Relativistic Treatment of the DC Conducting Straight Wire

[vanhees71]

Table of Contents
ToggleIntroductionAnalytical treatmentAlternative argument for $\rho_{\text{wire}}’=0$References
Introduction
The direct-current-conducting infinitely long wire is often discussed in the context of relativistic electrodynamics. It is of course a completely academic discussion since for the typical household currents the drift velocity of the electrons in the wire, making up the conduction current, is tiny (of the order...

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[bhobba]

Excellent treatment. Better than Purcell which for me always had issues, but I do suggest it as a starting point in understanding Maxwell's equations from Coulombs Law. Then something like the following to derive Maxwell's equations:
http://cse.secs.oakland.edu/haskell/Special Relativity and Maxwells Equations.pdf
Then you can use it to see there is a myriad of ways why the constant c in Relativity is the speed of light.

But after that you really should go back and analyse what motivated it correctly. Then you can point out that sort of thing is done not infrequently in physics teaching. For example the standard beginning treatment of the double slit should be redone without recourse to wave-particle duality once the full machinery of QM is developed - but few do it. Note to those that know my posts I do often post an analysis of the double slit without wave-particle duality, but it is not really correct either, I use just for pedantic purposes.

Thanks
Bill

 

[vanhees71]

I don't think that one can derive all Maxwell equations from electrostatics. The problem with the "derivation" in the quoted paper is that he assumes that there's a reference frame, where the entire charge distribution is at rest and electrostatic can be applied. That's of course not generally true.

I don't like Berkeley Physics Course vol. 2 since my days when I was a student. I tried to make sense of the treatment of the straight wire at the time with no success. Now I know why: It's simply wrong (i.e., it's not using the assumption that the charge density in the rest frame of the conduction electron vanishes, which is consistent with the simple model of a metal as a system with positive ions with rigid charge distributions. It assumes wrongly that the charge density is 0 in the rest frame of the ions, which must be wrong due to the Hall effect, as discussed in my Insight (see also the paper by Peters cited there).

 

[Ibix]

I need to look at this with a larger screen than my phone to make substantive comment, but I caught one typo. In the first sentence of your third paragraph, one of the two current densities that form a four vector should be a charge density.

 

[vanhees71]

Do you mean Eq. (3)? I don't see a typo. Note that a four-current is of the form
$$j^{\mu}=\begin{pmatrix} j^0 \\ \vec{j} \end{pmatrix}=\begin{pmatrix} c \rho \\ \vec{j} \end{pmatrix},$$
where $\rho$ is the usual charge density (dimension charge per volume) and $\vec{j}$ the current density (dimension charge per time and per area). The additional factor $c$ in the time-like component must be there for dimensional reasons. The quantitites $n$ are particle-number densities in the (local) rest frames of the corresponding particles (dimension inverse volume).

 

[PAllen]

The following occurs right where @Ibix describes:Since the current density and the current density form a four-vector,

 

[Ibix]

Do you mean Eq. (3)?

No - it's in the text, first sentence of the third paragraph. @PAllen reproduces the exact wording.

 

[vanhees71]

Yes of course! Sorry. I'll correct it immediately. Thanks for pointing out the typo!

 

[Swamp Thing]

Insight article, in paragraph 2:

In the rest frame of the conduction electrons there is a current due to the moving positively charged background and a corresponding magnetic field. The charges within the positive background are however bound and can be considered not to move due to the electromagnetic field. Thus in this reference frame the charge density vanishes everywhere within the wire as if there were no current at all since the freely movable conduction electrons are at rest,

... the charge density in the rest frame of the conduction electron vanishes, which is consistent with the simple model of a metal as a system with positive ions with rigid charge distributions.

Consider a target audience (including me, to a large extent) to whom the above reasoning would not at all seem self-evident. And assume that the rest of the analysis in the insight article is above my pay grade and also this audience's.

Can we come up with a simple thought experiment that would make the above statements clearer? My specific sticking point is this: The fact that the positive ions are constrained to maintain a fixed lattice structure as they move along -- well, how exactly does that fact prevent them from sending out their own electric field? (*) When I think of a long, thin cylindrical insulator prepared with an embedded positive charge distribution, I think of it producing an electrostatic field. If I now move that dielectric rod at some speed of the same order as copper's drift velocity, the field doesn't vanish. Where is my error - is it in my description of the insulator, or is it in assuming that that description carries over to the metal wire?

(*) Or for that matter, their own magnetic field, if the dielectric is moving.

 

[vanhees71]

A dielectric is a completely different thing! For them you need other constitutive equations.

The point is that the conduction electrons can move quasi freely against the ions, which (for the very weak fields considered here) as rigid, which means that within the system of ions no shifts between charges occur, because all these charges are considered as bound. In the reference frame, where the conduction electrons are at rest there cannot be any force acting at them. There's no magnetic force, because they are at rest to begin with, and thus there can also be no electric field in this reference frame, because otherwise there's be an elecstrostatic force. The conclusion is, using $\vec{\nabla} \cdot \vec{E}=\rho$ that in the rest frame of the conduction electrons the charge density (i.e., ions + conduction electrons) must vanish, i.e., it's in this frame that (within this model) the wire as a whole is uncharged.

In the restframe of the wire (i.e., the ions) there acts a magnetic force perpendicular to the wire on the conduction electrons due to the current made up by them. Since the electrons by our ansatz move with constant velocity along the wire, there cannot be a net force on them, and thus there must be a radial electric field to compensate the radial magnetic force, and thus in the rest frame of the wire the wire must carry a (negative) net charge distribution. That's nothing else than the self-induce Hall effect.

All this is worked out analytically in the 2nd section of the article, but I think the qualitative arguments should be clear also without the math.

I hope these arguments make it clearer. If so, I'll also update the Insights text.

 

[Swamp Thing]

there must be a radial electric field to compensate the radial magnetic force, and thus in the rest frame of the wire the wire must carry a (negative) net charge distribution.

Please correct me if I'm getting this wrong... This is the sense I'm getting:-
In the wire's rest frame:-
[1] There is a radial re-distribution of electrons, bringing more electrons out to the surface and leaving fewer in the interior. Meanwhile, the radial distribution of (positive) ions can't change due to the current.

[2] If we count the total charge in any small segment/slice of wire, there will still be a net charge of zero. The pluses (ions) and the minuses (free electons) will still cancel out over the volume of the slice. Electron and ion count per unit segment volume would be an invariant across reference frames.

[3] To a test electron outside the wire, the negative "skin" charge due to radial redistribution will look like a negative charge which would repel the test electron away from the wire.

Hope the above is correct.

Another point: If we calculate / predict / measure any physical parameter that is a function $f(r)$ of radial distance $r$ measured orthogonally to the relative motion of the two reference frames, but independent of position $x$ along the relative motion, then that function $f(r)$ should be invariant across the reference frames. (*) True? Now the radial distribution of charge is such a thing, so analyses in both reference frames should predict the same distribution. Is this correct?

(*) Except for E and H fields, which "mix" with each other as a function of relative frame velocity?

 

[vanhees71]

Ad [1]-[3]: To the contrary the net charge of the wire in its rest frame (= rest frame of the ions) is negative due to the Hall effect. Note that the electrons are negatively charged. Perhaps I should draw a figure to make that clear (I'm admittedly always too lazy to draw figures, but of course, I should do it to make the whole thing better understandable). The battery must deliver some net negative charge (i.e., electrons).

Your argument concerning the invariance of $f(r)$ only holds for scalar fields $f$, but the charge and current densities build a four-vector. That's why the charge densities are different in the wire rest frame and the conduction-electron rest frame as demonstrated in my article.

 

[Swamp Thing]

Thanks. I still need to think that through and maybe get back.

One concern is that the self-capacitance (or maybe capacitance to nearby things) of the battery could be quite negligible compared to that of the wire, which we could make as large as we wanted. That would call for an arbitrarily large voltage across the battery's self capacitance in order to transfer the charge Q called for by the model.

Anyway, I'll have a little think first.

 

[vanhees71]

I've just added a section to the Insights article, proving the charge neutrality of the wire in the rest frame of the conduction electrons using a calculation entirely made in the rest frame of the wire.

 

[haushofer]

Nice! Just above eqn.10 a spacing is missing in the sentence (theCoulomb). If it's just my phone, never mind ;)