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Consider some positive charge that is distributed uniformly over a very long line along the z-axis.

If I am stationary with respect to the line then there is only static electric field which has cylindrical symmetry.

Assume now that I am moving with some constant velocity which has only a (positive) component along the z axis. With respect to me there is now current running down the (negative) z-axis, hence I expect to find a static magnetic field in my moving inertial frame.

My question is, do I still find (a static) electric field as well in this case? [Because on one hand I find the answer to be NO, since static electric fields should come only from static charges, and the charges here will not be static. On the other hand, although the line carries current, the line is still charged (which is different from currents running in conducting wires where the wires are neutral at all times). That means the answer is YES and one expects to see static electric field because the line is continuously charged.]

So is the answer yes or no (and why)?

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2 Answers 2

up vote 7 down vote accepted

By taking the contraction of the electromagnetic field tensor with itself, you can obtain the following Lorentz invariant quantity: $$B^2 - \frac{E^2}{c^2}$$

Now, in the original frame we have only an electric field and no magnetic field. Thus, the quantity above is negative.

In the other frame, we know that we have a magnetic field. If we had no electric field, the resulting quantity $B^2 - E^2/c$ would be positive. But that cannot be, because if a quantity is invariant, it must stay the same after a Lorentz transformation, and it can surely not simply change sign!

So there must still be a static electric field!

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Wonderful. That answers automatically the question in the title of my post as well. So an electric field CANNOT turn into a magnetic field and vice versa, right? By the way, that Lorentz invariant quantity is true regardless of the fields are time dependent or not, right? –  Revo Mar 11 '12 at 0:07
    
Well, it's true that the quantity is Lorentz invariant, but you might have to be careful about at what time you evaluate the fields, e.g. $B(t)$ in one frame and $B'(t')$ in another frame. –  Lagerbaer Mar 11 '12 at 0:32
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Won't the linear charge density simply increase (for the moving versus at-rest observer) according to the $\gamma$ factor? I thought charge itself was taken to be Lorentz invariant, since if you had a spaceship with a given charge all observers would agree on that charge. Obviously it's a little different for a line charge. –  AlanSE Mar 11 '12 at 1:04
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You can define the Lorentz covariant four-density $(c\rho, \vec{j})$, and that already suggests that if you have $(c\rho, \vec{0})$ in one system, you can get a non-zero current density in a system in relation to that. –  Lagerbaer Mar 11 '12 at 15:33
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Lagerbaer's answer is great, and convincingly demonstrates that it is impossible to eliminate a static electric field through a Lorentz transformation. (In fact, you only make it stronger!)

Another argument uses Gauss's law. No matter how you Lorentz-transform your configuration of static charges, you cannot eliminate the charge; all you do is scale its density to account for the change in volume caused by length contraction. And if there is any charge at all in some frame, Gauss's law tells you there must be electric field in that frame. So: no way to get rid of the electric field.


In your specific case:

If you are travelling parallel to the line of charge, the line of charge will be contracted by the factor $\gamma$, so the density of charge will increase by the factor $\gamma$. Then, if you draw a cylinder around some length $L$ of charge, Gauss's law tells you that there must be net electric flux through this cylinder; in particular, $\gamma$ times the electric flux through a cylinder of length $L$ in the rest frame.

If you knew that the electric field in the moving frame were perfectly radial, you could infer that the electric field in the moving frame would have to be $\gamma$ times the electric field in the rest frame. This is a little dicey, though -- since the charges are moving in a certain direction, you've lost a reflective symmetry which you used to ensure that the field was radial in the rest frame.

As it turns out, the electric field is radial, even in the moving frame, so you get $E'=\gamma E$. Off the top of my head, I can't think of an easy way to see this, though.

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