# When Does an Observer Become an Inertial Observer?

Discussion in 'Physics & Math' started by Mike_Fontenot, Aug 2, 2019.

1. ### Mike_FontenotRegistered Senior Member

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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? I.e., is it required that you must NEVER have accelerated in the past, and that you can guarantee that you will NEVER accelerate in the future?

Or, can you be an inertial observer if it has been long enough since you stopped accelerating, and if you can guarantee that you will not accelerate for some period of time into the future?

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.

What say you?

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3. ### SeattleValued Senior Member

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You have to not be accelerating at the moment that you call yourself an inertial observer.

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5. ### DaveC426913Valued Senior Member

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This is needlessly complicated.
If you are in free fall - i.e. not experiencing a force (acceleration or gravity) - then you are an inertial observer. Full stop.

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7. ### Q-reeusValued Senior Member

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Nowhere have you specified the spacetime metric. Must we assume it is Minkowski and not say exterior Schwarzschild? Because in the latter bodies can have zero proper acceleration but non-zero coordinate accelerations.
Assuming it is Minkowski, I'd say the only requirement for SR inertial frame formulae to be strictly valid, as 'snapshot' determinations, is both parties are inertial over a round trip radar duration. Which is not endorsing the notion of distant 'now' having any general usefulness.

Last edited: Aug 3, 2019
8. ### James RJust this guy, you know?Staff Member

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No. The only requirement is that you not be accelerating right now.

I don't really understand what you're talking about when you say "simultaneity at a distance". Can you explain?

Why don't they just say an observer is inertial if he is not accelerating now?

He can use any law of physics at all in the period of time in which he is not accelerating. That's a feature of inertial frames of reference.

9. ### Mike_FontenotRegistered Senior Member

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That's my contention also. But some physicists disagree.

When the traveling twin asks, at some instant in his life (for example, when he is 20 years old), "How old is my home twin RIGHT NOW?", he is asking a question about simultaneity at a distance.

In my opinion, physicists who put other restrictions on what is required for an observer to be an inertial observer, do so because they know that the MCMIFM assumption leads to the conclusion that, if the traveling twin suddenly increases his velocity in the direction AWAY from the home twin, that the home twin will suddenly get YOUNGER (according to the traveler). They find that result so ABHORRENT that they impose other restrictions on the definition of an inertial observer, in order to eliminate the possibility of negative ageing.

10. ### James RJust this guy, you know?Staff Member

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Mike_Fontenot:

Okay. I'm not familiar with these physicists who want to alter the definition of what it means to be an inertial observer, away from the usual definition used in general relativity. I'm also not aware of any need to do that.

It appears that you and I are in agreement.

11. ### QuarkHeadRemedial Math StudentValued Senior Member

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An imprecise statement, if I may be so rude!

Look - since acceleration is the second derivative of position with respect to time, and since the existence of a second derivative mandates the existence of the first derivative, and since the first derivative of position with respect to time is unaccelerated motion, then, at least in principle and provided times and distances are infinitesimal, then at that instant an accelerating body can be considered as inertial.

12. ### James RJust this guy, you know?Staff Member

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Well, yes, it's imprecise. I was trying to stick with the terms that were already being used in the discussion.

If you want more precision, just stick with the original definition: an inertial frame is one in which Newton's first law of motion (i.e. the law of inertia) holds.

The first derivative of position with respect to time is velocity, which can either be constant (unaccelerated) or changing (accelerated).

I don't think so.

13. ### QuarkHeadRemedial Math StudentValued Senior Member

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Huh? Acceleration as a first derivative? I think not

14. ### arfa branecall me arfValued Senior Member

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Velocity is an 'instantaneous' derivative; I think that means it pertains only to infinitesimal intervals of time.

So that an accelerating object also has instantaneous velocities (at each instant!); therefore it seems to follow that an object (or observer) is inertial at every instant.

Or something.

15. ### Mike_FontenotRegistered Senior Member

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An accelerating observer does have a velocity at each instant, but he can't use the time-dilation result for any finite interval of time that contains any acceleration during that interval. You can see this from two facts: First, the time dilation result says that the traveler will say that the home twin is ageing more slowly than he is. But while he is accelerating, the traveler will conclude that his home twin is ageing much faster than he is (either getting rapidly older, or getting rapidly younger). The traveler CAN, however, use the time-dilation result right up until the instant that he starts accelerating, and he can also use it immediately after he stops accelerating.

Another way to see that he can't use the time-dilation result during the entire trip is via the twin "paradox" itself. The home twin can use the time-dilation result for the entire round trip, and she knows that he will be the younger at the reunion. But if the traveler tries to do that, he will conclude that the she will be the younger one at the reunion, and they can't both be right. That's the apparent paradox. The resolution of the apparent paradox is that the traveler must conclude that the home twin instantly ages by a large amount during the instantaneous turnaround, just enough so that they will both agree at the reunion that he is the younger.

16. ### James RJust this guy, you know?Staff Member

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Please pay attention to what was actually written, which in this case was "the first derivative of position with respect to time is velocity,..."

If you have questions, I'm happy to answer them.

arfa brane:

More precisely, instantaneous velocity is the change in position over a time interval, in the limit as the time interval goes to zero. In other words, a standard derivative of position with respect to time. If you want to visualise it, it is the gradient at any point on a position vs time graph.

No. Go back to the basic definition of inertial frame, which I gave above. Since Newton's first law does not work for an accelerating observer, an accelerating observer is not an inertial observer.

17. ### arfa branecall me arfValued Senior Member

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I'd quibble with an instantaneous derivative corresponding to a change in position. I know that velocity is a gradient at a point (something I learned about when I did calculus and physics at uni).
But the first law must hold "instantaneously". It's only when velocities (and the gradients) change from point to point that a second derivative exists.
Since a gradient at a point does not correspond to a change in position, I think my work here is done . . .

18. ### James RJust this guy, you know?Staff Member

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I didn't mention anything like that. Here's the maths:

$\vec{v} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{x}}{\Delta t}$

Newton's first law is about motion. Motion doesn't take place "instantaneously".

See the mathematical definition above. That $\Delta \vec{x}$ means a change in position.

19. ### arfa branecall me arfValued Senior Member

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Hmm. Newton's calculus only makes sense if there are points in spacetime. Physically, a point in time does not make sense, only intervals do. Calculus seems to be able to overcome this by allowing an interval to get very small. If an object is in motion, it can't possibly ever "be at" a point.

But we have this instantaneous velocity thing--a gradient defined mathematically at a point. Do the points exist in reality? Does it matter? According to Zeno, it does (but his paradox has been resolved, with infinitesimals).

20. ### James RJust this guy, you know?Staff Member

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I don't see why there's any problem with points in time or space.

We need intervals when we start talking about any kind of rate of change (such as velocity = rate of change of position with respect to time). A rate of change says if the change one thing (like time) by a little bit, then some other thing (like position) changes by a certain amount, too.

Why not? Doesn't it have to pass through all points on its trajectory? (Restricting ourselves to classical physics for now.)

You tell me. Is it possible to specify position coordinates?

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..........

22. ### QuarkHeadRemedial Math StudentValued Senior Member

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Oh I did James. It was

23. ### arfa branecall me arfValued Senior Member

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And yet, velocity is an instantaneous derivative.

Although motion isn't instantaneous, velocity is; since velocity is instantaneous, so is acceleration.