Suppose we are dealing with electrostatics for this question. A physicist carries out experiments with static charges and determines that, the electric field $\vec { E } (\vec { r } )$ is a quantity which behaves as,

$$\nabla .\vec { E } =\frac { \rho }{ { \epsilon }_{ 0 } } $$

Further he notices that, this quantity drops to zero as $\vec { r } \rightarrow \infty$

Is it possible to deduce the field from these conditions alone?

Posed mathematically, the question is that the divergence equation is a first order PDE so by giving enough boundary conditions we should be able to determine the field right? If this was so then why do we need the curl equation? And again if one were to use the curl equation then we would have 3-unknowns and 4 equations, so some of them must be redundant right?

Note:Assume that there is no magnetic field present for the purpose of this question.

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    $\begingroup$ If you follow the logic of the Helmholtz decomposition, even if you know the divergence and the boundary conditions, you still can't fully determine the value of the field. $\endgroup$ – secavara Jan 14 '18 at 16:16
  • $\begingroup$ If you use the curl equation, then you would have four equations and 3-unknowns, so some equation must be redundant right? $\endgroup$ – Abhikumbale Jan 14 '18 at 16:19
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    $\begingroup$ The basic idea is that, if we are no longer in an electrostatic case, then the electric field could have a contribution which has no divergence (that is, in the kernel of the divergence operator). It's the same as when you're solving an inhomogenous differential equation. The particular solution (what you deduce above) can be supplemented with a homogenous solution. Maxwell's other equations allow you to solve for the homogenous part. $\endgroup$ – Bob Knighton Jan 14 '18 at 16:48
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    $\begingroup$ In my answer, I added an alternative mathematical solution to your partial differential equation for $E$ which doesn't explicitely use the $curl (\vec E) =0$ $\endgroup$ – freecharly Jan 14 '18 at 17:40
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    $\begingroup$ This is an ambiguous question hinging upon what "dealing with electrostatics" means. Normally it means that you are assuming that the Maxwell equations are all valid but the charges are stationary and the magnetic fields are not time-changing, so as to force $\nabla\times E=0.$ But inside the question proper you seem to be asking about a world where perhaps this latter equation does not hold at all, and all we know is that $\nabla\cdot E = \rho/\epsilon_0.$ Which is it? $\endgroup$ – CR Drost Jan 14 '18 at 20:59

Your statement is correct within your stipulation of stationary charges and no time-varying magnetic fields.

Outside of the restricted cases where a) there are no time varying magnetic fields present and b) the electric field is conservative, i.e. is the gradient of a scalar potential, , we need the curl equation

$$\nabla \times \vec E = - \frac {\partial \vec B} {\partial t}$$

to explain the results of additional experiments (starting with Faraday), namely those involving electric fields resulting from electromagnetic induction in a time varying magnetic field.

  • $\begingroup$ For an electrostatic problem $$\nabla \times \vec E = 0$$ Thus you don't have any time dependent electric and magnetic fields. $\endgroup$ – freecharly Jan 14 '18 at 16:48
  • $\begingroup$ I'm well aware of that. I interpreted OP's question as asserting the curl equation in Maxwell's equations was unnecessary. Given their comments on your answer, so did the OP. $\endgroup$ – paisanco Jan 14 '18 at 16:50
  • $\begingroup$ It was ambiguous. The OP added later that there is no magnetic field. $\endgroup$ – freecharly Jan 14 '18 at 17:31
  • $\begingroup$ The question is poorly titled and was ambiguous. $\endgroup$ – paisanco Jan 14 '18 at 17:38
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    $\begingroup$ Would whoever is downvoting please explain yourself? I'd be more motivated to improve the answer if its deficiencies are explained. $\endgroup$ – paisanco Jan 14 '18 at 18:48

This is an electrostatic problem. This means, as you mentioned, that $$\nabla \times{\vec E}=0$$ which is equivalent to the existence of an electrostatic potential $\Phi(\vec r)$ so that $$\vec E(\vec r)=-\nabla \Phi(\vec r)$$ Inserting this into $$\nabla \vec { E } =\frac { \rho }{ {\epsilon }_{ 0 } } \tag{I}$$ yields $$\Delta \Phi =-\frac { \rho }{ { \epsilon }_{ 0 } } $$ which is the Poisson equation for the electrostatic potential. This has a unique solution for $\vec E=-\nabla {\Phi}\rightarrow 0$ when $\vec { r } \rightarrow \infty$.

Another point of view is that, mathematically, using the Green's function of the equation (I), which corresponds to Coulomb's law, a solution to equation (I) with the boundary condition $\vec E\rightarrow 0$ when $\vec { r } \rightarrow \infty$ is $$\vec E(\vec r)= \int {\frac{\rho(\vec r') (\vec r - \vec r')d^3r'}{4 \pi \epsilon_0|\vec r - \vec r'|^3}}$$

Note: Equation (I) is also known as the differential form of Gauss's Law. Gauss Law is equivalent to Coulomb's Law. Gauss's law follows from Coulomb's Law and vice versa.

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    $\begingroup$ I'm supposing that we do not know the curl relation. $\endgroup$ – Abhikumbale Jan 14 '18 at 16:52
  • $\begingroup$ @Abhikumbale - So what do you mean by "Suppose we are dealing with electrostatics for this question?" $\endgroup$ – freecharly Jan 14 '18 at 16:56
  • $\begingroup$ I assume that there are no-time varying magnetic fields present in the region and the charges are stationary. $\endgroup$ – Abhikumbale Jan 14 '18 at 17:00
  • $\begingroup$ @Abhikumale - You only need Coulomb's Law and the boundary condition to find the solution. $\endgroup$ – freecharly Jan 14 '18 at 18:45
  • $\begingroup$ @Abhikumbale - You only need Coulomb's Law and the boundary conditions to find the solution. $\endgroup$ – freecharly Jan 14 '18 at 18:58

".... the question is that the divergence equation is a first order PDE so by giving enough boundary conditions we should be able to determine the field right?"

Not so. As mentioned in the comments the answer to this question is essentially about the Helmholtz decomposition, but let's actually partly delve into a certain proof of this decomposition that shows very clearly, geometrically and intuitively what the problem is, at least for a large class of vector fields, namely, those that have Fourier transform, as discussed in my answer here.

Imagine a vector field's Fourier decomposition $\vec{F}(\vec{k})$, a function of the plane wave wavevector $\vec{k}$, i.e. we decompose a vector valued function $\vec{f}(\vec{r})$ of position $\vec{r}$ into a superposition of plane wave vector fields of the form $\vec{F}(\vec{k})\,\exp(i\,\vec{k}\cdot\vec{r})$.

Now, how do divergence and curl look in Fourier space? $\nabla\cdot \vec{f}$ has the Fourier transform $\vec{k}\cdot\vec{F}$ and $\nabla\times \vec{f}$ has the transform $\vec{k}\times\vec{F}$; you should be able to prove this fairly straightforwardly.

So now, ask your question in Fourier space terms. It is, "why can we determine the vector $\vec{F}$ from $\vec{k}\cdot\vec{F}$ alone?". It should be very clear that this can't be done; we need to know the components of $\vec{F}$ that are orthogonal to $\vec{k}$ and these can be assigned essentially independently, since the divergence of a vector field everywhere orthogonal to $\vec{k}$ vanishes.

In general, one can assign a smooth scalar field in Fourier space $g(\vec{k})$ and a second smooth vector field $\vec{H}(\vec{k})$ that is everywhere orthogonal to $\vec{k}$, but otherwise arbitrary. As I discuss in this answer here and here, the information $g(\vec{k})$ and $\vec{H}(\vec{k})$ are exactly the information to determine a vector field $\vec{F}$ such that:

$$g(\vec{k}) = \vec{k}\cdot\vec{F}(\vec{k})$$ $$\vec{H}(\vec{k})= \vec{k}\times \vec{F}(\vec{k})$$

So the answer to your question is essentially that the divergence condition tells you only the component of the vector field that is along the wavevector; the solenoidal part, that orthogonal to the wavevector, is missing (it has zero divergence) and can be assigned independently.


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