# Action at a Distance in Electromagnetism?

1. Mar 6, 2009

### Phrak

Maxwell's equations in integral form are not obviously local.

Faraday's Law, a good enough example, in differental form is

$$\partial_i B_j - \partial _t E_k = J_k$$

where (i,j,k) are cyclic permutations of (x,y,z).

In integral form,

$$\frac{d\Phi}{dt} = \oint E \cdot dl$$

The potential around a closed loop is equal to the time rate change of a nonlocal magnetic flux.

Of course, as the integral form is equivalent to the differential form it must be local. What am I missing?? What connects the nonlocal \Phi to the local B?

2. Mar 7, 2009

### Phrak

Phi is nonlocal to E. How is this

$$\frac{d\Phi}{dt} = \oint E \cdot dl$$

not action at a distance?

3. Mar 7, 2009

### clem

When you postulate that the contour can be shrunk to infinitesimal size, it becomes local.

4. Mar 7, 2009

### Phrak

But do you find this somewhat bothersome when the contour is not shrunk to infintesimal size?

The action at a distance between charged particles is eliminated with the introduction of electric and magnetic fields that act as intermediaries. But here, it appears that the very fields that cure the pathology are themselves pathological.

5. Mar 8, 2009

### Phrak

Does this has something to do with the fundemental theorem of calculus?

6. Mar 8, 2009

### atyy

What if you considered Faraday's law together with the law containing the displacement current? I don't know how that will actually help, but I'm guessing that the "locality" in EM has to do with the finite propagation speed of light, and it's Faraday law and D-current law together that give the wave equation (maybe plus some of the others, I don't remember).

7. Mar 8, 2009

### Phrak

I looked up displacement current, just to be sure. No displacement current in Faraday's law, so that's not a part of it. And it required a dialelectric medium too; otherwise D=E.

As long as the change in magnetic flux is constant over time, we don't have to worry about including Ampere's law, and the speed of light, so that doesn't seem like it.

Man, you guys are going to make me do all the hard work, and actually think 'n stuff.

8. Mar 8, 2009

### atyy

Yes, no displacement current in Faraday's law. It's in another law. I meant the pair of laws together give the wave equation which gives the finite speed of propagation.

9. Mar 8, 2009

### atyy

Yes. Since I don't know the answer!

Anyway, another reason to try considering a pair (or more) of laws is that in the Coulomb gauge you get an "action at a distance" equation (Eq 397), but it isn't once you consider more than one Maxwell equation.

http://farside.ph.utexas.edu/teaching/em/lectures/node45.html

10. Mar 8, 2009

### Phrak

That's wild, atyy, and thanks for checking around the internet for this. The phi in equation 397 is the scalar potential in this case, whereas the Phi in Faraday's law is the integral of the magnetic field over an area.

I think the article basically says that changes in the scalar potential may propagate at faster than the speed of light, but it's OK with us, because it can't be measured, only the electric fields.

But the vector potential, 'A' has been measured (up to some spatial integral so this is probably the critical point that prevents action at a distance) in experiements that verify the Bohm-Aharanov effect. But I wonder if something couldn't be discovered from a variation of the experiment involving dA/dt. I really don't know what I'm talking about yet.

Last edited: Mar 8, 2009
11. Mar 8, 2009

### atyy

Yes, I know that, just an analogy where you get rid of action at a distance by considering more than one Maxwell equation. I don't think the Aharonov-Bohm is relevant, at least not if you're asking about classical EM.

To be a little more explicit, I am suggesting that you try considering these two equations together:
(i) rate of change in B flux = line integral of E
(ii) rate of change in E flux = line integral of B

12. Mar 8, 2009

### Phrak

Sorry. Just to be sure.

You're right, Arharonov-Bohm isn't relevant, but it does bring up a whole new can of worms I was suprised to see. The more I know, the more I don't know... Oh well.

I think there is no action at a distance, but it just looks like it. Consider the problem statement:

"Show that Faraday's law in integral form does not imply that the magnetic fields act at a distance. Consider the case where $d \Phi_B /dt$ is constant."

The constant part of it ensures that we can ignore Ampere's circuital law 'cause E is constant.

Where are the smart guy around here that can tell me how pea-brained I am?

Last edited: Mar 8, 2009
13. Mar 8, 2009

### Hans de Vries

In Classical EM terms:

The flux $$\Phi$$ through the surface is associated with magnetic field lines, which,
since they are closed, have to circle around outside of the wire loop to
enter the surface at the same place they left.

This means that any magnetic disturbance has to have spread beyond the loop to
take effect. Any small localized disturbance which not has spread beyond the loop
has the same number of magnetic field lines going both "up and down" through the
surface resulting in a zero net flux.

Regards, Hans

Last edited: Mar 8, 2009
14. Mar 8, 2009

### Phrak

Thanks for the response Hans.

Say we had an long solenoid with radius r, where the interior magnetic field points upward. The exterior return magnetic field would point downward. The interior flux changes at a constant rate. There's a potential, integral E dot d lambda, around a loop at some radius R>r.

I don't want to say an infinitely long solenoid, incase it results in a non-physical answer. The interior flux is nonlocal to the loop R. There's an exterior return flux between the radii r and R. The magnetic field strength around the loop can be made arbitrarily small as the length of the solenoid is made arbitrarily long, so long as each remain finite.

The equation might be written like this.

$$\frac{d}{dt} (\Phi_r - \Phi_{R-r}) = \oint_{\partial M} E \cdot d\lambda$$

All of $\Phi_r$ is spatially remote from the loop. Some of the exterior field, $\Phi_{R-r}$, isn't. It still appears to be nonlocal.

15. Mar 8, 2009

### atyy

Well, if this is correct, I can provide a wrong argument as to why it's wrong.

line integral of E = rate of change of B flux

Change variable names:
line integral of B = rate of change of E flux

This looks like Ampere's law + Displacements current:
line integral of B = current flux + displacement current flux

If displacement current is constant, then why not just treat it like a current in magnetostatics. In which case the solution is the Biot-Savart law, which looks like Coulomb's law, which is an action at a distance law.

16. Mar 8, 2009

### Phrak

I've considered that non-local is a fuzzy word. Is there any action at a distance? You both, atyy and Hans were alluding to this, I think.

If we put a test wire loop around a long solenoid we'll measure a constant potential for a constant $d \Phi_{B} / dt$. There's no 'action' in this case; no information is being transmitted; things stay constant.

But we can take energy out of the wire loop by attaching a load resistor. After everything settles down to some steady state condition, it appears that energy is being transmitted by some unknown action acting at a distance.

I'm vacillating over this issue. I'm beginning to think that the electric and magnetic fields alone cannot account for local action. But the vector potential can.

Even though the magnetic field outside of an infinitely long solenoid is zero, the vector potential is not.

$$\int_{\partial M} A \cdot d \lambda = \int_{M} B \cdot d \sigma = \Phi_B$$

The condition of interest has been a constant time rate change in magnetic flux, so

$$\int_{\partial M} \partial_{t}A \cdot d \lambda = \int_{M} \partial_{t}B \cdot d \sigma = \partial_{t}\Phi_B = constant$$

Last edited: Mar 8, 2009
17. Mar 9, 2009

### Hans de Vries

Hi Phrak.

It's still local, notwithstanding the more complex geometry.

A solenoid still has the same number of (returning) field lines at the outside as at the
inside. They are only located further away because the A field close to the solenoid
goes with 1/r which has a zero B=curl A.

$$\vec{B} ~=~ \mbox{curl}\left\{~ -\frac{y}{r^2}~,~ \frac{x}{r^2} ~,~ 0 ~ \right\} ~=~0$$

The field A loops around the solenoid. An increasing current through the solenoid
generates a constant B (instead of a zero B) at its near outside area in the form of:

$$\vec{B} ~=~ \mbox{curl}\left\{~ -\frac{(ct-r)y}{r^2}~,~ \frac{(ct-r)x}{r^2} ~,~ 0 ~ \right\} ~=~ c\left\{~ 0~,~0~,~ -\frac{1}{r}~ \right\}$$

There is also a circular E field:

$$\vec{E} ~=~ -\frac{\partial}{\partial t}\left\{~ -\frac{(ct-r)y}{r^2}~,~ \frac{(ct-r)x}{r^2} ~,~ 0 ~ \right\} ~=~ c\left\{~ \frac{y}{r^2}~,~ -\frac{x}{r^2} ~,~ 0 ~ \right\}$$

Plus a radially outwards directing Poynting vector:

$$\vec{P} ~=~ \vec{E}\times\vec{H} ~=~ \frac{c^2}{\mu_o}\left\{~ \frac{x}{r^3}~,~ \frac{y}{r^3} ~,~ 0 ~ \right\}$$

Note that div P = 0 so the energy-density in the near outside field of the solenoid
remains constant.

A sudden increase in the current in the solenoid will cause a sudden increase in A
looping around the solenoid which spreads outwards with the lightspeed c. It only
acts on the loop when it reaches it (via E = -dA/dt).

There's absolutely nothing non-local.

Regards, Hans

Last edited: Mar 9, 2009
18. Mar 10, 2009

### fizzle

It's an identity, there is no cause-and-effect implied so there's no action at a distance. See Stoke's Theorem relating the surface integral to the line integral of a vector field.

19. Mar 10, 2009

### Phrak

Thanks Hans. Those equations are an end result. I'll see if I get them for a solenoid--but in cylindrical coordinates.

@fizzle. Most equations are identities. So you're saying transformers don't work, or if they do, classical electromagnetism doesn't tell us how they work?

20. Mar 10, 2009

### fizzle

No, I'm saying that there is no cause and effect implied in that equation. The line integral and surface integral change simultaneously. They are two descriptions of the same thing.