physics_fun
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There is nowone who can help me on the homework forum, so i'll try it here...
physics_fun said:I'll try a little LATEX to make the problem more clear:
<br /> p_{1} + p_{2} = p_{3} + p_{4}<br />
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: <br /> p_{1} + p_{2} = p_{3} + p_{4}<br />
square this: (p_{1})^(2) + (p_{2})^(2) + 2p_{1}p_{2}= (p_{3})^(2) + (p_{4})^(2) + 2p_{3}p_{4}<br /> //<br /> (p_{1})^(2)=(p_{3})^(2)=m (electron mass)//<br /> (p_{2})^(2)=(p_{4})^(2)=M (positron mass)<br />
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?