-mv_x^2/(2kT)}$$ If you replace $v_x$ by $-v_x$ what happens to the overall sign of the integrand?

 

[vela]

If i integrate over v_x: [-inf to inf]C∫∫[-KTv_x/m*exp(-m(vx^2+vy^2+vz^2)/(2KT))](-inf to inf] + KT/m [-inf to inf]∫exp(-m((vx^2+vy^2+vz^2)/(2KT)) dvy dvz. The integral [-inf to inf] Is odd right ?. But i don't know if vx is even or odd

Since you're having problems evaluating the integral, it would help if you showed what you're doing rather than just saying you got the wrong answer.

I don't know if it's just a typo, but what's up with the red part?

[-inf to inf]C∫∫[-KTv_x/m*exp(-m(vx^2+vy^2+vz^2)/(2KT))](-inf to inf] + KT/m [-inf to inf]∫exp(-m((vx^2+vy^2+vz^2)/(2KT)) dvy dvz

 

[Firben]

I got the right integral to be equal to C[-inf to inf on all integrals]∫v_xe^{((-m*v_x2))/(2KT)}dv_x∫e^{((-m*v_y2))/(2KT)}dv_y*((2πKT)/(m)) = C((2πKT)/(m))^(1/2)∫v_xe^{((-m*v_x2))/(2KT)}dv_x*((2πKT)/(m)) = C((2πKT)/(m))^(1/2) ([-/KTv_x)/m*e^-mv_x2/(2KT)]+ KT/m∫e^{-mv_x2/(2KT) dv_x = the first is equal to zero and the last is (2πKT/m)^(1/2) =C(2πKT/m)^(3/2)*(KT/m)}

If i replace v_x with -v_x in the integral then i got it to be equal to C[inf to -inf]v_xe^{-(mv_x2)/(2KT)}, when i switch -inf to inf

 

[TSny]

The integral over $v_x$ has an odd integrand, so the integral must be zero. There is then no need to evaluate the other integrals over $v_y$ and $v_z$.

 

[vela]

I got the right integral to be equal to C[-inf to inf on all integrals]∫v_xe^{((-m*v_x2))/(2KT)}dv_x∫e^{((-m*v_y2))/(2KT)}dv_y*((2πKT)/(m)) = C((2πKT)/(m))^(1/2)∫v_xe^{((-m*v_x2))/(2KT)}dv_x*((2πKT)/(m)) = C((2πKT)/(m))^(1/2) ([-/KTv_x)/m*e^-mv_x2/(2KT)]+ KT/m∫e^{-mv_x2/(2KT) dv_x = the first is equal to zero and the last is (2πKT/m)^(1/2) =C(2πKT/m)^(3/2)*(KT/m)}

If i replace v_x with -v_x in the integral then i got it to be equal to C[inf to -inf]v_xe^{-(mv_x2)/(2KT)}, when i switch -inf to inf

The problem is that you're not integrating by parts correctly. In fact, integration by parts doesn't work with this integral; the substitution $u=v_x^2$ would work. The easiest route, of course, is what TSny has said. The integrand is odd, so the integral must vanish.