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Edit: Can someone check my answer and possibly complete my task at the end?

The helmholtz theorem states that any vector field can be decomposed into a purely divergent part, and a purely solenoidal part.

What is this decomposition for $\vec{E}$, in order to find the field produced by its divergence, and the induced $\vec{E}$ field caused by changing magnetic fields.

The Potential Formulation:

$$\vec{E} = -\nabla V - \frac{\partial \vec{A}}{\partial t}$$

Is often transformed as $$\vec{E} = - \frac{\partial \vec{A}}{\partial t}$$

For showing induced fields, where charge density is not important ( and subsequently the scalar potential is zero).

There are a number of issues with this however using current density without including charge density violates $\vec{J} = \rho \vec{V}$

with that being said, in general, although it is a good approximation for the induced part of the field.

When not modeling $\rho$ as zero, From the lorenz gauge condition

$\nabla \cdot \vec{A} = - \mu_0 \epsilon_0 \frac{\partial V }{\partial t}$

We know the divergence of

$- \frac{\partial \vec{A}}{\partial t}$

Is non zero.

And thus that component of the $\vec{E}$ field cannot "just" be caused by the induced part, it is caused by the $\vec{E}$ fields divergence.

So what is an expression for the purely solenoidal part of the E field?

Edit:

artificially removing $V$ and then choosing the coulomb gauge to show its zero divergence is also incorrect as artifically removing $V$ with approximations should remove $A$ as well. Instead do the same with the full equation. Also, the components each terms represents is gauge dependent as well.

Perhaps the helmholtz decomposition is :

$$\vec{E} = -\nabla V - \frac{\partial \vec{A}}{\partial t}$$

Only under the coulomb gauge. As in the coulomb gauge, the first term has zero curl, but has divergence. But the second term has curl, but zero divergence?

What each part represents is gauge dependant, the full E field however being gauge independant.

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2 Answers 2

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For showing induced fields, where charge density is not important ( and subsequently the scalar potential is zero).

This seems confused. We may choose a gauge in which $V=0$, in which case $\vec E = -\frac{\partial}{\partial t} \vec A$. This is called the Weyl gauge, though this is a bit of a misnomer because there is still some residual gauge freedom left; as a result, this defines a family of gauges.

For example, when we have a single point charge $q$ at the coordinate origin, the electric field is given by $$\vec E = \frac{q \vec r}{4\pi \epsilon_0 |\vec r|^3}$$ In electrostatics, we typically choose the potentials $V= \frac{q}{4\pi \epsilon_0 |\vec r|}, \vec A=0$. However, we can perform a change of gauge $$\matrix{V \mapsto V- \frac{\partial}{\partial t}\chi\\ \vec A \mapsto \vec A +\nabla \chi}, \qquad \chi(\vec r,t) := \frac{q t}{4\pi \epsilon_0 |\vec r|}$$ in which case we obtain the potentials $V=0$, $\vec A = -\frac{qt \vec r}{4\pi \epsilon_0 |\vec r|^3}$. This can always be done, and is not a statement about the presence or absence of charge density.

There are a number of issue with this however as using current density without including charge density violates $\vec J = \rho \vec V$.

This is only true if there's only one species of particle in your system. In general, one has that $\rho = \sum_i \rho_i$ and $\vec J = \sum_i \rho_i \vec v_i$. It's entirely possible that $\rho=0$ and $\vec J\neq 0$; for example, if you have electrons moving through a background of stationary positively-charged ions.

From the lorenz gauge condition $\nabla \cdot \vec{A} = - \mu_0 \epsilon_0 \frac{\partial V }{\partial t}$, we know the divergence of $-\frac{\partial \vec{A}}{\partial t}$ is non zero.

Unless of course $\frac{\partial V}{\partial t}=0$. But note that in the Lorenz gauge the electric field is generally not given by $\vec E = -\frac{\partial}{\partial t} \vec A$, because the Lorenz gauge is not generally a Weyl gauge.


The helmholtz theorem states that any vector field can be decomposed into a purely divergent part, and a purely solenoidal part [...] So what is an expression for the purely solenoidal part of the E field?

The Helmholtz decomposition theorem applies to (twice-differentiable) vector fields which are defined on a simply-connected bounded subset $V\subset \mathbb R^3$. In that case, we may define

$$ \vec Q(\vec r,t) \equiv \frac{1}{4\pi} \int_V \frac{\nabla' \times \vec E(\vec r',t)}{|\vec r-\vec r'|} \mathrm d^3r' - \frac{1}{4\pi} \int_{\partial V} \hat n \times \frac{\vec E(\vec r',t)}{|\vec r-\vec r'|} \mathrm dS$$

where $\nabla'$ refers to differentiation with respect to $\vec r'$, $\partial V$ is the boundary of $V$, and $\vec n$ is the normal vector to $\partial V$. From there, $\nabla \times \vec Q$ is the purely solenoidal part of $\vec E$. If $\vec E$ goes to zero as $r\rightarrow \infty$, we may remove the requirement that $V$ be bounded and obtain $$ \vec Q(\vec r,t) \equiv \frac{1}{4\pi}\int_{\mathbb R^3} \frac{\nabla ' \times \vec E(\vec r',t)}{|\vec r-\vec r'|} \mathrm d^3 r'$$

Note also that if we choose the Coulomb gauge, then by definition $-\frac{\partial}{\partial t}\vec A$ is solenoidal. If the Helmholtz decomposition theorem applies (e.g. on the domain $\mathbb R^3$ where $\vec E\rightarrow 0$ as $r\rightarrow \infty$), then $\nabla \times \vec Q=-\frac{\partial }{\partial t} \vec A $ is the unique purely solenoidal component of $\vec E$ (by which I mean, the vector field obtained when the irrotational component has been subtracted).


My assertion with this modified formula was in relation to griffiths stating this, in the lorenz gauge, that because a wire was electrically neutral, $\rho=0$, and subsequently the scalar potential was zero.

If $\rho=0$, then the electric field is solenoidal as per Gauss's law. In the Coulomb gauge, this implies that $V=0$ and that

$$\nabla \times \vec B = -\nabla^2 \vec A = \mu_0 \vec J - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \vec A$$ $$\implies \left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2\right)\vec A = \mu_0 \vec J$$

The solution to this equation is given by $$\vec A = \int \frac{\mu_0 \vec J(\vec r', t_r)}{4\pi |\vec r-\vec r'|} \mathrm d^3 r'$$

as per the solution given in Griffiths' text. Note that this choice of gauge also satisfies the Lorenz gauge condition in this particular case.

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  • $\begingroup$ First part of your answer: the Weyl gauge is very interesting, I have never come across that before and will definitely look into it, and I definitely get that when choosing this gauge, it has nothing to do with the presence of some charge density. My assertion with this modified formula was in relation to griffiths stating this, in the lorenz gauge, that because a wire was electrically neutral, $\rho$ = 0, and subsequently the scalar potential was zero. ( which led me away from this specifically to learn about how to write down the purely solenoidal part of the field). -- $\endgroup$ Commented Apr 18, 2022 at 20:23
  • $\begingroup$ My question was more focusing on "IF I choose to ignore the scalar potential, as Griffiths has done. What does this actually represent?" As my aim goal with this helmholtz question was finding an expression for the purely solenoidal component of the E Field. In writing my own answer, you have also been writing. In my answer I eventually figured out on my own, that $-\frac{\partial \vec{A}}{\partial t}$ represents the induced electric field as a result of a changing B field, however it only represents this in THE COULOMB GAUGE. --- $\endgroup$ Commented Apr 18, 2022 at 20:27
  • $\begingroup$ Would it also be in your opinion, that griffiths ignoring the scalar potential, and then calculating the induced electric field in the LORENZ gauge, was a mistake? Unless like you said, his assumption that $\rho = 0$ does not contradict the fact that $\vec{J} = \rho \vec{V}$ is non zero. And subsequently $(\rho = 0,V= 0)$ and then $\nabla \cdot \vec{A} = 0$, ( so da/dt is purely solenoidal for the special case $\rho=0$. $\endgroup$ Commented Apr 18, 2022 at 20:32
  • $\begingroup$ Although I guess he got lucky as when $\rho ≠ 0$, the formula he used in the lorenz gauge, Does not represent the induced part only. $\endgroup$ Commented Apr 18, 2022 at 20:33
  • $\begingroup$ @jensenpaull I have edited my answer to address your question. If $\rho=0$ everywhere, then the Coulomb gauge is a Lorenz gauge because $\nabla^2 V=0 \implies V=0$ (assuming we require that $V\rightarrow 0$ at infinity) and $\nabla \cdot A=0$ by definition of the Coulomb gauge; as a result, $\nabla \cdot \vec A + \epsilon_0 \mu_0 \partial V/\partial t$ vanishes because each term vanishes individually. $\endgroup$
    – J. Murray
    Commented Apr 18, 2022 at 20:47
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This question was inspired by finding the E fields from the induced part only of a wire with increasing current.

$$\vec{E} = -\nabla V - \frac{\partial \vec{A}}{\partial t}$$

Subsequently the scalar potential V is set to zero and the E field takes the form

$$\vec{E} = - \frac{\partial \vec{A}}{\partial t}$$

This approximation, of removing V, isn't due to the fact that $\rho = 0$ everywhere, as if this were the case, $\vec{J}= \rho \vec{V}$ would be equal to zero aswell, and subsequently $\frac{\partial \vec{A}}{\partial t} = 0$ as there is no current density.

Removing $\nabla V$ is done to remove this contribution of the field, to find the induced part only.

Gauge dependance:

However, we will see that this naive approach is wrong, as what each term represents, is gauge dependant.

$$\vec{E} = -\nabla V - \frac{\partial \vec{A}}{\partial t}$$

$$\nabla \cdot \vec{E} = -\nabla^2 V - \frac{\partial}{\partial t}\nabla \cdot \vec{A}$$

$\nabla \cdot \vec{A}$ is gauge dependant.

Depending on the gauge choice , $-\frac{\partial \vec{A}}{\partial t}$ Could be purely solenoidal, or could infact have divergence.

Let's assume we choose the coulomb gauge to work with.

Clearly by definition, $\nabla \cdot \vec{A} = 0$

In this gauge, the term $-\nabla V$ is purely divergent, and the term $-\frac{\partial \vec{A}}{\partial t}$ is purely solenoidal.

The induced E field is due to the purely solenoidal part.

Meaning ONLY in the coulomb gauge,

Does the equation

$$\vec{E} = - \frac{\partial \vec{A}}{\partial t}$$

Make sense, to represent the induced electric field.

In the lorenz Gauge:

$\nabla \cdot \vec{A} = -\mu_0\epsilon_0\frac{\partial V}{\partial t}$

Meaning the although the term $\nabla V$ is purely divergent, the term - $\frac{\partial \vec{A}}{\partial t}$ is no longer purely solenoidal.

So in the lorenz gauge, the equation

$$\vec{E} = - \frac{\partial \vec{A}}{\partial t}$$

Does NOT represent the induced electric field as a result of the changing magnetic flux.

Misconception:

Although you could be tempted to say that $\nabla \cdot \vec{A}$ is related to V, and if $\rho$ = 0, then V is zero ( and subsequently the second term is purely solenoidal). The reason that setting $\nabla V = 0$ supposedly finds the solenoidal part isn't because $\rho = 0$ its because we choose to ignore it. (As we only want the solenoidal part) As otherwise $\vec{J}$ must be zero and then $\vec{A}$ must be zero.

This misconception could also be invalidated by considering situations where the body IS charged, but I only want to find the magnetic contribution.

**I would be extremely interested if someone could find the "induced" field, using the above equations, using the coulomb gauge, and in the lorenz gauge, to put my logic to the test, solving this question: **

enter image description here

I am aware of his logic to set the scalar potential to be zero, however in reality, it isn't because $\rho = 0$, otherwise there'd be no current. What he should have said, is that he is ignoring it purposely. He performed his calculations with a non zero $\vec{J}$(so he MUST have been ignoring it) in the lorenz gauge.

I have shown in this gauge, it has divergence, so it can't be just the induced part, it must also represent some fields produced by a charge density (from which he supposedly says was neutral, this contradicts this)

Can someone answer this in the coulomb gauge and compare it to his answer?

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  • $\begingroup$ $\rho = 0$ does not imply there is no current. If the wire is made of perfect conductor, there need not be charges anywhere in/on the wire for the current to flow. Only for wires with non-zero resistance, some surface charges usually have to be present to maintain electric field and current inside. $\endgroup$ Commented Apr 18, 2022 at 21:10
  • $\begingroup$ Electric current density is not, in general, expressible as $\rho \vec{V}$ where $\rho$ is total charge density. For example, DC current density in a wire can't be expressed using total charge density $\rho$, because when there is constant DC current, $\rho$ vanishes inside the wire. $\endgroup$ Commented Apr 18, 2022 at 21:16

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