# Relativistic Treatment of the DC Conducting Straight Wire

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## 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 ##\mathcal{O}(1\,\text{mm}/\text{s})##!). Nevertheless it is unfortunately only quite confusingly discussed in the literature. So here is my attempt for a more consistent description using a naive classical model of a metal as consisting of a continuum of effectively positive bound charges (consisting of atoms and the bound electrons making up the lattice forming the metal, in the following called “ions”) and negative freely moving conduction electrons, treated as a freely moving fluid subject to some friction when moving against the positive charged rigid background.

The confusion starts with the fact that usually it is not carefully discussed 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 rest frame of the conduction electrons [1]. The qualitative argument is simple: We consider a straight wire with a constant current. 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, and there is thus no net force acting on them and thus in this reference frame no charge separation occurs.

Since the charge density (times ##c##) and the current density form a four-vector, consequently in the rest frame of the wire (i.e., the rest frame of the ions) there must be a non-vanishing charge density within the wire due to the Lorentz-transformation properties of vector components. This is also easily explained dynamically: In this reference frame the conduction electrons move along the wire with constant velocity, and thus a magnetic field is present, which causes a radial force on the conduction electrons, which consequently arrange such that an electric field is built up which exactly compensates this magnetic force. This is, of course, nothing else than the  “self-induced” Hall effect.

This is taken into account automatically when the correct relativistic version of Ohm’s Law is considered, which is sometimes approximated with the simple non-relativistic form, consequently leading to non-covariant approximations of the fields.


## Analytical treatment

As a usual material for wires we can consider copper, which in the static case has permittivity and permeability very close to 1 (in the here used Heavyside-Lorentz units). Thus we make the simplifying assumption that, using the usual electromagnetic field components wrt. an arbitrary inertial reference frame,
$$\label{1} \vec{E}=\vec{D}, \quad \vec{B}=\vec{H}.$$
Now we use our above defined very simplified classical model of a metal to describe the wire, which we consider to be a very long cylinder along the ##3##”=axis of an inertial reference frame. It consists of a positively charged continuum of ions with only bound charges of proper density ##n_+(x)##, which denotes the number-density of positively charged ions and quasi-freely movable conduction electrons with proper density ##n_-(x)##, where ##x=(c t,\vec{x})## is the space-time four-vector. Further let ##u_+(x)## and ##u_-(x)## be the four-velocity fields of the positive ions and negative conduction electrons respectively, which are normalized such that ##u_{\pm} \cdot u_{\pm}=1##, with the Minkowski-metric components ##\eta_{\mu \nu}=\eta^{\mu \nu}=\diag(1,-1,-1,-1)##. In our arbitrary frame the four-current density field thus reads
$$\label{2} j=e c (n_+ u_+-n_- u_-).$$
In the following we assume that we can find a solution of the magnetostatic Maxwell equations thus that in the rest-frame of the wire (i.e., the restframe of the positive ions) the conduction electrons move with a constant velocity ##\vec{v}=c \vec{\beta} \vec{e}_3=c \vec{u}_-/u_-^0=\text{const}## along the wire.

As we argued above, we assume that the charge density in the rest frame of the electrons can be considered as vanishing. In this frame, ##\Sigma’##, we have ##u_+’=(\gamma,0,0,-\gamma \beta)## and ##u_-‘=(1,0,0,0)##, where ##\gamma=1/\sqrt{1-\beta^2}##, which implies that
$$\label{3} j^{\prime 0}=c e (\gamma n_+-n_-) \stackrel{!}{=}0 \; \Rightarrow \; n_-=\gamma n_+.$$
Of course, if both the ions and the conduction electrons are at rest, i.e., if no conduction current is flowing, ##n_-=n_+##. This implies that the battery connected to the wire must deliver net negative charge!

In the restframe of the wire we have ##u_+=(1,0,0,0)## and ##u_-=(\gamma,0,0,\gamma \beta)##, and the four-current density reads
$$\label{4} (j^{\mu}) = \begin{pmatrix} c e (n_+-\gamma n_-) \\ -c e \gamma \beta n_- \vec{e}_3 \end{pmatrix} = c e n_+ \begin{pmatrix} 1-\gamma^2 \\ -\gamma^2 \beta \vec{e}_3 \end{pmatrix} =-c e n_+ \beta \gamma^2 \begin{pmatrix} \beta \\ \vec{e}_3 \end{pmatrix}.$$
As expected, for an observer in the rest frame of the wire the cable carries a negative charge density ##\rho_{\text{wire}}=j^0/c=-e n_+ \beta^2 \gamma^2##. The conduction-electron charge density in this frame is ##\rho_{\text{cond}}=j^3/(\beta c)=-e n_+ \gamma^2##, i.e., ##\rho_{\text{wire}}=\rho_{\text{cond}} \beta^2##.

The Maxwell equations are easy to solve, given the charge and current density along the wire. To that end we introduce the electromagnetic potentials ##A^{\mu}=(\Phi,\vec{A})## which are related to the fields via
$$\label{5} \vec{E}=-\nabla \Phi-\frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
As is well known, with this ansatz the homogeneous Maxwell equations,
$$\label{6} \vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0,$$
are identically fulfilled, as can be easily checked by inserting (\ref{5}) on the left-hand sides of Eqs. (\ref{6}). The inhomogeneous Maxwell equations read
$$\label{7} \vec{\nabla} \times \vec{B} – \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho_{\text{wire}}.$$
Here we insert (\ref{5}), leading to
$$\label{8} \vec{\nabla} (\vec{\nabla} \cdot \vec{A})+\left (\frac{1}{c^2} \partial_t^2 -\Delta \vec{A} \right)+\frac{1}{c} \partial_t \vec{\nabla} \Phi=\frac{1}{c} \vec{j}, \quad \Delta \phi + \frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{A}=-\rho_{\text{wire}}.$$
Since any other potentials derived from ##A^{\mu}## via a gauge transformation,
$$\label{9} \tilde{A}^{\mu}=A^{\mu} + \partial^{\mu} \chi \, \Rightarrow \, \tilde{\Phi}=\Phi+\frac{1}{c} \partial_t \chi, \quad \vec{\tilde{A}}=\vec{A}-\vec{\nabla} \chi,$$
do not change the fields given by (\ref{5}), where ##\chi## is an arbitrary field, we can impose one gauge constraint on the potentials. In our static case the most simple choice is the Coulomb gauge,
$$\label{10} \vec{\nabla} \cdot \vec{A}=0.$$
Also, since ##\rho_{\text{wire}}## and ##\vec{j}## are time-independent we can also assume that the potentials are time-indpendent in this gauge. Then (\ref{8}) simplifies to
$$\label{11} \Delta \vec{A}=-\frac{1}{c} \vec{j}, \quad \Delta \Phi=-\rho_{\text{wire}}.$$
The vector and scalar potential decouple completely, and due to the cylindrical symmetry we introduce cylinder coordinates,
$$\label{12} \vec{x}=\vvv{R \cos \varphi}{R \sin \varphi}{z}$$
and make the ansatz
$$\label{13} \Phi=\Phi(R) -E_z z, \quad \vec{A}=A(R) \vec{e}_z$$
with ##E_z=\text{const}##. The latter part we need for the electric field driving the current along the ##3##”=axis. Then the equations (\ref{11}) read
$$\label{14} \frac{1}{R} [R A'(R)]’=-\frac{1}{c} j_3, \quad \frac{1}{R} [R \Phi'(R)]’=-\rho_{\text{wire}}.$$
Let the radius of the wire be ##a##. Then for ##R>a## we have ##j_3=\rho_{\text{wire}}=0##, and the solutions are found by simple integrations:
$$\label{15} A(R)=A_{1>} \ln \left (\frac{R}{a} \right) + A_{2>}, \quad \Phi(R)=\Phi_{1>} \ln \left (\frac{R}{a} \right) + \phi_{2>},$$
and for ##R<a##
$$\label{16} A(R)=A_{1<} \ln \left (\frac{R}{a} \right) + A_{2<}-\frac{j_3 R^2}{4 c}, \quad \Phi(R)=\Phi_{1<} \ln \left (\frac{R}{a} \right) + \phi_{2<}-\frac{\rho_{\text{wire}} R^2}{4},$$ To determine the integration constants we note that we only need ##A_{1>}## and ##\Phi_{1>}## since for ##R<a## there are no singularities along the ##3##-axis and thus ##A_{1<}=\Phi_{1<}=0##, and the constants ##A_2## and ##\Phi_2## do not enter the fields,
$$\label{17} \vec{E}=-\vec{\nabla} \Phi=E_z \vec{e}_z+\left [ \frac{\rho_{\text{wire}} R}{2} \Theta(a-R) -\frac{\Phi_{1<}}{R} \Theta(R-a) \right] \vec{e}_R$$ and $$\label{18} \vec{B}=\vec{\nabla} \times \vec{A} = \left [\frac{j_3 R}{2c} \Theta(R-a) + \frac{A_{1>}}{R} \Theta(a-R) \right] \vec{e}_{\varphi}.$$
The constant ##A_{1>}## is determined by the demand that ##\vec{B}## must be continuous at the boundary of the wire, i.e., ##A_{1>}=j_3 a^2/(2c)##. Now the total current, ##I##, through the wire is ##I=\pi a^2 j_3## and thus
$$\label{19} \vec{B}=\frac{I}{2 \pi c a^2} \left [R \Theta(a-R)+\frac{a^2}{R} \Theta(R-a) \right] \vec{e}_{\varphi}.$$
The same continuity argument holds for the electric field, leading to ##\Phi_{1<}=-\rho_{\text{wire}} a^2/2##, i.e., $$$$\label{20} \vec{E}=E_z \vec{e}_z + \frac{\rho_{\text{wire}}}{2} \left[R \Theta(a-R)+\frac{a^2}{R} \Theta(R-a) \right] \vec{e}_R.$$$$ To also determine the constant ##E_z## we need Ohm’s Law. This we get most easily in covariant form: The electromagnetic four-force on a conduction electron is ##K_{\text{el}}^{\mu}=-e F^{\mu \nu} u_{\nu}## [2], i.e., its spatial components are $$$$\label{21} \vec{K}_{\text{el}}=-e \gamma \left (\vec{E}+\vec{\beta} \times \vec{B} \right).$$$$ Then there’s a friction force, which we assume to be given by $$$$\label{22} \vec{K}_{\text{fric}}=-\alpha \vec{p}=-m \gamma \frac{\alpha}{\rho_{\text{cond}}} \vec{j},$$$$ where ##\vec{p}=m c \gamma \vec{\beta}## is the electron’s momentum. Since the conduction electrons move with constant velocity the total force must vanish, and thus we have $$$$\label{23} \vec{j}=\gamma \sigma \left (\vec{E} + \vec{\beta} \times \vec{B} \right)$$$$ with $$\label{24} \sigma=-\frac{\rho_{\text{cond}} e}{\alpha \gamma}>0.$$
From (\ref{23}) we see that ##\sigma## is the electric conductivity since for ##\beta=|\vec{v}|/c \ll 1## we get ##\vec{j} \simeq \sigma \vec{E}##. By defining the covariant Ohm’s Law with the Lorentz ##\gamma##-factor as in (\ref{23}), we define ##\sigma## as a Lorentz scalar, as it should be for transport coefficients since we can write (\ref{23}) in Lorentz-covariant form as
$$\label{25} j^{\mu}=\sigma F^{\mu \nu} u_{\nu} = \sigma \gamma \begin{pmatrix} \vec{\beta} \cdot \vec{E} \\ \vec{E}+\vec{\beta} \times \vec{B} \end{pmatrix}.$$
To check the consistency of this classical model of conduction currents we note that
$$\label{26} j^0=\vec{\beta} \cdot \vec{j}=\rho_{\text{cond}} c \beta^2 = \rho_{\text{wire}} c,$$
as discussed above right after Eq. (\ref{4}). From this equation we also get, using (\ref{20}),
$$\label{27} E_z=\frac{\rho_{\text{wire}}c}{\sigma \gamma \beta}=\frac{\rho_{\text{cond}} \beta c}{\sigma \gamma}.$$
Now we use (\ref{23}) to get
$$\label{28} \vec{j} = \rho_{\text{cond}} \beta c \vec{e}_z=\gamma \sigma E_z \vec{e}_z,$$
which implies that, using (\ref{17})
$$\label{29} \vec{\beta} \times \vec{B}=-\beta B_{\varphi} \vec{e}_R \stackrel{!}{=} -E_R \; \Rightarrow\; \rho_{\text{wire}}=\frac{I}{\pi c a^2}.$$

## Alternative argument for ##\rho_{\text{wire}}’=0##

We can also derive the correct ansatz that the charge density of the wire must vanish in the rest frame of the conduction electron by first wirking entirely in the rest frame of the conduction electrons. We use the same notation as above but now start from the magnetostatic Maxwell equations written in this frame. The crucial point is, of course, to use the correct relativistic form of Ohm’s Law:
\begin{alignat}{2} \label{30} &\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0,\\ \label{31} &\vec{\nabla} \cdot \vec{E}= \rho_{\text{wire}}, \\ \label{32} & \vec{\nabla} \times \vec{B}=-\rho_{\text{cond}} \vec{\beta} \\ \label{33} &\vec{j}_{\text{cond}}=-\rho_{\text{cond}} \vec{v}=\sigma (\vec{E} + \vec{\beta} \times \vec{B}). \end{alignat}
Now we make the ansatz ##\vec{v}=c \vec{\beta}=v \vec{e}_3=\text{const}.## Taking the divergence of (\ref{33}) then yields
$$\label{34} \sigma [\vec{\nabla} \cdot \vec{E} + \vec{\nabla} \cdot (\vec{\beta} \times \vec{B})]=0$$
or
$$\label{35} \rho_{\text{wire}} + \partial_i \epsilon_{ijk} \beta_j B_k=\rho_{\text{wire}} -\vec{\beta} \cdot (\vec{\nabla} \times \vec{B}) = \rho_{\text{wire}}-\rho_{\text{cond}} \beta^2=0.$$
From this we get the four-current,
$$\label{36} (j^{\mu})=\vv{c \rho_{\text{wire}}}{\rho_{\text{cond}} \vec{v}} = \rho_{\text{cond}} \vv{\beta^2 c}{\vec{v}}.$$
Boosting to the rest frame of the conduction electrons leads to

$$\label{37} \vv{j^{\prime 0}}{j^{\prime 3}} = \beta c \rho_{\text{cond}} \gamma \begin{pmatrix} 1 & -\beta \\ -\beta & 1 \end{pmatrix} \vv{\beta}{1} = \beta c \gamma \rho_{\text{cond}} \vv{0}{1-\beta^2} =\frac{\beta c}{\gamma} \rho_{\text{cond}} \vv{0}{1}.$$
This shows that indeed in the rest frame of the conduction electrons the wire is uncharged, and in this reference frame the positive ions have a charge density ##\rho_{+}’=-\rho_{\text{cond}}/\gamma## (since the velocity of the ions in this frame is ##\vec{v}_+’=-\beta c \vec{e}_3’##.

The charge densities in terms of the manifestly covariant quantities ##n_+##, ##n_-##, and ##u^{\mu}## then follow as argued starting from Eq.  (\ref{2}).

## References

[1] H. v. Hees, Special Relativity, Physics Forums FAQ (2019).
https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

[2] P. C. Peters, In what frame is a current-conducting wire neutral, Am. J. Phys. 53 (1985) 1156.
https://doi.org/10.1119/1.14075

Tags:
11 replies
1. bhobba says:
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:

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

2. vanhees71 says:
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).

3. Ibix says:
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.
4. vanhees71 says:
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).
5. P
PAllen says:
The following occurs right where @Ibix describes:

Since the current density and the current density form a four-vector,

6. vanhees71 says:
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.

7. vanhees71 says:
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.

8. vanhees71 says:
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.