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So I understand that we select the divergence of A (magnetic vector potential) to be:

$$\frac{1}{c^2}\dot{\phi} + \nabla\cdot\vec{A} = 0.$$

The Lorenz gauge (1).

because of the symmetries in Maxwell's equations. Also, I know that the A and the scalar potential ($\varphi$) are not unique. We can write:

$$\phi \rightarrow \phi' = \phi-\dot{\Psi},$$

$$ \vec{A} \rightarrow \vec{A}' = \vec{A} +\nabla\Psi.$$

Which give the same E and B fields. Until here everything is general. So we have some freedom to select A and $\varphi$.

The question is: how does this imply the Lorenz gauge (Equation 1)? If there is further information needed to prove Equation 1 please let me know!

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  • $\begingroup$ For arbitrary potentials, the Lorenz gauge condition is not necessarily true. But you can always do a gauge transformation to make it true. $\endgroup$ – Javier Oct 26 '19 at 17:07
  • $\begingroup$ I think I got the point. It is like a constant of integration, we can fix it to satisfy the condition. Thanks! $\endgroup$ – AHMED KRS Oct 29 '19 at 14:11
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I don't think it does imply the Lorenz gauge. You can take advantage fof the gauge freedoms to specify any scalar relationship between the magnetic vector and scalar electric potentials, which is equivalent to choosing a form for the function $\Psi$.

For instance, the Lorenz gauge implies that $\Psi$ is any function that satisfies $$\nabla^2 \Psi = \frac{1}{c^2}\frac{\partial^2 \Psi}{\partial t^2}$$

The Lorenz gauge is often used because it is Lorentz invariant and is useful in deriving symmetric, inhomogeneous wave equations of the form $$\nabla^2 \vec{A} - \frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2} = -\mu_0 \vec{J}$$ $$\nabla^2 \phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = -\frac{\rho}{\epsilon_0} $$ that can then be used to solve problems in electrodynamics using retarded potentials.

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  • $\begingroup$ So we do not need a formal proof for the Lorenz Gauge. It is just true because A is not unique. Therefore, we can fix it such that it satisfies the gauge condition. Thank you so much! $\endgroup$ – AHMED KRS Oct 29 '19 at 14:15

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