Action at a Distance in Electromagnetism?

[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?

 

[Phrak]

Phi is nonlocal to E. How is this

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

not action at a distance?

 

[Meir Achuz]

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

 

[Phrak]

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

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.

 

[Phrak]

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

 

[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).

 

[Phrak]

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).

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.

 

[atyy]

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.

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.

 

[atyy]

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

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

 

[Phrak]

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

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.

 

[atyy]

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.

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

 

[Phrak]

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.

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.

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

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?

 

[Hans de Vries]

Phi is nonlocal to E. How is this

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

not action at a distance?

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

 

[Phrak]

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

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.

 

[atyy]

"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."

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

Faraday:
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.

 

[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

 

[Hans de Vries]

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.

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

 

[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.

 

[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?

 

[fizzle]

@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?

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.

 

[Phrak]

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.

Yes. That's roughly the point of this thread.

 

[fizzle]

Yes. That's roughly the point of this thread.

I guess I'm missing your point. From your first two posts, it seems like the problem is reconciling relationship between the line integral of E and the time rate of change of phi (surface integral of B). Remember that the time derivative of B is related to the curl of E. These equations aren't independent. This substitution reduces the equation to Stoke's Theorem, which is a mathematical identity for any vector field. There's no cause-and-effect so there's no problem with action-at-a-distance.

The cause-and-effect topic has dogged em theory for a long time. There doesn't seem to be any cause-and-effect regarding E and H implied in Maxwell's Equations according to Oliver Heaviside. We've all heard the "rolling wave" explanation of the propagation of EM waves, where a change in E with time causes a change in H with distance, and vice-versa. As Heaviside notes in "Electrical Papers" vol I, p. 490, way back in 1888, this explanation fundamentally involves action-at-a-distance, even if it's only over a very small distance. We all know that action-at-a-distance is impossible, so the explanation is not correct. I think this becomes clear for TEM waves when you substitute E = cB into the time derivative equations. They reduce to a mathematical identity for an object moving at a constant speed => d/dt (E) = -c * d/dx (E).

 

[Phrak]

...reduces the equation to Stoke's Theorem, which is a mathematical identity for any vector field.

But not for every field. Though this may actually be more to the general issue than anything yet.

In one dimension, to make things easy,

\int_{a,b}f(x)dx = F(a) - F(b) ,

if f is a continuous function.

It seems that for the integral form to be equivalent to the differential form (of Maxwell's equations) ,B and E must be continuous, integrable fields regardless of the boundaries of materials with different permeability or permittivity. Hmm.

 

[fizzle]

But not for every field. Though this may actually be more to the general issue than anything yet.

In one dimension, to make things easy,

\int_{a,b}f(x)dx = F(a) - F(b) ,

if f is a continuous function.

It seems that for the integral form to be equivalent to the differential form (of Maxwell's equations) ,B and E must be continuous, integrable fields regardless of the boundaries of materials with different permeability or permittivity. Hmm.

I wouldn't waste my time chasing down exceptions to Stoke's Theorem. My guess is that you'll find it applies to almost all situations. Regardless, just stick to the differential forms since they're points. There is no action-at-a-distance (tired of typing that, let's just use AAAD from now on) and, anyway, there are much more interesting em topics to explore. For example, if an electron is moving with or against a TEM wave, why does the Lorentz force reduce to a simple Doppler Shift of the electric field intensity? You can dispense with the troublesome magnetic field, which doesn't seem to be a real thing in the first place. Is there any physical insight given by this result? Another example, take a standard point electron and excite it with a TEM wave. Look at the em wave emitted by the point electron. Why does the emitted field's magnitude equal the incident wave's magnitude at exactly the classical electron radius? It doesn't seem to make sense that the emitted wave's magnitude could ever be stronger than the incident wave's magnitude (ie. inside the classical electron radius). What does the Poynting vector say about energy flow in this situation?

BTW, you can solve the transformer problem by reducing it to one that involves two coupled transmission lines. This simplifies the physical picture immensely. The input loop couples to the output loop using the standard crosstalk equations (even and odd modes). The input loop is terminated with a short and the output loop starts with a short. That's all you need to know to see the fundamentals of a transformer in action. It also applies to a two-turn inductor.

 

[Phrak]

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

Hans, thanks for the kick in the right direction.

So just to be sure, I finally noticed that in the absense of charge and current, in free space the wave equation is true for all solutions around a long solenoid where r is << L, where L = the length of the solenoid,

\partial_{r}{}^{2}E_\theta = \frac{1}{c^2} \partial_{t}{}^{2} E_\theta

So as it turns out,

E=E_\theta = E_0 \ exp(\pm kr \mp \omega t)
and
E=E_\theta = E_0 \ cos(\pm kr \mp \omega t + \phi)

are solutions in E_theta, and likewise for B_z. I was so sure that making d\Phi / dt=0 would simplify everything so that Ampere's circuital law could be ingnored. That was misleading.

For a current, forever increasing through the solenoid,

E= E_0 \ exp[(-r+ct)/R] .

It seems the world is in order again and Maxwell's equations in differential form are locally causal. The interesting thing remaining is that the integral forms don't seem to be compete without the addition of one or more continuity conditions on E and B that would ensure that Stoke's Theorem is satisfied.