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For gravity, it is at rest relative to the Earth. For an accelerating charge, it is also at rest relative to the Earth. I think there is an assumption here that if radiation is seen in any reference frame then it is seen in all reference frames. Is this reasonable?

A.T.
For gravity, it is at rest relative to the Earth. For an accelerating charge, it is also at rest relative to the Earth.
If the antenna is at rest relative to the charge in one case, then it also must be at rest relative to the charge in the other case, to have equivalence.

Dale
Mentor
For an accelerating charge, it is also at rest relative to the Earth.
The equivalence principle relates behavior at rest in gravity to behavior accelerating in the absence of gravity. So there cannot be any earth in the accelerating case. That is why the motion must be specified relative to the charge.

I think there is an assumption here that if radiation is seen in any reference frame then it is seen in all reference frames. Is this reasonable?
Not without specifying how the radiation is seen. It matters substantially.

In the gravitational case, if the antenna is at rest it detects no radiation, but if the antenna is in free fall it detects radiation. In the accelerating case, if the antenna is co-accelerating then it detects no radiation, but if the antenna is inertial it detects radiation.

In the gravitational case, if the antenna is at rest it detects no radiation, but if the antenna is in free fall it detects radiation.
What happens as time goes on? If the antenna is at rest then there is no problem - the situation can go on forever. If the antenna is in free fall and it detects radiation, can this situation go on forever? Wont the charged matter run out of energy after a while?

In the accelerating case, if the antenna is co-accelerating then it detects no radiation, but if the antenna is inertial it detects radiation.
Same question, almost. If the antenna is co-accelerating, then the situation can go on forever. If the antenna is inertial, can this situation go on forever? Won't the charged matter run out of energy after a while?

I think there is an assumption here that if radiation is seen in any reference frame then it is seen in all reference frames. Is this reasonable?
I asked if this was reasonable. I now claim that it is in fact reasonable.

A.T.
If the antenna is inertial, can this situation go on forever? Won't the charged matter run out of energy after a while?
Since the charged matter is accelerating in the inertial frame of the antenna, something must be doing work on the charged matter in that frame.

Dale
Mentor
What happens as time goes on? If the antenna is at rest then there is no problem - the situation can go on forever. If the antenna is in free fall and it detects radiation, can this situation go on forever? Wont the charged matter run out of energy after a while?
The equivalence principle does not apply forever. It only applies for a small region of space and time where the spacetime curvature is negligible. The curvature is not negligible forever.

If the antenna is co-accelerating, then the situation can go on forever. If the antenna is inertial, can this situation go on forever? Won't the charged matter run out of energy after a while?
No, if the charge is being accelerated forever then it is being given an infinite amount of energy. However, the amount of energy captured by the antenna is not that large, although that is a calculation that I will leave to the interested reader.

I asked if this was reasonable. I now claim that it is in fact reasonable.
I am not sure how you came to the sudden conclusion that a false claim is reasonable: https://en.wikipedia.org/wiki/Unruh_effect

But the important question in terms of the equivalence principle is not whether or not radiation exists in the frame, but whether or not the detector detects it.

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Help me to understand how you can have radiation in one frame of reference but not in another, or how you can detect it in one frame, but not in another.

Dale
Mentor
Help me to understand how you can have radiation in one frame of reference but not in another, or how you can detect it in one frame, but not in another.
The detectors will have the same results in all frames, regardless of whether or not there is radiation in that frame.

Consider the accelerating charge and an inertial antenna. In the antenna’s inertial frame the accelerating charge emits radiation which is measured by the antenna as a time varying voltage. In the charge’s accelerating frame the charge does not emit any radiation and the time varying voltage is a result of the antenna moving through the static field

Help me to understand how you can have radiation in one frame of reference but not in another, or how you can detect it in one frame, but not in another.
This may be relevant

https://arxiv.org/pdf/physics/0506049‎

Yeah it looks relevant

PeterDonis
Mentor
2019 Award
Help me to understand how you can have radiation in one frame of reference but not in another, or how you can detect it in one frame, but not in another.
Talking about "frames" just confuses the issue. The relevant factor is the relative motion of the source and the detector. That should be obvious from the examples @Dale gave.

The detectors will have the same results in all frames, regardless of whether or not there is radiation in that frame.

Consider the accelerating charge and an inertial antenna. In the antenna’s inertial frame the accelerating charge emits radiation which is measured by the antenna as a time varying voltage. In the charge’s accelerating frame the charge does not emit any radiation and the time varying voltage is a result of the antenna moving through the static field
Talking about "frames" just confuses the issue. The relevant factor is the relative motion of the source and the detector. That should be obvious from the examples @Dale gave.
Not to messy the waters, but I have a slightly different question: Assuming each observer has their own local detector, and one observer is comoving with the charge, shouldn't that comoving observer see an electrostatic field, because any radiation an inertial observer sees (with respect to the accelerated charge) would be forever inaccessible to the comoving observer due to a kind of event horizon (basically for the observer at rest with respect to the charge, any radiation would be outside of that person's light cone)?

Or am I way off base here?

Talking about "frames" just confuses the issue. The relevant factor is the relative motion of the source and the detector. That should be obvious from the examples @Dale gave.
So we are at the point where, if the detector is comoving with the "accelerated" charge (or the charge in a gravitational field) then it doesn't detect radiation because of the speed of light or because of some reason, but if the detector is not comoving with the charge then it can detect radiation? That seems like it would resolve the "paradox" (apologies to those who don't like that word).

Staff Emeritus
2019 Award
Not to messy the waters, but I have a slightly different question
Maybe that better fits its own thread, as the OP misunderstands at least two things, so muddier waters is not what we need.

PeterDonis
Mentor
2019 Award
So we are at the point where, if the detector is comoving with the "accelerated" charge (or the charge in a gravitational field) then it doesn't detect radiation because of the speed of light or because of some reason, but if the detector is not comoving with the charge then it can detect radiation?
That's basically what @Dale was saying in post #28.

This paradox may have come from Feynman's Lectures on Physics, ...
And Feynman says in the lectures that accelerated charges don't radiate. It is the 3rd derivative of position (change in acceleration, not just acceleration alone) that is responsible for the radiation of charged particles.

...But according to the Principle of Equivalence,...
You could in principle construct, instead of an accelerating elevator in flatspace, an elevator which not only accelerates but additionally changes acceleration. Then you could produce radiation from a charged particle, but that would not be an equivalence with the charged particle on earth which does not radiate.

Staff Emeritus
2019 Award
And Feynman says in the lectures that accelerated charges don't radiate.
Where?

And Feynman says in the lectures that accelerated charges don't radiate. It is the 3rd derivative of position (change in acceleration, not just acceleration alone) that is responsible for the radiation of charged particles.
After all that has gone through this thread, I would very much like to see where anyone says that a charge undergoing constant acceleration doesn't radiate. I think I will dig through Jackson on this subject, though I am not really at Jackson's level in my review of physics.

After all that has gone through this thread, I would very much like to see where anyone says that a charge undergoing constant acceleration doesn't radiate. I think I will dig through Jackson on this subject, though I am not really at Jackson's level in my review of physics.
Again, Feynman said it.

But your first reference says that the radiated power is proportional to $a^2$, not $\frac{d\vec{a}}{dt}$

But your first reference says that the radiated power is proportional to $a^2$, not $\frac{d\vec{a}}{dt}$
No. Equation 9.1.1 is the equation that Feynman says has led us astray in our thinking.

Dale
Mentor
Note, things like energy and power are frame variant. What is invariant is the outcome of measurements. The measurements we discussed above were simply about constant or time varying voltages, regardless of the power or other frame variant considerations.

Note, things like energy and power are frame variant. What is invariant is the outcome of measurements. The measurements we discussed above were simply about constant or time varying voltages, regardless of the power or other frame variant considerations.
Good point.