# Is cross section is proportional

1. Oct 22, 2006

### alfredbester

If I show something that a cross section is proportional to:

$$1 / (16p^2 sin^4( x) + 8mpsin^2 (x) + m^2)$$

does it imply that the cross section is proportional to

$$1 / sin^4 (x)$$ as well?

Last edited: Oct 22, 2006
2. Oct 22, 2006

### OlderDan

No. But can you factor the first denominator?

3. Oct 22, 2006

### alfredbester

$$1 / (4psin^2(x) + m)(4psin^2(x) + m)$$

Not sure what to do from there.

4. Oct 22, 2006

### OlderDan

The two factors in the denomnator are identical. What you have shown is that your thing is inversely proportional to the square of (4psin^2(x) + m). It is inversely proportiona to sin^4(x) only if m = 0. You could reach the same conclusion from your original expression, but it is a bit more evident when you factor it into a product of identical terms.

5. Oct 22, 2006

### alfredbester

That cause a problem then I was the scattering cross section is proportional to Mfi and

$$|M_{fi}|^2 /propto 1 / (q^2 + m^2)^2$$

Starting from the definition of q = $$p_{final} - p_{initial}.$$

Show that the angular dependance of the scattering is then given simply by the Rutherford formula:CS= scattering cross secion

$$CS /propto 1 /(sin^4(\theta /2)$$

I found $$q = 2psin(\theta /2)$$
Since Mfi is proportional to the scattering CS I just tried sticking q into into the expression for $$|M_{fi}|^2$$ as above but that clearly didn't work. I guess the wording angular dependance is key here.

Last edited: Oct 22, 2006
6. Oct 22, 2006

### OlderDan

I had no context for your question, and I'm not completely following what you are trying to do. I have to log off now. See if this helps you any

http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html

7. Oct 22, 2006

### alfredbester

Thanks, I've got it now m=0 because the exchange boson is a photon :).