Maxwell's equations and gravitation

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  • #1
naima
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Hi PF
I am reading Feynman's lectures on gravitation.
The equivalence principle says that no local measurement (it includes measurement of the electromagnetic field) can tell you if you are accelerated or in gravity.
Feynman agrees and writes that we have then a problem. Accelerated charges are the sources of radiations but what about motionless charges in gravity emitting photons?
He writes that Maxwell's equation will have to be rewritten to be coherent with this principle.
I have two questions.
Is there a mainsream theory for these new equations?
Did experiments show that motionless charges can emit in gravity?
 

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  • #2
stevendaryl
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Hi PF
I am reading Feynman's lectures on gravitation.
The equivalence principle says that no local measurement (it includes measurement of the electromagnetic field) can tell you if you are accelerated or in gravity.
Feynman agrees and writes that we have then a problem. Accelerated charges are the sources of radiations but what about motionless charges in gravity emitting photons?
He writes that Maxwell's equation will have to be rewritten to be coherent with this principle.
I have two questions.
Is there a mainsream theory for these new equations?
Did experiments show that motionless charges can emit in gravity?

This issue is discussed in Wikipedia here: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field

As far as mainstream theory, the changes to Maxwell's equations to accommodate curved spacetime (gravity) are pretty well understood.

https://en.wikipedia.org/wiki/Maxwell's_equations_in_curved_spacetime

(Some people object to using Wikipedia as a reference, but my experience is that it is often, if not always, pretty good.)
 
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  • #3
naima
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great!
I ignored that feynman proposed that a linear uniform acceleration of a charged particle would produce no radiation
 
  • #4
naima
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Rohrlich's argument looks like the Unruh effect. A radiation is
seen by an observer but not by another. But here we have two
fields.
Is it an analogy or is it the same with different ground states
and Bogoliubov transformations?
I understand Rohrlich until he talks about a supported charge
in a supported frame in gravity. Is it a motionless charge on earth?
What is the equivalent with Unruh?
 
  • #5
naima
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There is a classical way to illustrate the equivalence principle.
an accelerated elevator moves upward. there is a hole in the wall. A stone follows an inertial path outside and then crosses the hole. the observer
inside sees a parabolic orbit.
suppose that this stone is charged and made in iron
the observer will see a radiating particle and an outside physicist will say that it does not.
I can accept that this radiation is frame dependent but what about its measurement by the inside observer? the outside observer can look at
the devices in the elevator. was it a paradox for Einstein?
 
  • #6
Vanadium 50
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First, radiation is a far field effect. That is the definition of what is called radiation and what is called some other part of electromagnetism.

Second, the equivalence principal is a nearby effect. It is only true so far as the distances are small. Over large distances it doesn't work (over the size of the earth, the acceleration is not only not constant, it even changes sign!)

As a general rule, taking two approximations that not only apply to different situations but opposite situations is unlikely to work.

It is possible to do these calculations in an intermediate region where neither approximation is very good, but not at the B level. Or the I level. Or, for most people, the A level. What you will find is that different observers agree on what they see, but not necessarily why they see it. For instance, they may all agree that a radiation detector went off, but not everyone will agree it was radiation that did it - some might observe that it was the near fields from moving charges that did it.
 
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  • #7
naima
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According to Rohlrich, a free falling detector would detect radiation from a supported charged system (on the earth) but no radiation from another free falling chage.
A supported detector (on the ground of the earth) would detect
radiations from a free falling charge but not from a charged
metalic sphere on the ground.
Is it mainstream?
 
  • #8
Vanadium 50
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What was unclear about my message #6?
 
  • #9
naima
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Nothing.
I summarized.
 
  • #10
Vanadium 50
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So why are you still trying to match approximations with not only different assumptions but opposite assumptions?
 
  • #11
naima
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I am not interessed here in antennas and telecommunications.
The equivalence principle is about charge and uncharged particles falling together in in a labotatory on earth or in an ac accelerated elebator with detectors aboard.
Stevendaryl gave me a good link to the charged particle's paradox. I bought Rohlrich's book on the subject.
What you wrote was not wrong or unclear.
Rohlrich try to solve the paradox, do you?
The main point is that for him Maxwell equations are only valid for free falling detectors. They are not valid on the surface
of the earth.
If it is true experiments should show it with a 1 g gravity.
 
  • #12
PeterDonis
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Did experiments show that motionless charges can emit in gravity?

Have you tried to calculate the intensity of radiation from a charge at rest on the surface of the Earth, as seen by an observer free-falling past the charge, using the formula given in the Wikipedia page that stevendaryl linked to? Have you noticed how small the result is?

Have you also noted that, according to Rohrlich, a charge at rest on the surface of the Earth will not appear to radiate to an observer who is also at rest on the surface of the Earth?

Answering these two items answers the question you pose in the quote above, which as far as I can tell is the only actual question you have asked in this thread.
 
  • #13
PeterDonis
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Thread closed as the question posed in the OP has been answered.
 

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