# Can $\mathbf{D}$ field only exploit symmetry for linear?

1. Oct 18, 2015

### davidbenari

There are problems in classical electromagnetism where they ask you to find the electrical displacement given some geometry (like a sphere or a cylinder) and the dielectric constant $\epsilon_r$.

The solution to these problems typically employs symmetry arguments along with Gauss' laws for the D field. However, it got me thinking: Is this only valid in the case when the dielectric is linear?

What justifies symmetry arguments for the D field in general? I have a vague intuition about the validity of this, but these arguments aren't as obvious as the case for the E field.

I wouldn't like to say much more because I want a general argument that explains why exploiting symmetry works for the D field.

You could take as an example the case of an infinite cylindrical capacitor with a dielectric in between. (Maybe its linear, maybe not)

Why should the D field be radial in this case? I agree with you it has to look the same at whichever point separated a distance $s$ from the cylinder since the cylinder is infinite, but how would you counter someone who says that it's not perfectly radial but has some inclination as well?

Thanks!

Last edited: Oct 18, 2015
2. Oct 18, 2015

### nasu

Ask him which way would it be inclined.

3. Oct 18, 2015

### davidbenari

Well the field due to dipoles has some curvature associated to it. So if you're calculating the field between,e.g., two charged plates and considering a linear dielectric then I don't see why the point in question wouldn't inherit some of the curvature of the dipoles and thus have some inclination...

Also I ask about linearity since in general terms the susceptibility $\chi_e$ is a tensor and thus would make the D and E field not necessarily collinear , so its not immediately obvious symmetry works here.

4. Oct 18, 2015

### davidbenari

Also if the dielectric is nonlinear the curl doesn't necessarily vanish and we could consider some inclination.

5. Oct 18, 2015

### nasu

You mentioned an infinite capacitor with cylindrical symmetry.
Of course there are non-symmetric cases (all real cases pretty much) and the field is curved in various directions, depending on the specific geometry.

6. Oct 18, 2015

### davidbenari

The problem of two charged plates also exploits symmetry by the way. The cylindrical capacitor too.

How do you validate exploiting symmetry in the typical ways?

My argument for electric fields was that any charge had its complementary which cancelled some component , blah blah blah... But maybe there is some mathematical way of seeing symmetry that I don't know of? Something along the lines of a unique solution or something like that.

7. Oct 18, 2015

### davidbenari

Many times when we solve Laplace's equation we can see that some point has the same background regardless of translations in the $z$ coordinate and we make the ansatz that our potential doesn't depend on $z$, which happens to work.

I guess this could to be it. You consider certain translations of a point by varying certain coordinates, and if you're point has the same background each time then it necessarily means it has no dependence on those coordinates. My problem though is that this wouldn't eliminate dependence on the unit vectors, only on the scalar valued coordinate.

How can I be sure that dependence on the unit vectors would also disappear?

8. Oct 18, 2015

### nasu

The two charged plates can exploit symmetry if they are infinite plates. In this case the field has to be perpendicular to the plates. Not because of blah,blah but just there is no way to specify another direction towards the field will be inclined. For finite plates the field curves towards the edges.

9. Oct 18, 2015

### davidbenari

So why can't we specify another direction? Countering with "which one would you suggest" doesn't really prove anything IMHO.

I want a proof for this.

You've seen the field caused by dipoles, why couldn't there be a slight inclination inherited in the field?

10. Oct 18, 2015

### davidbenari

The fact that the susceptibility is a tensor can introduce these components in my opinion. I don't see how you can simply exclude them with symmetry arguments.

11. Oct 18, 2015

### nasu

Of course you can. Only that you must have the symmetry for the argument to hold. The tensor has nothing to do with it. It works as well with air or vacuum as dielectric.

I suspect you don't grasp yet the meaning of infinite planes or cylinders.
Let's take the example of an infinite cylindrical capacitor, with uniform charge. If the field is not perpendicular to the axis, it must have an inclination towards one of the "ends".
For a finite cylinder this is OK. It may be inclined towards the end that is closer or maybe the one which is farther away. Only in the middle the field must be perpendicular to the axis.
Now consider an infinitely long cylinder. Every point is equally far from either "end".

12. Oct 18, 2015

### davidbenari

Take as an example an infinite cylindrical capacitor with uniform charge. And lets consider some field such that it satisfies $\int \mathbf{D} \cdot d\mathbf{a}= Q_f$ (this is the D field, but lets not stick to its meaning). The fact that the capacitor is infinite implies that the solution will take the same values for all points a distance $s$ from the central axis. It doesn't imply anything else. It doesn't seem to negate the possibility of a non-radial component. (please keep reading even if you disagree with this).

In the case of the electric field you then have to employ other arguments like "it has to be radial since if you take consider some patch of charge, then there is another patch of charge somewhere which cancels any non-radial component". But the D field has no coulomb law associated to it so you can't resort to charges or whatnot.

So I guess this is really about what mathematics has to say about solutions when there are symmetries; and not specifically about physics. It could be related to what I said about Laplace's equation above.

What you said about inclining towards the ends sounds like you're resorting to charges. And again, the D field has no coulomb law associated so I don't see how you could say such a thing.

If the dielectric is nonlinear then the curl could be nonzero and indeed we could be dealing with another beast here even if we have symmetry.

So I sincerely don't agree with you. Where do you suggest I am going wrong?