This question is motivated by a student of mine. Is there a general characterization, or classification, of electrostatic fields whose field lines are straight? There are some obvious examples:

  • Uniform field; immediately generalizes to the field of any charge distribution that varies only in one direction, but is constant in the other two directions.
  • Field of a point charge; immediately generalizes to the field of any spherically symmetric charge distribution.
  • Field of a straight charged wire; immediately generalizes to the field of any axially symmetric charge distribution.

An alternative formulation of the question might be: find all (Newtonian) gravitational fields such that any particle, initially at rest, will freely fall along a straight trajectory.

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    $\begingroup$ The plane perpendicular to the line between two equal charges and halfway between them has only straight field lines. The same is true if the "charges" are masses and the field is gravitational. The analogous situation for two infinite wires would hold true. $\endgroup$
    – S. McGrew
    Feb 19, 2021 at 0:16
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    $\begingroup$ Moreover, any configuration of matter or charge that has either rotational symmetry or plane symmetry will have at least one straight field line. $\endgroup$
    – S. McGrew
    Feb 19, 2021 at 0:42
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    $\begingroup$ @S.McGrew What I had in mind were electric fields such that all field lines, not just one or two, are straight. Sorry if that wasn't clear from the question and the given examples. $\endgroup$ Feb 19, 2021 at 13:05

6 Answers 6


I do not know about existing classifications. But it seems that the field should satisfy in general the following equation (btw it reminds me of the conformal Killing equation): \begin{equation} \vec E(\vec x + \epsilon \vec E) = F(x) \vec E(\vec x), \end{equation} with $F(x)$ being a scalar function. The infinitesimal form is \begin{equation} f(x) E_i(x) = E_j (x) \partial_j E_i(x). \end{equation} Multiplying by $E_i$ both sides fixes $f(x)$ as \begin{equation} f(x) = \frac{1}{2E^2(x)} \partial _j \left ( E^2(x) E_j(x) \right ), \end{equation} where I assume that $\partial _i E_i(x)=0$. As a result we get \begin{equation} E_i(x) E_j (x)\partial_j E^2(x) = 2 E^2(x) E_j(x)\partial_j E_i(x). \end{equation} Maybe it would be easier to analyze this equation using scalar potential \begin{equation} E_i(x)=-\partial _i \varphi(x). \end{equation}

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    $\begingroup$ It might be worth noting that for an electrostatic field, $\partial_i E_j = \partial_j E_i$ (this is the curl-free condition in disguise.) $\endgroup$ Jan 29, 2021 at 15:13
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    $\begingroup$ @ Michael Seifert thanks, I think that indeed might be useful. Using this we get $E_i E_j \partial _j E^2 = E^2 \partial _i E^2$, which implies $\partial_i E^2(x) =c(x)E_i(x) $ $\endgroup$
    – nwolijin
    Jan 29, 2021 at 15:40
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    $\begingroup$ Thanks @nwolijin. This might perhaps be more easily formulated using vectors. The straight field line condition is equivalent to requiring that the directional derivative of $\vec E$ along $\vec E$ itself, that is $(\vec E\cdot\vec\nabla)\vec E$, is parallel to $\vec E$. Thanks to the fact that $\vec E$ is conservative and so curl-free, one has that $(\vec E\cdot\vec\nabla)\vec E=\frac12\vec\nabla(\vec E^2)$. So one ends up with your condition that $\vec\nabla(\vec E^2)$ is parallel to $\vec E$. It is not necessary to assume that $\vec\nabla\cdot\vec E=0$. $\endgroup$ Jan 29, 2021 at 19:40
  • $\begingroup$ What is meant by "multiplying by E_i both sides fixes f(x)"? Isn't f(x) arbitrary? If you simply rearrange one equation to "solve for f(x)" and then substitute back in to the same equation the result should be trivial like 1=1. Am I missing something? $\endgroup$
    – hft
    Feb 20, 2021 at 0:22

As a partial answer, it's true that there exist charge distributions with straight field lines that aren't symmetric in any way. For example, suppose a charge is placed outside a grounded conducting sphere; then charges are induced on its surface. We know from the method of images that the electric field produced by those induced charges, outside of the sphere, is precisely the same as the electric field produced by an "image" charge. Therefore, the field of the induced charges has perfectly straight field lines, even though the induced charges don't have any symmetry.

This makes it unlikely that there's a simple general way to describe the charge distributions. But the equipotential surfaces in this case are still spheres; you might have better luck constraining those.

  • $\begingroup$ That's a very nice observation! It shows that in my own partial answer to the question, I have forgotten to include the possibility that the $\vec E$ field may be discontinuous. (I assumed that the direction of $\vec E$ is well-defined everywhere except for points where $\vec E$ vanishes.) I agree that it might be easier to characterize equipotential surfaces. $\endgroup$ Feb 20, 2021 at 21:37
  • $\begingroup$ @asking_anonymously Here is yet another geometric reformulation of the problem using equipotential surfaces. Once the E-field has been normalized to unity, the potential difference between two nearby equipotential surfaces equals simply their distance measured along a joint normal vector. So the question is: given two surfaces that share normal vectors, and whose distance is a constant (independent of the choice of normal vector along which we measure the distance), are these surfaces necessarily a pair of parallel planes, coaxial cylinders or concentric spheres? $\endgroup$ Feb 22, 2021 at 7:24
  • $\begingroup$ @TomášBrauner I gave it another shot, but I'm not at all confident my second answer is right; take a look! $\endgroup$
    – knzhou
    Feb 26, 2021 at 23:59

Here is an attempt at a partial answer, extending the argument in the previous answer by nwolijin. Let us start with some vector calculus. For any vector fields $\vec u,\vec v$ and scalar field $\phi$ we have the identities \begin{align} \tag{1} \vec\nabla(\vec u\cdot\vec v)&=(\vec u\cdot\vec\nabla)\vec v+(\vec v\cdot\vec\nabla)\vec u+\vec u\times(\vec\nabla\times\vec v)+\vec v\times(\vec\nabla\times\vec u),\\ \tag{2} \vec\nabla\times(\phi\vec v)&=\phi\vec\nabla\times\vec v-\vec v\times\vec\nabla\phi. \end{align} Now the condition that the field lines of $\vec E$ are straight is equivalent to the condition that the directional derivative of $\vec E$ along itself, that is $(\vec E\cdot\vec\nabla)\vec E$, is parallel to $\vec E$. Since electrostatic field is conservative, it follows from (1) that $(\vec E\cdot\vec\nabla)\vec E=\frac12\vec\nabla(\vec E^2)$. Hence $\vec\nabla(\vec E^2)$ must be parallel to $\vec E$.

As the next step, use (2) with $\vec E$ in place of $\vec v$ and any function $f(\vec E^2)$ in place of $\phi$. Since $\vec\nabla(\vec E^2)$ is parallel to $\vec E$, so is $\vec\nabla f(\vec E^2)$, and (2) then tells us that $\vec E'\equiv f(\vec E^2)\vec E$ is a conservative field. At the same time, $\vec E'$ is parallel to $\vec E$ everywhere, and thus also has straight field lines. We arrive at the conclusion that the electric field maintains the straight field line property (and remains conservative) if we rescale it by an arbitrary function of $\vec E^2$. We can in particular replace the field $\vec E$ with a unit vector of the same direction everywhere in space, except for points where $\vec E=\vec 0$; at such points the direction of the field is ill-defined.

So if we are only interested in the topology of the field lines, not in the magnitude of the electric field, we may as well assume that the field is a unit vector everywhere where it is nonzero. It appears that it should be possible to characterize such unit vector fields just by their singularities, i.e. by specifying the set of points in space where the electric field vanishes.

Here is where some hand-waving starts. The singularities can be either isolated points, curves, surfaces or three-dimensional domains. In case of point singularities, the field $\vec E'$ will point radially outwards (or inwards) in their immediate neighborhood. In the other cases, the field lines will enter the singularity from directions perpendicular to it. This also applies to the surface of three-dimensional domains. We have thus reformulated our problem in geometric terms. We look for sets of points, curves, surfaces and domains in three-dimensional Euclidean space such that any straight line starting from a point, or any straight line starting along a normal direction to a curve or surface, will either stretch to infinity or end up on another singularity, entering it along a normal direction. Any two such lines are not allowed to cross anywhere outside of the given set of singularities.

Some obvious possibilities for the set of singularities are:

  • A set of parallel planes. This corresponds to a one-dimensional electric field.
  • A set of concentric spherical surfaces. This corresponds to a spherically symmetric electric field.
  • A set of coaxial cylindric surfaces. This corresponds to an axially symmetric electric field.
  • Special limits thereof, including a sphere of zero radius (point singularity) or a cylinder of zero radius (straight line singularity), a slab (stack of planes with infinitesimal spacing), a solid cylinder or sphere (the same for cylinders and spheres), and the like.

Is there some elegant way to solve this geometric reformulation of the problem?

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    $\begingroup$ The divergence of your unit normal field n is the mean curvature H. Maybe you can prove that H is constant on an equipotential, and also that the Gaussian curvature K is constant. If so, then your equipotential must be either a plane, a cylinder, or a sphere. For a field of the form you want, I suspect that the Hessian of the potential only contains three rotationally invariant pieces of information: H, K, and the charge density. $\endgroup$ Feb 19, 2021 at 16:24
  • $\begingroup$ @asking_anonymously Thanks for the interesting tip! I'll have to think about this more carefully... $\endgroup$ Feb 19, 2021 at 22:38
  • $\begingroup$ @TomášBrauner - I figured out that if you use the unit vector instead of the full vector then the curvature approach works for all the cases. $\endgroup$ Feb 20, 2021 at 22:25

What is commong among all the three cases is the presence of a symmetry:

  • reflection symmetry in the case of a charged plane
  • spherical symmetry in the case of a point charge
  • a symmetry axis in the case of a straight wire or cylinder

The electric field is then directed along the coordinate curves of the coordinate system natural for each symmetry.

So there are two main ingredients for a generalization are:

Note that in general case these coordinate curves are not straight lines in the sense implied in the question - i.e., being straight lines in Euclidean coordinates.

Finally, similar claims could be made about the magnetic field.

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    $\begingroup$ Thanks for the response. I agree that all the examples I list have special symmetry, that is rather obvious. But it's not clear to me how the suggested generalization goes. Do you have any example that is different from those already listed in the question? Note also that I don't require the potential of the electric field to be harmonic. There can be an arbitrary charge distribution so that the potential satisfies the Poisson equation. All I assume is that the electric field is static and conservative. $\endgroup$ Feb 19, 2021 at 13:15
  • $\begingroup$ Regarding the harmonic functions - I think what is important here is that the field is twice differentiable and that its lines do not cross, which allows for its use as coordinate lines probably for an arbitrary charge distribution. What seems to me too vaguely defined in your question is straight line... $\endgroup$
    – Roger V.
    Feb 19, 2021 at 13:38
  • $\begingroup$ Do you have in mind some kind of curved space on which the electric field would live? Or are you seriously asking for a definition of a straight line in Euclidean space? $\endgroup$ Feb 19, 2021 at 16:38
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    $\begingroup$ Yes, that's what I meant. This post is based on a question I received from a second-year student of a basic course on electromagnetism. It's a simple question that is meant to be simple, there are no hidden tricks. But it is a very good example of a simple question that is not easy to answer. $\endgroup$ Feb 19, 2021 at 19:17
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    $\begingroup$ If we limit ourselves to the straight lines in the Euclidean space, then we should consider only the coordinate systems that has some of their coordinate lines straight - this probably limits us to the three cases that are already mentioned. I will think about it. $\endgroup$
    – Roger V.
    Feb 19, 2021 at 20:19

This is my second attempt at an answer; criticism is welcomed. This could all be incredibly off base, because I've never taken a differential geometry class, and I strongly suspect the problem has a simple formulation in that language.

To incorporate the curl-free condition, we characterize the field in terms of its equipotential surfaces. Consider a specific equipotential surface $S$. The electric field on this surface must all be pointing directly inward or outward; let's suppose it's outward for concreteness, and also that it has a single connected component.

I have a hunch (but no proof) that in order for the field lines to be straight, the electric field magnitude on each surface $S$ must be a constant $E_S$. This is at least true in two dimensions, but I can't really tell in three.

Suppose this is true. Then consider two infinitesimally distant surfaces $S$ and $S'$. By the divergence-free condition, we can apply Gauss's law to the volume bounded by $S$ and $S'$, getting $E_S A_S = E_{S'} A_{S'}$ where $A_S$ is the area of $S$. On the other hand, we can consider a small patch $dA_S$ of $S$, which expands to a small patch $dA_{S'}$ of $S'$. Applying Gauss's law to the volume bounded by these surfaces and the field lines gives $E_S dA_S = E_{S'} dA_{S'}$. Therefore, $$\frac{dA_S}{dA_{S'}} = \frac{A_S}{A_{S'}} = \text{const.}$$ In other words, as the equipotential surfaces expand outward uniformly, the area of each piece of the surface has to change at the same rate.

Now consider a patch of the surface $S$, and place it at the origin with the normal vector pointing along the $\hat{\mathbf{z}}$ direction. If we orient the $xy$ axes appropriately, the equation of the surface in a small neighborhood is $$z \approx \frac{a}{2} x^2 + \frac{b}{2} y^2$$ where $a$ and $b$ are the principal curvatures. The normal vector of the surface is $\hat{\mathbf{n}} \propto (ax, yb, 1)$. Thus, after transporting a distance $\epsilon$ along the normal vector, any patch on the initial surface is stretched by a factor $1 + \epsilon a$ in the $x$ direction, and $1 + \epsilon b$ in the $y$ direction, so the area scales as $$\frac{dA_{S'}}{dA_S} = (1 + \epsilon a)(1 + \epsilon b) = 1 + \epsilon (a + b) + O(\epsilon^2).$$ That is, the rate of increase of area is characterized entirely by the sum of the principal curvatures. This implies the equipotential surfaces must be constant mean curvature surfaces, and there are many examples besides the plane, cylinder, and sphere. The question reduces then to whether these other surfaces continue to have constant mean curvature as they are expanded outward.

  • $\begingroup$ Very nice! This goes a long way towards a complete answer to the question. Just some extra comments. First, for a conservative field $\vec E$, the field lines are straight if and only if $(\vec E\cdot\vec\nabla)\vec E=\frac12\vec\nabla(\vec E^2)$ is parallel to $\vec E$. Hence $\vec E^2$ must be constant on any equipotential surface, which proves your assumption. Second, in the absence of electric charge, the equipotential surfaces should form a family of constant mean curvature surfaces. I believe this is only possible for planes, cylinders and spheres, although I don't have a proof. $\endgroup$ Feb 27, 2021 at 22:00

Is there a general characterization, or classification, of electrostatic fields whose field lines are straight?

I had a previous answer (and some other still present answers) that related to the curvature with the expression depending upon something like $\left( \mathbf{E} \cdot \nabla \right) \mathbf{E}$ which only works for divergenceless fields.

The part I had missed is that I left the expression in terms of the magnitude of the field instead of using the unit vectors. So let's define a unit vector as $\hat{u} = \tfrac{ \mathbf{V} }{ \lvert \mathbf{V} \rvert }$ then we can define curvature in terms of this as: $$ \boldsymbol{\kappa}_{c} = \left( \hat{u} \cdot \nabla \right) \hat{u} \tag{0} $$

The radius of curvature is just the inverse of the magnitude of the curvature given by: $$ \rho_{c} = \frac{ 1 }{ \lvert \boldsymbol{\kappa}_{c} \rvert } \tag{1} $$

A field line that is straight will satisfy $\rho_{c} \rightarrow \infty$, i.e., a zero magnitude curvature.

Example: Point Charge
For a point charge, the electric field $\hat{u} = (1,0,0)$ in spherical coordinates and so that Equation 0 goes to zero. If we use the full electric field vector $\mathbf{E} \propto r^{-2} \hat{r}$ in place of $\hat{u}$ then Equation 0 would not be zero, which was my the with my previous answer and the issue with some of the other answers herein.

Example: Linear Gradient
Suppose we a linearly increasing (in magnitude) electric field given by: $$ \mathbf{E} = E_{o} \left( 1 + z \right) \hat{x} \tag{2} $$

Again if we use $\mathbf{E}$ instead of its unit vector, Equation 0 will not have a zero magnitude but it will if we use the unit vector.

Example: Dipole
Suppose we have a dipole made of two point charges separated by some vector $\mathbf{d}$ and we observe said field at some displacement $\mathbf{r}$ in the $\phi = 0$ plane. Then the electric field will have an $r$ and $\theta$ component given by: $$ \mathbf{E} \simeq \frac{ k \ d }{ r^{3} } \left[ \left( 2 \ \cos{\theta} \right) \hat{r} + \sin{\theta} \ \hat{\theta} \right] \tag{3} $$

We know the fields are not straight here and indeed the expression for Equation 0 is not zero but a complex function of $r$, $d$, and $\theta$ given by: $$ \begin{align} \left( \left( \hat{e} \cdot \nabla \right) \hat{e} \right)_{r} & = - \frac{ 8 \sin^{2}{\theta} }{ r \left( 5 + 3 \cos{2 \theta} \right)^{2} } \tag{4a} \\ \left( \left( \hat{e} \cdot \nabla \right) \hat{e} \right)_{\theta} & = \frac{ 8 \sin{2 \theta} }{ r \left( 5 + 3 \cos{2 \theta} \right)^{2} } \tag{4b} \\ \left( \left( \hat{e} \cdot \nabla \right) \hat{e} \right)_{\phi} & = 0 \tag{4c} \end{align} $$ where $\hat{e} = \tfrac{ \mathbf{E} }{ \lvert \mathbf{E} \rvert }$ is the electric field unit vector given by: $$ \begin{align} e_{r} & = \frac{ 2 \cos{\theta} }{ \sqrt{ 4 \cos^{2}{\theta} + \sin^{2}{\theta} } } \tag{5a} \\ e_{\theta} & = \frac{ \sin{\theta} }{ \sqrt{ 4 \cos^{2}{\theta} + \sin^{2}{\theta} } } \tag{5b} \\ e_{\phi} & = 0 \tag{5c} \end{align} $$

Therefore, the field line curvature must be calculated in terms of the unit vector of the field, not the field with its magnitude included. In the limit of $\rho_{c} \rightarrow \infty$, the field lines will always be straight (ignoring critical points where the magnitude is zero).

  • $\begingroup$ This doesn't seem to say anything specific about electric fields... of course, a line is straight if its curvature is zero, but that doesn't help us with the present question. $\endgroup$
    – knzhou
    Feb 20, 2021 at 18:07
  • $\begingroup$ @knzhou - As you may or may not have seen my previous answer, I tried to do this in general starting from the actual fields. I came to realize that there is no analytical generalization of what I was trying to do. In the case of electrostatic fields, this would just be a special limit for my $\mathbf{V}$. What I show here is as close to a general solution as can be determined, I think. $\endgroup$ Feb 20, 2021 at 18:19
  • $\begingroup$ @knzhou - I updated the answer and figured out why my previous answer was wrong, i.e., I wasn't using the unit vectors. $\endgroup$ Feb 20, 2021 at 20:30
  • $\begingroup$ @knzhou - The OP asks about how to determine whether field lines are straight, do they not? If so, then my answer is correct now (after a few stumbles, admittedly). Yes it starts specific to electric fields but then they ask about gravitational fields later in the question, thus I tried to generalize it. The curvature is the key and the simplest approach here. $\endgroup$ Feb 20, 2021 at 21:28
  • $\begingroup$ Thanks for your effort. Unfortunately I still don't see how this adds anything to the already existing answers. First, note that your straight field line criterion of vanishing of (0) is just a special of the criterion valid for fields of arbitrary magnitude, present in the answers by nwolijin and myself, that is, $(\vec E\cdot\vec\nabla)\vec E$ must be parallel to $\vec E$. Second, as pointed out by @knzhou, your answer does not address electrostatic fields. These have the important property that they are conservative. It is preserving this property that makes my question nontrivial. $\endgroup$ Feb 21, 2021 at 20:04

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