Lorenz and Coulomb gauge-fixing conditions. What is physical difference between these two gauge-fixing conditions? Mathematical expression are clear but how to we choose one of these means what they really means.

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    $\begingroup$ The "physical" difference is nonexistent, that's the whole point of gauge theory. It's simply a choice that makes solving certain problems easier; i.e. you have a degree of freedom in choosing one of your variables and hence you should choose the one that simplifies the math! $\endgroup$ – Chris Gerig Aug 6 '13 at 6:16

The fields are not affected by this gauge transformations, and only those quantities have physical meaning that are invariant under the transformations. So essentially the physics is the same in these two gauges. But if you want some of the implications of choosing one of these gauges, so you can choose the appropriate one, the following may be of help:

Lorenz Gauge is commonly used. Because leads to similar wave equations for both $\Phi$ and $\mathbf{A}$: $$\nabla .\mathbf{A}+{1\over c}{\partial \Phi\over \partial t}=0\to\cases{ {1\over c^2}{\partial^2 \Phi\over \partial t^2}-\nabla^2\Phi={4\pi \rho} \\ {1\over c^2}{\partial^2 \mathbf{A}\over \partial t^2}-\nabla^2\mathbf{A}={4\pi\over{c}}\mathbf{J}}$$ and so fits well into the considerations of special relativity:

The D'Alembert operator $\Box ={1\over c^2}{\partial^2 \over \partial t^2}-\nabla^2$ in the above equations is the invariant 4D Laplacian and using this, the above equations can be written in covariant forms $\partial_{\alpha}A^{\alpha}=0$ and $\Box A^{\alpha}={4\pi\over c}J^{\alpha}$.

Coulomb Gauge leads to a Poisson equation $\nabla^2\Phi=-{\rho\over \epsilon_0}$ for $\Phi$ ,as in electrostatics, and the vector potential can be shown to satisfy a wave equation with just the divergenceless component of $\mathbf{J}$ as it's source (SI units): $$\nabla^2\mathbf{A}-{1\over c^2}{\partial ^2\mathbf{A}\over \partial t^2}=-\mu_0\mathbf{J}+{1\over c^2}\nabla {\partial \Phi\over \partial t}=-\mu_0\mathbf{J}_t$$

So, it will have a simple form when no sources are present at all. More importantly, this gauge is useful and simple for far field radiation problems. The asymptotic behavior of $\Phi$ is as $Q\over r$ ($Q$ is the total charge), and because it's gradient behaves as: $$-\nabla \Phi\sim {Q\over r^3}\mathbf{r}$$ we can neglect $\Phi$ when calculating radiation fields and the equations will become: $$\mathbf{E}\sim-{\partial \mathbf{A}\over \partial t}$$ $$\mathbf{B}=\nabla \times \mathbf{A}$$ for far fields.

It can be shown that the Lorenz Gauge leads to exactly the same fields for radiation as the Coulomb gauge.

  • $\begingroup$ So can we still formulate the co variant form of electromagnetism using coulomb's gauge? $\endgroup$ – Naman Agarwal Apr 25 '18 at 12:19

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