Redundant degrees of freedom in EM fields?

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If we consider E and B individually, there are 6 total degrees of freedom.

But they are actually related to each other by Maxwell's equations.

So we can find potentials and reduce dof to 4; 3 in vector potential and 1 in scalar potential.



Thus, there remain 2 redundant dof.

This is the problem...

I think one of them gives the polarization of EM wave and the other gives the gauge symmetry.

For example, the vector potential has gauge symmetry, A' = A + del (f)

f is scalar and has one dof.

Am I understanding right ?
 
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  • #2
vanhees71
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There are 6 physical degrees of freedom, ##\vec{E}## and ##\vec{B}##. Indeed, the homogeneous Maxwell equations,
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}$$
lead us to introduce the scalar and vector potentials ##\Phi## and ##\vec{A}##. The relation to the physical field is
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
The gauge symmetry makes everything invariant under the gauge transformations,
$$\vec{A}'=\vec{A} - \vec{\nabla} \chi, \quad \Phi'=\Phi+\frac{1}{c} \partial_t \chi$$
with an arbitrary scalar field ##\chi##. Thus we can pose one constraint in addition to the inhomogeneous Maxwell equations. If we choose the Lorenz-gauge condition,
$$\frac{1}{c} \partial_t \Phi + \vec{\nabla} \cdot \vec{A}=0,$$
the inhomogeneous Maxwell equations become separated
$$\Box \Phi=\rho, \quad \Box \vec{A}=\frac{1}{c} \vec{j}, \quad \frac{1}{c^2} \partial_t^2 - \Delta.$$
The entire system of Maxwell equations are only consistent, if electric charge is conserved,
$$\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0.$$
It turns out that this fixing of the gauge with the Lorenz gauge makes the solutions to the wave equations unique if charge and current densities are present.

For the free Maxwell field that's not the case, but even with the Lorenz-gauge constraint you still have the freedom to change the gauge via a field obeying the source-free wave equation. Thus you have to use another constraint to fix the gauge completely. One convenient possibility is to demand
$$\Phi=0$$
in addition to the Lorenz condition. This means you have
$$\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0,$$
i.e., you have only two transverse field-degrees of freedom left. There are only 2 polarization degrees of freedom (e.g., helicity +1 and helicity -1 states, which is the same as left- and right-circular polarized waves).
 

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