Solving a Relativistic Ideal Gas Momentum Probability Distribution

[Xerxes1986]

Homework Statement

For a relativistic ideal gas, the momentum probability distribution is given by

where [URL]http://latex.codecogs.com/gif.latex?\epsilon_p=\sqrt[]{m^2c^4+c^2p^2}.[/URL] Determine A

Homework Equations

The Attempt at a Solution

I know that:
[URL]http://latex.codecogs.com/gif.latex?\int_{-\infty}^{\infty}Ae^{-\frac{\epsilon_p}{k_bT}}dp=1[/URL]

Which boils down to:
[URL]http://latex.codecogs.com/gif.latex?A\int_{-\infty}^{\infty}e^{-\frac{\sqrt[]{m^2c^4+c^2p^2}}{k_bT}}dp=1[/URL]


I have no idea how to integrate this function. I have tried substitution, integration by parts...everything. After researching on line I know that the solution is:

[URL]http://upload.wikimedia.org/math/4/e/6/4e6b1c37ebe913700d63984beaa7a429.png[/URL]


But I need help understanding how a modified Bessel function pops into the solution...I don't even really understand what the Bessel function is.

 

[Thaakisfox]

First of all the momentum density distribution you gave is not the distribution of the absolute value of the momentum vector. i.e. it is not the density of the probability of finding a particle between p and p+dp. Rather it is the density probability of finding a particle with momentum (p_x+dp_x,p_y+dp_y,p_z+dp_z) i.e. it is the distribution density of the momentum vector.

So to normalize we have to integrate over all three directions:

$$1=\int d^3p A\cdot \exp\left(\sqrt\left(m^2c^4+c^2(p_x^2+p_y^2+p_z^2\right)\right) = \int_{-\infty}^{\infty}dp_x\int_{-\infty}^{\infty}dp_y\int_{-\infty}^{\infty}dp_z A\exp\left(\sqrt\left(m^2c^4+c^2(p_x^2+p_y^2+p_z^2\right)\right)$$

Now this integral looks quite tough, hence let's change to spherical coordinates in the momentum space. We see that the integrand doesn't depend on the angles hence In this case we know that the volume element is simply $$4\pi p^2 dp$$ where obviously $$p^2=p_x^2+p_y^2+p_z^2$$.

So the integral will look like:

$$1=4\pi A\int_0^{\infty}dp p^2e^{-\frac{mc^2}{kT}\sqrt{1+\frac{p^2}{m^2c^2}}}$$

Now put $$\sinh u = \frac{p}{mc}$$
Then perform the partial integration (which is a bit tricky, but i will let you think about it). And you will recognize one of the integral representations of the modified Bessel function of the second kind with order 2.

 

[Thaakisfox]

The Bessel functions are basically solutions of the Bessel differential equation. The usually arise when writing the laplacian in cylindrical coordinates and separating the variables. The radial part will be the Bessel differential equation. The solutions to these are the Bessel functions.
In this case however we have modified Bessel functions. You can think of these just as of the usual Bessel functions, just with a purely imaginary argument.
They can be represented in many forms but not as a combination of elementary functions. Thats why they have a separate name and are called special functions. They have many power series representations, integral representations etc.
Usually as a rule of thumb when you see an integral which has an exponential term with an argument containing sines or cosines, hyperbolic functions , then it is most probably a representation of a Bessel function.
In our case as you will see the exponential term contains a hyperbolic cosine function, hence we can guess that it will be a modified Bessel function.
Here are some of the integral representations of the modified Bessel function we will obtain:

http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/07/01/01/

Out of these number 3 will be the one you need.