Difference between revisions of "Forest Bhabha Scattering"
Line 32: | Line 32: | ||
::<math> = m_1^2 + m_2^2- 2(E_1E_2- p_1p_2 \cos(\theta))</math> | ::<math> = m_1^2 + m_2^2- 2(E_1E_2- p_1p_2 \cos(\theta))</math> | ||
:: <math>m_{\gamma}^2=E_{\gamma}^2 - p_{\gamma}^2= m_1^2 + m_2^2 -4p_1p_2\sin^2(\theta/2)</math> | :: <math>m_{\gamma}^2=E_{\gamma}^2 - p_{\gamma}^2= m_1^2 + m_2^2 -4p_1p_2\sin^2(\theta/2)</math> | ||
+ | |||
+ | if | ||
+ | :E_{\gamma}^2 - p_{\gamma}^2= m_1^2 + m_2^2 -4p_1p_2\sin^2(\theta/2) \ne 0</math> | ||
+ | |||
+ | Then | ||
+ | :<math>m_{\gamma}^2 \ne 0 \Rightarrow</math> photon with mass | ||
for the case that the electron and positron have equal and opposite momentum (CM frame) the exchanged particle is clearly a virtual photon | for the case that the electron and positron have equal and opposite momentum (CM frame) the exchanged particle is clearly a virtual photon |
Revision as of 04:04, 15 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) (space-like)
If the electron and positron simply scatter off of one another via a coulomb interaction, then they exchange a 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.
- The 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). Also note that a virtual, neutral Z-boson may also be exchanged via the electro-weak interaction.
- Momentum conservation at the electron vertex
if
- E_{\gamma}^2 - p_{\gamma}^2= m_1^2 + m_2^2 -4p_1p_2\sin^2(\theta/2) \ne 0</math>
Then
- photon with mass
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) (time-like)
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 at the first vertex
- In reality the
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