What Happens to the Fields When a Charged Wire Moves Relativistically?

[eXorikos]

Homework Statement

We have a uniformly charged wire that is infinitly long. The charge density is $$\lambda$$. Calculate the fields when the wire is at rest. Somebody moves the wire with a constant speed, what happends to the fields?

Homework Equations

Gauss' equation for finding the electric field at rest. Transformation rules from special relativity.

The Attempt at a Solution

The electric field is obviously:
$$E=\frac{\lambda}{2\pi r}$$
The magnetic field is zero, because there is no current.

Now we move the wire. We chose the x-axis allong the wire. y and z are perpendicular to the wire ofcourse. I assume it is ok to calculate the fields first for the case of velocity in the x-direction and then in the r-direction. The fields tranform as follows.
$$\overline{E}_x=E_x=0$$
$$\overline{B}_x=B_x=0$$
$$\overline{E}_y=\gamma (E_y-vB_z)=\gamma\frac{\lambda}{2\pi y}$$
$$\overline{E}_z=\gamma (E_z-vB_y)=\gamma\frac{\lambda}{2\pi z}$$
$$\overline{B}_y=\gamma (B_y+\frac{v}{c^2}E_z)=\gamma \frac{v}{c^2}\frac{\lambda}{2\pi z}$$
$$\overline{B}_z=\gamma (B_z-\frac{v}{c^2}E_y)=-\gamma \frac{v}{c^2}-\frac{\lambda}{2\pi y}$$

Now here's the thing. Since the observer moving allong the wire clearly sees a magnetic field caused by the passing chargedensity, according to him there must be a current in the wire. But why is the electric field in the x-direction still zero in his referenceframe? Or is it the field at the point of the observer?

How do I calculate the field for a motion in a random direction? Is it just calculating the fields for the two possible motions and adding them up or is it not that simple?

 

[Jhero]

Thank you for your question. It is important to consider the effects of relativity when dealing with moving charged objects. In this case, we can use the Lorentz transformation equations to calculate the fields in the reference frame of the moving wire.

First, let's consider the electric field. In the reference frame of the moving wire, the electric field is given by:

E' = \gamma (E - vxB)

where E is the electric field in the rest frame and B is the magnetic field. In this case, since the wire is infinitely long and has a constant charge density, the electric field in the rest frame is given by:

E = \frac{\lambda}{2\pi r}

and the magnetic field is zero, as you correctly stated. Plugging these values into the Lorentz transformation equation, we get:

E' = \gamma \frac{\lambda}{2\pi r} - \gamma vB

Since B is zero, the second term also becomes zero and we are left with:

E' = \gamma \frac{\lambda}{2\pi r}

which is the same as the electric field in the rest frame. This means that the electric field does not change when the wire is moving at a constant speed.

Now let's consider the magnetic field. In the reference frame of the moving wire, the magnetic field is given by:

B' = \gamma (B + \frac{v}{c^2}xE)

Plugging in the values for B and E from the rest frame, we get:

B' = \gamma \frac{v}{c^2} \frac{\lambda}{2\pi r}

This means that in the reference frame of the moving wire, there is a non-zero magnetic field due to the motion of the charged wire. This is because the motion of the wire creates a current, which produces a magnetic field.

To calculate the fields for a motion in a random direction, you would need to use the full Lorentz transformation equations, which can be found in most textbooks on special relativity. It is not as simple as adding the fields from two different motions, as there are additional factors to consider.

I hope this helps answer your questions. Good luck with your calculations!