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It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...

11

Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ... 5 Regardless of renormalizability, the term that you wrote down (gA_\mu A^\mu \phi) does not describe photons because it is not gauge invariant. This would be a theory of a massless vector boson with three dynamical propagating degrees of freedom (two transverse and one longitudinal), which is inconsistent with Lorentz invariance and irrelevant to ... 3 The longitudinal mode decouples from all physical processes as a consequence of gauge invariance, which in turn forces the Ward identity$$ k^\mu \mathcal{M}_\mu = 0where the S-matrix element decomposition \mathcal{M}^\mu is obtained from the polarization vector \epsilon^\mu(k) by \mathcal{M} = \epsilon^\mu(k) \mathcal{M}_\mu. This decoupling (and, ... 3 It's a mixture of c_\infty = c_0 = c and "the question doesn't make sense". So, first, how it does not make sense: What's the "speed" of a quantum object? It has, in general, no well-defined position, so v = \frac{\mathrm{d}x}{\mathrm{d}t} is rather ill-defined. Instead, we should probably look at the mass of the photon, since all massless objects ... 2 Take a look at equation (3.65) on page 48. While the barred spinors \bar{u}^s are properly orthogonal to v^r \begin{align} \bar{u}^s(p)v^r(p) = u^{s\dagger}(p)\gamma^0v^r(p)=0 \end{align} the hermitian conjugated u^{s\dagger} obey u^{s\dagger}(\mathbf{p})v^r(\mathbf{-p})=0$$where (-\mathbf{p})=(E_p,-\mathbf{p}) is imho. a bit of an unfortunate ... 2 You might end up slapping yourself in the forehead because it's probably simpler than what you were thinking.$$\left(\frac{4\pi}{\Delta}\right)^{\epsilon/2} = \exp\left[\frac{\epsilon}{2}\log \left(\frac{4\pi}{\Delta}\right)\right] = 1 + \frac{\epsilon}{2}\log \left(\frac{4\pi}{\Delta}\right) +\mathcal{O}(\epsilon^2) whence the result follows.

1

If we write $A_\mu(x)=\varepsilon_\mu(p)e^{ipx}$, the polarization vector should satisfy $\varepsilon_\mu p^\mu=0$, which is a Lorentiz-invariant relation, and is necessary to make sure that we have an irreducible representation of the Lorentz group (actually, the little group that leaves the momentum invariant). This knocks down the number of D.O.F to 3. ...

1

The Coulomb "interaction" appears as the answer to a very specific question, even in QED, that is "what is the correction to the ground state energy of the electromagnetic field if two charges $q_1$ and $q_2$ are pinned at specific locations separated by a distance $r_{12}$?" . The answer to that question is exactly $\Delta E_0 = \frac{q_1q_2}{4 \pi ... 1 It is a standard exercise in most quantum field theory books that the 2-point function does not vanish outside the light-cone of a particle. Less technically; the probability that some particle$r$away from a source feels the effect of the source quicker than light to travel would$r$is non-zero. That is a good definition of superluminal motion. See for ... 1 A diagram which is first order in$\alpha_\text{EM}$would have to have one vertex, because$\alpha_\text{EM}\propto g^2$where$g\$ is the factor associated with each vertex (and the amplitude corresponding to the diagram gets squared). There's only one possible vertex in QED, namely the photon-electron-positron vertex, and it's impossible to arrange this in ...

1

What is potential energy truly? It depends on the circumstances. When you compress a spring it's stress in the bonds or electromagnetic field between the atoms. IMHO at the fundamental level it's essentially spatial stress. That might sound unfamiliar, but it shouldn't, because the stress-energy-momentum tensor "describes the density and flux of energy and ...

1

Jezstarski is mostly correct, The para-positronium (p-PS) state ends up being the main mode of annihilation of positronium (PS). Positrons can annihilate in at least eight different ways but once ortho-positronium (o-PS) forms in a void/vacuum, it has additional time to undergo another mode of annihilation. P-PS annihilates in under 125 picoseconds. O-PS ...

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