18
votes
Accepted
Trouble reconciling these two views on gauge theory
Gauge theory resembles general relativity applied to extra dimensions beyond the big four. Whether that's the true nature of gauge fields or just a convenient mental picture isn't clear, but it's at ...
- 28.7k
16
votes
What is the $\,\phi=0\,$ gauge called?
The gauge $\phi = A_0 = 0$ is called Weyl gauge or temporal gauge.
This gauge is incomplete, as one can see from the definition of a gauge transformation,
$$A_\mu \to A_\mu + \partial_\mu \alpha(x).$...
- 105k
16
votes
Accepted
Is special relativity falsified by the Aharonov-Bohm effect?
This is not the case. The Aharanov-Bohm effect yields an observable of the form
$$\oint_{\mathcal C} A,$$
where $\mathcal{C}$ is some circuit. This is however gauge invariant. A way to see this is by ...
- 3,985
15
votes
Accepted
Are Maxwell's equations "physical"?
Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-...
- 105k
14
votes
Can we do path integrals in gauge theories without fixing a gauge?
You are misunderstanding what a gauge theory is if you think we shouldn't get rid of the gauge symmetry at some point. A gauge symmetry is not like other symmetries, it does not relate configurations ...
- 129k
13
votes
Accepted
How many degrees of freedom in a massless $2$-form field?
It is natural to generalize to an Abelian $p$-form gauge field
$$A~=~\frac{1}{p!} A_{\mu_1\mu_2\ldots\mu_p} \mathrm{d}x^{\mu_1}\wedge\ldots\wedge \mathrm{d}x^{\mu_p}\tag{1}$$
with $\begin{pmatrix} D ...
- 213k
12
votes
Accepted
A question on gauge fixing
Yes, all those things are correct.
The equivalence class of potentials that are related by gauge transformation is called a gauge orbit, since it is an orbit for the action of the group of gauge ...
- 129k
11
votes
Why is nonzero net charge density incompatible with the cosmological principle?
The relevant difference between gravitational and electrostatic forces is that a gravitational field accelerates everything equally, whereas an electric field produces different accelerations on ...
- 6,987
10
votes
Gauge-fixing of an arbitrary field: off-shell & on-shell degrees of freedom
In this answer, we summarize the results. The analysis itself can be found in textbooks, see e.g. Refs. 1 & 2.
$\downarrow$ Table 1: Massless spin $j$ field in $D$ spacetime dimensions.
$$\...
- 213k
10
votes
Accepted
Can we choose the Coulomb gauge if we're in a gauge where the gradient of the scalar potential is zero?
You're asking whether we can impose both $\phi = 0$ and $\nabla \cdot \mathbf{A} = 0$ simultaneously. This will not be possible in any situation where $\rho \neq 0$, since if both conditions on the ...
- 51.6k
9
votes
Faddeev-Popov Gauge-Fixing in Electromagnetism
A gauge symmetry means that the equations of motion do not uniquely determine the evolution of all the configuration variables, i.e., that the Euler-Lagrange system is under-determined. The canonical ...
9
votes
Accepted
Gauge theory and eliminating unphysical degrees of freedom
Answer to the first question
You can fix the gauge doing the transformation $A^\mu \rightarrow {A^\prime}^\mu = A^\mu + \partial^{\,\mu} f$ and choosing an appropriate function $f$. You cannot fix it ...
- 643
9
votes
Accepted
Does Coulomb gauge imply constant density?
You secretly impose 2 gauge fixing conditions:
$$\nabla \cdot \vec{A} = 0$$
$$\nabla \cdot \vec{A} + \mu_0 \epsilon_0\frac{\partial \varphi}{\partial t} = 0$$
The latter coming from that fact you use ...
- 6,797
8
votes
Are Maxwell's equations "physical"?
You are correct that there are gauge degrees of freedom in the solution for $A_\mu$ - precisely the ordinary gauge transformations. But $A$ is not physical, the electromagnetic field strength $F$ is. ...
- 129k
8
votes
Accepted
Gauge symmetry of massive vector field
Symmetries of the action must be considered without use of the equations of motion. An on-shell symmetry is a vacuous notion - if you use the equations of motion, as you do when using eq. (5) to ...
- 129k
8
votes
Accepted
Quantum Theory of Radiation Enrico Fermi 1932
Fermi is working in the gauge $V =0$.
Suppose you have some choice of potentials $\mathbf{A}(\mathbf{x},t), V(\mathbf{x},t)$. Then define $\tilde{\mathbf{A}}(\mathbf{x},t) = \mathbf{A}(\mathbf{x},t) + ...
- 2,851
7
votes
Accepted
Radiation gauge and choice of the gauge function
$\nabla\cdot \mathbf{A} = 0$ is known as the Coulomb (or transverse) gauge. It's also called the radiation gauge. Note that, for boundary conditions at infinity, the Coulomb gauge is complete - there ...
- 22.8k
7
votes
Accepted
Can we choose other than Gaussian integral for Faddeev-Popov gauge fixing?
Any integrable function $f$ will in principle do. But the calculations may become more cumbersome.
It should be obvious why we normally choose the function $f$ to be Gaussian, because it is ...
- 213k
7
votes
Accepted
Counting massive degrees of freedom after gauge fixing
What you can remove with a field redefinition (of the form of a $U(1)$ gauge transformation) is the phase of $\phi$. But the total number of degrees of freedom (dof's) is not going to change since the ...
- 5,192
7
votes
Accepted
Confusion on Maxwells equations and Gauge Transformations
What is "real" in electrodynamics is what you measure. The fields can be measured without ambiguity due to their influence on charges and currents, and so these are the things that are real. Maxwell's ...
- 1,074
7
votes
What exactly are the sections in gauge theories?
The Principle fiber bundle can be thought of as an expansion of spacetime: Given a gauge group $G$ and a principal $G$ bundle $\pi: P \to M$ over spacetime $M$, we locally get $$\pi^{-1}(U) \cong U \...
- 554
7
votes
Why do we impose de Donder gauge?
Actually this is not possible.
To impose the gauge $h=0$, I have to solve
$$
2\partial\cdot \epsilon = -h\,,\qquad\qquad (1)
$$
which always admits solutions.
This partial gauge fixing leaves a ...
- 2,237
7
votes
Accepted
Why does Lorenz gauge condition $\partial_\mu A^\mu =0$ pick exactly one configuration from each gauge equivalence class?
The Lorenz gauge condition does not fix the gauge completely.
Let $A^\mu$ be a field satisfying the Lorenz gauge condition $\partial_\mu A^\mu = 0$. Given a scalar function $f$, let $B^\mu = A^\mu + \...
- 1,297
7
votes
Accepted
Why does Coulomb gauge condition $\partial_i A_i =0$ pick exactly one configuration from each gauge equivalence class?
The Coulomb gauge actually also leaves residual gauge freedom, just as the Lorentz gauge does. This is another example of the Gribov ambiguity mentioned in my answer to the other question. In general, ...
- 49.4k
7
votes
What does adding a gauge fixing term $-\frac{1}{2\xi}(\partial_\mu A^\mu)^2$ really mean?
There is a trick to derive this in path-integral formulation of QED by introducing an auxilliary field $\pi(x)$ (See Schwartz's QFT 14.5 on p. 267 for detailed explanation).
Consider following path-...
- 1,450
7
votes
What does adding a gauge fixing term $-\frac{1}{2\xi}(\partial_\mu A^\mu)^2$ really mean?
OP seems mainly concerned about what happens classically at the level of equations of motion when we alter the E&M Lagrangian density
$$ {\cal L}_0~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \mp j^{\mu}A_{...
- 213k
7
votes
Accepted
$R_\xi$ gauges and the EM-field
In the Lagrangian of the path integral, a popular class of gauge-fixing terms is of the form$^1$
$${\cal L}_{GF}~=~-\frac{\chi^2}{2\xi}.$$
The word "gauge" is here confusingly used in 2 ways:
The ...
- 213k
7
votes
QED scattering cross sections are independent of the gauge-fixing terms
Greiner is only considering a few specific Feynman diagrams at tree-level. For a systematic approach OP's question is best asked within the framework of BRST quantization. The scattering cross ...
- 213k
7
votes
Accepted
Why does $\mathbf{A}(x) = \frac{1}{2}(\mathbf{B}(x) \times \mathbf{x})$ work?
This identity holds only for a uniform magnetic field $\vec B$. Many terms cancel in your formula. It remains
$$\eqalign{
\vec\nabla\times\vec A
&={1\over 2}\big((\vec\nabla.\vec x)\vec B-(\...
- 3,708
6
votes
Can we do path integrals in gauge theories without fixing a gauge?
As of today, nobody knows how to canonically quantise a classical theory with gauge symmetries. The standard approach (Dirac's algorithm) where one replaces the canonical brackets by (anti)commutators ...
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