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In electromagnetism, we say that any conservative electric field $\vec{E}(\vec{r})$ is associated to a scalar potential $V(\vec{r})$ such that $\vec{E}(\vec{r}) = -\nabla V(\vec{r})$. If the electric field is continuous, the respective electric potential must be differentiable because, if not, its gradient could not be calculated everywhere.

There are some cases, though, in which the electric field is discontinuous, leading to a non-differentiable electric potential. The latter is, however, still continuous.

Why is this? Why is it that even when the electric field is discontinuous, the electric potential is not? Must the electric potential always be continuous everywhere? A mathematical approach (i.e. not just a qualitative insight) is what I'm looking for.

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No. For example, the potential of a point charge is discontinuous at the location of the point charge, where the potential becomes infinite.

Since all charges in nature seem to be point charges (elementary particles such as electrons and quarks), electric potential always has discontinuities somewhere. When we work with continuous charge distributions, we are simply using an approximation that averages over lots of point charges and smears out the discontinuities in their charge density, potential, field, field energy density, etc.

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  • $\begingroup$ And what about the first question? "Why is it that even when the electric field is discontinuous, the electric potential is not?" $\endgroup$
    – Tendero
    Sep 7, 2019 at 18:10
  • $\begingroup$ That isn’t true. In the case of the point charge that I mentioned, both the field and the potential are discontinuous. $\endgroup$
    – G. Smith
    Sep 7, 2019 at 21:44
  • $\begingroup$ Isn't the electric field infinite in that point? What I want to understand is why the electric potential is continuous when the electric field shows finite discontinuities. $\endgroup$
    – Tendero
    Sep 8, 2019 at 2:43
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    $\begingroup$ The change in potential between two points is the line integral of the field between those points. Suppose the field has a finite discontinuity somewhere. Take two points $\epsilon$ away from the discontinuity and calculate the line integral, which will have two finite pieces. As $\epsilon\to 0$, this line integral goes to zero, so the change in potential across the finite discontinuity is zero, so the potential is continuous there. $\endgroup$
    – G. Smith
    Sep 8, 2019 at 4:03
  • $\begingroup$ What definition of continuity are you using? With those normally used in mathematics, the potential of a point charge is not defined at the charge. Being not-defined is not the same as discontinuous. $\endgroup$ Jul 23, 2023 at 14:34
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Yes, it is. In electromagnetism, the electric field is defined in terms of the electric potential through the equation $\mathbf{E}=-\nabla \phi$, where $\nabla$ is the gradient operator. If the electric potential were to have a discontinuity or singularity at a certain point, it would lead to a divergence in the electric field at that point, which is physically unrealistic. In classical electromagnetism, such singularities or discontinuities are not allowed within the domain of valid solutions.

As for point charges, they are not physical objects, just mathematical models. The Maxwell equations don't have solutions that are discontinuous functions. On the other hand, singular potentials play a crucial role in defining Green's functions and simplifying various concepts through the introduction of analyticity. They provide formal solutions for non-homogeneous equations in integral form and serve as valuable mathematical models.

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  • $\begingroup$ If you have a step function (the Heaviside's theta function equal to 0 for negative argument and equal to 1 for positive, do you really say that this function has a derivative at x=0? If you are thinking in terms of Dirac's delta function, that is not right. Dirac's delta is not a function at all. It is just a short name for a singular distribution. Quite a different object. $\endgroup$ Jul 24, 2023 at 6:45
  • $\begingroup$ what is the value of the electric field in the point x=0? $\endgroup$
    – freude
    Jul 24, 2023 at 7:58
  • $\begingroup$ It is not defined. Only the right and left limits are defined (and are both finite). Within a macroscopic description, this is the only possibility mathematically consistent. By the way, this whole issue is well known and explained in a mathematically consistent way in every textbook on "Potential Theory". $\endgroup$ Jul 24, 2023 at 8:09
  • $\begingroup$ it is like a complex-valued amplitude, isn't it? It is mathematically consistent but is not directly related to physically observed quantities. $\endgroup$
    – freude
    Jul 24, 2023 at 12:51
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A real discontinuity in the mathematical sense (not a divergence) of the electric potential is only possible in the presence of a surface dipole distribution (see, for instance, this page ). It can be obtained starting with the expression of the potential due to a continuous dipole density ${\bf P(r)}$: $$ \phi({\bf r})=\frac{1}{4 \pi \epsilon_0}\int {\bf P(r')}\cdot\nabla'\left( \frac{1}{| {\bf r} - {\bf r'} |} \right) dV' $$ Analyzing the effect of a dipole density confined on a surface, it can be shown that the limit values of the potential on the two faces of the surface must be different.

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  • $\begingroup$ It is a simplification. Step-wise like potential implies infinite electric fields. $\endgroup$
    – freude
    Jul 24, 2023 at 5:49
  • $\begingroup$ @freude Not really. If math is used in a rigorous way, the gradient of the potential at the surface is defined only as a limit from one side or the other. Of course, the discontinuous jump is a mathematical idealization, but it is coherent with a macroscopic description. One could say that most of the mathematical applications to Physics are idealizations. Do sharp, smooth surfaces are the boundaries of solids in nature? Certainly not. But at the macroscopic level, everything works as if they exist. $\endgroup$ Jul 24, 2023 at 6:33
  • $\begingroup$ agree, and you can also say, that microscopic electric fields are infinitely large $\endgroup$
    – freude
    Jul 24, 2023 at 8:01

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