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What are the properties of the field lines of an irrotational vector field like electrostatic field $\bf{E}$?

Zero divergence fields like $\bf{B}$ have the property to be always closed field lines. But what about the zero curl?

The idea I get is that for an irrotational field there are only point sources, where field lines "born" and then "return", but I don't know if this is correct and if it is general.

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    $\begingroup$ Might Mathematics be better suited for this math question? $\endgroup$ – Kyle Kanos Jan 2 '17 at 21:42
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    $\begingroup$ @KyleKanos No. I think because this is related to the physical sense of a physicist of field lines. $\endgroup$ – AHB Jan 7 '17 at 9:45
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About divergence-free field lines

It isn't necessary for the lines of a divergence-free field to be closed. Zero divergence implies that the field lines can't end. Closed curves don't have endpoints, but there are other possibilities, such as lines that extend infinitely in both directions.

The relevant mathematical result here is Gauss's theorem. For any field, it says that the integral of its divergence in any region of space equals the integral of the field over the boundary of the region, the flux of field lines crossing the boundary surface. If you enclose in such a surface a point at which field lines start or end, the flux doesn't vanish, so the divergence of the field can't be zero.

For magnetic fields, the divergence is zero and then there aren't any ending points for its lines. For electric fields, the divergence equals the density of charge, and so its lines have ends at point with non-vanishing charge. The flux of lines across a closed surface is proportional to the charge enclosed by it.


A geometrical picture of irrotational fields

So we have a picture of electric field lines being born and dying at some (charged) points or regions of space. But we haven't used the property that it's irrotational.

If you're looking for some geometrical implication of zero rotational, a way of getting it is considering the surfaces that are orthogonal to field lines at every point. The existence of this surfaces, even locally, is not guaranteed if our field is not irrotational. However, they can be defined for some fields with non vanishing rotational. For example, for the magnetic field around an infinite straight wire, they would be the semiplanes limited by it. A surface defined in this way has an orientation: at each point one of its two orthogonal directions is determined by the direction of the field at that point. Although it's not a standard terminology, let me call them here "field surfaces". A nice analogy with field lines in the previous case will arise.

We can imagine field surfaces having boundaries; that is, there are curves in space at which these surface can end. For example, you can picture such a curve and a family of surfaces sharing it as a boundary, just as a family of field lines might end at a point. If we take an oriented closed curve wrapping around a boundary curve, the integral of the field over it will be proportional to the number of field surfaces crossing it, including a negative sign for each that has opposite orientation as the curve an a positive sign otherwise.

There's a generalization of Gauss's theorem that's useful in this case. It's called Stokes' theorem. The version we need now says that the integral over a surface of the rotational of a field equals the integral of the field over the boundary of the surface, a closed curve. Suppose that the surface over which we are integrating cuts a boundary curve. Therefore, by integrating the rotational we are counting the field surfaces that cut its boundary.

For electric fields (in the static case) the rotational is zero. Thus, following the reasoning above, we can conclude that electric field surfaces can't end.


A more precise formulation of the second part

The picture above can be formulated more precisely using differential forms and the Frobenius theorem. The idea is not to use the fact that a divergence-free field has a potential, but to work with geometrical objects which might be understood intuitively, as field lines are. In particular, we want to understand what theses field surfaces are and to learn something about their existence.

In general, any non-vanishing $1$-form $\omega$ on a $n$-dimensional manifold $M$ defines at each point $p\in M$ a $(n-1)$-dimensional subspace $S_p$ of the tangent space $T_pM=\mathbb{R}^3$ as $S_p=\{v\in T_pM:\;\omega(v)=0\}$. Such a smooth assignment of subspaces of the tangent space to every point is called a distribution.

One question we can ask about a distribution is whether we can find a family of $(n-1)$-dimensional submanifolds such that their tangent space at each $p$ is $S_p$. If we can, the distribution is said to be integrable. For our case, we will need a particular case of the Frobenius theorem: the statement that if a differential form is closed, then its associated distribution is integrable.

Now, the electric field can be seen as a $1$-form $E=E_xdx+E_ydy+E_zdz$ in the manifold $M=\mathbb{R}^3$. The distribution associated with $E$ is nothing more that the family of planes given by orthogonality to the electric field at a point. The condition that its rotational is zero is then just $dE=0$. So by the Frobenius theorem, the ($2$-dimensional) surfaces that are tangent to the planes $E$ defines exist. These are just the "field surfaces" that are used above.

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  • $\begingroup$ What @coconut calls "field surfaces" is probably what Ampere called the "directive plane". The magnetic field is not "really" a vector but rather a pseudo-vector or an axial vector, or even better: a skew-symmetric tensor and as such it is best imagined as a "planar magnitude" (Sommerfeld). $\endgroup$ – hyportnex Jan 10 '17 at 17:44
  • $\begingroup$ Your condition is necessary but it is by no means sufficient. $\vec{E}(x,y,z)= \exp(x) \vec{e}_z$. This field admits surfaces normal to it which never end, they are the planes normal to the $z$ axis. Hovever $rot \vec{E}= - \exp(x) \vec{e}_y \neq 0$. $\endgroup$ – Valter Moretti Jan 10 '17 at 18:40
  • $\begingroup$ @ValterMoretti: I see. I wasn't thinking it was a sufficient condition, though. Maybe I should state that clearly. Anyways, I'm realizing now I didn't think very much when I wrote this answer. I imagined that it would be easy to give the simple geometric picture that I tried to explain. As I have said, I'll try to make everything clearer if I have the time. $\endgroup$ – coconut Jan 10 '17 at 19:39
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jan 11 '17 at 17:07
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If you'd like the physical intuition on the flow, rather than the field lines, the Wikipedia article on vorticy has some nice pictures and gifs. The clearest (in my opinion) is this one.

irrotational flow vs rotational flow

Differential particles undergo only translation in irrotational flow, while they translate and rotate in rotational flow.

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The most important property for the study of electromagnetism is that an irrotational field is the gradient of a scalar potential function (which in the case of the electric field, is the negative of the electric potential).

It is also not true that an irrotational field can only be generated from point sources; consider, for example, a sphere of uniform charge density. All field lines produced are radially arranged, so the field here is also irrotational, but there were no point charges involved anywhere.

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As probably_someone pointed out, an irrotational field can locally be expressed as the gradient of a potential function. Geometrically what this means is there is a family of surfaces (of constant potential) such that the irrotational vector field is always perpendicular to them.

But for an arbitrary vector field it is not always possible to construct a family of surfaces with this property. As an example, consider the vector field $u(x,y,z)$ $$u_x = \cos z,\quad u_y= \sin z,\quad u_z = 0$$ you won't be able to construct surfaces orthogonal to it even in a small neighborhood.

Now this property of being able to construct orthogonal surfaces is not quite unique to irrotational fields. A vector field that is some arbitrary function times an irrotational field also has the property. This is the cost of talking about field lines which lose some information about the magnitude of the vector field. What is necessary though for this geometric property is the component of the curl that is in the direction of the field itself must vanish.

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We know these:

$$ \text{In free space:} \left\{ \begin{array}{l} \nabla\times\mathbf{E}=0 \\ \nabla\cdot\mathbf{E}=\rho/\epsilon_0 \\ \end{array} \right. \ \ \ \ \ \ \text{In matter:} \left\{ \begin{array}{l} \nabla\times\mathbf{D}=\nabla\times\mathbf{P} \\ \nabla\cdot\mathbf{D}=\rho_{f}/\epsilon_0 \\ \end{array} \right. $$

Where the electric displacement is defined by $\mathbf{D}:=\epsilon_0\mathbf{E}+\mathbf{P}$.


$$\nabla\times\mathbf{E}=0 \implies \mathbf{E}=-\nabla V$$

By definition,

$$dV=\nabla V\cdot d\mathbf{l}$$

So if $d\mathbf{l}$ is perpendicular to $\nabla V$, V will not change. This means:

In 2D, If you start from any point on the plane, moving in the direction perpendicular to field lines, you have to form a closed loop. Because the surface you are in, must intersect with the equi-potential surface associated with the potential at your position. And there must exist one closed loop on that plane on which all points share the same potential with that of your position. That's because equi-potential surfaces are closed ones, no matter at infinity or not.

In 3D, If you start from any point in space, sweeping any possible point on your perpendicular-to-field-lines path, you have to from a closed surface.

This doesn't hold for $\mathbf{D}$ unless you can prove $\mathbf{P}$ is irrotational, that is, to prove it has these properties. This is often done by symmetry.


By definition,

$$\nabla\cdot\mathbf{E}=\lim_{\Delta V\to 0}\left[\frac{\oint_\mathcal{S} \mathbf{E}\cdot d\mathbf{a}}{\Delta V}\right]=\rho/\epsilon_0$$

This means if you consider a very small closed surface, The lines which come in, are exactly the same amount as the lines which get out of, the surface; if there is no charge enclosed by the small surface. This means those lines must be continuous in charge-free space.

In fact, Divergence is a measure of sources. So if you have discontinuities in some points, in which, the outgoing lines are more than incoming lines, you must be on a positive source point.

This holds for $\mathbf{D}$ too, as long as you neglect bound charges, those which are not much real ones, but almost our creation.


By application of Stokes's law on $\nabla\times\mathbf{E}=0$, we also get:

$$\oint_\mathcal{P}\mathbf{E}\cdot d\mathbf{l}=0$$

But this is not any help to our sense one field lines. Instead, we can play with the gradient formula to achieve another not-much-different-from-the-first-rule rule. :D

We know, by $\mathbf{E}=-\nabla V$, field lines must point to the steepest local decrease in the amount of potential. To if you go where $\mathbf{E}$ tells you, your potential should decrease the fastest. So if it turned out that your potential hasn't changed, or has even increased, something must be wrong about the field lines.

Some tracing will help you make sure what is drawn in front of you exists there in the real world.


But the first and the third point assume a previous sense of potential.

Looking at this integral:

$$V(\mathbf{r})=\frac1{4\pi\epsilon_0}\int\rho(\mathbf{r'})\frac{\mathbf{r}-\mathbf{r'}}{\left|\mathbf{r}-\mathbf{r'}\right|^3}d\tau'$$

It seems that potential is like a measure of positiveness inheritance which decays with getting more and more distant sharply.

If you are near a positive charge, you will inherit more positiveness. If you get farther from it, your positiveness will decrease kinda fast.


I think this is the most I could provide. And that this is probably the end of it and you can't dig more to get more sense. Instead of this, one could stick to mathematics to discover new fact!

Good luck.

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With regard to the field (as opposed to the field lines, as you asked about), the following are equivalent (see D. J. Griffiths' E&M book, Ch. 1, "Theorem 1"):

  1. $\nabla \times \mathbf{F} = 0$ everywhere.
  2. $\int_\mathbf{a}^\mathbf{b} \mathbf{F}\cdot\text{d}\mathbf{l}$ is independent of path from $\mathbf{a}$ to $\mathbf{b}$.
  3. $\oint\mathbf{F}\cdot\text{d}\mathbf{l} = 0$ for any closed loop.
  4. $\mathbf{F}$ is the gradient of some scalar function.

In terms of the field lines, it follows from (3) that they can't close on themselves, since if they did you could make a closed loop that follows a closed field line; if you integrated $\oint\mathbf{F}\cdot\text{d}\mathbf{l}$ over such a loop you would get something nonzero. (The converse is NOT true - you can't infer that if the field lines never close on themselves then the field is irrotational.)

If, following on (4), the vector field is $\mathbf{E}$ and you define the scalar function to be the negative of the electrostatic potential $V$, i.e. $\mathbf{E} = -\mathbf{\nabla} V$, then you can infer information about the potential from the field lines, and conversely if you know the potential field you can infer what the field lines are doing. Specifically, the field lines always point in the steepest downhill direction, and how close together the field lines are tells you how steep the potential gradient is in that direction. Peaks in potential correspond to regions of positive charge, and valleys to negative charge, so the field lines diverge from positive charges and converge to negative ones.

The idea I get is that for an irrotational field there are only point sources, where field lines "born" and then "return", but I don't know if this is correct and if it is general.

This is not right. The sources need not be points but can be distributed charge density, and if the field lines are "born" at one source (a positive charge density) they don't "return" to it but either diverge to infinity or converge on another "negative source", or sink (a negative charge density).

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  • $\begingroup$ What does "everywhere" mean in 1? If "everywhere" means everywhere in $\mathbb R^n$, then (1) <=> (2), (3), (4) [which are always equivalent]. If instead "everywhere" means "where F is defined, then (2),(3),(4) => (1) but the converse is generally false. $\endgroup$ – Valter Moretti Jan 10 '17 at 13:11
  • $\begingroup$ @ValterMoretti As physicists (not mathematicians), let us not quibble about such things! I don't think the OP was concerned about singularities in the field or spaces that are not simply connected (and nor was David Griffiths, where I quoted that from). Let's presume everything is, as they say, "sufficiently nice," so "everywhere" means "everywhere in $\mathbb{R}^3$. $\endgroup$ – pwf Jan 10 '17 at 17:01
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    $\begingroup$ In my view singulatities are however important in physics, especially in this case as they have topological implications. I think that other requirements as regularity are negligible. However I do not want to insist. $\endgroup$ – Valter Moretti Jan 10 '17 at 17:43
  • $\begingroup$ Alternatively, the two way implication holds if $\mathbf{F}$ and its curl are defined everywhere in a simply connected region, and we confine our discussion to that region. Keep the simply connected in mind and you won't go wrong: you need it to make the theorem to work, for example, for the 2D potential flow whose complex potential is $\Omega = \log z$. This one illustrates Valter's point about topological implications nicely. $\endgroup$ – WetSavannaAnimal Jan 11 '17 at 23:21
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By Gauss' theorem: $$ \int_V \underbrace{\nabla \cdot \vec{B}}_{=0} ~ dV= \int_{\partial V} \vec{B} \cdot d\vec{A} = 0 $$

Thus for every compact volume $V$ and its closed surface $A = \partial V$, the "sum" of the projections of $\vec{B}$ onto the normals to the surface $d\vec{A}$ vanishes. E.g. there is no net flow of $\vec{B}$ out of the surface $A$. Thus there are no sources (= produce net flow out of a surface). Thus field lines are closed at finite points (lines that go to infinity can be disclosed, e.g. $\vec{B} = \vec{e}_z$).

Similar by Stokes' theorem: $$ \int_A \underbrace{\nabla \times \vec{E}}_{=0} \cdot d\vec{A} = \int_{\partial A} \vec{E} \cdot d\vec{l} = 0 $$

Thus for every compact plane $A$ and its closed boundary $\partial A$, the sum of the projection of $\vec{E}$ onto the tangent of the boundary $d\vec{l}$ vanishes. E.g. there are no curls present. Thus there can be no closed field line.

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