# How can a current carrying wire be electrically neutral if the wire at rest is already electrically neutral? (relativistically speaking)

I am stationary with respect to a conducting wire

This conducting wire is electrically neutral (From observation)

Current is now made to move within this wire.

The motion of charges will be seen as length contracted(from my stationary reference frame) and therefore the density of negative charges must increase giving the wire a net charge.

But the current carrying wire is found to be electrically neutral (From observation).

How does this happen? I have seen the videos from Veritasium and the Science asylum as well as many threads here that fail to explain this(adequately)? How can the wire be neutral in both scenarios?

• physics.stackexchange.com/q/327511/37364 Nov 1, 2019 at 16:15
• @mmesser314 your answer there was kinda vague. What can I read so that I can answer my own question? Nov 1, 2019 at 17:30
• Charges released simultenously in an uniform E-field do not get closer to each other. That follows from the uniformity of the field. Balls dropped simultaneously in an uniform gravity field do get closer to each other. That is caused by the non-uniform potential, which causes a non-uniform time dilation. Accelerating observer may think that there is a gravity field. Nov 2, 2019 at 1:15
• @underdog Find an answer to the question: A million light years long formation of spaceships changes its speed from 0 to 0.87c in one year. How much does it contract? The answer "half million light years" would be very wrong. But I guess you'll never understand why density of spaceships does not double, or why density of electrons does not increase when current is turned on. Sorry. Well, it seems to be impossible to understand to almost everyone. Nov 2, 2019 at 10:26
• @stuffu why not just point out where or how I can find the answer to your question instead of presuming I won't understand. Nov 2, 2019 at 15:22

How can the wire be neutral in both scenarios?

This question usually comes up in the context of relativity. It stems from an interesting but confusing explanation pioneered by Purcell:

http://physics.weber.edu/schroeder/mrr/MRR.html

However, despite the relativistic context this specific question is actually not related to relativity at all. Both the charge on and the current through a wire is under experimental control.

A wire has self capacitance and resistance. By raising the voltage on both ends of the wire we can take advantage of the self capacitance to control the net charge on the wire. By increasing the voltage difference across the wire we can take advantage of the resistance to control the current through the wire.

So the experimenter can make the wire have no charge and no current, or charge but no current, or current but no charge, or both current and charge. Relativity is not relevant to that at all. Relativity only comes in once you have fully specified the settings chosen by the experimenter in the lab frame and want to determine what the scenario looks like in another frame. So the lab values are arbitrary “given” quantities that are then used as inputs for the relativistic calculations.

• Very good explanation. Thanks! Nov 3, 2019 at 20:59
• But what if the return path is infinitely far away from the conductor under observation? In any case, my doubt was more about length contraction than anything else really. As someone who has never studied this formally, it's hard to wrap your head around.. Nov 4, 2019 at 3:03
• @underdog changing the location of the return path would change the value of the capacitance of the wire but would not otherwise alter my answer. The point is that turning on the current does not determine whether the wire is charged or not, that can be separately controlled.
– Dale
Nov 4, 2019 at 3:44

Suppose your wire is a straight one between the terminals of a battery. Electrons enter the wire at one end and leave at another. Let's assume the electrons are point particles and that wire is very narrow and that the current is such that one electron follows the next down the wire. Because the electrons are moving relative to you, the gap between them, as measured in their frame, is longer than the gap between them as measured in your frame.

This is just the 'pole in a barn paradox' in another guise. Simultaneity in the frame of the electrons will be different to simultaneity in yours, so what will happen is that in your frame extra electrons will seem to have entered the wire before all the others have left it, thus preserving the neutrality of the wire.

• But how can we disagree on an events occurrence? If extra electrons enter in my frame, must it not happen in all frames at some time? Also, how does the addition of extra electrons counteract the negative charge density at a part of the wire due to length contraction? Nov 1, 2019 at 14:33
• Sound reasonable for a straight wire, but what if a bend it so the "now" at the 2 terminals is the same for both the positive lattice and the moving electrons?
– JEB
Nov 1, 2019 at 15:49
• @JEB - Now you have the twin paradox in another guise. Nov 1, 2019 at 16:06
• Yes, and to make matters worse, from the point of view of the electrons, the lattice of ions is length contracted, so the charge density problem applies in the opposite way. What a muddle we're in! But never fear, Marco is here! I will give it more thought. Nov 1, 2019 at 16:09

Step one is simplify. So we have a 1 dimensional lattice of equally spaced ($$s$$)positive and negative charges, labeled by position $$n$$:

...0...1...2...3...4...5

The position of the n-th positive ion vs. time is:

$$x^+_n(t) = ns$$

The electrons all start moving simultaneously (in the lattice frame), and have positions:

$$x^-_n(t) = ns + vt$$

The difference in position gives the linear charge densities:

$$\Delta^+(t) = x^+_{n+1}(t) - x^+_n(t) = s$$ $$\Delta^-(t) = x^-_{n+1}(t) - x^-_n(t) = \Delta^+(t) + (vt-vt)=s$$

So they are equal, and the wire is neutral.

Boosting to the moving frame ($$v$$) with the electrons:

$$x'^+_n(t) = \gamma (ns -vt)$$

$$x'^-_n(t) = \gamma (ns +vt -vt) = \gamma ns$$

Now get the right time coordinate:

$$t = \gamma(t' + vx'/c^2)$$

so

$$x'^-_n(t) = \gamma ns$$ $$x'^+_n(t) = \gamma (ns -v\gamma(t' + vx'/c^2)) =\frac{ns}{\gamma}-vt'$$ are the world lines in the electron frame.

The electrons are stationary (and further apart, which is shown in the Science Asylum video at 6:03), while the protons move to the left, and are contracted.

Here is a plot of 5 ions and 5 electrons moving to the right at $$\gamma=2$$: Note that in the lab frame the densities are:

$$\rho_{\pm} = \pm 1$$

The magenta line is $$t'=0$$, aka the $$x'$$ axis, with uniform symbols at integer values of $$x'$$. It is easy to see that:

$$\rho_{\pm} = \gamma^{\pm 1} = 2^{\pm 1}$$

Now I can boost that plot to the electron frame (so the magenta line becomes the abscissa): That explicitly shows the electron density is reduced by $$\gamma$$, and the ion lattice contracted by $$\gamma$$.

• When you are finding the position of moving electrons from the lab frame shouldn't the term (ns+vt) be divided by γ ? Also what is the X' axis supposed to represent? Nov 3, 2019 at 11:53
• If you move at $v$ your displacement versus time is $vt$. A $\gamma$ would make your velocity $\gamma v$, and that would exceed $c$ at some point, and lead to the conclusion that another $\gamma' = 1/\sqrt{1-(\gamma v)^2}$ is in order, and that's no good. $X'$ is the hyperslice of simultaneity in the electron frame, with unit gradations (where a factor of $\gamma$ is appropriate). All you need to do to get the (inverse) density is count the number of red and blue lines crossing it between 2 magenta circles.
– JEB
Nov 3, 2019 at 16:18
• yes, I get it now, from the ion's perspective, the line of simultaneity are just parallel to the 'lab x' axis and thus the density is same. But in the electron frame the line of simultaneity is stretched out so it counts double ions. I guess, my only question that remains is why there are no relativistic effects when all the electrons start moving in the lab frame? Nov 3, 2019 at 16:49
• The electrons are not a rigid body. Each is on its own separate path. The $\gamma$ appears in the electron frame, where their density is reduced by $\gamma$, so transforming back to the lab frame, they pick up a $\gamma$ and get back to 1.
– JEB
Nov 3, 2019 at 16:58
• In relativity we must speak about 4-vectors. So, when electrons are stationary we have electrostatic field. When they start moving we have magnetic field also. So this is actually a relativistic effect. You could see it if you imagine a wire that is stationary and the wire that is moving. Moving wire actualy is length contracted and so we see change of density. This change of density could be called magnetic force.... Nov 3, 2019 at 19:03