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0

Hamilton's principal of least action would apply which is more general than Fermat's principle. If Energy remained constant then action would be proportional to time so probably something like Fermat' principle would apply, but this is just a guess.


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 ...


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

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 ...


0

I never did the quantitative calculation. But possibly it can be understand in this way. Note that a free electron cannot absorb a photon. Therefore, it is not so surprising that a loosely bound electron has less possibility to be kicked out by a high energy 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}} = ...


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.


0

Consider a correlator like $$ \frac{1}{Z}\int \phi(x)^4 \phi(y_1)\ldots\phi(y_n)\ e^{-\int \phi(z)^4 dz}\ d\mu(\phi) $$ where $d\mu$ is the perturbed free field measure. Both $\phi^4$'s are composite fields but very different ones. The $\phi(z)^4$ in the exponential is a composite field for the free field theory $d\mu$ whereas the $\phi(x)^4$ outside is a ...


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

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 ...


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 ...


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 ...


-4

This may follow Einstein's equation and may appear to fit into classical picture, but this is taking place between two particles obeying Fermi Dirac statistics. Hence, it is a quantum phenomenon, and the direction of photon emission is arbitrary, as required by the fundamental assumption of quantum mechanics.


20

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 ...


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 ...


0

First, note that you don't have a trace yet. You get a trace when you do the spin sum for the external fermion lines for the square of the amplitude; c.f. Peskin and Schroeder p. 132. Second, you need two photon propagators; one for each line in the diagram. One will couple to your $\gamma^\mu$ and to a new vertex with some $\gamma^\rho$ on the loop, and ...


0

How did he get it? How did he get what? Which part of the calculation is confusing you. You have listed multiple equalities, which one is at issue? Eq 3.55 of the pdf online version of Peskin and Schroder shows that: $$ u^\dagger u=2E_p\xi^\dagger \xi $$ and since $$ \gamma_0^2=1 $$ we know that $$ \bar u\gamma_0u=u^\dagger u $$ and, in the ...



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