Divergence of v x B = Divergence of E in the v=0 frame?

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Consider a scenario where in one frame R, I have a magnet at rest and a solid slab of charges with an arbitrarily large mass moving at velocity v. The overall acceleration of the slab is trivial, however, the v x B exerted on the slab is divergent, thus compressive/tensile stresses are exerted on the charges within the slab.

Now, let's change the frame of reference such that we are now in the frame of reference R' of the slab. There are compressive/tensile stresses on the charges are still exerted. However, in this rest frame, the average v' is zero, therefore v' x B' = 0. This leaves us only with the electric field E'. In the prior case (frame R), we had non-zero v x B which was perpendicular to the velocity of the slab v. Therefore, in the latter case (frame R'), we would expect an electric field exerted on the slab that is perpendicular to the velocity of the magnet in frame R'. This confines the electric field to planes whose normal is parallel with said velocity. But according to theory the distant magnet should not be able to create a divergent E-field at a distance from it. So somehow the compressive/tensile stresses in the slab in frame of reference R (magnet's rest frame) which are explained by a divergent v x B would somehow have to be explained by a non-divergent E-field in frame R' (slab's rest frame).

Before I should go any further on this line of reasoning, I would like to know what your thoughts are about this. Have I already made a serious or minor mistake somewhere?

Thanks in advance.

Sincerely

Kevin M.
 

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  • #2
Orodruin
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Apart from the issues we have talked about in other threads with taking care when switching inertial frames, you do realize that if ##\vec v## is constant then ##\nabla\cdot(\vec v\times\vec B) = -\vec v \cdot (\nabla\times \vec B) = 0## in the absence of currents just as ##\nabla\cdot\vec E=0## in the absence of charges?

Edit: Also, you seem to use "divergent" to mean "with non-zero divergence". This is not an appropriate nomenclature.
 
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Apart from the issues we have talked about in other threads with taking care when switching inertial frames, you do realize that if ##\vec v## is constant then ##\nabla\cdot(\vec v\times\vec B) = -\vec v \cdot (\nabla\times \vec B) = 0## in the absence of currents just as ##\nabla\cdot\vec E=0## in the absence of charges?
Indeed, ##\nabla\cdot\vec E=0## in the absence of charges, but ##\nabla\times \vec B=0## in the absence of both electric currents and displacement currents (or where their contributions cancel), and the absence of electric charge (density) at a point only guarantees the absence of electric current (density) at that point.
 
  • #4
BruceW
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I'm not 100% sure if I understood the question. But, in the magnet frame, the slab is moving, there is a force on the slab since it is moving in a magnetic field. In the slab frame, the magnet is moving, so there is a changing magnetic field that acts on the slab, so there is still a force on the slab even though the slab is not moving (initially).
 
  • #5
Dale
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compressive/tensile stresses are exerted on the charges
This seems like a good motivational problem for jumping into the full relativistic formulation of Maxwell's equations. Simply calculate the stress energy tensor and then transforming between frames is almost trivial.
 
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This seems like a good motivational problem for jumping into the full relativistic formulation of Maxwell's equations. Simply calculate the stress energy tensor and then transforming between frames is almost trivial.
I want to be able to put in the 3-vectors of the electric field and magnetic fields according to one rest frame and get 3-vectors electric and magnetic fields for the other rest frame. What I am thinking is that the 3-vector formulation should be sufficient for describing the fields acting on charges and currents. You seem to be suggesting that to properly address my problem, I should pursue the full relativistic formulations (i.e. with tensors not just vectors). Is this simply because it is easier to do it that way, or is this a way of saying that 3-vectors cannot do what I expect it to do, that is, explain how the compressive/tensile stress are affected under Lorentz boosts?

I looked at:

https://en.wikipedia.org/wiki/Stress–energy_tensor
https://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

In one of the paragraphs, I saw that the latter linked to:

https://en.wikipedia.org/wiki/Electromagnetic_tensor
https://en.wikipedia.org/wiki/Maxwell_stress_tensor

So if I defined my scenario initially in terms of source charges and currents, and then defined the corresponding fields E and B, I should be able to construct the "Electromagnetic tensor". I should also be able to construct the "Electromagnetic stress-energy tensor" with the explicit matrix form provided on the wiki which carries nine components from the "Maxwell stress tensor", three of which are diagonal terms for the pressure. I noticed that the divergence of the Maxwell stress tensor gives the momentum flux density, but this is different the Lorentz force density by a term proportional to the time-derivative of the Poynting vector.

So, now, if I understand correctly, whether there is or is not a compressive or tensile force acting on a charge density in its rest frame is not strictly based upon the divergence of the electric field E, but instead it appears to depend on "the divergence of the divergence" of the Maxwell stress tensor divided by the charge density. As the charge density contributes its own components of the E field used in the calculation of the Maxwell stress tensor, it follows that the divergence of the Maxwell stress tensor depends on the self-field of the charge, and the compressive/tensile stress components do not actually require that externally-applied E field be divergent like I was thinking. The divergence of the time-derivative of the Poynting vector would become the thing responsible for the pressure gradients within the slab itself, rather than a (non-existent) divergent E field generated at a distance by the external moving charge.

This is basically undermines a premise that I held in the back of my mind, the premise that the Lorentz force density should in the slab's rest frame be expected to describe the compressive/tensile forces acting on the slab when an external charge moves relative to it at some velocity.

To say it another way, the E field does not account for the entire compressive/tensile forces acting on the slab in its rest frame, but there is a term which contains as a factor the time-derivative of the Poynting vector power flux (W/s per m^2). The divergence of that time-derivative is essentially the time-derivative of the power flux to/from the charge density (W/s per m^3).... something that looks like mass density*(acceleration*acceleration + velocity*jerk) + rate change of the reactive power density in the E and B fields....

Hmmm.... Does this mean that, actually, that the stresses are divided between the charge and its fields and that the share between them is frame-dependent? Wouldn't that kind of mean that the charges and the field are part of the same "object" in regards to the stress-strain state?
 
  • #7
vanhees71
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What you need is simply the Lorentz transformation of the electromagnetic field components (not fields, which are as a 2nd-rank antisymmetric tensor Lorentz invariant!). Of course you can translate the elegant covariant Minkowski-space formalism also to the (1+3)-formulation. You find the formulae in Wikipedia

https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames

The more natural way to do this is to use the Riemann-Silberstein vector ##\vec{F}=\vec{E}+\mathrm{i} \vec{B}## (Heaviside-Lorentz or Gauss units), which provides a proper representation of the proper orthochronous Lorentz group in terms of ##\mathrm{SO}(3,\mathbb{C})##, which is precisely the transformation property of the Riemann-Silberstein vector, i.e., you can build up a manifestly covariant formalism based on this representation.

https://en.wikipedia.org/wiki/Riemann–Silberstein_vector

Note that Wikipedia uses the ugly SI of units, and thus you have ugly factors ##c## around :-(.
 
  • #8
Dale
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Is this simply because it is easier to do it that way, or is this a way of saying that 3-vectors cannot do what I expect it to do
It is just a matter of ease and convenience. 3 vectors can do this, but tensors just make it much easier. This is one of those problems that would be so simple in tensor notation and so cumbersome in 3 vector notation that the effort in learning the strange tensor notation almost "pays for itself" in a single problem.
 

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