# Relativistic explanation of attraction between two parallel currents

This has been answered a lot of times, and I have been reading about it on many websites, including PSE, but I still don't get it. So please don't mark it as a duplicate unless it really is one.

Two parallel wires at rest don't attract or repel each other because both of them have roughly the same amount of electrons and protons, so their net charges are mostly zero.

If there's a current (in both wires, same direction), the attraction between the wires can be calculated

1. using Ampère's force law
2. using the Maxwell equations (the current creates a magnetic field)
3. using Special Relativity

I want to understand the third case.

Take the reference frame of the moving electrons. Due to relativistic length contraction, the protons get closer together, so the number of protons per length unit is higher than that of the electrons. So far so good.

At this point, all answers to this question just go like

Because wire 2 has more protons than electrons (per length unit), the electrons in wire 1 are attracted to wire 2, so the two wires attract each other.

But wire 1 has more protons too! Both wires become effectively positively charged due to the movement of the electrons. Because both wires are positively charged, they should repel each other.

Yes the electrons in wire 1 are attracted to wire 2 because of wire 2's net positive charge. But the protons in wire 1 are repelled from wire 2 because of the same reason. And there are more protons (per unit length) in wire 1, so the repellent force is stronger than the attraction.

So, again: Why do the wires attract each other?

EDIT: Maybe it has to do with the fact that the electrons can "re-arrange" themselves to counterbalance the positive charge? But then both wires would again have zero net charge.

EDIT 2: Here are some PSE posts and websites that try to answer this question:

Relativity and Current in Wire

However, as I said, they don't go into my interpretation that both wires should have the same net charge and thus should repel each other.

• re: So please don't mark it as a duplicate unless it really is one. Just to head possible closures off at the pass, perhaps you should point out how your question is different from the others, if you yourself think there are duplicates. Best of luck with it.
– user81619
Commented Oct 22, 2015 at 11:52
• @count_to_10: I think I did that. If you read the quotation in the middle of my question, that is how all the other answers go. But as I write in the following paragraph, I can't see why this is true. How could I point out the difference better? Furthermore, I didn't say there were duplicates. I just think the likelihood that someone marks it as a duplicate before reading it in detail is high.
– Bass
Commented Oct 22, 2015 at 12:11
• @BastianTreichler - I recommend you include a link to the "this is not a duplicate" questions / answers. Commented Oct 22, 2015 at 12:13
• @Floris: See edits.
– Bass
Commented Oct 22, 2015 at 12:23
• Why do you say "But wire 1 has more protons too!"? What is your logic? Commented Oct 22, 2015 at 12:30

As far as your comment goes, you mean there is an absolute symmetry between the 2 wires. Maybe, but one thing I must tell you that when you are considering the electrons in WIRE 1, the relativistic effects will be as follows:

1. The electrons in their reference frame will consider the protons IN WIRE 2 to be in motion.
2. Then due to relativistic length contraction, the electrons will observe that WIRE 2 has a higher positive charge density. So the electrons will face more Coulombic attraction from the WIRE 2 than repulsion from the electrons in that wire.
3. Most importantly, where I think you are going wrong, the electrons in WIRE 1 will NEVER see the protons in WIRE 1 to be MORE as you say. That is, electrons in WIRE 1 will see that WIRE 2 has a higher positive charge density than WIRE 1 always due to special relativity and nothing else. On the other hand, in a similar fashion, electrons in WIRE 2 will see that WIRE 1 has a higher positive charge density than WIRE 2 always due to special relativity and nothing else.

Hope your doubt has been resolved.

• Well I think I'm starting to understand it. However, I'm not sure if I can follow your third point. Why don't the electrons in WIRE 1 see more protons in WIRE 1? In their reference frame, the protons in WIRE 1 are moving exactly the same way as they are in WIRE 2.
– Bass
Commented Oct 22, 2015 at 14:38
• I have a proposal for a new third point in your explanation: 3. The electrons in WIRE 1 see more protons in both wires. The coulomb forces from the protons in WIRE 1 cancel each other out, because the electrons in WIRE 1 are between the protons of WIRE 1, so their net force is zero. The coulomb force of the protons in WIRE 2 attract them. The thing I got wrong: The protons in WIRE 1 must be considered in their reference frame. There, the protons are at rest and the electrons are moving, so the protons see more electrons, so they are attracted too.
– Bass
Commented Oct 22, 2015 at 14:42
• How do you type so fast? I barely finished writing the same thing. Anyways your reasoning is partially right. However I couldn't get what you meant by "The coulomb forces from the protons in WIRE 1 cancel each other out, because the electrons in WIRE 1 are between the protons of WIRE 1, so their net force is zero." One point I would like to add is that if the electrons in WIRE 1 see more protons in WIRE 1,then it will seem the wire has length contracted implying higher positive charge density and similarly higher negative charge density too. Which is a contradiction. Commented Oct 22, 2015 at 14:50
• (Contd.) By relativity, electrons cant say from their own reference frame that a property due to them has increased without themselves increasing. Take an analogy. A train is going by and I am standing still near the rail tracks. A friend is on the train. From my ref frame, I am still and I have no K.E. But from my friend's ref frame, the train is still and I have K.E. But he will never be able to measure my velocity for which I have K.E. nor will I ever claim that yes I have K.E., the train is still. Commented Oct 22, 2015 at 14:56
• "The coulomb forces from the protons in WIRE 1 cancel each other out, because the electrons in WIRE 1 are between the protons of WIRE 1, so their net force is zero." I mean that as long as everything is distributed homogeneously in the wire, there are no forces on the electrons from the wire they're in. This is clear from simple symmetry arguments.
– Bass
Commented Oct 22, 2015 at 15:05

I realize this post is a bit old, but I had this exact same question for quite a long time and never dedicated any time to figuring it out until now. I'd like to take a stab at explaining the discrepancy you've identified (which is the same issue I'd been having). Disclaimer: I'm not a physicist but only an electrical engineer with limited education in physics.

First, the electric and magnetic fields are different sides of the same coin, as seen from different reference frames. I won't go into detail, but there is a "legit" reference here from the University of Virginia.

Perhaps the results in some examples can have the forces be cleanly divided by magnetic or electric fields based on the reference frame you're considering, while in other cases, such as the one you brought up, both forces are needed to explain the phenomenon in certain frames.

I realize this conclusion may conflict with @SchrodingersCat's, but as I understand it, if you want to look at the problem from the electron frame of wire 1, you will in fact see an increased positive charge density in wire 1 as well as wire 2, and the repulsive Coulombic electric field force applies.

If you then consider, in the same reference frame, the magnetic field observed by the motion of the positive charges in wire 1 and 2, you can explain an attractive force due to motion of charge coupled with magnetism. I assume the result of a net attractive force can be explained by the math, where the magnetic effects overcome the electric for the electron frame of reference - but alas that calculation is beyond me.

Again, I'm not an expert, so if someone sees a flaw in my explanation then please correct me!

• Welcome to Physics.SE! I suggest the following: 1) Take the tour! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. 3) If you have a good question, ask it! Just search for duplicates and follow the help center rules. Commented Aug 26, 2016 at 22:45

This s a very good question. A classical reference to this problem is Chapter 5.9 "Interaction between a moving charge and other moving charges" of Purcell: Electricity And Magnetism. Surprisingly he also doesn't discuss the forces on the protons although he is usually very explicit.

If you start looking at the protons in the moving frame, you will see their magnetic fields and resulting Lorentz forces. These are such, that what's actually happening in both frames is the same. So if there was no net force on the protons in their rest frame, there will also be no net force in the moving frame. So the forces from new $$E$$ fields must be canceled by Lorentz forces from new $$B$$ fields. Here the protons in one wire feel an attractive Lorentz force from the protons and electrons in the other wire. The result is the same as in their rest frame, were they see a neutral wire, which is neither attractive nor repulsive. So the protons don't contribute a net force between the wires.

"Riding on an electron of Wire 1" you see the prevalence of positive charge on Wire 2. More positive charge there, per unit length, due to Lorentz contraction. So YOUR electron, the one you ride on, get's attracted. This is as it should be.

You do not worry, riding on your electron, what the positive charges on your wire (Wire 1) see themselves. YOU are attracted to Wire 2. To check from OUTSIDE, moving frame of reference, what forces are "felt" by positive charges in your wire, you have to use full EM theory. ALTERNATIVELY, you switch to a point of view (frame of reference) of a positive charge on your wire. Then you feel the attraction again.

This argument is often used to support the idea that magnetism is a relativistic effect. However, it only works for parallel currents. The currents must also be carried by identical charges moving at the same as speed in the same direction. The positive background is a complication. My opinion is that this argument is only confusing.

@UtopianParadox answer seems correct to me, so I would like to accompany similar reasoning with some math (not rigorous, and I'm not sure about some of the steps, but at least these two wires attract each other in both inertial frames).

Also, a very similar situation is considered in this Feynman lecture (13–6The relativity of magnetic and electric fields), where the interaction of a charged particle and a current-carrying wire is described in two inertial frames.

According to the principle of relativity physical laws should be the same in every inertial frame of reference. It follows that if Ampère's force law is valid in one inertial frame it should remain valid in any other inertial frame. When we take an inertial frame in which the electrons are at rest, we still have two electric currents flowing in the same direction, thus the wires must attract each other according to Ampère's force law. It turns out that this attraction is stronger than in the inertial frame where the electrons are moving, and it is stronger for the same reason that these wires receive a net positive charge. So, in one inertial frame there is only an attractive force, and in another inertial frame there are both attractive and repulsive forces, but it is not clear which one is stronger.

Let's first consider an inertial frame $$S$$, in which the wires (their proton lattices) are at rest, and the electrons in both wires move in the same direction with speed $$v = \beta c$$. Then the electric current flowing through each wire is $$I = \frac{dq_{-}}{dt} = \lambda_{-} \frac{dl}{dt} = \lambda_{-}\beta c \ ,$$ where $$\lambda_{-} = \frac{dq_{-}}{dl}$$ is the linear charge density of electrons in each wire. Assuming that each wire has zero net charge in frame $$S$$, the net linear charge density is $$\lambda = \lambda_{+} + \lambda_{-} = 0$$. In frame $$S$$, according to Ampère's force law, the magnetic force per unit length is $$\frac{dF_B}{dl} = \frac{\mu_0}{2\pi} \frac{I^2}{r} = \frac{\mu_0}{2\pi} \frac{\lambda^2_{-} \beta^2 c^2}{r} \ .$$

Now consider the same wires in an inertial frame $$S'$$, in which the electrons are at rest. As correctly stated in the question, due to relativistic length contraction, the protons get closer together in frame $$S'$$ (protons move in frame $$S'$$ and are at rest in frame $$S$$). The opposite is true for electrons, since electrons move in frame $$S$$ and are at rest in frame $$S'$$. So in frame $$S'$$, the net linear charge density is $$\lambda' = \lambda'_{+} + \lambda'_{-} = \gamma \lambda_{+} + \frac{\lambda_{-}}{\gamma} = \lambda_{+} \left(\gamma - \frac{1}{\gamma} \right) = \lambda_{+} \beta^2 \gamma \ ,$$ where $$\gamma = \frac{1}{\sqrt{1 - \beta^2}} \ ,$$ and there is the electric field due to line charge $$E' = \frac{\lambda'}{2\pi \epsilon_0 r} = \frac{\mu_0}{2\pi} \frac{\lambda' c^2}{r}$$ (here we used the relation $$c^2 = \frac{1}{\mu_0 \epsilon_0}$$). The electric force per unit length is $$\frac{dF'_E}{dl'} = \lambda' E' = \frac{\mu_0}{2\pi} \frac{\lambda^2_{+} \beta^4 c^2 \gamma^2}{r} = \frac{dF_B}{dl} \beta^2 \gamma^2 \ .$$ Since in frame $$S'$$ the electric current in each wire is $$I' = \lambda'_{+} \beta c = \gamma \lambda_{+} \beta c$$, the magnetic force per unit length is $$\frac{dF'_B}{dl'} = \frac{dF_B}{dl} \gamma^2 \ .$$ In frame $$S'$$, the electric force is in the opposite direction to magnetic force, so the net force per unit length is $$\frac{dF'}{dl'} = \frac{dF'_B}{dl'} - \frac{dF'_E}{dl'} = \frac{dF_B}{dl} \gamma^2 (1 - \beta^2) = \frac{dF_B}{dl} \ .$$

This may be oversimplifying things... but if you are moving in the frame of reference of the electrons, the (counterbalancing) positive charges (lattice) will appear to be moving the other way. This is in effect a positive current in the opposite direction, providing the same force as before. No relativity required there.

You may need to adjust your question to address "electrons in free space" if you want to get past that point.

• I think you are indeed oversimplifying. I'm asking about the attraction of the wires, which can be explained using special relativity. It does not work for electrons in free space.
– Bass
Commented Oct 22, 2015 at 13:13