If an observer is stationary relative to a current-carrying wire in which electrons are moving, why does the observer measure the density of moving electrons to be the same as the density of electrons if there were no current in the wire?

I read the explanation of magnetic force as a consequence of special relativity. That is, when the observer moves with respect to a current carrying wire in the same direction as the flowing electrons, then s/he observes the density to decrease due to Lorentz expansion and observes the density of positive ions to increase due to Lorentz contraction. The imbalance of charges results in a force that can be explained by Coulomb's law.

That explanation mentions that when the observer is stationary, the density of positive ions and the density of electrons appears to be the same. That last point is what I am stuck on.

Why does the density of moving electrons appear to be the same as the density of stationary positive ions?

Clarification: I know that when the electrons are not moving, their charge density cancels out that of the stationary positive ions. When the electrons are moving, how could the charge density still cancel out that of the stationary positive ions?

Important: My assumption is that a current carrying wire will neither attract nor repel a charge that is stationary with respect to the wire. I am now not sure if that is right :-(

  • $\begingroup$ This is just saying that there is no net charge within the wire. $\endgroup$
    – Will
    Jul 14, 2013 at 22:35
  • $\begingroup$ @Will the question is why does there not appear to be a net charge, given that the positive charges are stationary while the negative charges are moving? $\endgroup$ Jul 14, 2013 at 22:44
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    $\begingroup$ I see your problem. It seems that it is an assumption of the derivation that the total charge density is zero in the frame of the wire. I can write more on this if you like (but it'll have to be later) :) (note that this would mean that the density of electrons in the electron rest frame is different) $\endgroup$
    – Will
    Jul 15, 2013 at 0:22
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    $\begingroup$ @randomstring, a current carrying wire in a circuit may indeed have a net charge. But, in the context of this problem, this is a distraction. We're considering an ideal, zero resistance wire with a constant current and zero net charge density in the reference frame of the wire. $\endgroup$ Jul 15, 2013 at 1:50
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    $\begingroup$ A good question, and the reason I think using neutral wires can be a distraction. <shameless self-promotion> I wrote an explanation of the same phenomenon that does away with the complication of having two charges here. </shameless self promotion> $\endgroup$
    – user10851
    Jul 15, 2013 at 2:51

7 Answers 7


Model: Let's simplify the model of a current in a wire, so we can be definite about what we are talking about. Take a wire (in the wire's frame) to have fixed positive charge density $\rho_{+}$ and assume the electrons at rest w.r.t the wire, with electron density $\rho_{-}$.

Introducing a current sets these electrons moving at some speed $v_\rm{drift}$ w.r.t wire, but leaves the positive charge fixed. We ask the following question:

What is the relationship between $\rho_{-}$ (the electron charge density at rest), and the electron density with current?

Answer: The density measured by the observer stationary w.r.t a current carrying wire is not the same as if the charges were stationary. They are related by a Lorentz transformation. Let's write the 4-current of the electrons when at rest, and when moving (with $c=1$): $$J^\mu_\textrm{rest} = (\rho_{-},\vec{0})^\mu,\,\,\,\,\,J^\mu_\textrm{moving} = (\tilde{\rho},\vec{j})^\mu = {\Lambda(v)^\mu}_\nu J^\nu_\rm{rest}$$ where $\Lambda(v)$ is the Lorentz Boost between these two frames. Note in particular that $\boxed{\rho_{-} \neq \tilde{\rho}_{-}}$ because $$J^2_\textrm{rest} = J^2_\textrm{moving}~~\implies ~~ \rho_{-}^2 = \tilde{\rho}^2_{-}-\vec{j}.\vec{j},$$ and $~\vec{j}\neq \vec{0}$.

This means that when you set up your problem, we have two possible scenarios:

$(i)$ $\rho_{-}+\rho_{+} = 0$, that is we ask that the electron density in the electrons rest frame has the same magnitude as the positive charge density in the stationary wire.

$(ii)$ $\tilde{\rho_{-}}+\rho_{+} = 0$, that is we ask that the electron density in the wire's rest frame has the same magnitude as the positive charge density in the stationary wire. This is the situation of zero force on a stationary external charge you talked about in your edit.

So the question you have to ask yourself, is what situation do you want to deal with? It seems that for the "explanation of magnetic force as a consequence of special relativity" you are interested in, one should consider case $(ii)$ as this allows you to see how a test charge, moving parallel to the wire with velocity $v$, experiencing a force due to a pure magnetic force in one frame (wire rest frame) $F = q v\times B$, is the same force experienced by the charge in its rest frame, effected only by the electric force, $F = q E$, in that frame (as in this frame it isn't moving).

I hope this helps. If you need further explanation, don't hesitate to ask.

  • $\begingroup$ Thanks Will. I can't see how (ii) could happen, because a potential difference would squeeze electrons together in the electron frame of reference, which in turn would make them appear to be denser than postive ions in the wire frame, resulting in a force even though test charge is not moving. It is difficult to "assume away" the compression (of moving electrons in electron frame due to potential difference). ... For details, please see 2nd and 3rd comments on @AlfredCentauri answer. $\endgroup$ Jul 15, 2013 at 22:00
  • $\begingroup$ I'm explaining what we do under a simplified model. If you want to get technical, there is no "electron frame", as the electrons are all jittering about throughout the metal with some average drift velocity. I believe it is likely that the book you were reading was using the simplified model I am talking about. The whole point of the model was to see how a magnetic field produced by a current in one frame is equivalent to an electric field in another frame. Which is what I believe you were asking about? Correct me if I'm wrong. $\endgroup$
    – Will
    Jul 15, 2013 at 22:29
  • $\begingroup$ electron frame = drift velocity frame. Your most recent comment above sounds like it is right. In lab frame the electrons density appears greater than positive ion density, so test charge experiences force. In drift frame, the positive ion density exceeds electron density, so test charge experiences force. Somewhere in the middle, as you say, the densities are equal and we can think of this as the original reference frame to understand where magnetism comes from. As we start drifting from this frame we start feeling a force, which is what elementary textbooks call magnetic force. Correct? $\endgroup$ Jul 15, 2013 at 22:55
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    $\begingroup$ I am still confused because the answers seem to want me to "just assume something" and may be I cannot properly express why it is difficult to just assume it. I'm still figuring this out. $\endgroup$ Jul 19, 2013 at 22:54
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    $\begingroup$ @randomstring My understanding is that for the purpose of this problem, the battery creates an electric field in the wire, so all electrons get pushed/pulled equally, so it's kinda like a train rather than a queue of people moving, if you know what I mean. As for your other comment, yes, I am trying to understand what everyone else is saying...could take a while haha $\endgroup$
    – mehfoos
    Jul 19, 2013 at 23:05

When the electrons are moving, how could the charge density still cancel out that of the stationary positive ions?

Recognize that the fixed charge is, well, fixed, while the mobile electrons are, well, mobile. This is the crucial difference here. The mobile electrons are free to "squeeze together" or "push apart" but the fixed charge is not.

Regardless of whether the electrons have an average drift velocity or not, with respect to the fixed charge of the wire, the mobile electron density may be such that the wire has net charge or not.

For example, connect a wire to one terminal of a battery. The wire is unconnected at one end so, in steady state, there is no current through the wire. But, it is certainly the case that the wire has net electric charge.

So, for this problem:

we choose the condition to be that, in the reference frame of the lab in which the constant current carrying wire is at rest, the wire is electrically neutral.

Then, the question is, in another reference frame relatively moving with respect to the lab (and thus, the wire), is the wire electrically neutral?

The answer is: in general, no.

In the lab frame, the wire is neutral, we necessarily have that the fixed (positive) charge density equals the moving (negative) charge density.

Now, for example, in the frame of reference in which the negative charge moving in the lab frame is stationary, the density of the moving charge is decreased. This is because, in the lab frame, lengths in the reference frame of the moving charge are contracted. So the negative charge density decreases.

Also, in the same frame, the fixed charge density is increased since, in this frame, the wire is contracted. So, the positive charge density increases.

It follows that, in this frame, the wire has a net positive charge density.

Now, we don't have to choose the wire to be electrically neutral in the lab frame. But then the question becomes: in another reference frame relatively moving with respect to the lab (and thus, the wire), is the wire charge density the same?

Again, the answer is, in general, no.

Imagine two identical, ideal wires with identical constant currents. In the lab frame in which both of these wires are at rest, one of these wires has zero net charge and the other wire has non-zero net charge. Do you accept or reject that this is possible?

I think this is not possible, but I will try to see why you might suggest this is possible.

Consider the following simple circuit at rest in the lab frame.

enter image description here

The "top" wire has a net charge density. The "bottom" wire does not. The current through each wire is identical.

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    $\begingroup$ +1 thank you. i will think this over. it sounds like a promising explanation. $\endgroup$ Jul 15, 2013 at 2:29
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    $\begingroup$ @randomstring, see the addendum to my answer. $\endgroup$ Jul 15, 2013 at 23:52
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    $\begingroup$ I see. In one wire the charges are backed up because of the battery push and the resistance bottleneck. In the other wire they are spread apart further. Thank you so much for the kind diagram + explanation. I think I am on the verge of enlightenment. Any moment now! :D $\endgroup$ Jul 15, 2013 at 23:56
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    $\begingroup$ @randomstring, I think you're making this far too hard. There is a frame, relatively moving with respect to the lab, in which the top wire is neutral and the bottom wire has a net charge density. I honestly am having a difficult time "seeing" what it is you're not comfortable with. I am now baffled that you're still not putting the pieces together. $\endgroup$ Jul 16, 2013 at 0:41
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    $\begingroup$ @randomstring, I confess I don't really understand your comment. If a test charge "feels" a force in the lab frame, it "feels" a force according to relatively moving frames. What changes is the mixture of the electric force and magnetic force. In one frame, it may be all electric. In another, all magnetic. In others, some mixture. I'm sorry that I haven't been able to pinpoint the precise misconception that is leading you astray here. $\endgroup$ Jul 16, 2013 at 0:55

Re your statement:

"That explanation mentions that when the observer is stationary, the density of positive ions and the density of electrons appears to be the same."

This part of the explanation looks to be incorrect (or at best misleading).

The problem is most easily set up by assuming a neutral wire (canceling + and - charge densities) with current in a "lab" frame, and an "observer" charge moving parallel to it. Once you boost to the observer's frame, the wire charge densities no longer cancel, and the observer feels a Coulomb force (contra that statement I quoted).

I think demonstrating that effect is the point of the other answers here: if you start with a neutral current-carrying wire and then boost, the wire is no longer neutral.

Now you could start with a charged wire with current, such that the wire looks neutral in the observer's frame, (like the statement says). But then there's no force on the observer charge: In the lab frame, the magnetic force cancels the coulomb force, and in the observer's frame, there's no net charge to attract it. (i.e. Your important assumption is an important correct assumption.)

  • $\begingroup$ +1 (congrats on 2k!) digesting your answer. Is this comment on @Will's answer consistent with yours "In lab frame the electron density appears > than positive ion density, so test charge experiences force. In drift frame, the positive ion density exceeds electron density, so test charge experiences force. Somewhere in the middle, as you say, the densities are equal and we can think of this as the original reference frame to understand where magnetism comes from. As we start drifting from this frame we start feeling a force, which is what elementary textbooks call magnetic force. Correct?" $\endgroup$ Jul 15, 2013 at 23:00
  • $\begingroup$ Thanks for the congrats. In the lab frame, the simplest thing is to assume the charges cancel, so there's only a current (i.e. your "original" frame). After all, the point is to see that a moving charge is attracted to a current, even if the current has no net charge. You can analyze (and set up) other possibilities with non-zero charge in the lab frame (i.e. putting the wire at +100KV or some such), but that just complicates things, since you have to tease out the effects of the non-zero charge from the current. The simplest case is to have the lab wire uncharged (lab= "original"). $\endgroup$
    – Art Brown
    Jul 15, 2013 at 23:07
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    $\begingroup$ If the observer is moving in the "original reference" frame, there's a magnetic force on it (but no electrostatic force). As you boost away from that frame, the mix of magnetic and electrostatic forces changes. $\endgroup$
    – Art Brown
    Jul 15, 2013 at 23:16

I think a good deal of confusion here comes from the fact that the wire is infinite, and so in some real sense the charge doesn't have to be conserved. You have an infinite number of negative charges and an infinite number of positive charges. The net charge on the wire is $\infty-\infty$, which is an indeterminate form that can equal any value.

So why then do we assume that a wire that is neutral before is neutral after? Because an infinite wire is not a physical configuration. Ultimately any real circuit is a finite loop. And since a finite loop has a finite number of charges, the net charge has to remain zero (in every reference frame, in fact, although the rest frame has the additional condition that the charge density is uniformly zero in the steady state).

So how is it possible that in the infinite case the charge density remains zero in one frane and not another? As with so many things in relativity, it comes down to simultaneity.

To change the current from zero to $I$, we apply a uniform voltage gradient simultaneously across the whole wire and remove it when the current reaches the desired value. But of course in another frame this gradient is not simultaneous, so the current is greater in parts of the wire than others. This causes a build up of charge across the wire.

Note that conservation of charge was not violated! In a global sense, the charge was initially in the indeterminate form $\infty-\infty$, and it still is afterwards. In a local sense, the charge buildup was because of conservation of charge: the continuity equation gives $${d\lambda\over dt}={dI\over dx}$$

Or, in words, the gradient in the current causes a change in the charge density of the wire.

  • $\begingroup$ thanks. i still don't understand this, and so haven't accepted an answer. but that's not because the answers are not good but because i haven't set aside the time to grok this. as soon as i do so (or if some answer gets highly upvoted) i'll accept. thanks for your clarification about the simultaneity; i kinda feel skeptical about \inf - \inf because instead of integrating positive and negative separately i could integrate the net charge over small segments of the wire. ignore me if i got it wrong. i'll fully understand this. it's on my to-do. $\endgroup$ Apr 22, 2018 at 10:54
  • $\begingroup$ It's right to be skeptical- in this case, simply integrating over the overall charge density seems to give a physically reasonable result. Infinities in math are weird, though. Even if initially there's one electron per proton in a volume element, just by rearranging them you can make it so there's twice as many electrons as protons. It's related to the Infinite Hotel paradox. It's basically just a consequence of the fact that $2\infty=\infty $. $\endgroup$
    – Chris
    Apr 22, 2018 at 17:53
  • $\begingroup$ @chris There is a current carrying wire. Test charges $q_1$, $q_2$ where $q_1$ sees the density of electrons in wire to be $\lambda_1$ and is stationary relative to wire, and $q_2$ sees electrons in wire to have density $\lambda_2$ and is moving relative to wire. We have $\lambda_1 \unequal \lambda_2$ and the reason for this is given like, relative velocity of electrons in wire causes length contraction. But if we measure electron density in wire for different currents relative $q_1$, we cannot use the reason of length contraction to say that electron density is reduced or increased. $\endgroup$
    – James Webb
    Mar 27, 2022 at 12:01
  • $\begingroup$ @JamesWebb I'm not sure what your point is. Of course you can't derive the results for two wires of unequal currents using just length contraction. Two wires with different amounts of current in them are entirely different physical situations, not related to one another by a simple Lorentz boost. $\endgroup$
    – Chris
    Mar 27, 2022 at 21:13

Simplify the problem. Start with the effect of a moving negative charge on a stationary negative charge. We know the effect of a stationary positive charge on the stationary negative charge.

We have a correct formula for this, which is compatible with relativity.

In the frame where the target charge is stationary, and there is no acceleration, and the source velocity is perpendicular to the direction to the target, the formula simplifies considerably.

$E(r,t) = K\frac{(1-\beta^2)(n-\beta)}{R^2} $

where $K$ is a few lumped fudge factors,

$n$ is the unit vector in the direction from the source when the force left it, and the target when the force arrives,

$R$ is the distance between the source when the force left it, and the target when the force arrives, and

$\beta$ is the source velocity when the force left it, divided by lightspeed.

The force is in a different direction! $n-\beta$ instead of $n$.

Also the vector sum $n-\beta$ has magnitude $\sqrt{1+\beta^2}$ so the total magnitude of the force is $\frac{(1-\beta^2)\sqrt{1+\beta^2}}{R^2}$ compared to $\frac{1}{R^2}$ when both are stationary.

So the total force is larger, but the force in the original direction is smaller.

Why do people not usually measure this? Because individual electrons may move very fast, but mostly they move in random directions and the effects cancel. The average drift velocity is very small, so the change in direction is very small and the change in magnitude also.


EDIT: I guess the reason I got down-voted was because the answer seemed to imply that any current carrying conductor would have net charge 0. I do not claim that. I presume that we start off with a neutral wire, because having it charged does not help illustrate anything relevant for this question. Also: Readers might get the impression from the answers that Charge Conservation does not apply here. Well, I think that it does. The Current density would have to be replaced by the 4-current version.

EDIT2: Does not apply to either wire in Alfred Cnaturi's diagram. This answer applies to a current carrying wire that is neutral in lab/conductor's frame of reference. I think that that is a fair assumption to answer this question.

If I understand your question correctly, then I would have to repeat Will's answer: There is not net charge in the wire.

Why? Because although the electrons (negative charges) are moving whilst the positive kernels are fixed, since there is a closed circuit in place, they are being replenished by new electrons, regardless of the section you are observing. (Just realised that bit is wrong if you ignore EDIT2 assumption, thanks to Alfred Centauri's excellent link)

Therefore the charge density cancels out that of the positive ions. This much does not require any knowledge of special relativity, and I am sure that this is not solely what you are after.

So, in the case of a moving frame (that of the electron, for instance), the charge density does change to that of rest frame (rest with respect to conductor). But then there has to be conservation of charge! Well, this change in charge density is due not due to accumulation of charge but rather the relativity of simultaneity affecting the measurement of charge density in the moving frame. It is also important to note that the distance between the charge particles that expand or contract does not change, unlike the popular illustrations in books, instead the charge particles undergo the change in dimension.

  • $\begingroup$ Confused... I understand charge density is zero when the electrons+observer are stationary (<-- case 1), and somehow it is also zero when the electrons are in motion but observer is stationary (<-- case 2). BUT, when the observer moves with the electrons, then the density appears changed (<-- case 3). I cannot reconcile the last case with the first two. ..... i. e. why is observer motion not symmetric to electron motion? $\endgroup$ Jul 15, 2013 at 0:44
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    $\begingroup$ 1) The fact that, in steady-state, there is 0 net flux of electrons into a section does not imply that the net charge in that section is 0. It could be anything. (The wire could be sitting at 100kV with respect to ground, for example.) 2) Conservation of charge means that the measured individual electron charge does not change with its velocity. $\endgroup$
    – Art Brown
    Jul 15, 2013 at 2:06
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    $\begingroup$ I could be wrong, but I'd like to clarify things for the benefit of others and myself: 1) I understand that charge of the wire depends on what we started off with. But assuming we started off with neutral wire: it is actually carrying current, and so the analogy to 100kV wrt ground does not apply here, right? If the conductor had a net charge, then a test charge (positive or negative) next to it will not experience a force due to the wire. 2)Agreed. I did not mean to imply otherwise. OP, can you please clarify what you mean by 'observer motion is not symmetric to electron motion'? $\endgroup$
    – mehfoos
    Jul 15, 2013 at 13:10
  • $\begingroup$ @randomstring Sorry, I keep pressing enter instead of 'shift'+enter. Anyway, just realised that I did not respond to your comment here. So you are happy with case 1 and case 2? case 3: From electron reference, there is length contraction on the positive charges, which causes change in the charge density of electrons wrt charge density of positive charges. Does that answer your question? $\endgroup$
    – mehfoos
    Jul 20, 2013 at 0:09

The following is a relativistic take on a current in a wire. And it might not be necessarily an answer to your question but I'm adding it because everyone else seems to think it's relevant.

If you do an experiment where you measure the force on a stationary charge in the vicinity of a current-carrying wire you will measure that there is none. From this you can deduce that the net charge in the wire is zero.

In a wire carrying current the positive ions are bound and stationary so we'll take their frame as our stationary frame of reference. The linear charge density in the wire in the ion frame is then: $$ \lambda=\lambda_-+\lambda_+=0$$ and the linear electron charge density is: $$ \lambda_-=-\lambda_0$$ Now let's move to the electron frame of reference. In this frame the linear electron density must be: $$ \lambda'_-=-\lambda_0/\gamma$$ This is simply follows from what we observe in the ion frame. If it were any different the linear electron charge density in the ion frame would be different: $$ \lambda_-=\gamma\lambda'_-=\gamma(-\lambda_0/\gamma)=-\lambda_0$$ Additionally, when we move to the electron frame ions are no longer stationary and move. So their linear charge density changes: $$ \lambda'_+=\gamma\lambda_0$$ Notice that this operation is inverse to that of linear electron charge density as here we are going from stationary to moving ions. The total linear charge density in the electron frame is then: $$\lambda'=\lambda'_++\lambda'_-−=\gamma\lambda_0−\lambda_0/\gamma$$ and this is not equal to zero.

If we have a stationary single electron in the ion frame beside the wire the sum of forces is experimentally shown to be zero. It follows that the same must hold in the electron frame.

In the electron frame the now "new" Coulomb force acting on the single electron has to be counteracted for the above to hold. The solution lies in the now moving ion charges. In this frame they cause the current which produces a magnetic field which now acts on our moving single electron (it was stationary in the ion frame). The forces cancel each other out exactly.

I've included a simplified graphic trying to demonstrate this.Relativistic current analysis

This is the source(1) I was using to answer this question. It's where you will also find actual calculation of the forces in the electron frame and a bit more.

However, your question seems to be about electron density when there is and isn't current flowing.

Unfortunately I do not have a source for this but I will try to answer with what I think is going on. I think the crux of the situation lies in the first 4 equations above. We start with some linear electron charge density: $$ \lambda^{v=0}_-=-\lambda_0$$ which we then speed up when we turn on the power and the current starts flowing. Now because of relativity the distance between electrons shortens resulting in a higher linear electron density from our perspective and the ions': $$ \lambda^{v\neq0}_-=-\gamma\lambda_0.$$ Now the electrons that are stationary and want to join the current see it as too negative for what their potential is. Therefor they don't join in as fast and this results in the whole thing normalizing to: $$\lambda^{v\neq0}_-=-\lambda_0.$$ So while we see the same linear electron density the actual number of electrons is smaller and it only appears not to be because of relativistic effects.

From the classical perspective, you need to look into what current is. Given that experiments show zero net charge, higher current means faster charge movement and not "more-charge" movement. At the moment I don't have the answer to microscopic reasons why this is the way it is. Most likely something to do with: more charge -> higher field -> higher force -> higher terminal velocity -> less charge.

(1) Current-Carrying Wires and Special Relativity by Paul van Kampen
Centre for the Advancement of Science and Mathematics Teaching and Learning &School of Physical Sciences
Dublin City University


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