Stationary charge next to a current-carrying wire

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The discussion centers on the behavior of stationary charges near current-carrying wires, specifically addressing the implications of special relativity on charge distribution. Participants reference the textbooks "Special Relativity" by Helliwell and "Electrodynamics" by Griffiths to explain that while a wire appears neutral in its rest frame, it can exhibit a net positive charge in the electron rest frame due to length contraction effects. The conversation highlights the importance of understanding charge density and the Hall effect in analyzing the behavior of charged wires.

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  • Knowledge of the Hall effect and its implications in electrodynamics.
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  • Study the implications of the Hall effect in current-carrying conductors.
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  • Review the derivations in "Electrodynamics" by Griffiths regarding charge distribution in wires.
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Physicists, electrical engineers, and students studying electromagnetism who seek to deepen their understanding of charge behavior in current-carrying wires and the effects of special relativity on electric fields.

  • #61
As I said, I don't understand this point in the IEEE paper. I think they want to put a short circuit at ##z=\ell/2##, but then the ansatz, assuming full cylindrical symmetry is not applicable anymore, and you have to solve a much more complicated boundary-value problem for a cylinder of finite length. I don't see why their Eq. (6) is justified. It's just assuming that you have everywhere the cylinder symmetric solution for the induced surface charges, but that breaks down with the short-circuit-boundary condition at ##z=\ell/2##. You cannot have a vanishing surface charge at two values of ##z## with a potential linear in ##z##!
 
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  • #62
vanhees71 said:
I think they want to put a short circuit at ##z=\ell/2##
I don't think so. But it would anyway make no difference, because with even with a short circuit at ##z=\ell/2##, for symmetry-reasons no current would flow between the inner and outer conductor.

Edit: Sorry, I misunderstood you. Correction: A short circuit at ##z=0## would make no difference.
 
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  • #63
Of course it would, because of the voltage source at the other end. This paper is really hard to read, but it could also be that they mean to have two voltage sources, one at ##z=-\ell/2## and one at ##z=+\ell/2##. Nevertheless also then my arguments holds, i.e., you cannot fulfill the boundary conditions with the simple separation ansatz of the potential, leading inevitably to a solution linear in ##z## (see my manuscript), which does not allow to fulfill homogeneous boundary conditions at two different ##z## values. That's of course, because the finite cylinder conditions break the corresponding cylindrical symmetry.
 
  • #64
vanhees71 said:
Of course it would, because of the voltage source at the other end. This paper is really hard to read, but it could also be that they mean to have two voltage sources, one at ##z=-\ell/2## and one at ##z=+\ell/2##. Nevertheless also then my arguments holds, i.e., you cannot fulfill the boundary conditions with the simple separation ansatz of the potential, leading inevitably to a solution linear in ##z## (see my manuscript), which does not allow to fulfill homogeneous boundary conditions at two different ##z## values. That's of course, because the finite cylinder conditions break the corresponding cylindrical symmetry.
See my edit/correction in posting #62.
Of course they have two voltage sources in the symmetrical case, one with "+" connected to the outer conductor, the other with "+" connected to the inner conductor.

ieee3.png


Coaxial-symm-case.png
 
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  • #65
vanhees71 said:
Nevertheless also then my arguments holds, i.e., you cannot fulfill the boundary conditions with the simple separation ansatz of the potential, leading inevitably to a solution linear in ##z## (see my manuscript), which does not allow to fulfill homogeneous boundary conditions at two different ##z## values. That's of course, because the finite cylinder conditions break the corresponding cylindrical symmetry.
In their symmetrical case with the two batteries, which can be constructed completely cylindrical symmetric, I don't see this problem.

ieee5.png

The potential at ##\rho=c## and outer surface charge density would vanish only at ##z=0##.
 
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  • #66
Since an ideal voltage source has 0 resistance, shouldn't ##E_R## be 0 at the place of these voltage sources? The more I read the paper the more I'm puzzled. The math is ok, but which physics situation does it describe?
 
  • #67
vanhees71 said:
Since an ideal voltage source has 0 resistance, shouldn't ##E_R## be 0 at the place of these voltage sources? The more I read the paper the more I'm puzzled. The math is ok, but which physics situation does it describe?
They don't specify that the batteries are ideal voltage sources and they calculate at ##z## coordinates, were both batteries are "far away".

In a practical setup the current must anyway be limited to protect the batteries, because real coaxial cables have a very low resistance.

Practical applications are more like their asymmetrical case.
ieee6.png
The asymmetrical case is contained in the book, but not in the IEEE paper.
 
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  • #68
I find suspicious, that Assis lists on page iii near the beginning of the book under "Acknowledgments" also Hartwig Thim, at PF and in Germany known as an anti-relativist.

Hartwig Thim did an experiment and wrongly claims, that it disproved relativistic time-dilation, see the last sentence of his abstract at IEEE.

See also at the end of my posting #45 information about the publisher of this book.
 
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  • #69
Note that there's a very convincing counter-argument against Thim's interpretation. As expected, nothing's wrong with (special) relativity:

https://doi.org/10.1109/TIM.2009.2034324

It's really amazing what gets published in the peer-reviewed literature ;-)).
 
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