# Feynmann diagram

1. Jun 9, 2007

### physics_fun

1. The problem statement, all variables and given/known data
2. Relevant equations
Hi, I'm studying Feynmann diagrams; electron-positron scattering.
If the 4-momentum of the incoming electron (positron) is p1 (p2) and of the outgoing electron (positron) is p3 (p4), momentum conservation gives
p1 + p2 = p3 + p4

If you now consider the scattering in the frame where the threemomentum of the incoming positron = 0. Why does then defining threemomentum p4=p3 imply that they equal 0?
And why does defining p3*p4=0 (three vectors) imply that p3(threevector)=0 or p4=0?

3. The attempt at a solution

I tried: 4-vectors: p1+p2=p3+p4
square this: (p1)^2 + (p2)^2 + 2*p1*p2= (p3)^2 + (p4)^2 + 2*p3*p4

(p1)^2=(p3)^2=m (electron mass)
(p2)^2=(p4)^2=M (positron mass)

So: p1*p2=p3*p4
The lab frame condition gives:
p1(0)*p2(0)=p3(0)*p4(0)-p3*p4(three-vectors)

But what are the next steps????
(I hope I made clear the problem, it isn't very readable I'm afraid....)

2. Jun 10, 2007

### physics_fun

I'll try a little LATEX to make the problem more clear:

$$p_{1} + p_{2} = p_{3} + p_{4}$$
These are four vectors of the in- and going momenta

You take the frame where the threemomentum $$p_{2}=0$$

Questions:
1) Why does then defining threemomentum $$p_{4}=p_{3}$$ imply that $$p_{4}=p_{3}=0$$? (threemomenta!)
And why does defining $$p_{3}*p_{4}=0$$ (three vectors) imply that $$p_{3}=0 or p_{4}=0$$?

3. The attempt at a solution

I tried: 4-vectors: $$p_{1} + p_{2} = p_{3} + p_{4}$$
square this: $$(p_{1})^(2) + (p_{2})^(2) + 2p_{1}p_{2}= (p_{3})^(2) + (p_{4})^(2) + 2p_{3}p_{4} // (p_{1})^(2)=(p_{3})^(2)=m (electron mass) (p_{2})^(2)=(p_{4})^(2)=M (positron mass)$$
So: $$p_{1}p_{2}=p_{3}*p_{4}$$
The lab frame condition gives:
$$p_{1}^{0}p_{2}^{0}=p_{3}^{0}p_{4}^{0}-p_{3}*p_{4}(three-vectors)$$

But what are the next steps????
(I hope I made clear the problem, it isn't very readable I'm afraid....)

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