What are the details behind causality and Maxwell's equations?

Question

Maxwell's equations are \begin{align} \nabla\cdot\mathbf{E} & = \frac{\rho}{\epsilon_0} & \nabla\times\mathbf{B} &= \mu_0\epsilon_0 \frac{\partial\mathbf{E}}{\partial t} + \mu_0 \mathbf{J} \\ \nabla\cdot\mathbf{B} & = 0 & \nabla\times\mathbf{E} &=- \frac{\partial \mathbf{B}}{\partial t}. \end{align} Viewed in light of Helmholtz decomposition these equations can be viewed as fixing independent parts of the fields, with the $\nabla\cdot$ column fixing the divergent (irrotational) parts of $\mathbf{E}$ and $\mathbf{B}$ and the $\nabla\times$ equations fixing the solenoidal (divergenceless) parts.

As suggested by the formula for Helmholtz decomposition, the divergent part of $\mathbf{E}$ is given by $$\mathbf{E}_{\mathrm{div}}(\mathbf{x},t) = \frac{1}{4\pi\epsilon_0}\int \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3} \rho(\mathbf{x}',t) \operatorname{d}^3\mathbf{x}' \tag1$$ even when $\rho$ depends on $t$, as written above.

The statement in $(1)$ seems to violate causality. It shouldn't matter since $\mathbf{E}_{\mathrm{sol}}$ should also violate causality in such a way as to make the total electric field obey causality. My question is: what are the details that show how the acausal parts of the Helmholtz decomposed parts of $\mathbf{E}$ and $\mathbf{B}$ cancel (esp. does it require charge conservation)?

Put another way, starting from these equations \begin{align} \mathbf{E}_{\mathrm{div}}(\mathbf{x},t) & = \frac{1}{4\pi\epsilon_0}\int \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3} \rho(\mathbf{x}',t) \operatorname{d}^3\mathbf{x}' \tag2 \\ \mathbf{E}_{\mathrm{sol}}(\mathbf{x},t) & = \frac{1}{4\pi} \int \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3} \times \left(\frac{\partial \mathbf{B}(\mathbf{x}',t)}{\partial t}\right) \operatorname{d}^3 x' \tag3\\ \mathbf{B}_{\mathrm{sol}}(\mathbf{x},t) & = - \frac{1}{4\pi} \int \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3} \times \left(\mu_0\epsilon_0\frac{\partial \mathbf{E}(\mathbf{x}',t)}{\partial t} + \mu_0 \mathbf{J}(\mathbf{x}',t)\right) \operatorname{d}^3 x' \tag4 \end{align} what is the process of transitioning to a manifestly causal form of $\mathbf{E}$ and $\mathbf{B}$ (e.g. Jefimenko's equations), and what parts of the two parts of $\mathbf{E}$ above cancel out in the process?

Answer 1

I would like to point out in more detail the problems in the question which compares the purely mathematical Helmholtz decomposition of an (almost) arbitrary vector field $\mathbf{E(\mathbf{r})}$ to the Coulomb solution of a static charge density distribution of the Maxwell equations.

According to the Helmholtz theorem, any vector field $\mathbf{E(\mathbf{r})}$ con be decomposed into a curl-free (irrotational) vector field $\mathbf{a(\mathbf{r})}$ and a divergence-free (solenoid) field $\mathbf{b(\mathbf{r})}:$ $$\mathbf{E(\mathbf{r})}=\mathbf{a(\mathbf{r})}+\mathbf{b(\mathbf{r})}=-\nabla\phi(\mathbf{r})+\nabla\times \mathbf{A(\mathbf{r})} \tag 1$$ where $\phi(\mathbf{r})$ is a scalar potential and $\mathbf{A(\mathbf{r})}$ is a vector potential, which are given by: $$\phi(\mathbf{r})=\frac{1}{4\pi}\int_V \frac{\nabla \mathbf{E(\mathbf{r'})}d^3r'}{|\mathbf{r}-\mathbf{r'}|} \tag 2$$ and $$\mathbf{A(\mathbf{r})}=\frac{1}{4\pi}\int_V \frac{\nabla \times \mathbf{E(\mathbf{r'})}d^3r'}{|\mathbf{r}-\mathbf{r'}|} \tag 3$$ This decomposition can also be applied to the electric field solution of Maxwell's equation at any chosen time $t$. Thus you could formally enter the time as a parameter in the electric field decomposition, as has been done in the question. It is intriguing that the curl-free part of the decomposition eq. (2), which is equivalent to eq. (1) of the question, looks like the Coulomb solution of Maxwell's equations for a static charge distribution $$\frac{\rho(\mathbf{r})}{\epsilon_0}=\nabla\mathbf{E(\mathbf{r})}$$ and it is indeed the static, time independent field solution of Maxwell's equations for a static charge density distribution. But this does not imply any violation of causality, as suggested, because there is no evolution in time following from this mathematical formula as opposed to the solution of Maxwell's equations, which of course, obey causality. Actually, you need the causal, time dependent solutions of the Maxwell equations to Helmholtz-decompose them at a chosen time $t$ according to eqs. (1),(2), and (3), which are equivalent to the formulae (2) and (3) of the question (and (4) if you also do the analogous decomposition for the magnetic field).

The "transitioning to the manifestly causal form of $\mathbf{E}$ and $\mathbf{B}$" is achieved by using the field solutions of Maxwell's equation in the decomposition. This not only guarantees the causality in the decomposed fields but also ensures the consistence with additional physically necessary conditions, like charge conservation.

Answer 2

I think that you have the right suspicion. In assuming an arbitrary time dependence of the charge distribution $\rho(\vec r, t)$, you contradict the law of charge conservation $$div \vec J=-\frac {\partial \rho}{\partial t}$$ For example, you cannot assume that a point charge pops up out of nothing at a certain location.

Therefore, your Coulomb law electric field solution corresponding to a charge distribution can only hold, in principle, for an eternally static (i.e. time independent) charge distribution. It also holds approximately at limited distances for slow changes of the charge distribution that are compatible with charge conservation as it follows from the known retarded solutions with finite propagation times of the electric field. When you have time dependent charges compatible with the continuity equation, you also get (time changing) currents and magnetic fields which you have to consider.

Answer 3

The supposition in the original question, that there is some cancellation between the divergenceful part of $\mathbf{E}$ and its solenoidal part is false. It was based on doing the kind of construction forbade in the questions/clarifications (writing the solenoidal field as total field minus the divergeneful part). Instead, the expression for the divergenceful part of the field is not unique, and a causal construction is possible using the continuity equation for the 4-current, $$\frac{\partial \rho}{\partial t} + \nabla \cdot\mathbf{J} =0. \tag1$$
The easiest way to see that there is no cancellation is to show that the equations governing the the solenoidal and divergenceful part of the fields decouple. Start by taking the time derivative of the Coulomb's law Maxwell's equation to get $\nabla \cdot \partial_t \mathbf{E} = \frac{1}{\epsilon_0} \partial_t \rho$. By the continuity equation the right hand side becomes $\nabla\cdot \partial_t\mathbf{E} = -\frac{1}{\epsilon_0} \nabla\cdot\mathbf{J}$. Now this is interesting, because it implies that the divergenceful parts of $\partial_t\mathbf{E}$ and $\mathbf{J}$ are proportional, i.e. \begin{align} \frac{\partial \mathbf{E}_{\mathrm{div}}}{\partial t}& = -\frac{1}{\epsilon_0} \mathbf{J}_{\mathrm{div}}. \tag2 \end{align}

Equation (2) changes the form of the curl based Maxwell's equations to read \begin{equation}\begin{aligned} \nabla \times \mathbf{E}_{\mathrm{sol}} & = - \frac{\partial \mathbf{B}_{\mathrm{sol}}}{\partial t} \\ \nabla \times \mathbf{B}_{\mathrm{sol}} & = \mu_0 \mathbf{J}_{\mathrm{sol}} + \mu_0\epsilon_0 \frac{\partial \mathbf{E}_{\mathrm{sol}}}{\partial t}. \end{aligned} \tag3 \end{equation} As is familiar from undergraduate level physics courses, the Equations in (3) can be combined to form wave equations for $\mathbf{E}_{\mathrm{sol}}$ and $\mathbf{B}_{\mathrm{sol}}$ with standard causal solutions in terms of the Green's function for the same.

So, what happens to the divergenceful part of the electric field? Its defining equation, $$\nabla \cdot \mathbf{E}_{\mathrm{div}} = \frac{\rho}{\epsilon_0},$$ does not seem to allow for dynamics at all, just a non-local relationship that fixes $\mathbf{E}_{\mathrm{div}}$ in terms of $\rho$. The key is that the dynamics come in from Equation (2). That can be solved, explicitly, as $$\mathbf{E}_{\mathrm{div}}(\mathbf{x},t) = \mathbf{E}_{\mathrm{div}}(\mathbf{x},0) - \frac{1}{\epsilon_0} \int_0^t \mathbf{J}_{\mathrm{div}}(\mathbf{x},t') \operatorname{d}t'. \tag4$$ Equation (4) is causal: $\mathbf{E}_{\mathrm{div}}(\mathbf{x},t)$ is fixed in terms of a set of initial values within its backward facing light cone on a space-like hypersurface at the base of the cone, and influences within the cone. This satisfies the spirit of the question, if not the letter.

That $\mathbf{E}_{\mathrm{div}}(\mathbf{x},t)$ can be expressed in terms of Equation (2) from the original post or Equation (4) from this one suggests that it should be possible to continuously deform between the two - from a space-like cone, through light-like, and time-like until the radius of its base drops to zero, reducing a volume integral to a line integral. Showing that is left for another question.

What of the wave equation $\mathbf{E}$ obeys, implying that the divergenceful part of $\mathbf{E}$ obeys one, too. That can be derived from taking the time derivative of Equation (2) and adding that to the divergence of the gradient of (4), keeping track of the continuity equation, produces the wave equation from the question linked in this paragraph. Similarly, the divergence of the gradient of Equation (2) from the original post works, as long as surface terms vanish.

Note that the linear partial differential equations that the divergenceful part of $\mathbf{E}$ obeys are more strict than the wave equation, so the former may allow some solutions that the latter forbids.

P.S. This also answers the original second part of the question that was ruled overly broad. Since there is no cancellation, just an equivalent causal expression for $\mathbf{E}_{\mathrm{div}}(\mathbf{x},t)$, there are no QFT consequences from a cancellation between $\mathbf{E}_{\mathrm{div}}$ and $\mathbf{E}_{\mathrm{sol}}$.