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The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...


4

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2 $$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ...


3

If your electron is very fast you can track its trajectory in a ionization chamber. But if the electron is slow, its wave-like (quantum) character enters into the scene. About absorption and re-emission of a photon, please check if such things are possible, i.e. write the laws of conservation of energy and momentum as if these two bodies are billiard balls, ...


3

[PDG] quoted something like $(q_p+q_e)/e$ without defining these quantities, That is exactly what you asked for. Recall that the charge on the electron $q_e$ is negative and that on the proton $q_p$ is positive, so the sum there is exactly the difference in their magnitudes. Taking it as a fraction of the defined base charge $e$ makes it a dimensionless ...


2

So as you say, the denominator is OK. For the numerator we use $$\tag{1}\not a\gamma_{\mu} = 2a_{\mu}-\gamma_\mu\not a, (\text{follows from } \{{\mu,\nu \} = 2\eta_{\mu\nu}}). $$ Then the numerator can be written $$Num = 2p_\mu-k_\mu-\zeta_\mu$$ where $$\tag{2}\zeta_\mu:=-k_\mu+\not k\gamma_\mu+\gamma_\mu\underbrace{(\not p-m)}_{=0 \text{ when hitting ...


2

You don't need to introduce photon's fictive mass: dimensional regularization can be used for IR divergences as well as for UV divergences. First, use gamma-matrices identities, $$ \gamma_{\mu}\gamma^{\alpha}\gamma^{\mu} = (2 - d)\gamma^{\alpha} ,\quad [\gamma_{\mu}, \gamma_{\nu}]_{+} = 2g_{\mu \nu}, \quad g_{\mu}^{\mu} = d $$ and do simple that you've ...


2

In a given volume, we can have light throughout, such that there is no space with no light in it (with the electron which is to be seen). Note that in this view you can hardly talk of photons as particles localized somewhere and somehow bouncing around. If you consider a given volume with a given amount of electromagnetic radiation in it you are ...


1

It shouldn't matter: all methods of regularization are equally applicable if they have a correct limit at $\mu \rightarrow 0$. It is like asking whether we should keep the $(2\pi)^4$ in dimensional regularization or promote it to $(2\pi)^n$. Doesn't matter. All physical results are the same. Again, every method which makes the integral converge at finite ...


1

When you change the free field $A_\mu$ by means of a gauge transformation, you can easily see that it affects longitudinal and timelike degrees of feedom. Since observables are gauge invariant, those degrees of freedom cannot be physical.


1

I read what is written in the physicsforums.com/threads/ that you indicate. So, you speak of the electron as of a QUANTUM particle. That means that it has a linear momentum of the same order of magnitude as that of the particles with which you want to test its movement. If the linear momentum of those particles were much smaller, s.t. the collision with ...



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