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The Lorenz gauge condition does not fix the gauge completely. Let $A^\mu$ be a field satisfying the Lorenz gauge condition $\partial_\mu A^\mu = 0$. Given a scalar function $f$, let $B^\mu = A^\mu + \partial^\mu f$. $B^\mu$ can also satisfy the Lorenz gauge condition if $$\partial_\mu B^\mu = \partial_\mu\partial^\mu f = 0,$$ i.e. if $f$ is the Minkowski ...

3

Let me begin by answering the simpler question "What is $U(1)$ as a manifold?" Recall that the elements of $U(1)$ are $1\times1$, complex, unitary matrices. If $g$ is a generic $1\times1$, complex matrix, then we can write \begin{align} g = \begin{pmatrix} r e^{i\varphi} \end{pmatrix} \end{align} for real numbers $r$, $\varphi$. A unitary matrix obeys $U^{\... 3 As the other answers point out, Lorentz gauge is actually just a partial gauge fixing that leaves residual degrees of freedom. The remaining gauge freedom is what is known as a Gribov ambiguity. To fully specify a gauge, you need to additionally specify enough boundary conditions to fix a particular solution to the wave equation for your transition function$...

3

There isn't exactly true: the condition you state doesn't uniquely fixes the vector potential, in fact you have what's called residual gauge freedom, which means that with the condition $\partial_\mu A^\mu = 0$ you haven't completely fixed the gauge. Let's prove this. Suppose you have $A^\mu$ such that $\partial_\mu A^\mu=0$ then you have infinite Vector ...

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In this answer I work in natural units, and deal mostly with the Maxwell Lagrangian density $$\mathcal{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu\nu} .$$ Canonical momentum The canonical momentum conjugate to the field $A_\mu$ is \Pi^\mu = \frac{\partial\mathcal{L}}{\partial (\partial_0a_\mu)} = (0,E^i). \end{...

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The Coulomb gauge actually also leaves residual gauge freedom, just as the Lorentz gauge does. This is another example of the Gribov ambiguity mentioned in my answer to the other question. In general, any gauge-fixing condition defined by a linear partial differential equation will have a Gribov ambiguity that corresponds to the kernel of the differential ...

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How can a diffeomorphism take a spacetime to a physically inequivalent one? This happens when a diffeomorphism acts non-trivially on the boundary data. … is it just a subset of the diffeomorphism group that should be regarded as identifying physically equivalent spacetimes? Yes. These are trivial or gauge diffeomorphisms. The theory of non-trivial ...

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After you've fixed a complete gauge (i.e one that really picks precisely one representant of every gauge orbit), the gauge symmetry is gone. If your gauge condition is incomplete, then some residual gauge symmetry might remain (e.g. the Lorenz gauge condition leaves a residual gauge symmetry for harmonic functions). Gauge symmetries cannot "turn into global ...

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Gauge-fixing of gaugesymmetry is necessary to avoid singular volume factors in the path integral, cf. e.g. my Phys.SE answer here. The gauge-fixing condition must as a minimum accomplish this task. The global part of gaugesymmetry (and other global symmetries) should not be gauge-fixed as they don't render the path integral ill-defined. It seems relevant to ...

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OP's first Lagrangian density is $${\cal L}~=~\frac{1}{2}(\partial_t \phi)^2 \tag{1a}$$ with a propagating field $\phi$, i.e. with the presence of time derivatives. The complete solution to the EL eq. reads $$\phi(x,t)~\approx~f_1(x)t+f_2(x).\tag{1b}$$ Given an initial time surface $\{t=t_i\}$, the functions $f_1(x)$ and $f_2(x)$ are Cauchy data that ...

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You may well be stumbling on unfortunate notation. Your first action, $$S_1=\int \left(\frac{\partial}{\partial t}\phi(x,t)\right)^2 dx^3 dt$$ symmetric under $\phi(x,t)\rightarrow \phi(x,t)+\chi(x)$, is but a decoupled sum of your second one, with the unfortunate variable conflation excised, $$S_2= \int \dot{\phi}(t)^2 dt$$ invariant under a translation $... 1 Leaving technical issues like Gribov copies and residual gauge freedom aside, how do gauge fixing conditions like the Coulomb condition$∂_𝑖 𝐴_𝑖=0$or the axial condition$𝐴_3=0$help in getting rid of the gauge redundancy? Because while there are a huge number of fields$A_\mu$that satisfy the defining equation$\partial_\mu A_\nu - \partial_\nu A_\mu ...

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Diffeomorphism invariance is a nearly vacous statement in and by itself. The thing that distinguishes GR from a field theory formulated in a fixed background is the lack of background structures. Consider for example a field theory in Minkowski spacetime $(M,\eta)$, where $\eta$ is the flat metric. In this case there is a background structure, namely the ...

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