# Spontaneous Random Neutron Generation in a Spher

1. Sep 14, 2008

### Enjolras1789

This question concerns a problem in Arfken and Weber (from the infinite series chapter, after the power series section). I went to the homework section, and the titles beneath each section specifically imply that a question from a graduate book is inappropriate for that section. I thus post it here. I apologize if this is the wrong place.

The problem reads,"Neutrons are created by a nuclear reaction inside a hollow sphere of radius R. The newly created neutrons are uniformly distributed over the spherical volume. Assuming that all directions are equally probable, what is the average distance a neutron will travel before striking the surface of the sphere? Assume straight line motion, no collisions." It then goes on to give steps on the way of the answer, one stating that the result is that

mean distance = 3/2 R integral( 0 to 1) integral (0 to pi) square root [(1-K*K sin(theta)*sin(theta)] K*K* dk sin(theta) d(theta)

No, I have no idea what K is physically, except by the nature of of what looks like the differential element at the end (but I am confused as to how one might get a distance variable inside a square root times sine of the angle).

Although help in working toward this answer would be appreciated, my request is more meager. I don't understand why the answer isn't simply R/2. If particles are spontaneously forming uniformly in a sphere, and there is total isotropy in direction, and no collisions, I would think that the mean distance traveled by a particle until colliding with the surface would be simply R/2. Why isn't it that simple?

2. Sep 14, 2008

### CompuChip

Well the equation you gave looks like an integration in spherical coordinates.
Isn't k^2 sin^2(theta) = x^2 + y^2 (in Cartesian coordinates) ?

3. Sep 14, 2008

### Enjolras1789

Perhaps you mean a LaPlacian in spherical coordinates? That was my thought, that somehow the 1/(r*r sin(theta) sin(theta)) term in front of the partial derivative of the function with respect to phi. However, it's not obvious to me why I would take a LaPlacian of something, seeing as I physically do not understand why the problem isn't very simple to just being R/2

4. Sep 14, 2008

### CompuChip

No, I meant an integration:

$$\iiint_V f(x, y, z) dx dy dz = \iiint f(r \cos\phi \sin\theta, r \sin\phi \sin\theta, r \cos\theta) r^2 \sin^2\theta dr d\theta d\phi$$

In this case, you would have f(x, y, z) = 3/2 R sqroot(1 - x^2 - y^2)
If (x, y, z) is on a sphere of radius R, that's just 3/2 R sqroot(z^2).

I'm not really into this material, but that's the mathematics I see in there; probably you can relate it to something physical more easily than I can...

5. Sep 14, 2008

### Enjolras1789

You are good; thank you very much for your insight in seeing that form. I will try to understand why it is that this functional form of f is the case. PS, if you are more the mathematician type, I posed a question in the Analysis section of PF that I am very curious about concerning the "best" convergence tests to use.