Difference between revisions of "Forest Bhabha Scattering"
Line 29: | Line 29: | ||
::<math> p^{\mu}_{out}(\gamma^*) = (2m_e,\vec{0})</math> | ::<math> p^{\mu}_{out}(\gamma^*) = (2m_e,\vec{0})</math> | ||
− | ==e+e- | + | ==e+e- Annihilation (s-channel)== |
− | If the electron and positron | + | If the electron and positron form an intermediate state which then decays back to an electron and positron. This is a "s-channel" process in which |
− | [[File: | + | [[File:TF_BhaBha_s-chanFeynmanDiag.png | 200 px]] |
;Momentum conservation | ;Momentum conservation | ||
− | :<math>p^{\mu}_{in}(e^-) + p^{\mu}_{in}(e^+) = p^{\mu}_{out}(\gamma^*) | + | :<math>p^{\mu}_{in}(e^-) + p^{\mu}_{in}(e^+) = 0 = p^{\mu}_{out}(\gamma^*) </math> |
=Step 2 identify 4-Momentum conservation= | =Step 2 identify 4-Momentum conservation= |
Revision as of 21:23, 14 April 2012
Bhabha (electron -positron) Scattering
Bhabha scattering identifies the scatterng of an electron and positron (particle and anti-particle). There are two processes that can occur
1.) scattering via the "instantaneous" exchange of a virtual photon
2.) annihilation in which the e+ and e- spend some time as a photon which then reconverts back to an e+e- pair
Step 1 Draw the Feynman Diagram
The Feynman diagram is a space-time description of the interaction where the horizontal axis (abscissa) is used to denote time and the vertical axis (ordinate) is 3-D space.
A particle which travels only along the horizontal time axis is not moving in space while a particle traveling only along the vertical axis is not moving in time (within the uncertainty principle).
e+e- scattering (t-channel)
If the electron and positron simply scatter off of one another via a coulomb interaction, then they exchange a virtual photon along the space axis. You start with an external line from the left to represent the electron. This is a "t-channel" process in which one of the particles emits a virtual photon that is absorbed by the other particle. You can tell the exchanged particle is virtual if it is drawn parallel to the time axis in the Feynman diagram.
Time axis is from left to right so the Virtual particle is along the space axis (in some books the diagram has the space axis horizontal)
- Momentum conservation
for the case that the electron and positron have equal and opposite momentum (CM frame) the exchanged particle is clearly a virtual photon
e+e- Annihilation (s-channel)
If the electron and positron form an intermediate state which then decays back to an electron and positron. This is a "s-channel" process in which
- Momentum conservation
Step 2 identify 4-Momentum conservation
Let:
- initial electron 4-momentum
- initial electron spinor
- final electron 4-momentum
- final electron spinor
- initial positron 4-momentum
- initial positron spinor
- finial positron 4-momentum
- finial positron spinor
Step 3 Determine Matrix element for each vertex
Step 4 Find total amplitude
Matrix element for scattering
According to the Feynman RUles for QED:
the term
is used at the vertex to describe the Quantum electrodynamic (electromagneticc) interaction between the two fermion spinor states entering the vertex and forming a photon which will "connect" this vertex with the next one.
- The QED interaction Lagrangian is
Matrix element for annihilation
Radiative Bhabha Scattering to measure running of alpha