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While solving the Maxwell's equation we often use the Lorenz or Coulomb gauge, but why is that? Are the equations unsolvable if the gauge is not fixed? Or is it just for the simplicity?

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    $\begingroup$ Related: physics.stackexchange.com/q/22747/2451 and links therein. $\endgroup$ – Qmechanic May 19 '16 at 13:53
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    $\begingroup$ This is rather a lot like asking "Why do we have to chose a zero of height when using gravitational potential energy in a problem?" $\endgroup$ – dmckee May 19 '16 at 15:22
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First, the gauge invariance means that the solutions $A_\mu(x^\alpha)$ are not unique. For every solution, the gauge transformations of it are solutions, too. That may be a problem because sometimes we want to have specific values of $A_\mu(x^\alpha)$ that answer a physical question.

Second, we sometimes gauge fix because the equations simplify. For example, when we discuss radiation (and antennas), even in classical electromagnetism, gauge-fixing may conveniently bring Maxwell's equations to the form $\square A^\mu = j^\mu$ which may be solved like other equations with the box operator.

Third, most importantly, in quantum field theory, gauge invariance means that some polarizations of $A_\mu$ create particle states with zero norm. Those mean that the differential operator as a matrix $M_{\mu\nu}$ acting on $A_\lambda$ is not invertible etc. Gauge-fixing gives a nonzero norm to such states, makes the matrices invertible etc., but we must still remember to deal with the unphysical (time-like and longitudinal) polarizations of the photon at the end. It's probably not a good idea to try to explain these things outside the context of the quantization of the electromagnetic field. One learns those things simultaneously with other issues of the quantization of the electromagnetic field. One doesn't need to know in advance – and in general, cannot know in advance – all the situations in which a procedure such as gauge-fixing may be helpful. But the list above is enough to understand that it is sometimes helpful.

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  • $\begingroup$ I got some idea. But the potentials are not physically measurable. Only fields are. And the fields are unique.. So why do we need an unique potential? $\endgroup$ – Siddhartha Dam May 19 '16 at 16:23
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    $\begingroup$ Generally, we don't need it. Gauge-fixing is helpful, it's not "mandatory". $\endgroup$ – Luboš Motl May 19 '16 at 17:46
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The problem is that they have too many solutions if the gauge is not fixed.

Imagine you have some initial values, and want to solve it on the computer. Then you have to solve the equations for the next time step given the values for the previous one. But you have to compute the values for, say, four variables but have only three equations. You somehow have to tell the computer what to do, what to choose among the many solutions which are allowed by the equations. You choose a gauge condition, and obtain equations which have only a single solution, which you can compute.

Once you have computed one solution, you can later compute a lot of other ones, which do not fulfill you gauge condition. Choose a completely arbitrary gauge transformation and apply it, that's all. The result will be another solution of the original equations without gauge condition. (But there would be no point of really doing this, this is simply for understanding the point).

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If we speak about classical physics, then Maxwell equations are written to find out $F(\mu,\nu)$ field from $J(\mu)$ - current. Solve it if you can and don't think about gauge. The only situation when you need to use potential $A(\mu)$ is non-trivial topology (your space-time isn't $R(4)$, but has singular areas). Maxwell equations are:

  1. $dF=J$

  2. $dF=0$

with restriction $dJ=0$

$A$ can be found from equation: $dA=F$. It has infinite number of solutions.

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