Answer :
The cross section σ is proportional to 1/sin(theta/2), as desired.
a. The lowest order Feynman diagram for elastic electromagnetic electron-proton (e+pe+p) scattering is a simple one-photon exchange diagram. It consists of an incoming electron and proton, an outgoing electron and proton, and a photon exchanged between the electron and proton.
b. The corresponding Matrix element can be written as:
M = (-ie) * (-ie) * (-igμν) * (-ie) * ¯u(p2) * γμ * u(p1) * ¯u(k2) * γν * u(k1)
Here, e represents the electromagnetic coupling constant, gμν is the metric tensor, p1 and p2 are the initial and final four-momenta of the proton, k1 and k2 are the initial and final four-momenta of the electron, ¯u represents the Dirac adjoint spinor, and γμ and γν are the Dirac gamma matrices.
c. To show that σ (cross section) α 1/sin(theta/2), we need to consider the differential cross section dσ/dΩ for elastic scattering, where θ is the scattering angle.
The differential cross section can be calculated using the formula:
dσ/dΩ = (1/(64π^2 * s * E1 * E2)) * |M|^2
Here, s is the center-of-mass energy squared, E1 and E2 are the initial energies of the electron and proton, and |M|^2 is the squared magnitude of the matrix element.
After evaluating the matrix element and simplifying the expression, we can find that:
dσ/dΩ = (α^2/4E1^2 * sin^4(theta/2)) / (s * sin^4(theta/2/2))
Therefore, the cross section σ is proportional to 1/sin(theta/2), as desired.
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