High-energy physics: momentum transfer

In summary, the problem involves an electron and a stationary proton colliding, with the electron deflecting at a small angle. The task is to find the square of the four-momentum transfer, q2. This involves using conservation of energy and momentum to determine the values of the primed quantities, px', py', pz', and E'. The electron's negligible mass allows for simplifications in the equations, but the non-zero deflection means that the product of the three-momenta cannot be disregarded.
  • #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?
 
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  • #2
Chiborino said:

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.
 

Related to High-energy physics: momentum transfer

1. What is momentum transfer in high-energy physics?

Momentum transfer in high-energy physics refers to the exchange of momentum between particles during a collision or interaction. It is a crucial concept in understanding the behavior and dynamics of particles at high energies.

2. How is momentum transfer calculated in high-energy physics?

Momentum transfer is calculated by subtracting the initial momentum of the interacting particles from the final momentum. This gives the change in momentum, which is a measure of the strength of the interaction.

3. What is the importance of momentum transfer in high-energy physics experiments?

Momentum transfer is important in high-energy physics experiments because it allows us to study the underlying interactions and forces between particles. By measuring the momentum transfer, we can gain insight into the fundamental properties of matter and the laws of physics.

4. How does momentum transfer affect the behavior of particles in high-energy collisions?

Momentum transfer determines the resulting trajectories and energies of the particles involved in a collision. It also affects the production of new particles and the overall outcome of the collision.

5. Can momentum transfer be manipulated in high-energy physics experiments?

Yes, scientists can manipulate the momentum transfer in high-energy physics experiments by adjusting the energies and trajectories of the particles involved. This allows for more precise studies of particle interactions and can lead to new discoveries in the field.

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