We know that the gradient of the electric potential function $V(x,y,z)$ is the electric field. But not all vector fields are gradients, for example $y\hat{i}-x\hat{j}$ is not a gradient. Does this mean that an electric field given by $\vec{E}=y\hat{i}-x\hat{j}$ cannot exist? In particular does this mean there is no arrangement of charges which could create such a field? What is the correct interpretation here?

Note: the example I have given here is a bit stupid, because the field is two dimensional and this is probably impossible anyways. But there are of course three dimensional fields which are not gradients, I just can't think of one at the moment. If someone can please let me know.

Edit: to answer the comment below I will show that $y\hat{i}-x\hat{j}$ is indeed not a gradient. If it is, then there exists a function $f$ such that $∇f(x, y) = yi − xj$. Obviously $\frac{∂f}{∂x}(x, y) = y$, $\frac{∂f}{∂y}(x, y) = −x$ so $\frac{∂^2f}{∂y∂x}(x, y) = 1$, $\frac{∂f}{∂x∂y} (x, y) = −1$ and thus $\frac{∂^2f}{∂y∂x}(x, y) \neq \frac{∂f}{∂x∂y} (x, y)$.

This contradicts the mixed partial derivative theorem: the four partial derivatives under consideration are everywhere continuous and thus, according to the mixed partial derivative theorem, we must have

$\frac{∂^2f}{∂y∂x}(x, y) = \frac{∂f}{∂x∂y}(x,y)$.

  • $\begingroup$ why isn't yi−xj a gradient? $\endgroup$ – TheQuantumMan Mar 22 '15 at 15:51
  • $\begingroup$ yes you are right.I deleted my answer. $\endgroup$ – TheQuantumMan Mar 22 '15 at 16:02
  • $\begingroup$ I would suggest looking at Maxwell's equations. The electric field can also be produced by a time varying vector potential which can give you that kind of field. $\endgroup$ – ajctt Mar 22 '15 at 16:09

In electrostatics, there is always the scalar potential (voltage) such that $\vec\nabla\phi = -\vec E$.

This does not hold true in general electromagnetism, which is described by the electromagnetic four potential $A$ consisting of the electrostatic potential $\phi$ and the magnetic vector potential $\vec A$ such that $$ \vec E = -\vec\nabla\phi - \partial_t \vec A$$ $$ \vec B = \vec\nabla\times\vec A$$

So, not being a gradient is not sufficient to conclude that a given vector field cannot be the electric field, since for non-zero $\vec A$, i.e. in the presence of currents/moving charges/magnetic fields, we do not expect $\vec E$ to be the gradient of a scalar field.

An electric field would be impossible if it cannot arise as the solution to Maxwell's equations, but as long as your magnetic field is unconstrained, you can obviously just set $\partial_t \vec B$ and $\rho$ such that the given field fulfills them, so you cannot from the form of $\vec E$ alone judge whether it is physically permissible.

  • $\begingroup$ How exactly does this answer my question? $\endgroup$ – Joshua Benabou Mar 22 '15 at 16:05
  • $\begingroup$ @JoshuaBenabou: I expanded the answer to more clearly address your question. $\endgroup$ – ACuriousMind Mar 22 '15 at 16:11
  • $\begingroup$ Ok now I understand. So a couple questions: (1) so you are saying in an electrostatic situation, the field I described is impossible. (2) Is there an "easy" example of a field which is not a solution to Maxwell's equations (something non-pathological if possible). (3) I am trying to think of a charge distribution which would produce a two-dimensional field (all the field vectors lie in the same plane). Is there an "easy" example? $\endgroup$ – Joshua Benabou Mar 22 '15 at 16:20
  • $\begingroup$ @JoshuaBenabou: 1. Yes, exactly. 2. Well, your field together with $\vec B = 0$ is impossible. As I say, there is no $\vec E$, however pathological, that cannot be created by choosing $\rho$ and $\vec B$ appropriately. 3. Infinitely big charged sheet, perhaps? $\endgroup$ – ACuriousMind Mar 22 '15 at 16:25
  • $\begingroup$ for #2 I am asking is there a field $\vec{E}$ for which for which there is no choice of $\vec{B}$ that gives no solution to Maxwell's equation. In other words a field which cannot exist in any situation, not just in an electrostatic situation. $\endgroup$ – Joshua Benabou Mar 22 '15 at 16:34

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