High-energy physics: momentum transfer

  • Thread starter Chiborino
  • Start date
  • #1
Chiborino
21
0

Homework Statement


I have an electron of 20 GeV and negligible mass that collides with a stationary proton (mc^2 = 9.38 GeV) and deflects at an angle of 5°. I'm asked to find the square of the four-momentum transfer, q2


Homework Equations


q = P - P', where P/P' is a 4-momentum vector <px, py, pz, iE>
a "primed" quantity represents a value after the collision with the proton.

The Attempt at a Solution


I took the quantity P-P' and squared it:
q2 = (P-P')*(P-P') = P2 + P'2 -2P*P'
I'm told the first two terms are negligible due to the electron's mass being negligible, but I'm not sure I see the sense in that.
Anyways, continuing on, I then have:
q2 = -2P*P' = -2(px*px' + py*py'+pz*pz' -E*E')
or q2 = -2p*p' + 2E*E'
This is where I'm stuck. Should I also assume the product of the 3-momenta is 0 and carry on with the +2EE' term I'm left with? And what do I even do about the E' since I don't know that quantity?
 
Last edited:

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,566
2,204

Homework Statement


I have an electron of 20 GeV and negligible mass that collides with a stationary proton (mc^2 = 9.38 GeV) and deflects at an angle of 5°. I'm asked to find the square of the four-momentum transfer, q2
That's a pretty massive proton!


Homework Equations


q = P - P', where P/P' is a 4-momentum vector <px, py, pz, iE>
a "primed" quantity represents a value after the collision with the proton.

The Attempt at a Solution


I took the quantity P-P' and squared it:
q2 = (P-P')*(P-P') = P2 + P'2 -2P*P'
I'm told the first two terms are negligible due to the electron's mass being negligible, but I'm not sure I see the sense in that.
Anyways, continuing on, I then have:
q2 = -2P*P' = -2(px*px' + py*py'+pz*pz' -E*E')
or q2 = -2p*p' + 2E*E'
This is where I'm stuck. Should I also assume the product of the 3-momenta is 0 and carry on with the +2EE' term I'm left with? And what do I even do about the E' since I don't know that quantity?
With the small deflection, you would expect that the electron didn't give up much of its energy, so you wouldn't expect the product of the three-momenta to vanish. Unfortunately, you're not going to be able to just erase terms you don't want to deal with.

You need to use conservation of energy and momentum to figure out ##\vec{p}'## and E'. Note that because the electron is essentially massless in this problem, the situation is essentially the same as Compton scattering.
 

Suggested for: High-energy physics: momentum transfer

Replies
10
Views
737
Replies
9
Views
436
Replies
16
Views
850
Replies
40
Views
2K
Replies
0
Views
364
Replies
6
Views
822
Replies
0
Views
287
Replies
8
Views
2K
Replies
0
Views
585
Top