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The electric field inside a charged spherical shell moving inertially is, per Gauss's law, zero.

If the spherical shell is accelerated, the field inside is not zero anymore, but it gains a non-null component along the direction of the acceleration, as mentioned, for example, in this paper.

The following picture from the above paper shows the field lines in the xy-plane in the instantaneous rest frame. The shell is undergoing rigid hyperbolic motion along the x-axis (toward the right)

enter image description here

The question I have is the following:

Assuming that a charged spherical shell is moving inertially, would an accelerated observer (test charge) inside of it detect an electric field like in the image above?

This question is also equivalent to asking the following ones:

Is there an electric field inside the shell if it is accelerating in an homogenous gravitational field? Will an observer (test charge) in the center of the shell that is not falling along with it, detect an electric field?

I read in a couple of papers that there won't be any field detected by such observer, but this is not demonstrated and sounds strange to me. The reasoning of these papers is that the electromagnetic tensor is invariant, so if it is zero in the inertial frame it will be zero also in the non-inertial one.

But here we are talking about the electric field only, which is a component of the electromagnetic tensor, so I don't see any a priori reason why it could not be made non-zero thanks to a frame transformation

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  • $\begingroup$ My reasoning for being dubious about this is that, intuitively, if one zooms out from the shell and looks at it from the distance, it will look like a point particle, and its field will also look like that of a point particle. As it has been detailed in other answers here and numerous papers throughout the years (spanning over half a century), an accelerated observer looking at a point charge that moves inertially will detect radiation coming from it. (An accelerated observer at rest relative to a co-accelerated point charge sees instead only a static Coulomb field) $\endgroup$
    – Povel
    Commented Nov 6, 2022 at 17:43
  • $\begingroup$ This then seems to imply to me that, if we consider that point charge to be the charged shell as seen from the distance, then for such distant observer the exterior electric field will look just like in the picture above, and therefore I would also expect the interior one to look the same. $\endgroup$
    – Povel
    Commented Nov 6, 2022 at 17:52
  • $\begingroup$ An accelerated observer in a field free region does not see a field. $\endgroup$
    – FlatterMann
    Commented Nov 6, 2022 at 18:24
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    $\begingroup$ Your question is very interesting. The invariance of the electromagnetic tensor under diffeomorphism is a fact, but a the same time electrons are subjected to a fictious force in the reference of the sphere, altering the initial equilibrium condition. Consider that electrons will start to emit radiation all around the direction of the acceleration, and that means also inside the sphere: the question is also if you can have destructive interference inside the sphere by all the photons emitted by all points of the surface. Anyway, I'm just going around, I don't have an answer for you now sorry. $\endgroup$
    – Rob Tan
    Commented Nov 6, 2022 at 18:24
  • $\begingroup$ @RobTan Accelerated motion was always different from inertial motion. That doesn't change in electromagnetism. $\endgroup$
    – FlatterMann
    Commented Nov 6, 2022 at 18:26

2 Answers 2

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I read in a couple of papers that there won't be any field detected by such observer, but this is not demonstrated and sounds strange to me. The reasoning of these papers is that the electromagnetic tensor is invariant, so if it is zero in the inertial frame it will be zero also in the non-inertial one.

But here we are talking about the electric field only, which is a component of the electromagnetic tensor, so I don't see any a priori reason why it could not be made non-zero thanks to a frame transformation

Right, the correct argument is as follows: if all components of the electromagnetic tensor are zero in one frame, then the same will be true in any other frame. Also, it does not matter whether one frame or the other or both is non-inertial.

The nonzero components of the electromagnetic tensor are the 3 components of the electric field, the 3 components of the magnetic field, and the same with their signs flipped. So if the electric and magnetic fields both vanish in the first frame, then we can be assured that an accelerated observer at the same location will also agree that they both vanish.

A stationary, uniformly charged spherical shell in an inertial frame has no interior electric field and no interior magnetic field, as measured in such frame. So, an accelerated observer would agree.

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  • $\begingroup$ Thanks for your answer, it is very interesting. So this also means that a supported observer/test particle inside a charged shell that is falling around it in a (homogeneous) gravitational field will not detect any electric field, right? $\endgroup$
    – Povel
    Commented Nov 6, 2022 at 21:45
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    $\begingroup$ @Povel I don't know the answer to that. I seem to recall that there are difficult issues involved with such questions (e.g. the ongoing controversy about whether freely falling charges radiate). $\endgroup$
    – Brian Bi
    Commented Nov 6, 2022 at 21:48
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    $\begingroup$ As far as I know, it seems the consensus is that the falling charge, which is inertial, is seen radiating by a supported, non-inertial observer. The situation I described above with the charged shell falling around the supported observer is a similar situation, and analogous to having the shell inertial and the observer/test charge accelerate inside it, so I suppose the answer should be the same. I still find this strange though. If instead of falling the shell was getting accelerated by an homogeneous external electric field, the same supported observer would detect an electric field. $\endgroup$
    – Povel
    Commented Nov 6, 2022 at 22:01
  • $\begingroup$ @Povel exacty, that is what I was thinking: in an electric field forces are proportional to the charges and so acceleration of protons in nuclei and of electrons are very different due to their very different mass: this creates a capacitor effect. Also, if you push in some way your sphere you should have the same effect. In the gravitational field is quite tricky, because there is the equivalence principle that I think Brian cited; but at the same time from a dinstant observer the sphere will have velocity and acceleration and will emit just from Lienard-Wiechert potential (check wiki page) $\endgroup$
    – Rob Tan
    Commented Nov 7, 2022 at 9:10
  • $\begingroup$ I reflected a bit more on your reply, but there's one thing I don't understand: the electric and magnetic fields inside the shell are both null only for that family of inertial frames that are stationary relative to the shell. If I transform to an inertial frame in relative motion with the shell and inside of it, there should be a magnetic field in such frame, no? I didn't try doing the calculation, but it seems like there should be one. $\endgroup$
    – Povel
    Commented Nov 10, 2022 at 19:56
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After thinking a bit now I think there should be an electric field inside: electromagnetic tensor depends on the charge density, and this changes in the reference frame of the sphere due to the fictious force drive, so should not be invariant: hope I'm not saying blasphemies. Anyway, for a mechanical reasoning I should expect that electrons are driven in the direction opposite to the acceleration until they reach a point in which electrostatic force equals $m\boldsymbol{a}$. You fundamentally have now a capacitor and the figure you posted seems to confirm this. You also have emission by Larmor formula due to low-mass charged electrons, while nuclei don't almost emit, but I really can't tell if the phases of all photon emitted by all points of the sphere make electrodynamic field cancel out inside the sphere: electrostatic field should be present for the capacitor effect I told you before. I really can't imagine more than this; if I didn't say anything wrong I'll wait for a more conscious and claryfing answer since it's an interesting question.

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  • $\begingroup$ I did not consider how the charge density changes when seen from an accelerated reference frame. I guess that this should happen, since a distant non-inertial observer will see the shell as a continuously radiating point charge, and such field is not symmetric. Not sure if you can attribute this to fictitious forces though. In its own frame, the shell is inertial afterall. $\endgroup$
    – Povel
    Commented Nov 6, 2022 at 19:52

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