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Inertial observers can legitimately use the famous time-dilation result of special relativity to determine simultaneity at a distance. Observers who are currently accelerating can't.

To be an inertial observer during some period of your life, do you have to be a PERPETUALLY inertial observer?

Or, can you be an inertial observer for some period of time, provided that you don't accelerate during that period?

The question matters, because the answer specifies WHO is entitled to use the famous time-dilation result, and WHEN can they use it, in order to determine simultaneity at a distance.

Different answers to that question have produced several different published procedures for answering the question, "How old is that particular distant person, who is moving with respect to me, RIGHT NOW?".

Dolby and Gull, in their "Radar Simultaneity", say that an observer is an inertial observer if he has not accelerated too recently, and will not accelerate too far into the future (and they exactly specify how much is too much). Dolby and Gull's method is clearly non-causal. Minguzzi says that an observer is an inertial observer if he hasn't accelerated too recently, but there is no requirement that he can't accelerate at any time in the future.

The "momentarily co-moving inertial frames montage" (MCMIFM) says that an observer is an inertial observer if he isn't CURRENTLY accelerating, even if he has accelerated infinitesimally-recently in the past, or will accelerate infinitesimally-soon in the future ... i.e., he can use the time dilatation result throughout any period of time in which he is not accelerating.

[...]I don't know these references.[...]

Maybe it would be more clear if you spelled out exactly which 'famous time-dilation result' you're talking about.

To be an inertial observer during some period of your life...

Quote from: HalcMaybe it would be more clear if you spelled out exactly which 'famous time-dilation result' you're talking about. I assume Mike is talking about different published solutions to the twins paradox, with one stay-at-home twin, and one travelling twin?

The other part of the problem is to obtain a common time reference between different observers, who may be traveling at different speeds, in different parts of the universe.As I understand it:- Two observers in the same inertial frame are able to synchronise their clocks accurately

- If you have two observers in two inertial frames who pass the same point at the same time, they are able to synchronise their clocks accurately.

- For most other cases, the clock setting is ambiguous- Throw in movement at varying speeds through different gravitational fields, and it is very hard to unambiguously synchronise clocks (and they won't stay synchronised!)

An example is me looking at Andromeda when it appears at night. I can see it without aid from here on a clear night. In my CADO reference frame, time is almost always moving backwards (quite quickly actually) up there at Andromeda.

Quote from: jeffreyH on 04/08/2019 12:51:28If this at all relates to the twins paradox then acceleration is unimportant. Twin A stays at home twin B flies off at relativistic speed to planet Z. As he passes planet Z he signals a ground station how long his trip took according to his clock.The ground station transmits back this information to twin A.Twin B carries on the same amount of distance again to space station X. As he passes X he again signals what his clock reads. X transmits this information back to twin A.Just a side note: Twin B is quite capable of sending signals to A directly without involving these relays at Z and X. The result is the same assuming these ground stations have no turnaround time for forwarding a message. So Z and X might as well be just milepost rocks sitting there.QuoteEven though the information takes time to get back to twin A this can be included in his calculations since the speed of light is known. His twin will be determined to be younger.Twin A could also be transmitting his clock readings at regular intervals like that, and since twin B knows the distance between them, he can similarly compensate for that distance when he gets the signals and compute that twin A is indeed younger than twin B. In other words, the situation is entirely symmetrical, and neither can deduce from these messages which is the one moving.

If this at all relates to the twins paradox then acceleration is unimportant. Twin A stays at home twin B flies off at relativistic speed to planet Z. As he passes planet Z he signals a ground station how long his trip took according to his clock.The ground station transmits back this information to twin A.Twin B carries on the same amount of distance again to space station X. As he passes X he again signals what his clock reads. X transmits this information back to twin A.

Even though the information takes time to get back to twin A this can be included in his calculations since the speed of light is known. His twin will be determined to be younger.

Quote from: MikeFontenot on 04/08/2019 15:50:25In the CADO reference frame, the Andromedons, according to us, would be rapidly getting younger ONLY if we were accelerating in a direction away from them.Exactly, and since my acceleration vector is more or less downward at any time I'm looking at the sky, Andomedons are always getting younger as I watch them. They're up and my downward acceleration makes me accelerating away.

In the CADO reference frame, the Andromedons, according to us, would be rapidly getting younger ONLY if we were accelerating in a direction away from them.

I'm not talking about gravity fields. If you like, pretend it's just the SR world and I'm attached to Earth with duct tape.