(Previous posts in this series: Part 1, Part 2, Part 3, Part 4, Part 5, Part 6, Part 7)

We have now acquired most of the background knowledge needed to start directly addressing the question that this series of posts started with, as to whether an electric charge and a neutral particle dropped from the ceiling will hit the ground at the same time. The first stage of that is what is known as the Principle of Equivalence.

Recall the Postulate #1 we started with that said that all objects falling freely in a gravitational field will fall at the same rate and hit the ground at the same time. We were able to explain this by saying that it followed from the fact that gravitational and inertial masses were equal. But we did not explain why they should be equal. The two masses were arrived at, after all, by distinct methods using different operational definitions. But by considering accelerating frames and the Principle of Equivalence, we have a simple explanation for it.

Consider two objects floating freely in space that are at rest with respect to each other and to an observer S, to signify space. No suppose we enclose the two objects in a closed room and accelerate the room ‘upwards’ (i.e., in the direction from the ‘floor’ to the ‘ceiling’) with a value g. We will call this frame E to signify an elevator or Earth. Then in the frame E, both objects will seem to be accelerated ‘downwards’ (i.e., towards the floor of the room) with a value g, just as if they were falling freely on Earth, and will both hit the floor at the same time. They will behave just as if they were falling down near the surface of the Earth and since they hit the floor at the same time, we can infer that their gravitational and inertial masses are equal, rather than tacitly assuming it to be the case as we did before.

Einstein’s insight was that the two situations are indistinguishable, that we can never tell if we are in a uniform gravitational field that is acting downwards (due to a large mass somewhere below us) or whether there is no mass anywhere near that is producing a gravitational field and instead we are in a frame that is accelerating upwards. There is no experiment that we can do within the enclosed room (i.e., within the frame E) that would enable us to distinguish between the two systems.

This insight was derived from the behavior of objects in a gravitational field and explains Postulate #1 and thus also the equivalence of gravitational and inertial masses. But crucially Einstein extended it into a general rule, that there is no experiment whatsoever, involving any and all forces, that will enable us to make the distinction between the two frames, a stationary frame in a uniform downward gravitational field or an upwardly accelerating frame in the absence of any gravity. That added level of generality has massive implications for all of physics and laid the foundation for the development of general relativity. This is called the Principle of Equivalence (PoE).

It should be noted that this equivalence principle as stated above says that we cannot distinguish between an accelerating frame and a uniform gravitational field, i.e., a field that has the same magnitude and direction everywhere. But real gravitational fields are not uniform because they are due to masses that have limited size. So for example, if we were to drop two objects in an enclosed room, then if we were in an upwardly accelerating frame and there were no gravitational field, the two objects would fall along parallel lines. But if the room were stationary on the surface of the Earth and the objects were falling due to its gravitational field, then the paths of the two objects would ever so slightly converge towards each other because each one is headed towards the same single point, that is the center of the Earth. In that case, we could distinguish the two situations. So the PoE is formulated either in terms of a hypothetical uniform field or by applying it only within a very small region of space, where the field is approximately uniform. To get a truly uniform gravitational field over a large volume one would need to hypothesize the existence of an infinitely large, uniform slab of matter, with the two sides being flat and parallel. This is of course, not physically possible. The closest one might get to it in the real world is close to the surface of the plane of stars in the spiral arms of galaxies.

With all this in hand, we can tackle Postulate #2, that asserts that an accelerating charge will radiate energy, thus reducing its kinetic energy. If so, an electric charge falling in a gravitational field will fall more slowly than a neutral charge that does not radiate and lose any kinetic energy, thus seeming to violate Postulate #1. Doing an actual experiment to test this would be extraordinarily difficult and has not been done as far as I am aware. But Einstein’s PoE gives us a way to address this question theoretically by allowing us to switch between a frame at rest in a gravitational field (which is our familiar world) and an accelerating frame in the absence of gravity, and requiring the observable results to be the same.

Consider an electric charge and a neutral particle both floating freely in space, far from any gravitational and other forces. If observe them to be at rest or moving with uniform velocity, then that means that we are in an inertial frame S. The electric charge will not radiate any energy because according to Maxwell’s equations, which are valid in inertial frames, a charge must have an acceleration in an inertial frame in order to radiate energy. Now consider another frame E that is moving with an acceleration g with respect to S. In the frame E, both the charged and neutral particles will have the same acceleration g but in the opposite direction to E. The situation in E is exactly as if the two particles are accelerating in a uniform gravitational field. But all we have done is change the observer from one in S to one in E. The state of motion of the charge is unchanged. By the PoE, observable results have to be the same in both frames and hence the charge still should not radiate in E. Hence it seems as if we can conclude that an electric charge will not radiate if its acceleration is entirely due to it falling freely in a gravitational field. Hence Postulate #1 is satisfied but Postulate #2 is found to be only valid under certain conditions, when a charge is accelerating in an inertial frame. We arrived at this conclusion purely theoretically, by invoking the PoE.

This appears to resolve the paradox that we started with but at the expense of Postulate #2, which has to be modified to say that an electric charge does not radiate when it is falling in a uniform gravitational field. This resolution requires us to go from a universal statement that an accelerating charge always radiates to having to first determine whether its acceleration is due to falling freely in a uniform gravitational field because then that situation is equivalent to the charge not accelerating in an inertial frame. Thus it seems as if a charge will radiate if it is accelerating in a horizontal direction on the Earth’s surface but not if it is falling in a vertical direction, because in that case it is not accelerating in an inertial frame. If it is falling at an angle, so that the acceleration has a vertical component and a horizontal component, that would presumably require appropriate adjustments.

All that is is a little awkward but some closed the book at this point, reconciled to the situation.

But wait, there’s more! That is by no means the end of the story. While we seem to have resolved this one issue, if we start looking a little more closely, we will find that that this resolution does not hold up and looking deeper leads to another paradox and resolving that leads to another, and so on.

To misquote Bette Davis from the film All About Eve, “Fasten your seatbelts. It is going to be a bumpy ride.”

(To be continued.)

1. GerrardOfTitanServer says

it seems as if we can conclude that an electric charge will not radiate if its acceleration is entirely due to it falling freely in a gravitational field.

I commented before, but this is just such an odd way of phrasing it. The particles might be accelerating due to gravity according to Newton, but they’re not accelerating. The accelerometer says that they’re not accelerating, and Einstein says that they’re not accelerating. It is the observer at ground level who is undergoing constant acceleration. The particles falling to the ground are not accelerating.

2. Mano Singham says

Gerrard,

Acceleration is not an invariant quantity. There is no such thing as the acceleration of an object. It is always measured with respect to some frame and in different accelerating frames it will have different values.

3. GerrardOfTitanServer says

Mano
Yes, but, whether an object is accelerating or not, that is an invariant property under GR, pretty sure.

4. GerrardOfTitanServer says

5. Rob Grigjanis says

Gerrard: Yes, if a body is accelerating relative to a particular inertial frame, it will of course be accelerating relative to any inertial frame. But the measured acceleration will be different in the two inertial frames -- generally, in magnitude and direction. The proper acceleration is measured in the particular inertial frame in which the accelerating object momentarily has zero velocity.

It’s also perfectly legitimate to say that, in a frame of reference accelerated relative to an inertial frame, a freely falling body is accelerating. Remember that one of the goals is to be able to relate arbitrary frames, not just inertial ones.

And if we’re talking about a curved spacetime, the word ‘local’ has to be sprinkled liberally in the foregoing.

6. GerrardOfTitanServer says

Rob,
Sure. Sounds good.

Under GR, one should not expect a falling charged particle to emit radiation because it’s not really accelerating (relative to the local same-velocity inertial frame). An accelerometer next to the particle moving with the particle says that it is not accelerating, and therefore it is not accelerating, and therefore it should not emit radiation. I think Mano disagrees with the obviousness of what I am saying here, and that confuses me.

In other words, Einstein taught us that gravity is not a force, and it’s not acceleration.

7. Rob Grigjanis says

Gerrard @6: So far, what seems obvious is that an inertial observer will not see radiation from a freely falling charged particle.

What’s not obvious is what an accelerated observer sees. If an inertial observer sees no radiation, does it necessarily follow that an accelerated observer sees no radiation? If an accelerated observer sees radiation, does that necessarily mean a reduction in the charged particle’s kinetic energy?

8. GerrardOfTitanServer says

Rob,
Thanks. I could easily be wrong, but my physics training is telling me this:

Where we can ignore large scale curvature of space time, we can use conservation of energy. Consider the case of a test particle and observer far from everything else, eg with flat space time. The test particle is at rest. The observer is undergoing constant acceleration from a rocket. Because of the conservation of energy, the observer cannot see the particle emitting radiation. Therefore, charged particles at rest in inertial frames do not emit radiation.

By Einsteins principle of equivalence, this must extend to “acceleration” from gravity. Ergo, charged particles in free fall in curved space time do not emit radiation.

I’m struggling to see how it might be some other way, and I’m struggling to see why this should even be controversial, given my understanding of the precepts and consequences of relativity. I only see a problem when we rely on the outdated Newtonian idea of gravity as a force. But where gravity is modeled not as a force, the charged particle in free fall cannot be described as accelerating.

I admit that I could easily be wrong. I admit I feel very weird coming in as an amateur and stating that the answer seems obvious when apparently many smart people think it’s not. So, clearly I must not understand something. (Or those smart people are talking about something else, eg a charged particle that is actually accelerating according to an accelerometer.)

9. Rob Grigjanis says

Gerrard @8:

The observer is undergoing constant acceleration from a rocket. Because of the conservation of energy, the observer cannot see the particle emitting radiation. Therefore, charged particles at rest in inertial frames do not emit radiation.

Conservation of energy can be subtle. The approach by Rohrlich was more straightforward. We know the em fields in the rest frame of the particle -- just a static Coulomb electric field. Then find the transformation (not a Lorentz transformation) which gives the em fields in the accelerated frame. When he did this, Rohrlich found that they correspond to a radiating particle. But the charged particle still falls at the same rate as the neutral particle.

How does that square with conservation of energy? I’m not sure.

10. Rob Grigjanis says

To be clear: The particle in your scenario is not radiating as seen from any inertial frame of reference.

I admit I feel very weird coming in as an amateur and stating that the answer seems obvious when apparently many smart people think it’s not

As someone who studied electromagnetism and GR for years, I feel very weird that there is stuff here that I don’t understand.

11. GerrardOfTitanServer says

The approach by Rohrlich was more straightforward.

After reading that, I am completely lost, and I no longer trust my physics education intuition at all. This in particular:

The radiation from the supported charge viewed in the freefalling frame (or vice versa) is something of a curiosity: where does it go? David G. Boulware (1980)[9] finds that the radiation goes into a region of spacetime inaccessible to the co-accelerating, supported observer. In effect, a uniformly accelerated observer has an event horizon, and there are regions of spacetime inaccessible to this observer. Camila de Almeida and Alberto Saa (2006)[10] have a more accessible treatment of the event horizon of the accelerated observer.

It just seems like it must be patent nonsense. However, this is way beyond my level of math, and so maybe it’s true, and I give up.

12. GerrardOfTitanServer says

Rob,
Here’s something else interesting to think about.
Just like my collision vs no-collision example, we can talk about a light bulb turning on, or not.

Place a charged test particle “at rest” on the surface of the Earth. If it’s constantly emitting radiation according to some observer, any observer, then that observer should be able to place a photovoltaic cell next to the particle and harness free energy. It’s a perpetual motion machine. We could have a globally flat spacetime that is entirely empty except for the Earth, and we could have a perpetual motion machine (albeit with a very low power density).

What does it mean for one observer to see radiation coming off the particle and another does not? What does it mean for one observer to have a perpetual motion machine and another does not? This is classical relativity, not quantum mechanics. Such weird questions should not arise.

Finally, after reading a few more articles, I think I’m finally starting to appreciate Feynman’s solution that uniformly accelerated charges do not radiate, regardless of the observer. His alternative mathematical approach leads to this conclusion, and this conclusion neatly solves our little paradox here.

13. Rob Grigjanis says

Gerrard @11: It’s certainly not nonsense.

I’m sure you’ve heard of light cones. If an inertial observer sends a signal, it cannot reach a spacetime point outside their future light cone. However, a signal sent from an arbitrary point outside the past or future light cone will eventually intersect with their future light cone.

For an accelerated observer, the situation is different. In this case, there are regions of spacetime from which signals will never reach them.

In particular, consider an observer accelerating with a charged particle. From any inertial frame, the particle is radiating; inertial observers will eventually see the radiation. But the observer accelerating with the charged particle will only see a static electric field, and no radiation. That’s because, after you do the daunting math, the radiation part of the field is outside the region of spacetime which can communicate with the accelerated observer.

14. Rob Grigjanis says

Gerrard @12: According to the photovoltaic cell sitting next to the charged particle (and thus accelerating with it), there is no radiation, as I explained in #13.

15. Rob Grigjanis says

Most of the work leading to the results I’ve described was done after Feynman expressed his views in the early 60s. I don’t know if he changed his mind, or even paid attention to those papers.

16. GerrardOfTitanServer says

Rob,
You’re still talking about Schodingers lightbulb, a lightbulb that is connected to the solar cell. The light bulb is on according to observer who is not-falling, and it’s off according to any observer who is falling (or vice versa?). We could then ask questions like temperature of the light bulb. It’s beyond weird. Surely we’re going to hit contradictions.

I’m familiar with light cones. However, in our example with roughly flat non-expanding space time, I fail to see how that radiation could be entirely outside of the light cone of the accelerating observer. That radiation is going at the speed of light according to all observers (who can see it), and given the assumption of no curved spacetime and non expanding spacetime (and non contracting aka static), it seems like that radiation will quickly intersect the light cone of the nearby observer assuming non pathological cases of extreme acceleration. The idea of an event horizon being relevant seems entirely implausible.

17. Rob Grigjanis says

Gerrard @16:

You’re still talking about Schodingers lightbulb

I don’t know what you mean. We’re talking about a flat relativistic spacetime, not quantum mechanics. The solar cell is accelerating with the charge, so collects no radiation, so can’t power the light bulb.

Implausible or not, an accelerated observer in a flat spacetime has an event horizon.

18. GerrardOfTitanServer says

The solar cell is accelerating with the charge, so collects no radiation, so can’t power the light bulb.

But an accelerating / falling observer sees the particle emit radiation, which is then absorbed by the solar cell, which then turns on the light bulb. If the falling observer sees it emit radiation, then it must happen, right?

19. Rob Grigjanis says

Gerrard @18: The freely falling observer’s detector (which is freely falling with the observer) sees radiation from the charged particle which is not freely falling. The accelerated observer’s detector (which is accelerating with the observer and the charged particle) sees no radiation, because to that observer (and detector) the radiation is unobservable.

The bottom line: what your detector (the one sitting next to you) sees depends on your frame of reference.

The radiation detected by one observer is not necessarily the same as the radiation detected by another detector

At this point, Gerrard Fatigue is setting in.

20. GerrardOfTitanServer says

Rob,

The radiation detected by one observer is not necessarily the same as the radiation detected by another detector.

How can we have consistent physics in light of such a wild claim? I’m assuming that there is a single external reality. I’m assuming we can ignore quantum mechanics for now. If the engineering and physics of the detector says that it should alert if it is hit with a certain kind of radiation, then it should alert. Observers don’t matter here. Yet, you’re saying we can tell a story where observer #1 sees no radiation hit the detector (and sees no alert), and at the same time we have observer #2 who sees radiation of the right kind hit the detector but also sees no alert. How do we explain what #2 sees? #2 sees radiation hit the detector, but also sees physics break down because the detector doesn’t alert.

21. John Morales says

I’m reminded of the utility of E-Prime:
(paraphrased)
“Rather than “Light is both a particle and a wave”, it’s better to say “Light behaves like a wave when measured using method X, and behaves like a particle when measured using method Y”.”

22. GerrardOfTitanServer says

John
And from those theories of seeming inconsistency, they are able to make wildly successful predictions about macroscopic reality that don’t involve any such macro superpositions. “Argument by analogy is fraud.”

23. Rob Grigjanis says

Gerrard @20:

Observers don’t matter here

Yes, they always do! I remember a comment from you, on a different topic, about “what the math tells us”. In this case, the math tells us that one observer sees radiation, and another observer doesn’t. It’s not a wild claim, and it’s not about “radiation of the right kind”. Just radiation. Observer #1 sees it, observer #2 doesn’t.

24. GerrardOfTitanServer says

Rob,
If observer #1 sees the radiation, how is it possible that they don’t see the detector issue an alert? From their perspective, they see radiation. They see the detector. They see the radiation strike the detector. They see all of the necessary and sufficient conditions for the detector to issue an alert. But they don’t see an alert? How is that possible?

25. Rob Grigjanis says

Bloody hell. Observer #1’s detector issues an alert (i.e. detects something). Observer #2’s detector doesn’t issue an alert (i.e. doesn’t detect anything). What part of this are you not getting? Is there some meaning of “alert” that differs from “detection”?

26. Rob Grigjanis says

Actually, never mind. I’m not wasting any more time on this.

27. GerrardOfTitanServer says

But why doesn’t observer #2’s detector issue an alert? Observer #1 sees radiation from the particle hit observer #2’s detector. Observer #1 should see observer #2’s detector alert.

28. Mano Singham says

Gerrard @#27,

I suggest you hang in there. These questions will be addressed in upcoming posts though you may of course not like the conclusions.

29. file thirteen says

Gerrard @27:

Insteas of imagining observers having detectors, imahine that the detectors are the observers. Observers don’t have to be people. And as already discussed, observer #2 doesn’t see the particle radiate. Does that help?

30. GerrardOfTitanServer says

File thirteen
I have seen the entire time what you’re getting at. I don’t think you see what I’m getting it. Let me put it in more opaque terms. If I shoot someone with a bullet from a gun, the guy getting severely injured is not just a matter of my perspective or not. There is a universal truth to the matter. If I am an observer, and if I see the bullet strike the target, then the bullet strikes the target. There should be no need for me to calculate what the target sees from the target’s perspective because I know the laws of physics in my (non-inertial) frame, and according to those laws of physics I know what bullets do to targets. And yet, I see the bullet magically disappear the moment that it strikes the target. That’s messed up. Worse, when I talk to the target, they say that they never saw any bullet in the first place.

I cannot imagine how there could be any relevant event horizon in play with basically zero spacetime curvature, assumed no expansion of spacetime, and relative speeds that are small relative to c. What does this event horizon business even mean? That the universe branches at the point of the experiment and that the two parts of the universe can never interact again? This cannot be. Not for classical physics in this scenario.

31. GerrardOfTitanServer says

>opaque
I meant clear.

32. Rob Grigjanis says

Gerrard @30: There’s nothing magically disappearing. Let’s define things carefully.

What follows all takes place in a flat spacetime; no QM, no curvature, no expansion.

Observer 1 (O1) and detector 1 (D1) are accelerating relative to any inertial frame, along with a charged particle.

Observer 2 (O2) and detector 2 (D2) are in an inertial frame.

D2 detects radiation from the particle, and emits a signal (like a flash). O1 (the accelerated observer) can even see that D2 has flashed. So there is no doubt, to anyone, that D2 has detected radiation. D1, on the other hand, has not detected radiation. And O2 can see that it hasn’t (no flash observed coming from D1). So again, no-one is in any doubt about D1. There is no contradiction about either event, or non-event. Everyone agrees that D2 flashed, and D1 didn’t.

Now, as to the existence of an event horizon. Imagine you are in deep space, in an inertial frame. A short distance away, on your z-axis, a spaceship begins accelerating away with a constant proper acceleration, along that same axis. You would see it accelerate, but in your frame, it’s acceleration would decrease, as it approaches the speed of light. It’s a simple consequence of the properties of a flat spacetime that no signal sent from a spacetime point with z < ct (in your frame) could ever reach the spaceship, as long as it kept up its proper acceleration. That’s all that is meant by ‘horizon’.

33. Rob Grigjanis says

Clarification: The scenario with the spaceship was assuming the use of a certain set of coordinates (Rindler) in which there is a relationship between the initial location of the ship in the inertial frame, and its proper acceleration. If the proper acceleration is α, the initial location is c²/α. Sorry for any confusion, but it merely amounts to a certain choice of origin for the inertial frame.

34. Rob Grigjanis says

…and in that case, if the proper acceleration was g, the initial distance to the ship would be about 1 light year, so not “a short distance”.

35. GerrardOfTitanServer says

Rob, we don’t need constant acceleration to the end of time. The observer just needs to accelerate for long enough for a radiation particle to be emitted from the charged test particle. Just accelerate long enough for the emission of a radiation particle, and then the acceleration of the observer can stop. The radiation particle doesn’t just disappear mid-flight, does it? We could time it so that the acceleration ceases before the radiation particle hits the detector according to all possible observers (eg part of the past light cone of the detector). The result is an inertial frame observer who sees a radiation particle hit a detector without the detector flashing. That makes no sense.

36. file thirteen says

Gerrard @35:

The result is an inertial frame observer who sees a radiation particle hit a detector without the detector flashing.

No. The inertial frame detector detects (“sees”) radiation hitting it and all observers see it flash. No observer ever “sees” radiation hit the other detector; it didn’t. If it had, the detector would have flashed.

Having said that, I don’t yet comprehend how there can be radiation that reaches the detector in the inertial frame, but no radiation locally within the accelerated frame to reach the detector there. Mano has hinted that this might be explained in a later post though.

37. Rob Grigjanis says

Gerrard @35: First of all, we’re not talking about radiation ‘particles’. We’re talking about classical em fields.

If O1 stops accelerating, the charged particle stops radiating. D1 still won’t detect radiation. And all inertial observers (now including O1) will agree that there is none.

We could time it so that the acceleration ceases before the radiation particle hits the detector

Which detector? No radiation is ever reaching D1, during or after acceleration.

38. Rob Grigjanis says

file thirteen @36: The classical em field of an accelerating charge, as seen in an inertial frame, has been known for over 100 years. The question is: what is the em field as seen by an accelerating (with the charge) observer? Various authors have calculated, by performing transformations on the em fields as seen by inertial observers, that the fields seen by the accelerating observer, are just static (non-radiating) ones.

At that point, one could say we’re finished. But yes, it is an apparent conundrum; there is radiation which inertial observers detect. From the accelerated observer’s point of view, where does it go?

Now I’m going to read Mano’s latest post on the topic!

39. Rob Grigjanis says

I’ll add that it’s tempting to think, these days, in terms of photons (“radiation particles”) rather than classical em waves. But quantum physics brings in additional subtleties, so we should stick to the classical theory for now. Looks like Mano will be doing that, at least for a bit. For what it’s worth:

In the classical theory, you calculate the em fields due to some distribution of charge, with the tacit assumption that you’re in an inertial frame, in which Maxwell’s equations hold. If you consider the fields (electric and magnetic) at a great distance R from the charges (say, where the distance is much greater than the size of the charge distribution), you can generally split them up into contributions with different R-dependence.

So there will generally be a part that falls off as 1/R², which reflects the net charge of the distribution, and higher powers, which reflect details of the distribution. But if there is a part that falls off as 1/R, that is the piece which corresponds to radiation. That part is generally due to accelerating charges.

An additional wrinkle is that the field you observe due to a particular charge depends, not on where it is now, but where it was roughly R/c seconds ago. It’s referred to as the retarded field, and is based on the notion that any field effects propagate at the speed of light.

40. file thirteen says

Thanks for another detailed explanation Rob. I said “I don’t comprehend how…”, and you’ve already got into the “how” when Mano’s latest post has only just got into the “why”!

However, if the detector does not detect anything, we cannot immediately infer that the charge did not radiate but need to look at the situation more closely to see if theory predicts that there should be radiation or not, since there may be other reasons for the lack of detection. The state of motion of the detector needs to be taken into account.

And that’s great pacing Mano, I’m keeping up with you… so far!

41. GerrardOfTitanServer says

Ok. New setup. Outer space far from any large masses. One charged test particle, one observer, and two detectors, all initially at rest relative to each other.

The observer fires his rocket for 1 sec of constant acceleration. The observer sees a disturbed portion of the EM field in the shape of a solid sphere around the test particle (eg a solid sphere of emotes photons). This sphere hits detector #1. Supposedly it does not flash. That makes no sense to me.

Then the observer drifts at constant velocity. Wait 1 second. The observer sees a disturbed region of the EM field in the shape of a hollow shell around the test particle (eg a hollow shell of photons). This doesn’t go away when the observer stops accelerating, right? Then the hollow shell strikes detector #2. Supposedly it also does not flash. That also makes no sense to me. Also, at this moment, we only have inertial frames. The inertial frame of the test particle and detectors, and the second inertial frame of the observer.

Then, the observer simply drifts until the end of time. Therefore, there are clearly no relevant event horizons.

42. Rob Grigjanis says

Gerrard, could you specify which of the things you’re talking about are accelerating with the observer?

What do you mean by “hollow shell of photons”? It’s as though you’re making things up to justify your conclusions. And we’re not talking QM here, just classical em theory.

If the particle is accelerating with the observer, the mathematics of the classical theory tells us that the observer (and their detector) sees no radiation, while the detector which remains at rest does see radiation for the 1 second duration of the acceleration.

The interpretation of this is not simple, and can’t be solved by making up “photon shells”. Yes, the observer will see that the detector at rest ‘flashed’ (if you like), and that their own detector didn’t. But you’re assuming that radiation is in some sense independent of frame of reference. That’s a dangerous assumption.

I’ll note here that the quantum physics version of this (i.e. in terms of photons) has been looked at, and also comes to the conclusion that the accelerated observer sees no photons. You could say that what you even mean by “a photon” is different between an inertial frame and an accelerated one.

43. GerrardOfTitanServer says

Rob
Let me try again.

New setup.

Observer, test charge particle, detector #1, and detector #2 begin stationary with respect to one another.

Observer begins constant acceleration at T=0 and ends constant acceleration at T_observer=1. Nothing else undergoes acceleration throughout the course of my thought experiment.

At T_observer=0.9, the observer sees a disturbed EM field (aka “emitted radiation”) in the shape of a solid sphere around the test particle. At T_observer=0.9, the observer sees the sphere of emitted radiation strike detector #1. At this time, the observer is undergoing constant acceleration.

AT T_observer=2, the observer sees a disturbed EM field (aka “emitted radiation”) in the shape of a hollow shell around the test particle. At T_observer=2, the observer sees the shell of emitted radiation strike detector #2. At this time, nothing is undergoing acceleration. Everything is in an inertial frame -- the test particle and detectors are in one inertial frame, and the observer is in a second inertial frame.

Because the acceleration happens only briefly (and not to time infinity), and because we’re assuming a flat space time without expansion, there are no (relevant) event horizons.

Does the “emitted radiation” completely disappear from the perspective of the observer as soon as the observer stops accelerating? That would be very weird.

Which detector(s) flash, if any? I think that you would argue that neither flash. I assume that they both must flash because of the physics of detectors according to the frame(s) of the observer, but then I would further argue that this leads to a contradiction because a hypothetical second observer who remained at rest with the test particle and detectors would see no such flash, and thus I would argue reductio ad absurdum for the conclusion that no observer sees the test particle emitting radiation. I would go further and say that it’s still an open question whether a uniformly accelerating charged particle emits radiation because the thought experiment that I set up involves a particle that does not accelerate according to an accelerometer.

44. Rob Grigjanis says

Observer begins constant acceleration at T=0 and ends constant acceleration at T_observer=1. Nothing else undergoes acceleration throughout the course of my thought experiment.

So the charged particle, and the two detectors, remain at rest? In that case, neither detector sees anything. If the (temporarily) accelerated observer has a detector, it will see radiation from the stationary particle, for the duration of the acceleration, according to Rohrlich.

You’re still bogged down by the notion that, if there is radiation in one frame, there must be radiation in every frame, accelerated or not. If you could just put aside that assumption for now, and follow Mano’s exposition, you’d be doing us all a favour.

45. GerrardOfTitanServer says

But then what of physics of the detector? What does it mean for the observer to see radiation strike the detector without a detector flash? How is that possible? In my setup, the observer who sees radiation is also in an inertial frame. This second inertial frame is just as special as the first inertial frame. By what magic can the detector know that it’s. It supposed to flash when hit by radiation?

46. GerrardOfTitanServer says

… it’s not* supposed to…

47. GerrardOfTitanServer says

Equivalently, what is different about the physics of the detector mechanism in the inertial frame of the observer so that the observer could determine in advance that the detector will not flash when hit by radiation? This is what I’m hung up on.

48. Rob Grigjanis says

What does it mean for the observer to see radiation strike the detector without a detector flash?

It doesn’t mean anything! The question is literally meaningless. If radiation hits it, it flashes. The observer can’t see radiation. They’re going by what the detector does. I suggest we follow file thirteen’s advice, and forget about the distinction between observer and detector. There are just detectors. They either go off or they don’t. If you like, you can imagine them recording their detections, so that a scientist can examine their history after their retrieval.

And there’s nothing “different about the physics of the detector mechanism in the inertial frame of the observer”. All that is different in different frames is the radiation.

49. GerrardOfTitanServer says