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.
Best Answer
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.
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.
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 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).
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.