Understanding Gauss's Law and Special Cases in Maxwell's Equations

[Researcher720]

I have an issue with Gauss's Law when it is considered one equation of Maxwell's complete system of equations. I don't have an issue with it when it is a standalone equation, but when it is one of several equations put together to form a complete system of equations, there is an issue.

How is one is to know that in Gauss's Law $$ \nabla \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho_f $$ that $$ \vec{E} = - \nabla \Phi $$ and is not $$ \vec{E} = -\nabla \Phi - \partial \vec{A} / \partial t $$?

Some people say that "this is a special case". Yes, exactly. But how in Maxwell's system of equations, where symbol "E" is shown in several different equations (Gauss's Law, Faraday's Law, Ampere's Law, Lorentz force equations) is one to know E is a special case and when E is not a special case?

Without explicit indication of when the special case is being invoked there is a mathematical problem.

Once a special case is invoked it must be stated explicitly so that it can be applied to all following equations. Otherwise, you have no idea if special conditions are being invoked or not. So, as a complete mathematical/logical system, not as independent equations, isn't some indication (some notation) required to indicate special cases?

If I have a system of equations that includes
$$ f = a x^2 $$
and
$$ f = a x^2 + b x^3 $$
where, in general, b is not 0, isn't this a problem, because this says, in this system of equations, that
$$ f = a x^2 \ne a x^2 + b x^3 = f $$
so then
$$ f \ne f $$
which is a contradiction for the general case b not equal to 0.

How is this handled in Maxwell's equations for E?

 

[Dale]

How is one to know that in Gauss's Law $$ \nabla \vec{E} = \dfrac{1}{\epsilon_0} \rho_f $$ that $$ \vec{E} = - \nabla \Phi $$ and is not $$ \vec{E} = -\nabla \Phi - \partial \vec{A} / \partial t $$?

The definition of the potentials are: $E=-\nabla \phi - \partial A/\partial t$ and $B=\nabla \times A$. The special case formula, $E=-\nabla \phi$ is not part of Maxwell's equations nor is it part of the definitions of any of the terms in Maxwell's equations. Any good electrodynamics textbook should be clear about that, for example section 12.1 here: http://web.mit.edu/6.013_book/www/book.html

But how in Maxwell's system of equations, where E is shown are we to know E is a special case and when E is not a special case. All that I am saying is that without explicit indication of when the special case is being invoked there is a mathematical problem.

The simplifying information does not and cannot come from Maxwell's equations. It comes from the physics or from the problem setup. If you physically know that you have a static scenario or if you are working a problem which tells you that it is a static scenario, then you are justified in simplifying the time derivatives to 0. Otherwise you must use the general equation. There is no mathematical problem because the simplification is not obtained mathematically.

 

[Researcher720]

This is wonderful input. Thank you.

So let's check Gauss's Law using the definition of E in terms of the scalar and vector potential functions:
$$ \nabla \bullet \vec{E} = -\nabla \bullet \nabla \Phi - \nabla \bullet \partial \vec{A} / \partial t \\
\nabla \bullet \vec{E} = -\nabla^2 \Phi - \nabla \bullet \partial \vec{A} / \partial t \\
\nabla \bullet \vec{E} = -\dfrac{1}{4 \pi \epsilon_0} \iiint \rho \nabla^2 \dfrac{1}{r_{TS}} dVol_S - \nabla \bullet \partial \vec{A} / \partial t \\
\nabla \bullet \vec{E} = -\dfrac{1}{4 \pi \epsilon_0} \iiint \rho \left( -4 \pi \delta(\vec{r}_T - \vec{r}_S) \right) dVol_S - \nabla \bullet \partial \vec{A} / \partial t \\
\nabla \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho(\vec{r}_T,t) - \nabla \bullet \partial \vec{A} / \partial t
$$
But this result is not Gauss's Law. There is a second term involving the vector potential.

Doesn't this, in general, show that there is a problem with Gauss's Law? I am confused.

It seems that, as a general system of equations, there is a problem with Maxwell's system of equations. Gauss's Law as
$$\nabla \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho(\vec{r}_T,t) $$
doesn't seem to be verifiable, in general. The left side doesn't equal the right side. What am I missing?

 

[Researcher720]

I have read the chapter 12.0 and 12.1 sections you pointed me to. http://web.mit.edu/6.013_book/www/book.html

The author writes: "In Chap. 4, we made $$ \vec{E} = - \nabla \Phi $$ ... Later he writes equation (3) as
$$ \vec{E} = -\nabla \Phi - \partial \vec{A} / \partial t $$.

But your comment says that $$ \vec{E} = - \nabla \Phi $$ is not part of Maxwell's equations. So, this is part of Maxwell's electrodynamics but is not part of Maxwell's equations? Is that what you mean? Clearly $$ \vec{E} = - \nabla \Phi $$ is being used in the textbook in some places. Then the question comes up again as to when to use one definition of E and when to use another definition of E. <personal speculation deleted>

 

[Dale]

What am I missing?

Your third step is wrong. For one thing the Laplacian should be outside the integral, and for two it seems like you are not correctly using the retarded time. There may be other mistakes too.

The expressions for $\phi$ and $A$ in terms of charge and current densities are designed so that they satisfy $\nabla^2 \phi +\frac{\partial}{\partial t} (\nabla \cdot A)=-\rho/\epsilon_0$ which is Gauss' law. If an equation purporting to calculate the potentials does not satisfy this equation then it is not a correct equation. In this case, that is your step 3.

he author writes: "In Chap. 4, ... So, this is part of Maxwell's electrodynamics but is not part of Maxwell's equations?

Chapter 4 is a chapter on electroquasistatics, not electrodynamics. The author is very explicit in that chapter about the simplifying assumptions being used in that chapter. In fact, he spends all of chapter 3 introducing the assumptions and their resulting simplifications.

Then the question comes up again as to when to use one definition of E and when to use another definition of E.

As I said, you only use the simplified equation when you have outside information that the simplification is appropriate. This textbook is very clear about that. I highly recommend that you read chapter 3.

 

[Researcher720]

O.k., the notation I used was not clear enough. Sorry. I was hurrying...

The Laplacian can go under the integrals because it operates on the test body's coordinates and the integral is with respect to the source body coordinates. Recall that in electrodynamics there are two different position parameters, test (or observation) and source positions. I use subscript "T" and subscript "S" to distinguish between the two. Often, these subscripts are omitted because it is understood that the nabla vector operator is with respect to the test (or observation) position parameters and not with respect to the source position parameters. For example,
$$ \vec{r}_{TS} \equiv \vec{r}_T - \vec{r}_S \\
\nabla_T \equiv \partial / \partial x_T \hat{x} + \partial / \partial y_T \hat{y} + \partial / \partial z_T \hat{z} \\
\nabla_T \bullet \vec{E} = - \nabla_T \bullet \nabla_T \Phi - \nabla_T \bullet \partial \vec{A} / \partial t \\
\nabla_T \bullet \vec{E} = - \dfrac{1}{4 \pi \epsilon_0}\nabla_T^2 \iiint \rho \dfrac{1}{r_{TS}} dVol_S - \nabla_T \bullet \partial \vec{A} / \partial t \\
\nabla_T \bullet \vec{E} = - \dfrac{1}{4 \pi \epsilon_0} \iiint \nabla_T^2 \rho \dfrac{1}{r_{TS}} dVol_S - \nabla_T \bullet \partial \vec{A} / \partial t \\
\nabla_T \bullet \vec{E} = - \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \nabla_T^2 \dfrac{1}{r_{TS}} dVol_S - \nabla_T \bullet \partial \vec{A} / \partial t \\
\nabla_T \bullet \vec{E} = - \dfrac{1}{4 \pi \epsilon_0} \iiint \rho (-4 \pi \delta(\vec{r}_T - \vec{r}_S)) dVol_S - \nabla_T \bullet \partial \vec{A} / \partial t \\
\nabla_T \bullet \vec{E} = \dfrac{1}{ \epsilon_0} \rho(\vec{r}_T,t ) - \nabla_T \bullet \partial \vec{A} / \partial t
$$
We can swap the Laplacian operator and the charge density function around because the charge density function rho does not depend on the test (observation) position.

This demonstrates that Gauss's Law is not mathematically correct in general. Do you agree? And that, as you say, more information is needed to make this mathematically correct. How would you specify this outside information?

You mentioned "retarded time". Again, sorry, not stating my assumptions clearly. I am working in the regime where retarded time can be ignored. I am also working in the regime where all speeds are (much) less than the speed of light, so special relativity doesn't come into it either. (Hope you are not suggesting that Maxwell's equations (Gauss's equation in particular) are only valid when time retardation is introduced. But maybe you are? I'd need more clarification as to why this would be.)

You wrote: "you only use the simplified equation when you have outside information". This sounds to me (I know you didn't say this explicitly) like you are saying that Maxwell's system of equations are not self-contained, and not consistent in general, but are only for special cases. I don't think that is what you mean, but that is what I am stuck on. Isn't Maxwell's electrodynamics, and Maxwell's differential equations in particular, mathematically consistent in general? If not, why not? And then why would anyone use a system of equations that is not mathematically consistent? For example, Gauss's Law, demonstrated above.

 

[Dale]

Recall that in electrodynamics there are two different position parameters, test (or observation) and source positions.

No, not really. There is only one set of coordinates. I think you are referring to the dummy variable used inside the integral. If so, then the only coordinates are what you call the test coordinates, and the dummy variable is what you call the source position.

I am working in the regime where retarded time can be ignored.

OK, but then you should not expect to recover Gauss' law exactly. Maxwell's equations only hold in general if you use the retarded time.

Isn't Maxwell's electrodynamics, and Maxwell's differential equations in particular, mathematically consistent in general?

Yes. But not every single equation that is ever used is one of those equations. $E=-\nabla \phi$ is not one of them generally.

 

[Dale]

Recall that in electrodynamics there are two different position parameters, test (or observation) and source positions. I use subscript "T" and subscript "S" to distinguish between the two.

OK, I was confused by both this notation and this description, both of which I had not seen previously. However, looking at the math your T subscript is just the standard coordinates and your S subscript is just the standard dummy variable. I am fine with your notation.

We can swap the Laplacian operator and the charge density function around because the charge density function rho does not depend on the test (observation) position.

This is not generally true. The retarded time introduces a dependence on both your T and S coordinates. See below for more details.

You mentioned "retarded time". Again, sorry, not stating my assumptions clearly. I am working in the regime where retarded time can be ignored.

You need to be careful here. You cannot simply ignore retarded time at will. If you are using the Lorenz gauge then the retarded time is unavoidable in general. You can get rid of the retarded time in two ways that I know.

The first is to work in a static scenario, so that everything is the same at the present time and the retarded time. If everything is static then you do recover Gauss’ law, but that seems to defeat the point of the whole exercise, so let’s skip that one.

The second is to use the Coulomb gauge. In that gauge the retarded time disappears from the scalar potential (but not the vector potential) and the scalar potential depends on the instantaneous charge density. In the Coulomb gauge $\nabla \cdot A=0$, so you recover Gauss’ law in the end.

The result of this is:
1) if you are using the Lorenz gauge then your math is wrong, you cannot switch the Laplacian around like that because of the retarded time
2) if you are using the Coulomb gauge your math is OK, and you are just missing one step to recover Gauss’ law.

Maxwell’s equations are mathematically consistent either way.