# Why do wires attract when they carry current in the same direction (in relativistic sense)?

I have seen the veritasium video on this subject (https://www.youtube.com/watch?v=1TKSfAkWWN0), but I have this question which am unable to find the answer for.

I have 2 wires carrying same current in the same direction.

When my 'wire 1' starts carrying current, I understand that the protons would undergo length contraction (riding on the electrons frame of reference) and thus making 'wire 1' positive overall.

When my 'wire 2' starts carrying current, I assume a similar effect would happen and thus making 'wire 2' positive overall.

So, I would expect both wires to repel each other. But since those are actually attracting each other, How is this explained in relativity?

This is actually a great way to look at the magnetic force on a moving charge from a current carrying wire as being a purely electric force on a stationary charge from a wire with a uniform charge density in a different reference frame.

Now when you complicate matters by adding another wire, you really have to keep track of the transformation of sources (charges and currents).

Your mistake was simple: You forgot to include the positive ions' current.

The following derivation relies strictly on the transformations of charges and currents, and at no times uses the Lorentz transform of $$F_{\mu\nu}$$(or $$E$$ and $$B$$ separately). Obviously, if you just Lorentz transform $$F_{\mu\nu}$$ and $$j_{\mu}$$, it is going to work: it is manifestly covariant. If you work in Minkowski space, there will be no apparent contradictions, and no insight into how things breakdown in $$3+1$$ will be gained.

The point of Veratasium's video is that a single configuration of charges and currents (and their fields) can appear very different in different inertial frames.

With $$S$$ ($$S'$$) being the wire (co-moving, at $$\beta$$) electron frame, and $$\rho_0$$ being the rest density of $$\pm$$ charges per unit length, you have in $$S$$:

$$\rho_+ = +\rho_0$$ $$\rho_- = -\rho_0$$ $$\rho \equiv (\rho_++\rho_-)= 0$$ and in $$S'$$:

$$\rho'_+ = \gamma \rho_0$$ $$\rho'_- = -\frac 1 {\gamma} \rho_0$$ $$\rho' = (\rho'_++\rho'_-) = \gamma\beta^2\rho_0$$

So the electric fields from Gauss's Law are:

$$E = 0$$ $$E' = \frac{\rho'}{2\pi\epsilon_0r} = \frac{\gamma\beta^2\rho_0 }{2\pi\epsilon_0r}$$

And the electric force per unit lengths are:

$$F_E = \rho E = 0$$

$$F'_E= \rho' E' = (\gamma\beta^2\rho_0)(\frac{\gamma\beta^2\rho_0}{2\pi\epsilon_0r}) = \frac{\gamma^2\beta^4\rho^2_0 }{2\pi\epsilon_0r} > 0$$

While there is no electric force between 2 stationary current carrying wires, in a frame moving against the current (with the electrons), there is an electric repulsion.

The current densities due to the charge carries in $$S$$ are:

$$j_+ = 0$$ $$j_- = \beta c \rho_-$$ $$j = (j_++j_-) = -\beta c\rho_0$$

while in $$S'$$:

$$j'_+ = -\gamma\beta c \rho_0$$ $$j'_- = 0$$ $$j' = (j'_++j'_-) = \gamma j$$

The magnetic fields from Ampere's Law in $$S$$ and $$S'$$, respectively are:

$$B = \frac{\mu_0 j}{2\pi r} = -\frac{\mu_0 }{2\pi r}\beta c\rho_0$$

$$B' = \frac{\mu_0 j'}{2\pi r} = \gamma B$$

The magnetic force (per unit length on an identical wire at $$r$$):

$$F_B = jB = -\frac{\mu_0 }{2\pi r}\beta^2 c^2\rho^2_0 < 0$$

$$F'_B = j'B' = \gamma^2 F_B$$

so that the magnetic forces are attractive, in both reference frames.

At this point, use $$\epsilon_0\mu_0 = 1/c^2$$ so that:

$$E' = \gamma\beta cB$$

so that

$$F'_E = \gamma^2\beta^3 \rho_0c B$$

The sum of electric and magnetic forces per unit length in $$S'$$ is:

$$F' = F'_E + F'_B$$ $$F' = \gamma^2\beta^3 \rho_0c B - \gamma^2\beta\rho_0 c B$$ $$F' = (\rho_0cB\gamma^2\beta) \times [\beta^2 - 1]$$

and of course $$[\beta^2-1] \lt 0$$ so the total force is attractive.

Addendum: Since Veritasium's video, there has been a lot of discussion of this problem "out there". General points of confusion are:

1. The electrons do not Lorentz contract, rather they dilate a la Bell's Space Ship Paradox (BSSP).

2. The wire remains neutral when current starts not just because of (1), but because it is stipulated in the problem, and the electrons must move on separate trajectories to achieve that. (a la BSSP).

3. The difference in charge density is not just because of Lorentz contraction: the relativity of simultaneity matters too. (See: BSSP).

4. Don't forget about the current. Since $$v'_e = 0$$, it doesn't matter for the original problem, but it does if charges are moving in $$S'$$. In either Science Asylum or Minute Physics video (I forget which), you can see the initial electron's trajectory from purely electric attraction to the wire in $$S'$$ start to bend once it is moving, because there is a magnetic field in $$S'$$.

5. When in doubt: draw a Minkowski diagram and do the Lorentz transformation.

That's why that's a bad description on the magnetic force between two wires. Simple electromagnetic theory, using Ampere's law to find the magnetic field due to one wire, and the law of Biot-Savart to find the force on the second wire gives the correct answer.

• OP's question is about special relativity, not 2 of Maxwell's equations.
– JEB
Commented Mar 17, 2023 at 23:36
• My first sentence is about the misuse of SR. Commented Mar 19, 2023 at 2:19
• What's the misuse?
– JEB
Commented Mar 19, 2023 at 21:31
• "So, I would expect both wires to repel each other. But since those are actually attracting each other," Commented Mar 21, 2023 at 2:05

So we have two positive proton lattices in two wires. There is a huge repulsive force between those lattices.

In the electron frame those lattices move, so the force is smaller, because of time dilation. It's 1/gamma times smaller.

Just like the repulsive force between the mirrors of Einstein light clock caused by the light is smaller when the clock moves.

Also in the electron frame the electron density is reduced so the repulsive force between electrons is smaller too.

These two reductions of repulsive forces are the reason why the wires attract each other.

The magnetic force is real and it isnt a result of Special Relativity.According to the 3rd equation of Maxwell $$\triangledown \times H = J$$,Both currents will create a magnetic field around them and will act as magnetic dipoles.

• They do not act as magnetic dipoles. Commented Mar 19, 2023 at 2:20

Let us stay in the wire frame. We just need to explain the one "magnetic force" that exists in this frame. Said "magnetic force" exists between the moving charges - electrons.

Said magnetic force is actually the time dilation of Coulomb interaction.

So now it is explained.

Explanation was simple, because no length contraction occurred, as another (otherwise really confused) answer explained.

Oh, I forgot one thing: The electric field of each electron length-contracts. But the formation formed by electrons does not length-contract. This causes an additional decrease of Coulomb force. So the Coulomb force gets divided by gamma squared. One gamma comes from time dilation and other gamma comes from length contraction of fields.

It was not quite as simple as I thought, but now it's correct. :)

• The question is about what's going on in the current co-moving frame in which there is clearly electric repulsion.
– JEB
Commented Mar 18, 2023 at 18:46
• @JEB In that frame protons' repelling of each other is reduced from the normal, and electrons' repelling of each other is reduced from normal. Here normal means "as in the wire frame". Right? : ) I mean, I can argue why it's that way, if there's a need. Commented Mar 18, 2023 at 19:31
• The proton-lattice is Lorentz contracted, so the charge density increases by $\gamma$, and the repulsion incearses as $\gamma^2$. When the electrons are included, the end results is an electric repulsive force scaling as $\gamma^2\beta^4$.
– JEB
Commented Mar 18, 2023 at 22:13
• @JEB That's wrong.Using the old theory: Protons start magnetically attracting when proton lattice is accelerated. At speed c magnetic attraction equals electric repulsion. At c we have infinite E-field, infinite current, infinite M-field and infinite time dilation ... Hmm maybe old theory is actually invalid here. So let's use the new theory which says: F'=F/gamma. That is the right formula to use when there is length contraction. When there isn't contraction it's F'=F/gamma^2. Commented Mar 19, 2023 at 0:06