# Do moving charges still generate an electrostatic field?

According to my EM professor, an infinite line of current $$I$$ creates only a magnetostatic field (the regular $$\vec B=(\mu_0 I\vec /2\pi r)\vec e_\varphi$$ you get from applying Ampère's law), and not an electric field. I don't understand this statement since if I place myself in the inertial frame of reference of the moving charges, then I wouldn't register any current, just static charges, which would of course give place to an electrostatic field: $$\vec E'=(\lambda'/2\pi\varepsilon_0 r')\vec{e}_r$$. Therefore, if my professor's statement were true, then the invariant $$E^2-c^2B^2$$ would be violated when placing myself in the frame of reference of the moving charges. Also, the quasistatic approximation would not hold even for small intensities. All this does not make sense to me. Is it really true that a line of current does not give place to an electric field?

• if you place your self in the inertial frame so the moving charges (electrons) are statioary, then you would note the previously stationary +ve charge would be moving
– jim
Commented May 28 at 18:20
• Your professor meant there is magnetostatic field and zero electric field in the frame of the wire. In the frame of the moving electrons, current is still non-zero, because of the opposite motion of the oppositely charged wire, and both electric and magnetic field is present. Commented May 30 at 8:20
• Commented May 30 at 11:59

When we speak of a current-carrying wire, we usually mean a wire that is electrically neutral (as most wires are). There are as many positive charges as there are negative charges. The net charge is zero and there is no electric field. The electrons are flowing so there is a magnetic field. However, if you only have a single line charge flowing, then the net charge density is nonzero and both fields will be present.

• Begs the question whether there still is a magnetic field for an an observer moving along the wire with half the electron speed: In that case the backwards "current" of the positive charges (the positively charged metal atoms in the crystal lattice) should cancel out the forward current of the electrons. Commented May 30 at 7:42
• @Peter-ReinstateMonica Please read the question again. That isn't what OP is asking. Also what you said is false. It is not true that there is no force. Commented May 30 at 8:42
• I didn't refer to (let alone try to answer) the original question with my remark. But the question of my remark stands: For an observer moving along the wire with half the average electron speed there is no net current, is there? And for an observer moving along with the average electron speed, there is an opposite current (and, as a consequence, opposite magnetic field), isn't there? (I'm not being facetious; this is an actual question.) -- Oh, I'm seeing my mistake -- the protons move in the opposite direction, therefore the current is the same. Any observer sees the delta. OK. Commented May 30 at 9:19
• @Peter-ReinstateMonica Oh I see. Some remarks: (1) Due to relativistic velocity addition, the frame with half the electron velocity will not have half the electron velocity forwards. (2) If you instead picked the frame in which the protons and electrons have exactly opposite velocities (which is a little faster than the half the electron velocity), they will not have the same spacing, so the net charge and current is nonzero, so there will be contributions from both electric and magnetic forces. Commented May 30 at 10:45
• @Peter-ReinstateMonica Otherwise, this question has been answered at physics.stackexchange.com/q/125932/174766, physics.stackexchange.com/q/694928/174766, physics.stackexchange.com/a/82844/174766, etc. Commented May 30 at 10:56

The phrasing of your professor's question is unclear, as it talks about an infinite line of current, which is, ofc, charged. So the implication in the question is:

There are two linear charge densities of $$\pm \lambda$$, with the positive one moving in the positive direction at $$\beta$$.

Note: please do not think of this thing as a real physical object, which in this case, would be a fixed lattice composed of antimatter copper ions and free positrons moving in the $$+x$$ direction...all that does is cause confusion. We are dealing with abstractions: co-located infinite linear charge densities.

So now we need to introduce observers and frames. They are Alice in the lab ($$S$$), and Bob, moving at $$+\beta$$ in $$S$$, which is his $$S'$$ frame.

(I am going to set $$c=\epsilon_0=\mu_0=1$$; feel free to add them--it's kinda cute how the mu's and epsilon's work out to be $$c$$).

In $$S$$, we have two charge distributions, which we can add:

$$\lambda = \lambda_+ + \lambda_- = \lambda((+1) + (-1)) = 0$$

And two currents, which we add.

$$j = j_+ + j_- = v_+\lambda_+ + (0)\lambda_- = \beta\lambda$$

Note that the conservation equation:

$$\frac{dj}{dx} - \frac{d\lambda}{dt} = 0 - 0 = 0$$

is trivially satisfied, making this both an electrostatic and magneto-static problem with:

$$E = 0$$ $$B = j = \beta\lambda$$

where both fields are the so-called perpendicular components, with $$E$$ being radial and $$B$$, azimuthal. I'm not worrying about the $$1/r$$ parts. Note:

$$(E^2-B^2) = -\beta^2\lambda^2$$

Ok, now transform the charges and currents to Bob's frame.

$$\lambda' = \lambda'_+ + \lambda'_-$$

The negative charge is contracted, while the positive charge is dilated (see Bell's Spaceship Paradox if that troubles you):

$$\lambda' = \lambda_+/\gamma + \gamma\lambda_-$$

$$\lambda' = \lambda/\gamma - \gamma\lambda$$

$$\lambda' = (\frac 1 {\gamma} - \gamma)\lambda = -\gamma(1 - \frac 1 {\gamma^2})= -\gamma\beta^2\lambda$$

So Bob sees a negative charge density, giving an electrostatic field:

$$E' = -\gamma\beta^2\lambda$$

For Bob, the current is:

$$j' = v'_+\lambda'_+ + v'_-\lambda'_-$$

and by definition, $$v'_+=0$$, and by symmetry, $$v'_- = -\beta$$, so:

$$j' = (-\beta)(\lambda'_-) = \beta\gamma\lambda$$

Bob sees a negative current going backwards, giving a positive magnetic field:

$$B' = \beta\gamma\lambda$$

so now the invariant is:

$$(E'^2 - B'^2) = \gamma^2\beta^4\lambda^2 - \beta^2\gamma^2\lambda^2$$

$$(E'^2 - B'^2) = \gamma^2\beta^2\lambda^2(\beta^2-1)$$

$$(E'^2 - B'^2) = -\beta^2\lambda^2 = (E^2-B^2)$$

QED.

Yes, a single moving charge does produce an electric field as well as a magnetic field (although it isn’t really correct to describe that electric field as “electrostatic”). But a typical current-carrying wire is electrically neutral - there is just as much positive as negative charge per unit length in the lab frame - so the electric field produced by the negative charges exactly cancels the electric field produced by the positive charges, and there is no net electric field.