# Integral of differential cross section over solid angle

1. Oct 7, 2012

### ck99

1. The problem statement, all variables and given/known data

Find σ , the differential cross section, starting from the expression below and integrating over solid angle Ω

2. Relevant equations

dσ/dΩ = r2sin2θ

3. The attempt at a solution

dσ = r2sin2θ dΩ

I remember that dΩ = sinθ dθ dμ

and doing the μ integral from 0 to 2∏ gives dΩ = 2∏ sinθ dθ

substitute in to get dσ = 2∏ r2sin2θ sinθ dθ

Now, the next step in the textbook is to substitute u = cos θ and write

σ = 2∏r2∫(1-u2)du with u = cos θ and integration limits from -1 to 1.

I have played around with this step for a while and I know the trig identity sin2 θ = 1 - cos2 θ, but I still don't get how to do this step properly. I think I'm suppose to put a du/dθ term in there, which is a -sinθ term, but I can't see how they get to the final expression in the book, which is

σ = 8∏/3

If anyone can help me work through these steps in detail I would really appreciate it; I should know how to do this by now (4th year physics MSc student!) but it's like my brain switches off every time I see the ∫ symbol and I have to learn everything from scratch every term . . . :(

2. Oct 8, 2012

### clamtrox

So when you substitute u = cos θ, you have to remember to change the integration measure as well. Using the chain rule, you have du = du/dθ dθ.

And yeah this is something you should really really really know. I suggest you go to the library, check out one of those massive analysis textbooks and go through it carefully until you at least understand the basics.

3. Oct 8, 2012

### ck99

Thanks clamtrox. I am sure I am on the right track, I just can't remember how to make it work properly. Using the trig identity from before, I can write

σ = 2∏r2 ∫(1-cos2 θ) sin θ dθ

by the chain rule, substitute dθ with (du/dθ)dθ

σ = 2∏r2 ∫(1-cos2 θ) sin θ (du/dθ)dθ

If u = cos θ then du/dθ = -sinθ

Substitute in u and cancel the dθ terms with each other (not sure I am doing this bit right!)

σ = 2∏r2 ∫(1-u2) sin θ (-sin θ) du

So I go wrong where I try to manipulate the (du/dθ)dθ part of the expression. I am trying to say that

(du/dθ)dθ = du (1/dθ) dθ = du

and cancel out the dθ terms to just leave du. But I know I am supposed to actually evaluate the du/dθ term!

They do it by saying

sin θ dθ = du where u = cos θ [A]

and it's this part I don't understand.

If I integrate the two parts of the expression in line [A] I get

∫sin θ dθ = -cos θ

∫du = u, or = cos θ

so ∫sin θ dθ ≠ ∫du

and I am even more confused!

4. Oct 8, 2012

### clamtrox

No! dθ≠(du/dθ)dθ!!

5. Oct 8, 2012

### ck99

Of course it isn't . . . .

If I try

du = du/dθ dθ

du = -sin θ dθ

dθ = 1/(-sin θ) du

Now the original expression

σ = 2∏r2 ∫(1-cos2 θ) sin θ dθ

becomes

σ = 2∏r2 ∫(1-u2 θ) sin θ/(-sin θ) du

σ = 2∏r2 (-∫(1-u2 θ) du)

And I think -∫ = ∫ with the limits swapped.

This is good! However I would just like to check that I have worked out the integral limits properly?

For the solid angle of a sphere, dμ is integrated from 0 to 2∏

and sinθ dθ is integrated from 0 to ∏. I take this to mean that the integral

σ = 2∏r2 ∫(1-cos2 θ) sin θ dθ

should also be evaluated from 0 to ∏ - is this correct?

These limits go from (0, ∏) to (-1, 1) because of the substitution, and to (1, -1) because I changed the sign of the integral. I evaluate the integral to get

- [(1/3)u3] (limits from 1,-1)

(1 - (-1)) - 1/3(1 - (-1))

2 - 2/3

4/3

So σ = 2∏r2 (4/3) = (8∏r2)/3 as required.

I don't know why I find it so difficult really; by the end of last year I was doing 4D GR calculus stuff with no trouble, but after a couple of months off it's like pulling teeth again and I am sub A-level standard!

6. Oct 8, 2012

### clamtrox

Yap now it looks pretty good.

I would still recommend you try to study analysis again. It's pretty easy and fast to learn, and it is something you have to be absolutely comfortable using if you plan on working in any job even slightly related to physics.

Often it takes some time to learn how to study stuff efficiently, and it's not uncommon to have to go back to the courses taught in the first years and go through the material again.