Accelerating Charged Particle: Does it Fall Slower?

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The discussion centers on whether a falling charged particle, like an electron, radiates energy as it falls under gravity. It is argued that in free fall, a particle follows the curvature of space-time and does not experience acceleration, thus it does not radiate. The conversation references Griffiths' question about potential energy loss and radiation, suggesting it may be misleading. The equivalence principle is highlighted, indicating that a coaccelerating observer would not detect radiation, while distant observers might. The complexity of the topic is acknowledged, with references to various perspectives, including those of Feynman and ongoing debates in the physics community.
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So I was doing a problem in Griffiths and it talked about an electron in free-fall towards earth.
And it would radiate. So does a falling charged object fall slower than an uncharged object.
Because some of its Gravitational potential energy goes into radiation and not all kinetic energy.
 
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It doesn't radiate until it hits the earth.
 
When a particle is in free fall, it is just following the space-time curvature. This comes from

\nabla\mathbf{u}=0

where \nabla is the covariant derivative. That is, the particle cannot "feel" acceleration. So no, it does not radiate.
 
Then why does Griffiths say : "An electron is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of the potential energy lost is radiated away? " Maybe I left out the fact that it started from rest.
 
Sounds like a trick question.

You can do a thought experiment to show this.

Imagine a metal box in free fall above the earth. In the box is a scientist and a large charge. If the radiative hypothesis were true, the box would feel a retarding force. As a result the scientist would feel a small amount of gravity, thus violating the equivalence principle.
 
Is this a valid perspective:

Relativity does not view free fall as acceleration...Is this exactly the same as post #5??
 
cragar said:
And it would radiate.

jfy4 said:
So no, it does not radiate.

This turns out to be a surprisingly tricky question which has stirred up controversy even among the professionals. There have been a number of threads about this on PF, for example:

https://www.physicsforums.com/showthread.php?t=160533
 
Naty1 said:
Is this a valid perspective:

Relativity does not view free fall as acceleration...Is this exactly the same as post #5??

I would put it like this; in GR, freefall is locally identical to no acceleration (and no gravity). Observers far away can tell the difference, but not the local observer.

I there's a lot of good discussion about this topic here. 5 years ago I posted about this explaining how a coaccelerating observer sees the static field of a charge while observers in inertial frames see radiation. There was also a paper published on this about 2 years ago. If you search for "Antiphon" and "uniformly accelerating charge" it should come up.
 
Interesting thanks for your answers.
 
  • #10
Yes, as JT says this is not an easy question to answer. Consider and electron stationary on a bench in a lab on the Earth's surface. According to General Relativity it is being accelerated, does it radiate? Feynman has an interesting perspective on this problem, as do many others.
 
  • #11
cosmik debris said:
Yes, as JT says this is not an easy question to answer. Consider and electron stationary on a bench in a lab on the Earth's surface. According to General Relativity it is being accelerated, does it radiate? Feynman has an interesting perspective on this problem, as do many others.

What does Feynman say about it .
 
  • #12
cragar said:
What does Feynman say about it .

I wouldn't start there, his explanations are pretty complicated. Read the thread that JT recommended first, to get a perspective on the problem.
 
  • #13
already read it
 
  • #14
cragar said:
already read it

OK good, now try this: http://http://www.mathpages.com/home/kmath528/kmath528.htm"
 
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  • #15
That link didn't work for me.
 
  • #16
cragar said:
That link didn't work for me.

I pasted an extra http bit at the front, it should be pretty obvious what it is meant to be.
 
  • #17
Thanks i didn't look that close at it .
 
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