Tag Info

New answers tagged

1

As in the case of Electromagnetism, in General Relativity you can perform two gauge transformations (diffeomorphisms). The first one being $x'^\mu(x)=\Lambda^\mu(x)$ you have four gauge functions which fix the gauge, for example the harmonic one $$\Gamma^\mu=\partial_\alpha(\sqrt{g}g^{\alpha\mu})=0.$$ Now it can be shown that exist a residual gauge ...


3

One has $ \dfrac{\delta \Gamma}{\delta K^\mu_a(x)} = sA_\mu^a(x)= \partial_\mu \omega^a(x) - g f^{abc} \omega^b(x)A_{\mu}^ c(x) $ We notice that : $$sA_\mu^a(x))|_{fields=0}= 0 \tag{1}$$ We notice that : $$\dfrac{\delta}{\delta \omega^b(y)}sA_\mu^a(x)|_{fields=0} = (\partial_\mu \delta^a_b \delta(x-y) - g f^{abc} \delta(x-y)A_{\mu}^ c(x)|_{fields=0} ...


5

The gauge connection is not unique, and this has nothing to do with the presence of matter fields. Let $\Sigma$ be our space-time, $P$ a principal $G$-bundle, and $\mathcal{A}$ the space of connections on $P$. Then, gauge transformations $t : P \to G$, forming the group of gauge transformations $\mathcal{G}$ have an action on $\mathcal{A}$ given by $$ A ...


6

Yes, one traditional alternative to the path integral formalism is the operator formalism. For QED with abelian gauge group, the old quantization formulation is the Gupta-Bleuler formulation. For QCD/Yang-Mills theory with non-abelian gauge group, the Gupta-Bleuler formulation is replaced by the BRST formulation. The BRST formulation exists in at least 3 ...


3

There exists an extensive literature for discretization of the abelian and the non-abelian gauge theories, known as lattice QED and lattice QCD, respectively. Here we will only sketch the main idea. Let us for simplicity use Euclidean signature $(+,+,+,+)$. A small Wilson-loop $$\tag{1} W~=~{\rm Tr}{\cal P}e^{ig\int_{\gamma}A}$$ lies approximately in a ...


2

The key point in all of this is that general relativity is a gauge theory, and, as the saying goes, "the gauge always hits twice" (apparently attributed to Claudio Teitelboim). What this means is that (1) you have an arbitrary freedom in defining your evolution, corresponding to the ability to make gauge transformations, and (2) some of the evolution ...


2

It is interesting to look at a linearized version of gravity, with $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ If you choose the Lorentz gauge : $$\partial^\mu \bar h_{\mu\nu}=0 \quad\quad \bar h_{\mu\nu} = h_{\mu\nu} - \frac{1}{2} h^i_i \,\eta_{\mu\nu} \tag{0}$$ the equations of movement in the vaccuum are simply : $$\square \bar h_{\mu\nu}=0 \tag{1}$$ ...


1

It isn't classical field theory, but there are a few features of using of 4-potential in QFT. The first one is that 4-potential as 4-vector can't be used for describing massless photons. It is because the fact that it must describe massless particles leads to its transformations not as 4-vector under the Lorentz group. Specifically, $$ A^{\mu} \to ...


2

If your question is asking whether the four-potential is more useful in classical electromagnetism from a purely computational standpoint, the answer would be no. It's not to say that it isn't useful, it's just that it only groups together two equations in the Lorentz gauge that are already useful themselves. The Lorentz gauge, $$ \Box\phi = ...


5

Why do we gauge-fix the path integral in the first place? If we were doing lattice gauge theory, we didn't need to gauge-fix. But in the continuum case, (the Hessian of) the action for a generalized$^1$ gauge theory has zero-directions that lead to infinite factors when performing the path integral over gauge orbits. In a BRST formulation (such as, e.g., the ...


4

We only have one contribution from each gauge-equivalent matter field configuration: Let $P$ be the principal $G$-bundle associated to our gauge theory on the spacetime $\mathcal{M}$ (for simplicity, assume it is $\mathcal{M} \times G$. The matter fields are constructed as sections of an associated vector bundle $P \times_G V_\rho$, where $V_\rho$ is a ...



Top 50 recent answers are included