Wednesday, May 15, 2019

An accelerating charged plane and an observer in the Minkowski space

NOTE May 19, 2019: the analysis on this page is incorrect. We must assume Gauss's law.

----

Our goal is to find out how the electric field of a charge behaves under a gravitational field.

We assume that the electric field is known in the case where the charge is static, or moves at a constant speed in the Minkowski space. Assuming that the Coulomb potential is 1/r for a static charge, Lorentz covariance dictates how the electric field must look like for a charge moving at a constant speed.

The problem becomes more complicated if the charge or the observer is in an accelerating motion.

Let the charge first be static. Then it is pushed rapidly, so that the speed is v. We suggested in our blog post a few days ago that the electric field should be formed by combining the old Coulomb field to the new Coulomb field and that Gauss's law does not hold.

The new Coulomb field is present in the circle of a radius ct around the original position of the charge.

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

It is the familiar diagram top right on the linked page.


A charged plane which is suddenly pushed away from the observer


        observer         push to speed v  --->
                              |
        o                    |
                              |
                              |
        <-- E

We have a static infinite plane with a uniform charge density per area. The electric field is a uniform E_0 at the observer. The plane is suddenly pushed to a relativistic speed v.

The observer will continue to see most of the plane at its original position because the speed of light is finite. But soon the observer will see a disk-like circular area of the plane opposite to him farther away than the rest of the plane. Furthermore, the observer will see the electric field of the circular disk flattened by length contraction.

The observer will initially see the electric field decrease, but it will start growing back towards E_0 as he sees the disk grow bigger and bigger.

How low can the electric field go? If v is very close to c, then the electric field of the disk is flattened extremely much.  Obviously, we can make the electric field E transiently as close to zero as we like.


An observer is suddenly pushed away from a static plane


In this case, the observer at a constant speed v will see the electric field E just as big as a static observer.


See Table 26-2 in the link. The Lorentz transformation of the electric field E in the direction of v is the identity mapping.


Both the observer and the plane are pushed to the same direction


           --> v
           |
           |
           |            o --> v
           |
           |
                       E -->

Suppose that the initial distance of the static plane from the static observer is 1 meter. Let us quickly push both to a relativistic speed v to the right. We want to keep the distance in the moving frame as 1 meter. To accomplish that, we first need to push the plane, then wait for a little while, and then push the observer. We need to wait because there is length contraction in the frame moving at a constant speed v relative to the original static frame. Without the wait, the distance would be bigger than 1 meter in the v frame.

The observer will see most of the plane at its original position and see that he is moving away from the plane. But soon he will see a circular disk of the plane close and static relative to him, at the the 1 meter distance from him. He will see the electric field E larger than the original field E_0.

         |
         |\
         |   \
         |      \  E -->
         |        o observer
         |      / the cone is 120 degrees wide
         |    /
         | /
         |

Half of the electric field E of a static plane at a static observer is generated by charges which are at most at an angle 60 degrees from the normal as seen by the observer.


The observer is behind the plane and both are pushed to the same direction


       --> v     --> v
                   |
      o           |
                   | 
                   |
           <--- E --->
                 
This is an interesting case. Let us assume that the distance of the observer and the plane is 1 meter both before and after the push, in the frame of the observer.

If the observer attains a relativistic speed, he will see most of the plane behind him. He sees a circular disk in front of him at the distance of 1 meter.

Suppose that v is almost the speed of light and the push is at t = 0. We work in the original static frame. When the observer has moved 2 meters to the right, he is receiving a ray of light sent from the plane sqrt(3) meters above him at t = 0. The observer sees a disk of an angular diameter 120 degrees 1 meter ahead of him. The rest of the plane he still sees at the original position behind him. The electric fields of these parts are opposite. He sees a total electric field E = 0.


An accelerating plane


We believe that in the case of smooth acceleration, the above sudden push examples give a qualitative understanding of what happens to the electric field.


A gravitational field


This is a much harder case. In the above example cases we worked in the flat Minkowski space. What happens when space is curved?

Hanni, Ruffini, Cohen, Wald, and Linet used Gauss's law to construct an electric potential around a black hole. But we saw that Gauss's law does not hold under acceleration. What to do?

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