B Stationary charge next to a current-carrying wire

[Sagittarius A-Star]

Yes, sorry, it's $l$ rather than $L$. I don't understand the physics of their "fourth boundary condition", i.e., $\Phi(\rho=l,\varphi,z)=0$. Where does this come from? What has the length of the wire to do with the radial boundary conditions?

They explain it under equation (5), but refer to Russell [8] to get an equation, that they use in equation (6) to get the justification for equation (5). Did you check their source Russell [8]?

In their Eq. (10) the electric field diverges logarithmically for $\rho \rightarrow \infty$, which I find even more worrysome than my constant electric field for $\rho>c$.

Their equation (10) is only defined between $a$ and $b$.

 

[A.T.]

To anticipate an objection about the creation of charge. Where did the charge in the box "come from"?

The answer to this objection is that current must flow in a closed loop. To find a "total charge", we must consider a closed loop.

Exactly, as also shown in the illustration by @DrGreg in post #40. And the explanation for why the charges are unevenly distributed along the loop in some frames is relativity of simultaneity: They cannot all start flowing simultaneously in all frames.

 

[vanhees71]

They explain it under equation (5), but refer to Russell [8] to get an equation, that they use in equation (6) to get the justification for equation (5). Did you check their source Russell [8]?Their equation (10) is only defined between $a$ and $b$.

I guess we don't refer to the same paper. I talk about:

Assis, Cisneros, The problem of surface charges and fields in coaxial cables and its importance for relativistic physics, in "Open questions of relativistic physics", ed. by A. Selleri, Apeiron (1998)

The inner conductor's radius is $a$ and the inner and outer radiuses of the outer condutor are $b$ and $c$. Here's the field he gives:

I checked Jefimenko's book. He simply doesn't discuss what happens outside the conductor ;-)). He discsusses indeed only Eq. (8) concerning the electric field.

The point indeed seems to be to have a "$z$-boundary condition" at the place of the battery (treated as an ideal voltage source), which I also have at $z=0$ but also at the place, $z=l$, where an additional resistor is placed and then neglecting edge effects. I think Assis's boundary condition at $z=l$ makes sense for a short-circuit there, but I still don't get, how the $l$ can appear in the logarithm in (10). For $\ell \rightarrow \infty$ one gets indeed my solution. Since $I/\sigma=(\phi_C-\phi_B)/\ell$.

I've to think, how to correctly implement the boundary condition at $z=\ell$ (with a finite or vanishing resistance between the inner and outer conductor). As I said, I'm a bit puzzled about the above copied solution by Assis. However that also seems not to solve my quibble with the electric field outside.

 

[Sagittarius A-Star]

I guess we don't refer to the same paper.

Then let's talk about the paper I was talking about. It contains at and behind equation (5) an answer to your question.

I don't understand the physics of their "fourth boundary condition", i.e., $\Phi(\rho=l,\varphi,z)=0$.

... is not a boundary condition at ...

I think Assis's boundary condition at $z=l$

Instead, it's a boundary condition at the surface of a thought cylinder with radius $l$.

 

[vanhees71]

But that's basically the same as the paper I've quoted. I still don't get the sense of this boundary condition. Also Russell's paper doesn't help. It seems as if they try to put two ideal voltage sources at the ends (one with potential difference $\Phi_A-\Phi_B$, but on the other hand that doesn't coincide with what they give for these surface charges in Eqs. (14-16) in the IEEE paper. It's very puzzling...

 

[Sagittarius A-Star]

It seems as if they try to put two ideal voltage sources at the ends (one with potential difference $\Phi_A-\Phi_B$, but on the other hand that doesn't coincide with what they give for these surface charges in Eqs. (14-16) in the IEEE paper. It's very puzzling...

I don't see the problem.

Consider $\text{radius} = l\gg c\gt b\gt a$:

 

[Sagittarius A-Star]

I found the English version of the book linked in posting #45.
https://www.ifi.unicamp.br/~assis/The-Electric-Force-of-a-Current.pdf

 

[Sagittarius A-Star]

Are their papers with a single wire (torus) around? Maybe that can be solved, but I guess it will involve some elliptic integrals or some other "higher functions", if it's solvable in closed form at all.

See free PDF in:
https://www.scielo.cl/scielo.php?pid=S0718-13372004000200003&script=sci_arttext

via:
https://www.ifi.unicamp.br/~assis/papers.htm

Another paper:
https://pubs.aip.org/aapt/ajp/artic...d-and-surface-charges?redirectedFrom=fulltext

Free version:
https://www.ifi.unicamp.br/~assis/Am-J-Phys-V71-p938-942(2003).pdf

 

[Sagittarius A-Star]

It seems as if they try to put two ideal voltage sources at the ends (one with potential difference $\Phi_A-\Phi_B$,

Visualization:

 

[Sagittarius A-Star]

I still don't get the sense of this boundary condition.

I think in the IEEE paper is not OK, that they don't provide their mathematical calculation as evidence for the claim under equation (6):

After solving these integrals we discovered that Φ went to zero not at infinity, but at ρ = ℓ.

 

[vanhees71]

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 $z$!

 

[Sagittarius A-Star]

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.

 

[vanhees71]

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.

 

[Sagittarius A-Star]

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.

 

[Sagittarius A-Star]

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.

The potential at $\rho=c$ and outer surface charge density would vanish only at $z=0$.

 

[vanhees71]

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?

 

[Sagittarius A-Star]

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.

The asymmetrical case is contained in the book, but not in the IEEE paper.

 

[Sagittarius A-Star]

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.

 

[vanhees71]

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 ;-)).