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I am trying to verify Gauss's law, differential form, SI units, non-special relativity regime, ignoring time retardation, by performing the differentiation on the left-side of the equations to see if the results match the right side of the equations. I can't get Gauss's law (electric) nor Ampere's law to match their left and right sides.
For example, I am using $$ \vec{E} = -\nabla \Phi -\partial \vec{A} / \partial t $$ Taking the divergence of both sides I get $$ \nabla \bullet \vec{E} = \dfrac{1}{\epsilon_0}\rho - \nabla \bullet \partial \vec{A} / \partial t $$ Some people have suggested that I now need to apply Coulomb gauge condition $$\nabla \bullet \vec{A} = 0$$ but 1) That is not indicated explicitly in Maxwell's equations. I only see $$ \nabla \bullet \vec{E} = \dfrac{1}{\epsilon_0}\rho . $$ Where is the Coulomb gauge requirement indicated?

And 2) doesn't selecting the Coulomb gauge fix the gauge, removing the gauge freedom from Maxwell's system of equations?

Does this mean that Maxwell's equations are only valid (mathematically consistent) in the Coulomb gauge? I thought Maxwell's equations were mathematically valid in general and for any gauge. But Gauss's Law doesn't seem to be valid for any gauge.

If you look at Jackson's book, for example, Gauss's Law is only derived in his Chapter 1 on Electrostatics.

Any comments on the following line of reasoning would be appreciated....

Let us do some expansion (also called a decomposition) of the electrodynamic field $\vec{E}$ $$ \vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + ... $$ The decomposition could be a Helmholtz decomposition (2 terms only) of the vector field. But in this case, I'll take the decomposition to be $$ \vec{E} = \vec{E}_1 + \vec{E}_2 = -\nabla \Phi -\partial \vec{A} / \partial t $$ where $$ \vec{E}_1 = -\nabla \Phi $$ and $$ \vec{E}_2 = -\partial \vec{A} / \partial t $$ Is there a mathematical problem in doing this? I can then write $$ \vec{E} = \vec{E}_1 -\partial \vec{A} / \partial t $$

If I understand what some people are telling me, maybe I misunderstand what they are saying, I must now relabel $ \vec{E}_1 \rightarrow \vec{E}$ (because in Jackson's book Chapter 1 $ \vec{E} = -\nabla \Phi$) and write $$ \vec{E} = \vec{E} -\partial \vec{A} / \partial t $$

I don't think this is mathematically correct.

Why isn't it valid to write Gauss's Law as $$ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$ And so, with this, Maxwell's Gauss's law is $$ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$ without any gauge fixing. Maxwell's system for electrodynamics then continues to have gauge freedom.

If this is not mathematically correct, can you point out what mathematically went wrong?

Again, I am trying to verify mathematically all of Maxwell's equations as a single system of differential equations and running into some issues.... Any help would be appreciated.

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  • $\begingroup$ If your aim is to verify if scalar and vector potential obey the Gauss law, you have to have expression of these potentials where there is some reference to charge density or charge. You need to take the functions $\phi(\mathbf x,t),\mathbf A(\mathbf x,t)$ in terms of $\rho,\mathbf j$ and calculate the divergence of $-\nabla \phi -\partial_t \mathbf A$. If you do not have such functions, there is no point in the exercise. Gauss law is stated for electric field and is a generalization of experience, you cannot "prove it" with the concept of potentials alone. $\endgroup$ – Ján Lalinský Jul 6 '18 at 11:45
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Your mistake is simply that $\nabla^2 \phi = -\rho/\epsilon_0$ only in Coulomb gauge. Otherwise you just have

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} = -\nabla^2 \phi - \frac{\partial \nabla\cdot\mathbf{A}}{\partial t}.$$

Maxwell's equations in terms of the fields are valid always, and so are the relations between the fields and potentials, $\mathbf{E} = - \nabla \phi - \partial \mathbf{A} / \partial t$ and $\mathbf{B} = \nabla \times \mathbf{A}$ together with the equations for the potentials:

$$\nabla^2 \phi + \frac{\partial \nabla \cdot \mathbf{A}}{\partial t} = - \frac{\rho}{\epsilon_0}$$

and

$$\frac{1}{c^2} \frac{\partial^2 \mathbf{A}}{\partial t^2} - \nabla^2 \mathbf{A} + \nabla(\nabla \cdot \mathbf{A}) + \frac{1}{c^2} \frac{\partial \nabla \phi}{\partial t} = \mu_0 \mathbf{J}.$$

Simplified versions of these only hold in certain gauges. For example, in Coulomb gauge, where $\nabla \cdot \mathbf{A} = 0$, we have $\nabla^2 \phi = - \rho/\epsilon_0$, but not in other gauges.

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  • $\begingroup$ Hi Javier, I think I am starting to see the light.... Are you saying (please correct me): 1) Maxwell's system in terms of fields is always correct because it is a set of equations that is postulated to be true. As a postulate, they are not subject to mathematical verification. 2) When we try to write the fields in terms of scalar and vector potential functions, then Maxwell's equations are not verifiable in general, but Gauss's Law is verifiable in the Coulomb gauge. Do I have that right? Is there a gauge where all the equations are mathematically verifiable? Thanks. Pleas corect. $\endgroup$ – Researcher720 Jul 6 '18 at 12:13
  • $\begingroup$ @Researcher720 1) I would say experimentally verified instead of postulated, but yes. 2) Maxwell's equations in terms of the potentials do hold: they're the last two I wrote in my answer. They do not include $\nabla^2 \phi = -\rho/\epsilon_0$, except in Coulomb gauge. Gauss' law is always true for $\mathbf{E}$, but not always true for $\phi$. The three big equations in my answer always hold. $\endgroup$ – Javier Jul 6 '18 at 12:26
  • $\begingroup$ Thank you for your input. It helps a lot. But I don't think your "I would say experimentally verified" is totally correct. Maybe mostly correct. For Gauss's Law, I can see how one might verify experimentally for electrostatic case, but not for electrodynamics case. Gauss's Law in Maxwell's equations is suppose to be for general electrodynamics case. How is that experimentally verified, and distinguished from electrostatic case? (Doesn't refer to the potential functions because talking experimental now.) $\endgroup$ – Researcher720 Jul 7 '18 at 13:34
  • $\begingroup$ @Researcher720 I don't know much about experiments, so I can't really help you there. But note that the four Maxwell's equations imply that the fields obey the wave equation, and these have definitely been confirmed many many times. A modification to Gauss' law would probably show up as a discrepancy somewhere, though I don't really know where. $\endgroup$ – Javier Jul 7 '18 at 13:53
  • $\begingroup$ Hi Javier, Yes, you are right. There is an issue with deriving the wave equations. This is solved by introducing retarded time. With time retardation, we can get an inhomogeneous wave equation. So, normal way to get homogeneous wave equation wont work, but you can get the inhomogeneous wave equation. Which I think is better because it identifies the wave sources explicitly. Thanks for your comment. $\endgroup$ – Researcher720 Jul 8 '18 at 14:42

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