Tag Info

New answers tagged

0

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 and irrelevant to considerations of ...


0

This gauge invariant vector potential increases without bound as long as there exists a static electric field. Indeed, even when the electric field is removed, there appears to be no mechanism by which the gauge invariant vector potential disappears. Static electric field has zero transversal component; entire field is longitudinal. The unbounded ...


1

Maxwell's equations and Lorentz force law do indeed summarize all electrodynamics. However there are physical situations in which you do not know the charge distribution a priori, but instead you specify some surfaces on which the electric field is always normal, which happens when you have metals around. In such situations you can in principle you can ...


1

From the Maxwell-Dirac Lagrangian $$ \mathcal L = -\frac{1}{2}F^2 + \overline{\psi}(i\gamma^\mu D_\mu +m) \psi $$ where $D_\mu$ is the gauge covariant derivative it is clear that the 4-current that acts as the source term in Maxwell's equations is $$ j^\mu_D = q\overline{\psi} \gamma^\mu \psi. $$ Using the Dirac equation it can be shown that (see, e.g., ...


2

I think you bring up an interesting question, I'm not sure why there were some derogatory comments to this... First off, the fermionic current doesn't couple to the gauge field due to its dimension. The Dirac field is dimension $ 3/2 $ and the current is dimension $ 3 $. Therefore, the coupling of the fermion to the gauge fields is of higher order. On the ...


3

The amplitude for emission of n photons with polarizations $\epsilon_{\mu_j}$ written as $\epsilon_{\mu_1}\ldots \epsilon_{\mu_n}\mathcal{M}^{\mu_1 \ldots \mu_n}$ satisfies $k_{\mu_1} \mathcal{M}^{\mu_1 \ldots \mu_n} = k_{\mu_2} \mathcal{M}^{\mu_1 \ldots \mu_n} = \ldots =0$ due to gauge invariance (do you know this fact? This is a consequence of ward ...



Top 50 recent answers are included