# Integrating a theta function/2d probability density function

1. Dec 22, 2015

### Thimble

1. The problem statement, all variables and given/known data
A two-dimensional circular region of radius a has a gas of particles with uniform
density all travelling at the same speed but with random directions. The wall of the
chamber is suddenly taken away and the probability density of the gas cloud subsequently
satisfies

$$\rho(r,\theta;t)=\frac{1}{\pi a^2}\int_0^{2\pi} \frac{d\phi}{2\pi} \Theta[a^2-r^2-(vt)^2+2rvtcos(\theta-\phi)]$$

where (r, θ) are the polar-coordinates of the point at issue, t is the time, v is the velocity
of the gas,Φ is the random orientation and Θ[x] vanishes if x is negative and is unity
if x is positive. Employing the equations below show that for each Φ and t the region mapped out by Θ is a circle and hence that the density is normalised.

2. Relevant equations
$$x=rcos(\theta)$$
$$y=rsin(\theta)$$
$$d_x=vtcos(\phi)$$
$$d_y=vtsin(\phi)$$

3. The attempt at a solution
Firstly I rearranged the theta expression to get
$$\Theta[a^2-(x-d_x)^2-(y-d_y)^2]$$ which I think satisfies the first part of the question with each Φ and t mapping out a circle with radius a and centre (dx,dy)

Now for the second part i'm thinking that i've got to show that the following statement is true
$$\int_{\phi=0}^{2\pi} \int_{r=r_{min}}^{r_{max}} \rho(r,\theta;t)rdrd\theta =1$$

where rmaxand rmin
are the maximum and minimum radii of the region the gas is in at that time which I think is a circle for a>vt and a doughnut shape for a<vt by thinking of the entire region as a series of circles of the same radius with centers moving away from the origin at the same speed in every possible direction.

The stumbling block i'm having is trying to integrate this theta function. My attempt at it was just treating the theta function as one as for all Φ there are points where the expression inside the theta is greater than zero and thus Θ=1, but this gives a constant probability density which doesn't make sense to me as the area over which the gas is in is expanding so the probability density must surely decrease. I've never come across this type of function before and all the examples i've seen on the internet for them all seem to be in one variable rather than 4 (r,θ,t and Φ)so could anyone point me towards any guidance on how to integrate these types of functions? Thanks in advance for any help.
EDIT: Changed upper limit in the first integral to 2π.

Last edited: Dec 22, 2015
2. Dec 22, 2015

### vela

Staff Emeritus
I think you mean $\theta=0$ on the lower limit, right? Use 0 and $\infty$ for the limits of the integral over $r$. For a fixed $\phi$ and $t$, the $\Theta$ function will only be equal to 1 when $r$ and $\theta$ correspond to a point inside the circle of radius $a$, so you can write down what the result of the double integral should be. Then you still have the integral over $\phi$ to take care of.

3. Dec 23, 2015

### Thimble

Yes I should have θ=0 on the bottom limit. I don't see why we use 0 and ∞ as the limits for r, surely the integral at r=∞ is 0 and after a certain time the integral at r=0 will be 0 as well (As Θ will be zero at these values)? Also i'm still struggling to figure out how to obtain what ρ(r,θ;t) is to be able to do the double integral.

4. Dec 23, 2015

### vela

Staff Emeritus
Generally, to show a probability density is normalized, you integrate over all space and show the result is equal to 1. You can certainly change the limits to omit the regions where the integrand vanishes, but why bother? It doesn't change the result and is an unnecessary complication.

You've been given $\rho$. What's there to obtain?

5. Dec 23, 2015

### Thimble

I've been given ρ in terms of that integral of Θ that i'm not sure how to calculate. Without having the actual value for ρ rather than in terms of that definite integral i'm not sure how to show that the integral of ρ is equal to 1, unless i'm missing something obvious.

6. Dec 23, 2015

### vela

Staff Emeritus
Keep $\rho$ in terms of the integral of $\Theta$, plug it into the double integral, and then change the order of integration so you integrate over $\phi$ last.

7. Dec 24, 2015

### Thimble

Ah, that makes sense. Just two more quick questions if you don't mind. As r would be treated as a constant in an integral with respect to Φ can i just take it inside the integral as follows to get
$$\int_{\theta=0}^{2\pi}\ \int_{r=0}^\infty \int_{\phi=0}^{2\pi} \frac{r \Theta[a^2-r^2-(vt)^2+2rvtcos(\theta-\phi)]}{2\pi^2a^2} d\phi dr d\theta$$
I'd then obviously take out the constants to make it look neater. Then am i able to just shift around the order that the integrals are performed in with no further changes? I've seen some places saying I can and some that I can't. Sorry for having so many questions and thanks for your help so far.