# When Does an Observer Become an Inertial Observer?

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

1. ### arfa branecall me arfValued Senior Member

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But that's avoided by instead showing that instants of acceleration are inertial, not periods or intervals.
QH has already posted evidence that Einstein addresses this back in 1912. But here we are.

3. ### exchemistValued Senior Member

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But actually what you say is interesting to me, inasmuch as there are (apparently) circumstances in which one can apply SR as if the system under consideration were not accelerating, even though it is.

5. ### exchemistValued Senior Member

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I can see where it is going, but none of it makes acceleration disappear and so, at least as far as I can see, the system is not inertial, even if only a single point on the circumference of the circle is considered.

However what I can also see, and maybe you can tell me if is this is relevant or not, is that the acceleration vector is perpendicular to the instantaneous tangential velocity, something that that would not be true in general for accelerated systems. Is that perhaps the key for why one can get away with the subterfuge of pretending the motion can be treated as inertial, when it comes to applying the Lorentz contraction?

7. ### Neddy BateValued Senior Member

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The acceleration vector does not have to be perpendicular to the instantaneous velocity. The Lorentz transformations only have velocity in them, (in addition to spacial and temporal coordinates), and so the velocity is all that is required.

For a rough example, imagine a rocket is blasting off a launch pad, accelerating constantly upward, and mission control is keeping track of its coordinates from earth. Mission control can calculate the rocket's Lorentz contraction at any time based solely on its instantaneous velocity, and in this case, the acceleration vector happens to point in the same direction as the instantaneous velocity. The same thing applies to the time-dilated tick-rate of the rocket's clock, its rate can be calculated based on its velocity.

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8. ### exchemistValued Senior Member

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Thanks for this.

So, if I understand what you are saying (you will have realised by now I am a neophyte where relativity is concerned), it is generally valid to apply the Lorentz contraction to systems that are in non-inertial frames of reference. Is that so?

9. ### Neddy BateValued Senior Member

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Special Relativity, or SR, (i.e. the Lorentz transformation equations) can be applied only when gravity is not present, or is small enough to be neglected. It does not matter how much coordinate acceleration might happen, even though an accelerometer would register that coordinate acceleration identically as it would if it were sitting at rest in a real gravitational field. Though I should add that SR calculations are much easier to do when the velocity is not constantly changing. This is somewhat analogous to algebra vs calculus. If you break down the accelerating rocket's trajectory into smaller and smaller sections, you can approximate them with small periods of constant velocity in SR.

However, if there is a real gravitational field, then General Relativity, or GR, equations apply. So my example of the rocket launching while mission control tracks it from earth was somewhat misleading, because the earth has a real gravitational field, though it might be small enough that it might be able to be neglected. I don't know whether its gravity would be significant or not in that case, because I don't know how to do GR, although I can usually do SR fairly well, for the simpler cases anyway.

Last edited: Aug 13, 2019
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10. ### exchemistValued Senior Member

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Aha, now we get to it. So you are saying the whole issue of inertial vs non-inertial frames is not relevant, regarding the application of Lorentz transformation equations.

The only thing now bothering me, from your reply, is that you seem to be saying there is a distinction between acceleration and "real" gravitation. I had thought the basis of GR was that the two are treated as indistinguishable.

11. ### exchemistValued Senior Member

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Now you are arguing a quite different point. Yes of course these instants of time, just like points in space, are subject to relativity and thus are perceived differently viewed from different frames of reference. But that does not mean they don't exist.

12. ### exchemistValued Senior Member

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So how about answering my question: what is the difference between your idea of an "instantaneous derivative" and any other derivative?

It is you that making this distinction.

13. ### Mike_FontenotRegistered Senior Member

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I thought exchemist was asking a slightly different question than what you answered. Here's what he asked:

"So, if I understand what you are saying [...], it is generally valid to apply the Lorentz contraction to systems that are in non-inertial frames of reference. Is that so?"

I THOUGHT he was asking if the TIME DILATION result can be applied when getting the accelerating observer's conclusion about the home twin's aging during the turnaround . (He used the term Lorentz CONTRACTION, which is normally used to describe LENGTH contraction, as opposed to TIME DILATION, but I think he was talking about time, not separation). If I'm right about what he was asking, the answer is obviously NO. When the traveler uses the famous time dilation result, he gets that the home twin (she) is ageing more slowly than he is. But during the turnaround, we know that he must conclude that she is ageing much faster than he is. (Because otherwise, the actual outcome at the reunion can't happen.) So the time dilation result CAN'T be used to determine what the accelerating traveler concludes during the turnaround.

14. ### exchemistValued Senior Member

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Thanks for the intervention but actually Neddy is right.

Quarkhead seemed to be arguing (in post 8) that one could treat accelerating frames of reference as inertial by considering infinitesimal intervals, and I thought this must be wrong. He then referred to Einstein's use of Lorentz contraction in rotating systems as evidence for this idea.

Neddy is now saying that the resolution of this argument is that yes, acceleration remains the same BUT that one can legitimately apply the Lorentz equations of SR to accelerating systems. It is only when gravitation is present that one cannot.

So all this is a bit of a side discussion from the main subject of the thread, I'm afraid, but I wanted to nail it down as it struck me as a bit crazy the way it was originally presented.

Last edited: Aug 13, 2019
15. ### Mike_FontenotRegistered Senior Member

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OK, I understand now what you meant. But just to be clear, you DO understand that the traveler in the twin "paradox" cannot use the famous time dilation result during his turnaround ... right?

16. ### arfa branecall me arfValued Senior Member

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What's the difference between the accepted (since Newton) idea of a time derivative being "instantaneous", and any other derivative--meaning I suppose a spatial derivative such as a gradient?
It's fairly obvious; it did occur to me some decades ago that's what my textbook and my lecturer meant, the difference is that first and second derivatives in physics usually mean time derivatives, which is why they're called instantaneous derivatives and have been called that for quite a long time.

I didn't make the distinction, everyone makes the distinction between time and space.
Jesus Christ this is fucking inane.

17. ### exchemistValued Senior Member

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That is what I have always understood, certainly.

In fact it came as something of a a surprise to me to learn that one can apply the Lorentz formula for length contraction to an accelerating frame of reference, such as a rotating one.

18. ### exchemistValued Senior Member

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I have never encountered the term instantaneous derivative, nor can I find a reference to it on the internet.

But now - at long last - you seem to suggest that what you mean by it is a derivative with respect to time, as opposed a derivative with respect to some other variable. If you'd made that clear several posts ago, we would not be having this fucking inane conversation.

19. ### Neddy BateValued Senior Member

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You are right that the founding basis of GR was the equivalence principle, which is the idea that coordinate acceleration in the absence of gravity would be locally indistinguishable from real gravity. This was the stepping stone Einstein used to take his special case (in which there was no gravity) and generalize it to the more general case which includes gravity. That is why SR it is called special relativity (as it is a special case) and GR is general relativity which is the general case that is supposed to cover everything.

If you and I were proficient in GR we would probably be saying right now that indeed there is no need to distinguish between acceleration and "real" gravitation. However, for those of us who are only proficient in SR, we look at it the other way around. SR is specifically not GR, and that is why we are supposed to exclude gravity from it.

Of course an accelerometer is going to register coordinate acceleration in the same way it would register gravity. But SR does not need to consider any accelerometers, because it is a kinetic theory which only relies upon instantaneous velocities. Changes in velocity will complicate things, but they can be dealt with in SR if one is very careful.

But gravity cannot simply be added in to SR as if it were just a regular change of velocity. On earth we experience 9.8m/s/s of acceleration, but if that were a coordinate acceleration going on constantly, it would only be a matter of time before our coordinate velocity would eventually surpass the speed of light. That is obviously completely the wrong way around.

The one thing I do know about GR is that the time dilation factor caused by actual gravity can be determined by the escape velocity. So, for example, on the surface of the earth the escape velocity is 11.2 km/s. Someone in deep space far away from any gravitational bodies would say that, because of our gravity, our clocks tick at a slower rate than their own, by the same factor as someone who was traveling at 11.2 km/s past them inertially, or even circling them at that speed, regardless of the radius. So once again we see that it is a velocity which determines the factor, not an acceleration.

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20. ### arfa branecall me arfValued Senior Member

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Is that because you haven't studied much physics? Perhaps it's because you couldn't type "what is an instantaneous derivative" in google's search bar?

Curiously, when I do that very thing I get more than 5,000,000 hits. That's pretty fucking inane, huh?
What I mean by that term is I'm quite sure, exactly what every textbook that uses it means too. Quite likely most of those > 5m links do as as well.
I'm reasonably sure Newton meant that same thing when he used it.

It is indeed also called the time rate of change. An instant of time is well-understood language; where the hell have you been? In inane land?

21. ### DaveC426913Valued Senior Member

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I didn't get 5,000,000 hits; I got zero. Google does not recognize the term.

I got lots of hits that say "Instantaneous Rate of Change: The Derivative" - but that's not the same thing at all.

(It's a bit of Googling "rate of speed" and getting lots of hits for "speed is a rate of change in distance over time". But "rate of speed" is nonsense.)

We're all on the same team here, guys.

Last edited: Aug 13, 2019
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22. ### arfa branecall me arfValued Senior Member

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Yes it is. It is the same thing and it always has been understood to be the same thing, in calculus.

You don't get that your quote is saying an instantaneous rate of change is also called an instantaneous derivative?
I'd say you're probably beyond help at this point.

Look, we're quibbling about language--language that has been used for a long time--and whether it makes sense to people who, it seems, haven't seen it before.
It's a bit like someone saying they've never heard of an electron or a proton before.
Seriously, you get introduced to the time derivative in physics fairly early on, like maybe in high school like I did. You come across this idea of an instant of time, "dt" as well; then it isn't too difficult (for most students) to come to grips with "instantaneous".

23. ### DaveC426913Valued Senior Member

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If you repeatedly and pointedly say "rate of speed" to your calculus people, you'll surely get a lot less polite responses than Exchemist gave - who simply asked for a reference, saying he (rightly) had never heard the phrase - nor has (at your derisive suggestion) Google.

I simply stated facts. Facts are not 'beyond help'.

I'm beginning to think you're not who I thought you were.

This would be a good time to say 'OK, maybe I didn't communicate my idea effectively. My bad. Let's start again.'

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