Equivalence principle: an electric charge and a coil

  • #1
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Main Question or Discussion Point

equiv2.jpg

The near-range magnetic field ##\vec{B}## of a point charge ##q## at distance ##\vec{r}##, moving at a non-relativistic velocity ##\vec{v}##, is given by
$$\vec{B}=\frac{q}{4\pi\epsilon_0c^2}\frac{\vec{v}\times\hat{r}}{r^2}.$$
Faraday's law of induction for the induced EMF ##V_c## in a coil, with area ##A## and turns ##N##, due to a changing magnetic field strength ##dB/dt## through its center is given by
$$V_c=-NA\frac{dB}{dt}.$$
In scenario ##A## the charge ##q## is falling downwards due to the gravitational force ##mg## and the coil is fixed at a distance ##r## so that the induced voltage ##V_c## measured by the coil is
$$V_c = - \frac{1}{4\pi\epsilon_0c^2}\frac{N\ A\ q\ g}{r^2}.$$
But according to the equivalence principle this is exactly the same as scenario ##B## in which the charge ##q## is fixed and the coil at distance ##r## is accelerating upwards with acceleration ##g##.

In scenario ##B## is there still an induced voltage ##V_c## across the moving coil?

P.S. I think one could actually do an experiment like this but one would replace the moving charge with a moving parallel plate capacitor which would produce a dipole magnetic field.
 
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  • #2
tech99
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View attachment 266748
The near-range magnetic field ##\vec{B}## of a point charge ##q## at distance ##\vec{r}##, moving at a non-relativistic velocity ##\vec{v}##, is given by
$$\vec{B}=\frac{q}{4\pi\epsilon_0c^2}\frac{\vec{v}\times\hat{r}}{r^2}.$$
Faraday's law of induction for the induced EMF ##V_c## in a coil, with area ##A## and turns ##N##, due to a changing magnetic field strength ##dB/dt## through its center is given by
$$V_c=-NA\frac{dB}{dt}.$$
In scenario ##A## the charge ##q## is falling downwards due to the gravitational force ##mg## and the coil is fixed at a distance ##r## so that the induced voltage ##V_c## measured by the coil is
$$V_c = - \frac{1}{4\pi\epsilon_0c^2}\frac{N\ A\ q\ g}{r^2}.$$
But according to the equivalence principle this is exactly the same as scenario ##B## in which the charge ##q## is fixed and the coil at distance ##r## is accelerating upwards with acceleration ##g##.

In scenario ##B## is there still an induced voltage ##V_c## across the moving coil?

P.S. I think one could actually do an experiment like this but one would replace the moving charge with a moving parallel plate capacitor which would produce a dipole magnetic field.
Should we be considering the relative acceleration here? When the charge and coil are the same height, there is zero relative acceleration so I would expect zero volts across the coil.
 
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In scenario ##A## I assume the charge is freely accelerating downwards due to the force of gravity whereas the coil is fixed to the Earth/observer (Newtonian viewpoint).

In scenario ##B## I assume the charge is at a rest whereas the coil (and the Earth/observer) is accelerating upwards (Einsteinian equivalent viewpoint).

Therefore I assume there is always a relative acceleration between them.

I think you get a voltage in the coil in both scenarios but as I understand it standard EM theory says you only get a voltage if the charge itself is accelerating.

One could argue that a falling object is not really accelerating as it does not experience proper acceleration. But it certainly looks to the observer on Earth that it is accelerating!
This common sense Newtonian view is backed up by the modern weak-field description of gravity as a standard force, acting in flat spacetime, mediated by gravitons.
 
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  • #4
tech99
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In scenario ##A## I assume the charge is freely accelerating downwards due to the force of gravity whereas the coil is fixed to the Earth/observer (Newtonian viewpoint).

In scenario ##B## I assume the charge is at a rest whereas the coil (and the Earth/observer) is accelerating upwards (Einsteinian equivalent viewpoint).

Therefore I assume there is always a relative acceleration between them.

I think you get a voltage in the coil in both scenarios but as I understand it standard EM theory says you only get a voltage if the charge itself is accelerating.

One could argue that a falling object is not really accelerating as it does not experience proper acceleration. But it certainly looks to the observer on Earth that it is accelerating!
This common sense Newtonian view is backed up by the modern weak-field description of gravity as a standard force, acting in flat spacetime, mediated by gravitons.
I believe Maxwell was aware of this reciprocal issue and started to touch on Relativity but did not manage to do that in his lifetime.
 
  • #5
tech99
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jcap, do you think that an EM wave is involved as the charge is accelerating? But in the second case it does not seem to work because the charge is not accelerating.
 
  • #6
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equivnn.jpg


One could imagine a Newtonian scenario where the charge is at a fixed position in the gravitational field and the coil is accelerating downwards under the force of gravity.

In the equivalent Einsteinian point of view the charge is experiencing an upwards proper acceleration to keep it at a fixed position whereas the coil feels no forces in an inertial frame. According to standard EM theory the accelerating charge should induce a voltage in the coil.

Therefore in the equivalent Newtonian point of view there should also be a voltage across the coil.

I guess an EM wave would also exist in the inertial frame of the coil although my argument just assumes local electromagnetic interactions between the charge and the electrons in the coil.
 
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  • #7
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I seem to recall that EM in curved spacetime is complicated, and the equivalence principle doesn't typically apply because the interaction cannot be described locally. And you can't really consider a Newtonian case because the fields are changing and the changes don't propagate slowly compared to the speed of light.

As I recall, @vanhees71 knew more about this kind of example.
 
  • #8
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Well I guess I'm just confining my argument to the local electromagnetic interactions between the charge ##q## and the electrons in the coil. Also I'm only assuming non-relativistic velocities between the charge and coil.
 
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I'm talking about the picture in post #1.

Let's say the q is a positive charge.

Case A: The q exerts an upwards force on the protons in the wire, downwards force on the electrons in the wire. As q exerts those forces, q feels two reaction forces that happen to cancel out.

Case B: Electrons in the wire exert an upwards force on q, protons in the wire exert a downwards force on q,
As electrons and protons exert those forces, they feel reaction forces. The voltmeter measures those forces.

Well, I hope Newtons third law applies here. "When a charge induces an electromotive force, it experiences an opposite electromotive force". If not, then all of the above is wrong.
 
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  • #10
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Well, I hope Newtons third law applies here. "When a charge induces an electromotive force, it experiences an opposite electromotive force". If not, then all of the above is wrong.
I agree with your analysis and your application of Newton's third law though many people only believe Newton's third law applies to equal and opposite contact forces.

I think Newton's third law does apply in the case of electromagnetic interactions between charged particles, and does obey special relativity, by the following mechanism:

Charged particle A emits positive energy virtual photons forwards in time that impart some energy-momentum to particle B. Simultaneously particle B emits negative energy virtual photons backwards in time that impart an equal amount of negative energy-momentum back to particle A.
 
  • #11
vanhees71
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I seem to recall that EM in curved spacetime is complicated, and the equivalence principle doesn't typically apply because the interaction cannot be described locally. And you can't really consider a Newtonian case because the fields are changing and the changes don't propagate slowly compared to the speed of light.

As I recall, @vanhees71 knew more about this kind of example.
Well, I'm not an expert in GR, but the usual argument from a typical theoretical high-energy-particle point of view uses gauge invariance. The free em. field can be entirely described with forms and thus does not rely on a Lorentzian pseudo metric, and that's the formulation for the em. field you usually assume when "translating" the Maxwell equations from flat Minkowski space to the curved Lorentzian manifold of GR. Another way to argue is that you need to keep electromagnetic gauge invariance and at the same time general covariance. Last but not least you only use the couplings with coupling constants with positive and 0 dimensions, because usually you expect the typical scale, where gravity comes into the game is at the huge Planck scale, i.e., in a contribution of the form ##\lambda R F_{\mu \nu} F^{\mu \nu}## in the Lagrangian for the coupling between the em. field ##F_{\mu \nu}## and gravity (##R## is the Ricci scalar) you expect the coupling constant ##\lambda##, which has dimension ##1/\text{energy}^2##, to be very small and thus you can neglect such a coupling.

There's a thorough discussion on how the laws of physics can be "translated" from "SR to GR", based on the equivalence principle in Carrol's lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019
 
  • #12
In scenario ##A## the charge ##q## is falling downwards due to the gravitational force ##mg## and the coil is fixed at a distance ##r## so that the induced voltage ##V_c## measured by the coil is
$$V_c = - \frac{1}{4\pi\epsilon_0c^2}\frac{N\ A\ q\ g}{r^2}.$$
Counter-argument in Wikipedia:
The key is to realize that the laws of electrodynamics, Maxwell's equations, hold only within an inertial frame
Source:
https://en.wikipedia.org/wiki/Parad..._a_gravitational_field#Resolution_by_Rohrlich

But according to the equivalence principle this is exactly the same as scenario ##B## in which the charge ##q## is fixed and the coil at distance ##r## is accelerating upwards with acceleration ##g##.

In scenario ##B## is there still an induced voltage ##V_c## across the moving coil?
No, according to the following paper.
7. Free falling charges

As an immediate consequence of Larmor’s formula (5) with the application note (7), a charged particle does not radiate if it is free falling in a gravitational field, because the gravitation is exerting the same acceleration on the particle and it’s field, and hence there is no stress between the particle and it’s field. We remark ...
Source:
https://arxiv.org/pdf/1509.08757.pdf
 
  • #13
PeterDonis
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Counter-argument in Wikipedia
The Wikipedia statement is false. It is perfectly possible to write Maxwell's Equations in generally covariant form, valid in any frame, in flat or curved spacetime, using differential forms, as @vanhees71 points out in post #11. A good textbook treatment of this is given in MTW.
 
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  • #14
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Charged particle A emits positive energy virtual photons forwards in time that impart some energy-momentum to particle B. Simultaneously particle B emits negative energy virtual photons backwards in time that impart an equal amount of negative energy-momentum back to particle A.
There is no need to introduce photons or any of the other frequently misunderstood apparatus of quantum electrodynamics here. Relativity is a classical theory and classical electrodynamics is sufficient to explain the forces between the charge and the coil.
 
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  • #15
vanhees71
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Let's first concentrate on the special-relativistic case. The problem concerning the radiation field of a uniformly accelerated charge (in the context of relativity a point charge moving in a constant direction) is notorious, and it's the only example I know of, where even the great Pauli made a mistake.

On the one hand the corresponding motion is among the most simple in special relativity. Just ignoring radiation it's pretty easy to solve as the motion of a point particle in a uniform electric field. What you get is hyperbolic motion, so named, because the world line in Minkowski space is a hyperbola, and this is already indicating the mathematical trouble: For ##\tau \infty## (##\tau## proper time of the particle) the speed of the particle asymptotically reaches the speed of light.

If you now consider the radiation field of this charge in hyperbolic motion and naively integrate the corresponding integrals giving the retarded (Lienard-Wiechert) potentials and the corresponding fields due to the asymptotic reaching of the speed of light you overlook an all-important ##\delta##-distribution like contribution, which however is all important, because it provides the contribution of the field that describes the radiation.

A clean way out is to get more physical in the sense that in reality there is no uniform electric field over all space. So just consider a capacitor with very large plates (large compared to the distance), so that you can treat the plates as infinitely extended, and then you have a uniform field only within the plates. You can consider a charged particle entering and leaving the capacitor through little holes. Then the speed of the particle always stays ##<c##, and there's no problem in evaluating the Lienard-Wiechert potentials, which of course inevitably leads to a radiation field (as well as the usual "bound field", i.e., the boosted Coulomb field dominating in the near zone around the charge). Then you can make the distance of the plates go to infinity holding the constant electric field between them fixed, and you get the correct limit for the field of the charge in hyperbolic motion, which of course radiates. Pauli was wrong in his famous masterpiece on the theory of relativity of 1921. If I remember right, already Sommerfeld got the issue right, but I'm not sure about that. There's a lot of literature about this issue. Among the most readable is the AJP article by Joel Franklin and Griffiths

J. Franklin, D. J. Griffths, The fields of a charged particle in hyperbolic motion
https://doi.org/10.1119/1.4875195
https://arxiv.org/abs/1405.7729

Erratum (concerning the Fig. not the calculation itself):

https://doi.org/10.1119/1.4906577

Another nice paper is by Jarrold Franklin:

J. Franklin, Electric field of a point charge in truncated hyperbolic motion
https://doi.org/10.1088/0143-0807/36/4/045015
https://arxiv.org/abs/1411.0640v3
 
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  • #16
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The key is to realize that the laws of electrodynamics, Maxwell's equations, hold only within an inertial frame
Ok - what do think about the following formulation of my argument?

equivnn.jpg


One could imagine a Newtonian scenario where the charge is at a fixed position in the gravitational field and the coil is accelerating downwards under the force of gravity.

In the equivalent Einsteinian point of view the charge is experiencing an upwards proper acceleration to keep it at a fixed position whereas the coil feels no forces in an inertial frame. According to standard EM theory the accelerating charge should induce a voltage in the coil.

Therefore in the equivalent Newtonian point of view there should also be a voltage across the coil.
 
  • #17
Therefore in the equivalent Newtonian point of view there should also be a voltage across the coil.
What you call "Newtonian point of view" is in SR treated as an accelerated rest frame.

The answer to this question is "no". Reason:
Nevertheless, comoving observers, that is, accelerated observers with respect to whom the charge is at rest, will not detect any radiation because the radiation field is confined to a spacetime region beyond a horizon that they cannot access.
Source:
https://arxiv.org/pdf/physics/0506049.pdf
 
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What you call "Newtonian point of view" is in SR treated as an accelerated rest frame.

The answer to this question is "no". Reason:

Nevertheless, comoving observers, that is, accelerated observers with respect to whom the charge is at rest, will not detect any radiation because the radiation field is confined to a spacetime region beyond a horizon that they cannot access.

Source:

https://arxiv.org/pdf/physics/0506049.pdf
Yes but in the equivalent Einsteinian view the coil is in an inertial frame and the charge is accelerating with respect to inertial frames. Therefore I think that standard EM implies that the charge should induce a voltage in the coil in such a situation.

Also in the scenario I describe above the coil's frame and the charge's frame are not comoving which is different from the situation you describe where the observer is accelerating along with the charge.
 
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  • #19
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There is no need to introduce photons or any of the other frequently misunderstood apparatus of quantum electrodynamics here. Relativity is a classical theory and classical electrodynamics is sufficient to explain the forces between the charge and the coil.
But even in classical electrodynamics proposals have been made to include both advanced and retarded solutions.

For example: Classical Electrodynamics in Terms of Direct Interparticle Action, Wheeler and Feynman (1949)

https://journals.aps.org/rmp/pdf/10.1103/RevModPhys.21.425

In the section Action and Reaction (P.430) the authors derive a relativistic generalization of Newton's third law of action and reaction using retarded and advanced electromagnetic forces between charged particles.
 
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  • #20
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But even in classical electrodynamics proposals have been made to include both advanced and retarded solutions.
Sure, but what does this have to do with virtual photons?
 
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Sure, but what does this have to do with virtual photons?
Ok one doesn’t need to use virtual photons but I find it interesting that Wheeler and Feynman showed how Newton’s third law could be understood in a relativistic way in classical electrodynamics using retarded and advanced interactions between charged particles.
 
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I agree with your analysis and your application of Newton's third law though many people only believe Newton's third law applies to equal and opposite contact forces.

I think Newton's third law does apply in the case of electromagnetic interactions between charged particles, and does obey special relativity, by the following mechanism:

Charged particle A emits positive energy virtual photons forwards in time that impart some energy-momentum to particle B. Simultaneously particle B emits negative energy virtual photons backwards in time that impart an equal amount of negative energy-momentum back to particle A.

Well, after thinking for a long time about induction and Newton's third law, I came to this conclusion:

Two charged particles A and B are side by side, and one of the charges accelerates.

The situation is symmetric, A sees B moving at speed v, B sees A moving at speed -v. A sees a magnetic field ## \vec {B} ## , B sees a magnetic field ## -\vec {B} ##, and so on... So the two particles experience opposite induced electromotive forces.

There is just a very small asymmetry: One charge accelerates and radiates, other one does not. But we can ignore that now, as the question was not about radiation, right?
 
  • #23
vanhees71
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You precisely can not ignore radiation, because only using the fields and the point particles together you get the "3rd law" right, i.e., momentum conservation. The field is a dynamical entity in the theory. Without it you cannot get the conservation laws right in a local form, and all successful theories in relativistic physics are local (quantum) field theories. Point particles are an exception, for which you have to pay the prize that there's up to now no consistent dynamical theory for them.
 
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You precisely can not ignore radiation, because only using the fields and the point particles together you get the "3rd law" right, i.e., momentum conservation. The field is a dynamical entity in the theory. Without it you cannot get the conservation laws right in a local form, and all successful theories in relativistic physics are local (quantum) field theories. Point particles are an exception, for which you have to pay the prize that there's up to now no consistent dynamical theory for them.
So, if we give an electric charge a very sharp vertical push, then the produced EM-wave propagates to all horizontal directions. But said wave contains vertical momentum. Right?


Or, if we first give an electric charge a very sharp upwards push and then a very sharp downwards push, then the produced EM-wave propagates to all horizontal directions. But the front half of said wave contains downwards momentum, while the rear part of said wave contains upwards momentum. Right?
 
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  • #25
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The em. wave propagates as given by the retarded propagator (Lienard-Wiechert potentials or equivalently Jefimenko's equation). The field momentum density is ##\propto \vec{E} \times \vec{B}##.
 

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