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0 votes
0 answers
61 views

Degree of freedom - Lorentz transfomation reduces it? [duplicate]

I am having a real difficult to counting degree of freedom. In fact, I notice that sometimes I am confused about what exactly we count as DoF, and what we do not count. See, for example, the ...
LSS's user avatar
  • 980
2 votes
0 answers
56 views

$R_\xi$ gauge and degrees of freedom counting

In the standard classical Maxwell theory, we use the following arguments to claim that there are only two propagating degrees of freedom $A_\mu$ has 4 components $A_0$ is non-dynamical (-1) $\...
Lelouch's user avatar
  • 669
2 votes
2 answers
207 views

"One-parameter" gauge transformation

In my advanced classical physics course, it was stated that the electromagnetic field strength tensor $F_{\mu\nu} = \partial_{\nu}A_{\mu} - \partial_{\mu}A_{\nu}$ is invariant under "one-...
Lili FN's user avatar
  • 234
1 vote
0 answers
173 views

Explicit counting of gauge field degrees of freedom

Consider a connection on a principal $U(1)$-bundle $A_\mu$ over the flat base manifold $M_4$. The action of the theory is described in terms of the curvatures of such connection coupled to some source ...
samario28's user avatar
  • 195
1 vote
3 answers
378 views

Why does gauge invariance in electrodynamics mean that there are redundant degrees of freedom? [closed]

It is possible to choose different gauges in electrodynamics. I am familiar with two of them: Coulomb gauge and Lorenz gauge. Let us stick to the Coulomb gauge. It sets $$\nabla\cdot\vec{A}=0.$$ The ...
Solidification's user avatar
10 votes
1 answer
3k views

Gauge theory and eliminating unphysical degrees of freedom

In free space we can express Maxwell's equations as \begin{align} \varepsilon^{abcd}\partial_bF_{cd}=0 ~~\text{ and }~~ \partial_aF^{ab}=0 \tag{1} \end{align} where $F^{ab}=-F^{ba}$. The most general ...
SigmaAlpha's user avatar
10 votes
2 answers
2k views

Question about physical degree of freedom in Maxwell Theory: Why Coulomb gauge can fix all redundant degree of freedom

Given $4$-potential $A^\mu(x)=(\phi(x),\mathbf{A}(x))$, the vacuum Maxwell equations: $$\nabla^2\phi+\frac{\partial}{\partial t}(\nabla\cdot \mathbf{A} )=0$$ $$\nabla^2 \mathbf{A} -\frac{\partial^2 \...
user avatar
2 votes
1 answer
2k views

Residual Gauge Freedom [closed]

How are we still left with one Residual Gauge Freedom in the choice of Electromagnetic Potential after having already exploited the Gauge Freedom once. As is mentioned in Halzen and Martin Section 6.9....
Gaurav Katoch's user avatar
18 votes
2 answers
3k views

Counting the number of propagating degrees of freedom in Lorenz Gauge Electrodynamics

How do I definitively show that there are only two propagating degrees of freedom in the Lorenz Gauge $\partial_\mu A^\mu=0$ in classical electrodynamics. I need an clear argument that involves the ...
QuantumDot's user avatar
  • 6,381
2 votes
1 answer
575 views

Coulomb gauge and two degrees of freedom of EM field

The EM field has two possible polarizations, which is caused by spin-one nature of field (leads to the Lorenz gauge) and massless of the field. Really, the Klein-Gordon equations for the EM field $$ \...
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