Stationary charge next to a current-carrying wire

vanhees71 said:

The only exception is Sommerfeld's Lectures of Theoretical Physics, vol. III, which I cited. What's now missing is the discussion of the Poynting vector, i.e., the energy transfer along the cable, which is interesting in its own right (and also discussed for the non-relativistic approximation by Sommerfeld).

Do you mean that Sommerfeld didn't consider the self-induced Hall effect?
I'm not sure why this is called "non-relativistic".

He assumed that there is no radial E-field in the wire volume, because the volume charge density in the wire is homogeneously zero.

If you calculate in Sommerfeld's book, §17, equation (9a), the ##\lim_{b \rightarrow \infty} ## of ##E_r## in the intermediate space ##a<r<b## of the coaxial cable, then you get ##E_r=0## and therefore no surface charge density on the inner wire.
Source:
https://archive.org/details/in.ernet.dli.2015.147970/page/n137/mode/2up

Calculation

I think this might be a good model for Purcell's infinite straight wire, when the outer return path is infinitely far away.

I can't find in your actual paper a proof for the following statement:

... in which reference frame the wire is uncharged. It is not the rest frame of the wire (i.e., the rest frame of the ions) but the restframe of the conduction electrons [Pet85].

In the AJP paper, which I linked in #18, a positive surface charge density compensates the negative volume charge density. Sommerfeld's assumption, that there is no radial E-field in the wire, is not fulfilled. But the positive surface charge density would compensate the ##E_r## outside of the surface to zero.

I see no indication in your paper (I don't have access to the full paper of Peters), what motivation the battery should have to provide extra electrons, if the wire is overall electrically neutral in it's rest frame.

Sagittarius A-Star said:

Do you mean that Sommerfeld didn't consider the self-induced Hall effect?
I'm not sure why this is called "non-relativistic".

What I mean with "non-relativistic" is the use of the standard "constituent equation" (Ohm's Law) as ##\vec{j}=\sigma \vec{E}##. This neglects the self-induced Hall effect and it's of course not relativistic, but nevertheless it's a damn good approximation. Note that the charge density inside the conductors due to the self-induced Hall effect is ##\rho_{\text{wire}}=\beta^2 \rho_{\text{cond}}##, and ##\beta=v/c## with ##v \simeq 1 \text{mm}/\text{s}##. For all practical purposes of electrical engineering the standard form of Ohm's Law is thus completely sufficient.

Sagittarius A-Star said:

He assumed that there is no radial E-field in the wire volume, because the volume charge density in the wire is homogeneously zero.

Exactly.

Sagittarius A-Star said:

If you calculate in Sommerfeld's book, §17, equation (9a), the ##\lim_{b \rightarrow \infty} ## of ##E_r## in the intermediate space ##a<r<b## of the coaxial cable, then you get ##E_r=0## and therefore no surface charge density on the inner wire.
Source:
https://archive.org/details/in.ernet.dli.2015.147970/page/n137/mode/2up

Calculation

I think this might be a good model for Purcell's infinite straight wire, when the outer return path is infinitely far away.

That's of course right, and then you end up with the solution of Gabuzda's paper. If didn't find this argument so convincing, but now given that it's found as this limit of the complete calculation, where only the usual physically well-motivated boundary conditions are used, it seems all right.

Sagittarius A-Star said:

I can't find in your actual paper a proof for the following statement:In the AJP paper, which I linked in #18, a positive surface charge density compensates the negative volume charge density. Sommerfeld's assumption, that there is no radial E-field in the wire, is not fulfilled. But the positive surface charge density would compensate the ##E_r## outside of the surface to zero.

But in my calculation, the overall charge neutrality comes out automatically from the standard boundary conditions, but that doesn't work with the single wire. That's why in this AJP paper Gabuzda adds the additional positive surface charge by hand. As the limit ##a_2 \rightarrow \infty##, i.e., putting the return path to infinity, it makes sense.

That there's no radial field in Sommerfeld's "non-relativistic" treatment follows immediately from ##\vec{j}=\sigma \vec{E}##, i.e., with ##\vec{j}=j \vec{e}_z## of course ##\vec{E}=\vec{j}/\sigma=E \vec{e}_z##. Gabuzda assumes the relativistic version of Ohm's Law and thus there must be a radial electric field compensating ##\vec{\beta} \times \vec{B}##. That's the same argument as in my writeup.

I'll add this limiting case as a model for a single DC conducing wire, including a citation of Gabuzda's paper.

Sagittarius A-Star said:

I see no indication in your paper (I don't have access to the full paper of Peters), what motivation the battery should have to provide extra electrons, if the wire is overall electrically neutral in it's rest frame.

Indeed, the charge-neutrality assumption isn't needed anymore with the full calculation of the complete coax cable including the return path, but it comes out as the said limit, ##a_2 \rightarrow \infty##. Everything that happens is a rearrangement of charges with the overall charge being 0 as it should be. That's, of course a frame-independent statement!

vanhees71 said:

That's of course right, and then you end up with the solution of Gabuzda's paper.

Yes. I saw, that you added his paper today to the references.

vanhees71 said:

But in my calculation, the overall charge neutrality comes out automatically from the standard boundary conditions, but that doesn't work with the single wire. That's why in this AJP paper Gabuzda adds the additional positive surface charge by hand. As the limit ##a_2 \rightarrow \infty##, i.e., putting the return path to infinity, it makes sense.

She didn't add the positive surface charge density by hand. The self-induced Hall effect shifts/compresses the electron current more towards the axis and then the surface has missing free electrons, which are not replaced from the battery.

Sommerfeld's calculation yields only an additional surface charge density.

vanhees71 said:

I'll add this limiting case as a model for a single DC conducing wire, including a citation of Gabuzda's paper.

vanhees71 said:

That's, of course a frame-independent statement!

Yes, if the "+" and "-" pole of the battery are defined to have the same ##z## coordinate, also while the ramp-up phase of the current.

The only quibble I still have with my calculation is the homogeneous part of the ##\vec{E}##-field in ##z## direction in region IV (outside the coax cable), but it seems to be unavoidable from the continuity condition. On the other hand, intuitively, there shouldn't be a field outside the cable. Maybe this has to do with the somewhat artificial assumption of an infinitely long cable and the ignorance of the fact that except the ideal voltage source at ##z=0## (which is indeed assumed to be infinitely thin too) that you have to close the circuit at some finite point ##z=L##. One could just put a conducting circle there, but then the problem seems not to be treatable analytically anymore.

Of course, I also didn't solve any transient state, i.e., "switching on the circuit" but only considered the final stationary DC state. That would also be an interesting task. One would have to solve the coax cable as a wave-guide problem. That's for sure much more complicated than the DC case, but it would also be worthwhile to study, because it would clearly show that the energy transfer is through the em. fields and not somehow along the wire as in this unfortunate analogy picture as if an electric circuit were analogous to a water pipe, where of course the energy (e.g., heat from hot water) is transferred through the flow of the water.

lightlightsup said:

Why doesn't the positive test charge (cat, here) experience an attraction to the wire if the electrons are getting length contracted?

As others explained, the distances between the electrons in the wire frame don't change when the current starts flowing, because unlike the positive ions, the electrons are not forced to keep constant proper distances. So the electron density in the wire frame doesn't change when the current starts flowing.

Here is a good explanation and diagram by @DrGreg :
https://www.physicsforums.com/threads/explanation-of-em-fields-using-sr.714635/post-4528480



This has been asked many times:
https://www.physicsforums.com/threads/electric-currents-and-length-contraction.930943/#post-5877453
https://www.physicsforums.com/threa...tion-of-electromagnetism.982883/#post-6284262
https://www.physicsforums.com/threa...magnetism-with-relativity.932270/post-5886984

vanhees71 said:

The only quibble I still have with my calculation is the homogeneous part of the ##\vec{E}##-field in ##z## direction in region IV (outside the coax cable), but it seems to be unavoidable from the continuity condition. On the other hand, intuitively, there shouldn't be a field outside the cable.

Sommerfeld avoids this problem in $17 by defining "outer radius ##c \rightarrow \infty##", from which follows his equation (3).

Yes, but that's also pretty artificial ;-).

I've uploaded a corrected version of the manuscript concerning the covariant treatment of Ohm's Law. I've gotten it wrong, because I've written down the equation too early in the special frame of reference, where the wire's at rest. Now everything is consistent in the sense that both transport coefficients, i.e., the Stokes-friction coefficient as well as the electric (or electrical?) conductivity are scalars and defined in terms of scalars (electron density as measured in their (local) rest frame):
$$\sigma=\frac{n_- e^2}{\alpha m},$$
where ##n_-## is the scalar number density of the electrons (i.e., the number density as measured in the rest frame of the electrons), ##e## the elementary charge, ##\alpha## the friction coefficient, and ##m## the electron mass.

I'm still somewhat puzzled by the result that there's a constant electric field along the wire outside of the coax cable, but it seems inevitable, if you don't make this somewhat artificial assumption as done in Sommerfeld by making ##a_3 \rightarrow \infty##.

vanhees71 said:

I'm still somewhat puzzled by the result that there's a constant electric field along the wire outside of the coax cable, but it seems inevitable, if you don't make this somewhat artificial assumption as done in Sommerfeld by making ##a_3 \rightarrow \infty##.

In the following paper, they don't make this somewhat artificial assumption. From their figure 3 follows the existence of longitudinal and radial components of the external electric field.

paper abstract said:

Surface charges and fields in a resistive coaxial cable carrying a constant current
Publisher: IEEE
A.K.T. Assis; J.I. Cisneros
...
Abstract:
We calculate the surface charges, potentials, and fields in a long cylindrical coaxial cable with inner and outer conductors of finite conductivities and finite areas carrying a constant current. It is shown that there is an electric field outside the return conductor.

Source:
https://ieeexplore.ieee.org/abstract/document/817391

From the same author A.K.T. Assis exists also the following German book.
See on page 94, chapter 6.4 "Radialer Halleffect" and on page 105, chapter 7 "Koaxialkabel":
https://www.ifi.unicamp.br/~assis/Kraft.pdf

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

Electrical field lines in the symmetrical case, according to the book:

Also, A.K.T. Assis mentions on page 8 of this book a here not yet discussed force of a wire on a charged test-particle at rest in the rest frame of the wire:

Das heißt, die externe Punktladung q induziert eine Verteilung von Ladungen längs der Oberfläche des Leiters und ergibt insgesamt eine elektrostatische Anziehung zwischen Leiter und q.

Google-translate:

That is, the external point charge q induces a distribution of charges along the surface of the conductor and results in an overall electrostatic attraction between the conductor and q.

From the publisher of this book, Apeiron, one can find also many dubious anti-SR books. However, that does not automatically mean, that every book of them is dubious.

This is not related to the offshoot question of what the charge distribution in a real wire is, but I hope it will be helpful to the OP if they are still around, or to other readers interested in the original question. It uses only Gauss' law to explore the issue.

A preliminary point is that Maxwell's equations are fully relativistic, so that we can use Gauss' law (one approach of one of Maxwell's equations) to determine the charge enclosed in a box (any box) at some instant in time. In this case, the box is a cylinder around an idealized wire with zero resistance.

While we have a pair of wires in a current loop, we only draw the Gauss' law cylinder around one wire.

We consider a pair of such cylindrical boxes, one that is stationary in the lab frame, one that is moving in the lab frame.

The charge enclosed by the box in the lab frame could, in principle, be anything. It's part of the experimental setup that current loop is at zero potential, and that there is no electric field around the wire. But we assume that the wire is uncharged, that it is an uncharged conductor in the lab frame, the experimental test of this is to measure the electric field around the neutral conductor in the lab frame and find no electric field. Then the Gauss' law surface integral is easy - the integral of zero is zero.

Now we consider a moving box around the wire. We know there is an induced electric field in the moving frame. Gauss' law now tells us, because of the induced electric field (proportional to v x B), that the material enclosed by the box contains a charge.

The next part I have to say is a little more advanced, but I don't know how to say it as simply as the first part. We don't expect that a stationary box will include no net charge, while a moving box will enclose net charge when we do a non-relativistic transformation. But when we do a proper relativistic transformation, we find that the box in the moving frame does contain a non-zero charge. The point is that for Maxwell's equations to work, the source must transform relativistically.

The result that the moving box contains a non-zero charge is unexpected, but it's not inconsistent, either internally or with the laws of relativity. The prediction that the moving box contains a net charge is not an error.

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.

When we do, we find in the moving frame that one leg of the loop has acquires a positive charge, while the other leg of the loop has a negative charge. The total net charge is zero in both the lab frame and the moving frame. I suggest again, without an overly detailed explanation, that we can explain this by the relativity of simultaneity. The total charge remains constant at zero, but the distribution of charge at any given "instant of time" varies with the choice of frame. In the moving frame, we see a positive charge density on one wire, and a negative charge density on the other, but the integral of the charge density around the loop remains zero.

A video of Veritasium about surface charges:

Sagittarius A-Star said:

In the following paper, they don't make this somewhat artificial assumption. From their figure 3 follows the existence of longitudinal and radial components of the external electric field.Source:
https://ieeexplore.ieee.org/abstract/document/817391

Yes, that's exactly the solution I consider in my writeup (in the non-relativistic approximation, i.e., ##\vec{j}=\sigma \vec{E}##. My quibble is with (12). I come to another solution though (my Eq. (41)) by assuming that there should be no divergent component for ##R \rightarrow \infty##. That leaves a homogeneous electric field outside of the conductor. The math seems to be correct, but I'm a bit puzzled, whether this is the right physics.

I didn't understand their boundary condition at ##\rho=L##. What has the longitudinal extent of a finite coax cable to do with the boundary conditions perpendicular to the conductors? There should be a boundary condition at ##\rho \rightarrow \infty## (##\rho## is my ##R##, i.e., the radial coordinate in cylinder coordinates).

Of course it would be desirable to have a solution for a conductor of finite length, but that seems to be too difficult for a closed, analytic solution.

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.

vanhees71 said:

I didn't understand their boundary condition at ##\rho=L##.

I can't find a capital ##L## in their equation (12), only a small ##l##. The ##E_z## fits there to a boundary condition at ##\rho = c##.

Source:
https://www.google.de/search?q=file...ive+coaxial+cable+carrying+a+constant+current

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?

All I use is the continuity of ##E_z## (the standard continuity condition of the parallel component of the electric field at boundaries). As my forth boundary condition I use that for ##\rho \rightarrow \infty## (using the paper's notation) the electric field must not diverge, and this leaves me with the constant electric field ##\vec{E}=-I/[\pi (c^2-b^2) \sigma]## (in the non-relativistic approximation, where ##\gamma=1+\mathcal{O}(\beta^2)##. 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##.

vanhees71 said:

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]?

vanhees71 said:

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##.

pervect said:

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.

Sagittarius A-Star said:

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.

vanhees71 said:

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.

vanhees71 said:

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 ...

vanhees71 said:

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##.

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...

vanhees71 said:

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##:

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

vanhees71 said:

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

vanhees71 said:

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

Visualization:

vanhees71 said:

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):

IEEE paper said:

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

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##!

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.

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.

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.

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.


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

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?

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

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.

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 ;-)).