0

If you have the typical infinite cylindrical coil and you pass current through it, you will get the typical $\bf B$ field pointing in the $z-$ direction. In this case I only have current so net charge should be zero so there is not electric field.

Now we know that special relativity give us the $P=E^{2}-B^{2}<0$ invariant. Now I would like to change to the charge reference frame, so in my mind in this would be equivalent to a frame where all the charges are still so I could model them as a cylindrical charge density and like there is no movement there will be not magnetic field in this frame (everything is super ideal but bear with me). But now I have $P>0$, which makes no sense.

Also if I use Lorentz transformations to go from the lab frame to the moving one I should get something like: $E'=\gamma(\beta \times B)$ , $B'=\gamma B$. Where $P<0$ as it should, but instead of having a vanishing $B$ it gets increased to compensate the $E$ that show up. For what I think $B$ points in $z$, $E$ in $r$ and $\beta$ should point in $\phi$ so the expression should make sense.

Where is the problem in starting from the charge reference?? Where is the big mistake?

Urb
  • 2,608
jonh_titor
  • 3
  • Would both frames you are thinking of be inertial? Otherwise, you are not talking about a Lorentz transformation. – secavara Dec 13 '20 at 11:31
  • Well the rotating frame is non.inertial whole meaning of the word. There is an implicit acceleration because you are rotating (with constant $\omega$), but i think that the expresion for the transformtaion should hold. I am thinking now that there should be equivalent virtual forces (like a coriollis like force) when I use the charge frame. – jonh_titor Dec 13 '20 at 11:47

1 Answers1

0

If you have the typical infinite cylindrical coil and you pass current through it [...]. In this case I only have current so net charge should be zero so there is not electric field.

Now I would like to change to the charge reference frame, so in my mind in this would be equivalent to a frame where all the charges are still so I could model them as a cylindrical charge density [...].

There are two things that stop you from doing this:

  • In a cylindrical coil, the charge carriers move in circular trajectories around the coil axis. If you want a reference frame where those charge carriers are stationary, your reference frame needs to be in uniform circular motion. There is nothing catastrophically wrong about this (i.e. you can still define such a frame) but it is not an inertial frame of reference, so the Lorentz transformations don't apply.

  • If you want to have a nonzero net current without any net charge, then you need charges of two opposite signs moving at different velocities. The most common way is to have a positive ionic lattice sitting stationary with negative electrons moving against that background (though it is important to remember that this is not the only way). If you have such a setup, there is no single reference frame in which all the charge carriers are stationary.

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
  • I would think that you could concatenate multiple rotations taking the Loretz transf in each point usign the tangential velocity to make the boost $v=\omega r$. – jonh_titor Dec 13 '20 at 12:18
  • No, that is wrong. It is possible to have local frames of reference that are (instantaneously) co-moving with individual sectors of the coil, but not with the whole thing. To be clear: there is no inertial reference frame with the properties that you want. – Emilio Pisanty Dec 13 '20 at 12:19
  • Ok, I was suspicious about that either way, thanks for the clarification. So in that case doesn't even make sense asking about the signature change of the invariant because the moving frame is out of the scope of lorentz transformations. – jonh_titor Dec 13 '20 at 12:36