Electron-Muon Scattering Cross Section

In summary, it is claimed that qμLμγ = qγLμγ = 0, which is fine because L is the product of 2 4-currents and qμjμe- = 0. However, according to the text that I am reading, the qμLμγ = qγLμγ condition implies that we can replace p' in the muon tensor with (p+q); ie, Meffective = 2[2pμpγ + (q2/2)gμγ.
  • #1
boltzman1969
7
1
Hi,

I am self-teaching Quantum Elctrodynamics, and have come across something which I do not understand. I would appreciate feedback from anyone on this specific issue from Atchison & Hey, "Guage Theories in Particle Physics" pg 238-239:

In calculating the u-channel electron-muon scattering amplitude at the one-photon exchange, one can simplify the calculation by introducing the electron and muon tensors:

LμγMμγ, where

Lμγ = 2[k'μkγ+k'γkμ+(q2/2)gμγ] (electron tensor) and

Mμγ = 2[p'μpγ+p'γpμ+(q2/2)gμγ] (muon tensor)

Now qμ = (k-k')μ = (p-p')μ is the 4-momentum of the exchanged photon; p, p' are the intial and final momenta of the muon; k, k' the initial and final 4-momenta of the electron.

It is claimed that qμLμγ = qγLμγ = 0, which is fine because L is the product of 2 4-currents and qμjμe- = 0. However, according to the text that I am reading, the qμLμγ = qγLμγ condition implies that we can replace p' in the muon tensor with (p+q); ie, Meffective = 2[2pμpγ + (q2/2)gμγ.

Does anyone know how to go from the condition qμLμγ = 0 to the constraint condition p' = (p+q)?

Thank you in advance for your assistance.
 
Physics news on Phys.org
  • #2
boltzman1969 said:
Hi,

Does anyone know how to go from the condition qμLμγ = 0 to the constraint condition p' = (p+q)?

Thank you in advance for your assistance.

The condition qμLμγ = 0 does not imply p' = (p + q). It is the opposite which is true.
There is only one diagram in the process, so the amplitude has two momentum-conserving delta functions: [itex]\delta^{ 4 } ( k - k' - q )[/itex] at the electron vertex and [itex]\delta^{ 4 } ( p - p' + q )[/itex] at the muon vertex. So when you do the integral [itex]\int d^{ 4 }q[/itex] you forced to put [itex]p' - p= q = k' - k[/itex]. After that you can use the Casimir's trick the trace theorems to obtain the [itex]| \langle \mathcal{ M } \rangle |^{ 2 }[/itex].
 
  • #3
Samalkhaiat,

Thank you for that reply. It is much clearer to me know how to obtain the scattering amplitude.

I was scratching my head trying to figure out the implication in the other direction.

The text was somewhat unclear in this regard.

Boltzman1969
 

1. What is electron-muon scattering cross section?

Electron-muon scattering cross section is a measure of the likelihood of an electron and a muon interacting with each other by exchanging a virtual photon.

2. How is the electron-muon scattering cross section calculated?

The electron-muon scattering cross section is calculated using the quantum field theory calculation known as the Feynman diagram, which takes into account the properties of the particles and their interactions.

3. What can the electron-muon scattering cross section tell us about the fundamental forces?

The electron-muon scattering cross section can provide information about the strength and nature of the fundamental forces, such as the electromagnetic force, the weak force, and the strong force.

4. What experimental techniques are used to measure the electron-muon scattering cross section?

Experimental techniques such as colliding beams of electrons and muons, and detecting the scattered particles and their properties, are used to measure the electron-muon scattering cross section.

5. Why is the electron-muon scattering cross section important in particle physics?

The electron-muon scattering cross section is important in particle physics because it helps to test and verify theoretical predictions and models, and provides insights into the fundamental properties of matter and the universe.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
0
Views
910
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
3
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
9
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
946
  • High Energy, Nuclear, Particle Physics
Replies
7
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
14
Views
1K
Back
Top