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In many textbooks, including Griffiths', they erroneously claim that a field is irrotational if and only if it is conservative (there exists a scalar potential).

This is true only if the domain of the field is simply connected, which is obviously not the case in electrostatics (we have $\mathbb{R}^3$ excluding the points where the charges are located).

Also, it seems like everyone assumes that $\vec{E}$ and $\vec{B}$ (electric and magnetic field) are continuously differentiable (which is not the case, just think about a conductor), so that they can use Stokes theorem, the divergence theorem and so on.

What am I missing?

I also have a bit of confusion about the definition of electrostatic field. By electric field I mean two things:

  1. The electric field generated by a finite number of charges $Q_0, \dots, Q_N$ located at $\vec{r_0}, \dots, \vec{r_N}$: $$ \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^{N}\frac{Q_i}{||\vec{r}-\vec{r_i}||^3} (\vec{r}-\vec{r_i}) $$
  2. The electric field generated by a continuous distribution of charge $\rho$: $$ \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \iiint_K \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||^3} (\vec{r}-\vec{r'}) \ d^3\vec{r'} $$

Griffiths proves that the electrostatic field is conservative by manually calculating the line integral of a single-charge electrostatic field over an arbitrary path (the same would be for a finite number of charges); the result depends only on the end points, so $\vec{E}$ is conservative. Does this prove the fact that the electrostatic potential produced by a continuous charge distribution is conservative? I don't think so, and I cannot find a proper proof of this.

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  • $\begingroup$ The definitions are identical one is simply the continous limit of the other and vice versa. If you truly wish to be technically correct, continuous charge distributions dont exist. Objects have charge that is distributed approximately continuously because of how small charges are both in size and value. In this case we just take the limit where charge is continuous throughout the continous (also a lie) matter. If you replace the distribution with a bunch of dirac deltas for each localized discrete charge you get the sum from the integral. $\endgroup$ – Craig Dec 22 '18 at 22:28
  • $\begingroup$ Related: physics.stackexchange.com/q/1324/2451 and links therein. $\endgroup$ – Qmechanic Dec 23 '18 at 7:13
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This is true only if the domain of the field is simply connected, which is obviously not the case in electrostatics (we have ℝ3 excluding the points where the charges are located).

Which definition of simply connected are you using? In $\mathbb{R}^3 $ the exclusion of isolated points does not destroy the simple connectivity. The reduction to one point of any closed curve is not hampered by isolated points.

..it seems like everyone assumes that E⃗ and B⃗ (electric and magnetic field) are continuously differentiable (which is not the case, just think about a conductor)...

In the case of a conductor, the fields E and B have discontinuities only at the surface. Elsewhere, not at the separation surfaces, they are modeled by smooth functions

Formulae [1] and [2] are equivalent. One is for isolated, finite charges, while the other is for continuous charge distribution

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  • $\begingroup$ Oh well, I just realized that in my analysis course we only defined simply connectedness for $\mathbb{R}^2$ sets, which is a bit different from $\mathbb{R}^3$ (the plane excluding a point is not simply connected, where evidently the space excluding a point is). Thanks! I get that the discontinuities are only at the surfaces, but still why can we apply Stokes and divergence theorem? I guess it's related to the fact that a surface has measure zero in $\mathbb{R}^3$ and so it doesn't really matter, but I can't really formally justify it. $\endgroup$ – Mattia F. Dec 23 '18 at 8:18
  • $\begingroup$ The reason we can apply stokes and divergence theorem is that we do it in every simply connected component $\mathbb{R}^3$ is partitioned by the discontinuity surfaces. Matching conditions at the discontinuities allow to "propagate" solutions from a component to the next. $\endgroup$ – GiorgioP Dec 23 '18 at 12:20
  • $\begingroup$ What do you mean by "matching conditions"? I don't really understand the last phrase. $\endgroup$ – Mattia F. Dec 23 '18 at 21:23
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    $\begingroup$ The conditions on the continuity of the tangential component of electric field and on the possible discontinuity of its normal component, realted to the surface density, for emplefor example. $\endgroup$ – GiorgioP Dec 23 '18 at 22:46

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