Electric field of a moving charge that's abruptly stopped

  • #1
vish22
34
1
Hello everyone,

This is in reference to fig 5.19 (screen shot attached - please read the paragraph which says "Figure 5.19 shows the...").
I don't get why the field outside of the sphere of radius ct acts as though the particle would have continued its motion. Author's words : "The field outside the sphere of radius R = ct must be that which would have prevailed if the electron had kept on moving at its original speed. That is why we see the “brush”
of field lines on the right in Fig. 5.19 pointing precisely down to the posi-
tion where the electron would be if it hadn’t stopped."

Why does it act this way? Why does the field outside R=ct assume an "apparent position" of the charge? Why does it think the charge is at x = v*t and that it remains in its state of motion?

In my opinion, it would make more sense if the field (outside R=ct) acted as though the particle is in a state of motion (with uniform velocity v) at x=0, ie. The field outside R=ct must belong to that of a charge moving (with a uniform velocity v along the x-axis and situated at x=0), before transforming into an electrostatic field belonging to a charge that's stationary at x=0. The field will transform once it knows the particle has stopped. But before such a transformation occurs, the field should reflect the actual state of the particle right before it's abruptly stopped, no (neglect any deceleration effects)? I'm not sure why the field assumes an "apparent position" in this case.This just doesn't seem right to me - I'm not sure what to make of it.
 

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  • #2
Orodruin
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Why does it act this way?
Because that is what comes out of Maxwell’s equations. What is outside of the sphere is not causally connected to any part of the particle worldline after the change in motion, only parts before the change. The field will therefore be the same as if the particle motion did not change because the change simply does not have enough time to propagate.

Why does the field outside R=ct assume an "apparent position" of the charge?
It doesn’t. It depends on where the particle was and how it was moving on the past lightcone of the event.

Why does it think the charge is at x = v*t and that it remains in its state of motion?
It doesn’t. See above.

In my opinion, it would make more sense if the field (outside R=ct) acted as though the particle is in a state of motion (with uniform velocity v) at x=0, ie.
Physics is not required to adhere to what you think would make sense. The theory of electromagnetism is based on Maxwell’s equations - a relativistic field theory. It behaves this way and has been very well tested experimentally.

This just doesn't seem right to me - I'm not sure what to make of it.
See above. Physics depends on experimental evidence, not on what you think seems right.
 
  • #3
vish22
34
1
See above. Physics depends on experimental evidence, not on what you think seems right.

Yes, yes absolutely.

For now I shall consider it as being experimentally proven and carry on my readings, thank you!
 

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