Maxwell-Boltzman distribution

[squenshl]

Homework Statement

1. According to the Maxwell-Boltzman law of theoretical physics, the probability density of a gas molecule is $f(v) = kv^2e^{-\beta v^2}$ if $v > 0$ and $f(v) = 0$ otherwise, where $\beta$ depends on its mass and absolute temperature, and $k$ is an appropriate constant. Show that the p.d.f. of the kinetic energy $E$, whose values are related to those of $V$ by means of the equation $E = \frac{1}{2}mV^2$ has a Gamma distribution.

2. Repeat the derivation when $V \sim N(0,\sigma^2)$.

Homework Equations

The Attempt at a Solution

For 1. I'm not too sure what exactly they are asking. Are they asking for pdf of E and how does that relate to the pdf $f(v)$?
For 2. I'm sure it just follows on from 1.

 

[Ray Vickson]

Homework Statement

1. According to the Maxwell-Boltzman law of theoretical physics, the probability density of a gas molecule is $f(v) = kv^2e^{-\beta v^2}$ if $v > 0$ and $f(v) = 0$ otherwise, where $\beta$ depends on its mass and absolute temperature, and $k$ is an appropriate constant. Show that the p.d.f. of the kinetic energy $E$, whose values are related to those of $V$ by means of the equation $E = \frac{1}{2}mV^2$ has a Gamma distribution.

2. Repeat the derivation when $V \sim N(0,\sigma^2)$.

Homework Equations

The Attempt at a Solution

For 1. I'm not too sure what exactly they are asking. Are they asking for pdf of E and how does that relate to the pdf $f(v)$?
For 2. I'm sure it just follows on from 1.

You ask: "Are they asking for pdf of E and how does that relate to the pdf $f(v)$?" Yes, that is exactly what the question says!

 

[squenshl]

So we apply the change of variable technique here then?

 

[Charles Link]

So we apply the change of variable technique here then?

This one I think is best done by using probability distribution functions which are integrals of the density functions from minus infinity to the value v or E. The derivative of the distribution function is the density function. The G(E)=F(v(E)) where the capital letters are distribution functions. (Note v=v(E) i.e. since E=E(v), we can also write v=v(E) ) Using the calculus chain rule, you should be able to get g(E) from F(v) and f(v), etc. (f(v)=density function for speed v). Question 2 isn't completely clear to me. It seems to be using a standard normal distribution with mean 0 and standard deviation $\sigma$, but will need to take another look.

 

[andrewkirk]

@squenshl There is a standard formula in probability theory (which can be found in most texts but also can be readily derived from first principles) for the distribution of $Y\equiv g(X)$ where $g$ is a monotonic function of a random variable $X$ that has a known distribution.

You know the distribution of $V$.

Is the function $f:v\mapsto \tfrac{1}{2}mv^2$ monotonic on the domain you have specified for $V$? If so then you can use that probability formula to find the distribution of $E=f(V)$.

 

[Ray Vickson]

So we apply the change of variable technique here then?

Well, you ARE changing variables, so what do you think?

 

[squenshl]

Right then.
I have $f(v) = kv^2e^{-\beta v^2}$ and we have $E(V) = \frac{1}{2}mV^2$ which is monotonically increasing for $V >0$. Hence, we can apply the change of variable formula $f(v(E))\left|\frac{dV}{dE}\right|$. Firstly, $V(E) = \sqrt{\frac{2E}{m}}$ and $\frac{dV}{dE} = \frac{1}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}$. So the new probability function is given by $$k\left(\frac{2E}{m}\right)e^{-\beta\left(\frac{2E}{m}\right)} \times \frac{1}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}} = \frac{k}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}e^{-\frac{2\beta E}{m}}$$.

Am I on the right track by comparing this to the Gamma distribution then just finding out the appropriate parameters to make this statement true?

 

[squenshl]

I'm sure I'm doing it wrong as I'm getting the same probability function

 

[Charles Link]

I'm sure I'm doing it wrong as I'm getting the same probability function

It does look like you got the correct result other than you are missing a minus sign in the exponent 1/2 when you did dV/dE in post #7. Perhaps you are also missing the concept that a probability distribution function is the integral of the probability density function, thereby giving you the gamma function. The probability density function is often loosely called a distribution function, but the distribution function is more formally the integral of the density function. The probability distribution function , e.g. F(x) gives the probability that a variable X is less than or equal to x. It's derivative $F'(x)=f(x)$ is the probability density function.

 

[Ray Vickson]

Right then.
I have $f(v) = kv^2e^{-\beta v^2}$ and we have $E(V) = \frac{1}{2}mV^2$ which is monotonically increasing for $V >0$. Hence, we can apply the change of variable formula $f(v(E))\left|\frac{dV}{dE}\right|$. Firstly, $V(E) = \sqrt{\frac{2E}{m}}$ and $\frac{dV}{dE} = \frac{1}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}$. So the new probability function is given by $$k\left(\frac{2E}{m}\right)e^{-\beta\left(\frac{2E}{m}\right)} \times \frac{1}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}} = \frac{k}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}e^{-\frac{2\beta E}{m}}$$.

Am I on the right track by comparing this to the Gamma distribution then just finding out the appropriate parameters to make this statement true?

I get exactly the same answer as you.

I don't see your problem; you DO have a gamma distribution with easily-determined parameters!

 

[squenshl]

We have $g(E) = \frac{k}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}e^{-\frac{2\beta E}{m}} = \sqrt{2}k\left(\frac{1}{m}\right)^{\frac{3}{2}}E^{\frac{1}{2}}e^{-\frac{2\beta E}{m}}$.

The PDF of the Gamma distribution is $g(E) = \frac{1}{a^p\Gamma(p)}E^{p-1}e^{-\frac{E}{a}}$ where $\Gamma(p) = (p-1)!$.

Comparing the 2 we get $p = \frac{3}{2}$ and $a = \frac{m}{2\beta}$ as our parameters but clearly this ain't the case so I'm a little stuck?

 

[Charles Link]

We have $g(E) = \frac{k}{m}\left(\frac{2E}{m}\right)^{\frac{1}{2}}e^{-\frac{2\beta E}{m}} = \sqrt{2}k\left(\frac{1}{m}\right)^{\frac{3}{2}}E^{\frac{1}{2}}e^{-\frac{2\beta E}{m}}$.

The PDF of the Gamma distribution is $g(E) = \frac{1}{a^p\Gamma(p)}E^{p-1}e^{-\frac{E}{a}}$ where $\Gamma(p) = (p-1)!$.

Comparing the 2 we get $p = \frac{1}{2}$ and $a = \frac{m}{2\beta}$ as our parameters but clearly this ain't the case so I'm a little stuck?

Try again. Don't you get p=3/2 ?

 

[squenshl]

Try again. Don't you get p=3/2 ?

Right I stuffed it up a bit.

 

[squenshl]

So $p = \frac{3}{2}$ but I think my $a$ is wrong. I actually think $p$ is wrong too because the 2 distributions are just not comparing right.

 

[squenshl]

I get exactly the same answer as you.

I don't see your problem; you DO have a gamma distribution with easily-determined parameters!

I get $p = \frac{3}{2}$ and $a = \frac{m}{2\beta}$.

 

[squenshl]

I guess that $p = \frac{3}{2}$ and $a = \frac{m}{2\beta}$ are the correct parameters.

 

[Ray Vickson]

I guess that $p = \frac{3}{2}$ and $a = \frac{m}{2\beta}$ are the correct parameters.

Or, maybe, $a = 2 \beta/m$, depending on which definition of the Gamma distribution you are using. (I have seen it done both ways in different sources.)

 

[squenshl]

Or, maybe, $a = 2 \beta/m$, depending on which definition of the Gamma distribution you are using. (I have seen it done both ways in different sources.)

Yes right thanks! I'm still not sure how these parameters give me a correct Gamma distribution vod plugging them into the Gamma dist doesn't give $G(E)$ for me

 

[Ray Vickson]

Yes right thanks! I'm still not sure how these parameters give me a correct Gamma distribution vod plugging them into the Gamma dist doesn't give $G(E)$ for me

Yes, it does, but you need to use the actual value of $k$, which you were not given in the original question. How would you determine $k$?