# Tag Info

## New answers tagged gauge-invariance

0

The kinetic term as well as any other terms containing only derivatives that are positive definite will just raise the energy of the system. So what one states first, is that the energy is lower for constant configurations (at tree level), this a first bound. Then one can say that within the constant configurations the ones that minimize the potential terms (...

2

Since OP does not state explicitly the definition of $h_{\mu\nu}$ I will guess it is defined by $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}.$$ If OP corrects this then I will update this answer to reflect the change. After expansion of the Einstein Hilbert action about $g_{\mu\nu} = \eta_{\mu\nu}$ all raising and lowering of indices is done by the flat space ...

5

To identify the Noether 4-current with the electric 4-current, one would in principle have to show that the Noether 4-current indeed appears as the source term in Maxwell's equations. The Maxwell equations with sources (Gauss's + Ampere's laws) are derived by adding the Maxwell Lagrangian $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ to a minimally coupled, gauge-...

2

Not sure if this is what you're looking for, but here is a exhaustive search for all gauge invariant objects in electromagnetism: Say $f[A_{\mu},\partial_{\mu} A_{\nu},\partial_{\mu}\partial_{\nu} A_{\rho},...]$ is a gauge invariant object. Here $f$ could itself have indices which have been suppressed. Under a gauge transform: $$A'_{\mu}(x)=A_{\mu}(x)+\... 3 It is up to a trivial constant the only gauge invariant Lagrangian that reproduces the Maxwell equations. Since the Maxwell equations are linear in F^{\mu\nu} only quadratic functions of F^{\mu\nu} are possible. There are two possibilities, namely F_{\mu\nu}F^{\mu\nu} and \epsilon_{\mu\nu\rho\sigma} F^{\mu\nu} F^{\rho\sigma}. The Maxwell equations ... 3 Let’s take quantum chromodynamics (QCD), the current theory of quark-gluon interactions, as a specific example of a Yang-Mills theory. And let’s avoid equations! There are three “colors” of quarks, usually called red, green, and blue. You can think of these colors as the QCD analog of electrical charge in QED. So there is a red quark field, a green quark ... 2 Building on the answer from @ACuriousMind I want to point out that the procedure being followed is the Faddeev-Popov gauge fixing. The symmetry transformations can be solved to put e(t) = T, a constant and \chi(t) = 0 (on the loop) or \chi(t) = \theta, constant (on the open line). T and \theta represent physically distinct configurations of the ... 0 The gravity gauge multiplet (e,\chi) enjoys a local supersymmetry with infinitesimal transformations$$ \delta e = -2\mathrm{i}\epsilon\chi \quad \delta \chi = \frac{\mathrm{d}\epsilon}{\mathrm{d}\tau}, \tag{1}$$where \epsilon is a fermionic parameter. Together with ordinary reparametrization symmetry by some rescaling f, this gives us two free gauge ... 1 Note that the notion of gauge theories is much more general than, say, Yang-Mills theory. There is of course a long list of gauge theories that have gauge potentials -- e.g. in the case of Yang-Mills theory, the A^a_{\mu}(x) field is the gauge potential -- but it is not a general requirement for a gauge theory. 2 The expression$$ \delta(A_\mu) = \epsilon^\alpha\partial_\alpha A_\mu$$is not gauge invariant in the usual, straightforward sense. Under a gauge transformation A\mapsto A' = A +\mathrm{d}f, it transforms exactly as you have shown,$$ \delta A'_\mu = \delta(A_\mu + \partial_\mu f) = \epsilon^\alpha\partial_\alpha(A_\mu + \partial_\mu f). Were the ...

1

Hamiltonian formalism also works for gauge systems, although one has to introduce constraints (and possibly the corresponding Lagrange multipliers), see e.g. Ref. 1. For the relativistic point particle, see e.g. this Phys.SE post. In fact the Hamilton-Jacobi equation is derived from the Hamiltonian formalism, not the other way around. The Hamiltonian $H$ is ...

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