# Physical reality EM potentials

1. Nov 21, 2004

### da_willem

I just learned the scalar and vector potentials arising from the electric and magnetic fields have a physical reality in relation to quantum mechanical effects, such as the Aharonov-Bohm effect. Contrary to what I've learned from my course on electrodynamics (Griffiths) this sure must mean they propagate with a finite speed.

Can you say that from maxwells equations they must have the same velocity as the fields?

Are there any other effects than a change in phase of a particles wavefunction observed or found?

And is the velocity at wich the potential propagate experimentally measured?

2. Nov 22, 2004

### speeding electron

The value of the electric and magnetic fields are determined by the value of the respective potentials, and vice versa. Therefore the fields must propagate at the same speed as the fields. This is pretty much by definition, and if the potentials travelled say faster than the fields, you would just have a different for the fields corresponding to those potentials. We would have two values for the field in the same place, an obvious contradiction.
In non-relativistic QM, electro-magnetic efects are brought in by their changing the phase of a wavefuction. So that is the one effect of them. I cannot speak for QFT.
In answer to the last question: Not that I know of - but again, we can measure the speed of the potentials by measuring the speed of the fields.

3. Nov 22, 2004

### hellfire

You can derive a wave equation for the electric and the magnetic fields fom Faraday’s and Ampere’s laws (no other equation is needed).

On the other hand, making use of the definitions of the vector and scalar potentials and substituting into Faraday’s and Ampere’s law, you can then either (a) impose the Lorenz gauge and obtain a wave equation for the scalar and vector potential or (b) impose the Coulomb gauge and get a Poisson equation for an ‘instantaneous’ scalar potential.

I am not sure how this has to be interpreted and I cannot answer the other questions.

[Sorry, I did not read the previous post of speeding_electron]

4. Nov 22, 2004

### dextercioby

That's right.

What do you mean??What did Griffiths say?From what u've written above,the author claims that QM effects (or is it the EM potentials) propagate with infinite speed.That's not right.Propagation of QM effects and EM potentials should be treated in a relativistic context.Either classical or quantum electrodynamics.
Either way,the best book on classical electrodynamics is J.D.Jackson's.I would advise you to read all of it.

Yes.It's an immediate result.Substiuing the definitions for the potentials in the original Maxwell's equations and chosing the covariant gauge Lorenz-Lorentz will give the wave equation for each of the potentials.

No,basically we can measure the fields (the electric and magnetic ones) only.The vector potential cannot be measured.I don't know about the scalar...I think not.
Either way,probably Jackson's book sets it straight.

5. Nov 22, 2004

### da_willem

I know the finite speed of the fields can be derived from two of the Maxwell equations. But my confusion arises from the statement Griffiths made that the velocity of propagation of the scalar potential is infinite. At least in the Coulomb gauge (as said by hellfire). This does not matter for the speed of the fields because these also depend on the magnetic vector potential wich does have a finite speed.

But the speed of the scalar potential cannot depend on the gauge. But in the formula for the electric Aharonov-Bohm effect (http://en.wikipedia.org/wiki/Aharonov-Bohm_effect) the potential appears, and causes a change in QM phase, so must be real. But how can something physically real have an infinite speed. The formula uses a difference in potential so does not depend on your gauge. And the speed of V sure does nor depend on your gauge, or does it....?

6. Nov 22, 2004

### speeding electron

Check again - I'm pretty sure the scalar potential propagates at a finite speed.

7. Nov 23, 2004

### da_willem

I quote from Griffiths "introduction to electrodynamics" paragraph 10.1.3 Coulomb and Lorentz Gauge.

How can this be true, in the light of the electrical Aharonov-Bohm effect? Is Griffiths just plain false? Is this only part of the story?

8. Nov 23, 2004

### dextercioby

Coulomb gauge (a.k.a. radiation gauge) appears in the consistent Hamiltonian analysis
of the free classical EM field and,hence,implies the inconsistency with the SR,as the Hamiltonian formalism implies breaking the equivalence between time and space coordinates as coordinates of a flat Riemannian manifold.
To avoid any weird results,you should be using the relativistic consistent Lorenz-Lorentz gauge.And Ahararonov-Bohm effect does not contradict the second principle of SR.

9. Nov 25, 2004

### Creator

Very interesting question, da willem.
I've always been fascinated with the AB effect, both the magnetic & electric versions; but are you aware that there is another Aharonov effect based on the vector potential?

It's called the Aharonov-Casher effect. Amazingly, it states that if a beam of
(polarized) neutrons is made to pass on both sides of a conductor of charge, the phase of the neutral particles are changed! Apparently, this is due to the participation of the vector potential in the Lagrangian of the neutrons.
Yikes :surprised

Creator

10. Nov 25, 2004

### pervect

Staff Emeritus
To expand just a bit on the answers already given, the Ahranov-Bohm effect, as is the case for any physical effect, does not depend on the gauge used. When one says that the Ahranov-bohm effect depends on the potentials, one does not mean that one can determine the gauge - this is impossible, the gauge degree of freedom remains.

Thus, given that there is obviously no faster-than-light propagation of physical effects in the Lorentz gauge, the same must be true in the Columb gauge, it's just not as obvious. Basically the Columb gauge is inconvenient for problems involving relativity.

Google finds

this archived usenet post

on the topic.

11. Nov 26, 2004

### da_willem

So some gauges in electrodynamics (e.g. the Coulomb gauge) do not describe the behaviour of the potentials correctly. Is the Lorentz gauge the only one that yields a velocity of c for the potentials? And how come the Coulomb gauge, wich is perfectly valid in electrodynamics, yields the unphysical result of a infinite velocity for V, where did it go wrong? Is this because we used the criterium that only the resulting fields matter?

12. Nov 26, 2004

### dextercioby

Gauges (all of them,including Coulomb and Lorenz-Lorentz) describe potential correctly.That is they insure that the only physical degrees of freedom appear in the correct description (classical/quantum) of the field.The reason why we chose Lorentz-Lorentz gauge is because it is consistent with the SR,as the equation (the wave equation) obtained is the same in all inertial frames,while chosing the Coulomb gauge to solve Maxwell equations for potentials leads to absurd results,like the quation for the magnetic potential is a Poisson one,but both the current density and the magnetic potential depend on time,yet derivatives wrt to time do not appear in the equation.But as i said,at hamiltonian leve,where relativistic covariance is lost (remember that hamiltonian dynamics takes place at equal times),Coulomb gauge appears naturally.And the physics is correct and and allows quantization.

13. Nov 27, 2004

### pervect

Staff Emeritus
No, all gauges describe the potentials "correctly", in the sense that they correctly predict the result of any experiment, including the "Arhanov-Bohm" (AB) experiments.

Thus the results of the AB experiment depend on some local function of the 4-potential, and cannot be expressed as some local function of the fields.

However, the AB experiment does not and cannot completely define or measure the 4-potential at any point - the potentials have a gauge degree of freedom that cannot be measured by any experiment.

The Lorentz gauge just makes certain theorems about the impossibility of transmitting information FTL more obvious than they would be in the columb gauge.

14. Nov 27, 2004

### da_willem

Although your post was very clear and helpful Pervect, I must admit it raised some more questions... The only way I can imagine that both the Coulomb gauge and the Lorentz gauge (in wich V propagates with a different speed) yield the same result for the Aharonov-Bohm effect is that A should also appear in the equation for the electrical AB effect. Is this what you mean by:

The formaulae found for both AB effects at wikipedia (http://wikipedia.org/Aharonov-Bohm_effect) can't be correct in the relativstic case can they? Are these the result from nonrelativistic QM?

And am I correct in saying there's no way to decide at wich speed V propagates, and this speed could even exceed the speed of light. But this is allowed by SR as this has no physical consequences, because all physical phenomena depend on A as well, wich does propagate with c?

And if A and V have a physical reality, why is there still a gauge freedom. In classical electrodynamics we used the argument that it arises from the fact that we can change A and V as ong as it does not change the fields. But now A and V have a physical reality themselves, so why can we still change them around by choosing a different gauge? Is this one of those things that just is\? Sot because all the equations we have in wich A and V appear we can change gauge and the results are the same?

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