Acceleration in SR makes it equivalent to GR?


by arindamsinha
Tags: acceleration, equivalent, makes
arindamsinha
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Dec19-12, 11:24 PM
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This thread is not about whether there is a difference between gravity and acceleration. I knew there are differences, even before starting this thread. However, if we extend such logic to extremes, we can falsely conclude that the equivalence principle itself is basically flawed and useless, because it treats two different phenomena as identical.

So when I am talking about the equivalence principle here, I am keeping within the same boundaries, where acceleration and gravity may be *considered* equivalent in terms of effects.

I will summarize my arguments in this thread below, but first need to respond to a few points in previous posts to set the context properly.

Quote Quote by tom.stoer View Post
This is a very problematic example. It's similar to purely accelerated motion w/o curvature, but it's not identical. The solution for the infinite plate is not identical to the Rindler coordinates for an accelerated observer in flat spacetime.

But in principle this doesn't matter b/c it's only one specific example; results based on this solution must not be generalized...
I am looking at it from the opposite direction. All I am arguing is that gravity can be non-tidal, so we cannot generalize the *usually* tidal nature of gravity as its differentiator from acceleration in the application of the equivalence principle. The equivalence principle holds, whether we are talking about tidal or non-tidal gravity. I am trying to eliminate the 'tidality' as a reason for refuting the logic I am putting forward.

Quote Quote by DaleSpam View Post
Suppose we were in a rocket and had a sphere of dust where each dust particle is a small accelerometer initially at rest wrt each other. Now, if the rocket were in free fall in the presence of tidal gravity the ball would stretch and distort shape while each accelerometer reads 0.

Now, according to you acceleration is equivalent to tidal gravity also. So, in flat spacetime, what acceleration profile would the rocket pilot use to make the sphere stretch while each accelerometer reads 0?


I am taking a non-tidal gravity example just to eliminate this type of complication. In the gravity field of an infinite plate, the gravitational acceleration and potential do not vary with distance from the plate. Therefore the ball of dust particles will not stretch as it accelerates.

Therefore, it should be possible to treat any acceleration as equivalent gravity, ignoring whether such gravity is tidal or not.

Quote Quote by PeterDonis View Post
...You can have "gravity" present in the sense of the equivalence principle, without having tidal gravity present.

The equivalence principle *does* hold "even if the gravity is not tidal". The EP doesn't talk about tidal gravity; it talks about felt acceleration. You can still feel acceleration in a spacetime that has no tidal gravity; that was my point.
Excellent. This is the point I was trying to make. Whether gravity is tidal or not doesn't matter in the equivalence principle. So let us eliminate tidality as a relevant factor when discussing the original question.

Quote Quote by PeterDonis View Post
Are you sure? Can you exhibit the solution of the Einstein Field Equation for this case that has a zero Riemann tensor?
The gravitational acceleration and potential caused by an infinite plate does not vary in magnitude with distance from the plate, and is therefore not tidal. As per my understanding, this means there is no spacetime curvature caused by such gravity. (You yourself stated in the previous post that spacetime curvature = *tidal* gravity.)

Quote Quote by PeterDonis View Post
Yes, please do.


OK, an infinite plate is not a realistic phenomenon in the Universe, so I will take a realistic example which we may consider an infinite plate approximation to a high degree of accuracy. The numbers used below are to give an idea of the magnitudes I imply, not to make it a mathematical exercise.

Take a large, high-density, flat disc-shaped galaxy, say 100,000 light years in diameter. We idealize it as a very large disc of uniform thickness and density (and no black holes present anywhere).

An observer 'A' is suspended say 1000 km away from the plane of the galaxy-size disc, and somewhere close to its centre/axis. 'A' can be considered to be in non-tidal gravity of an infinite plate (at least to a very large precision). The gravitational acceleration and potential at that location (and upto quite considerable distances in every direction) can be considered constants (> 0). This gravitational potential will cause A's time to dilate to a certain extent.

Now take observer 'B', who is far enough away, edge-wise from the galaxy disc, for gravitational potential to be reasonably considered zero. If B is at rest w.r.t. A, B's clocks will be faster.

To reach the same level of time dilation of A, B must travel and reach a particular velocity. An acceleration is necessary to reach that velocity from a rest state. This acceleration is what I am saying could also be considered as a gravity by the equivalence principle, even if of a non-tidal nature.


So, to summarize my logic on this thread, this is what I am saying:
  • SR, from its premises, deals with velocity (time dilation and other corresponding phenomena) only
  • In certain situations, acceleration is introduced in SR framework, either to resolve paradoxes or for many other purposes
  • By the equivalence principle, we can consider any such acceleration as being equivalent to gravity. Whether such acceleration is tidal or not does not come in the way, because even gravity can be non-tidal
  • In doing so, the distinction between SR and GR is blurred, and we are really dealing with the more general GR theory (which encompasses the velocity aspect of SR anyway)
What is wrong with this logic?
PeterDonis
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Dec20-12, 12:14 AM
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Quote Quote by arindamsinha View Post
This thread is not about whether there is a difference between gravity and acceleration.
That seems a little odd since your question in the OP was whether putting the concept of acceleration in SR makes it equivalent to GR. Since the only difference between SR and GR is that GR includes gravity while SR does not, it seems to me that the answer to your question in the OP hinges critically on whether there *is* a difference between gravity and acceleration. Of course, it's your thread and you're welcome to change the topic; is that your intent?

Quote Quote by arindamsinha View Post
So when I am talking about the equivalence principle here, I am keeping within the same boundaries, where acceleration and gravity may be *considered* equivalent in terms of effects.
Ok, but within these boundaries, we aren't using any of the features that make GR different from SR, so the answer to the question you posed in the OP would be simply "no".

Quote Quote by arindamsinha View Post
I am looking at it from the opposite direction. All I am arguing is that gravity can be non-tidal
The word "gravity" can have several meanings; but the only one that's relevant to GR, as opposed to SR, is the tidal one, since it's tidal gravity that makes the difference between flat spacetime and curved spacetime. So again, if you're using "gravity" in the non-tidal sense, GR is irrelevant to this discussion. However, that also creates problems for some of your scenarios; see next comment.

Quote Quote by arindamsinha View Post
I am taking a non-tidal gravity example just to eliminate this type of complication. In the gravity field of an infinite plate, the gravitational acceleration and potential do not vary with distance from the plate.
First, a clarification: if potential doesn't vary, then there is *no* acceleration, since gravitational acceleration is the gradient of the potential. I think you meant to say only that the acceleration doesn't vary with distance from the plate, and I'll assume that in what follows.

If we talk about the Newtonian solution for this scenario, yes, you're correct. But Newtonian gravity doesn't mix well with SR, so we don't really have a consistent framework for discussion.

If, OTOH, you want to talk about a GR solution, i.e., a solution to the Einstein Field Equation for a spacetime that is vacuum everywhere except on an infinite flat plate, I'm not sure that the gravitational acceleration is still independent of distance from the plate. But such a solution is the only way I'm aware of that we can have a consistent framework for discussion, so we would need to first find the appropriate solution to the EFE and agree on its properties.

Quote Quote by arindamsinha View Post
Excellent. This is the point I was trying to make. Whether gravity is tidal or not doesn't matter in the equivalence principle. So let us eliminate tidality as a relevant factor when discussing the original question.
But, as I noted above, if we do that there isn't anything left to discuss, at least not given your question in the OP. The answer is simply "no"; you can have acceleration without curved spacetime, so adding acceleration to SR does not make it equivalent to GR.

Quote Quote by arindamsinha View Post
The gravitational acceleration and potential caused by an infinite plate does not vary in magnitude with distance from the plate, and is therefore not tidal. As per my understanding, this means there is no spacetime curvature caused by such gravity.
See my comments above about whether this is really true of the appropriate solution to the EFE. Also see further comments below.

Quote Quote by arindamsinha View Post
OK, an infinite plate is not a realistic phenomenon in the Universe, so I will take a realistic example which we may consider an infinite plate approximation to a high degree of accuracy.
Fine, but you still need to exhibit an appropriate solution to the EFE and show that in fact there is zero spacetime curvature, at least in the vacuum region outside the plate. I'm assuming for the rest of this post that that can be done, but I'm not actually fully convinced it can be.

Quote Quote by arindamsinha View Post
To reach the same level of time dilation of A, B must travel and reach a particular velocity. An acceleration is necessary to reach that velocity from a rest state.
But not necessarily an acceleration that B will feel; he could just free-fall towards the plate. Is that what you have in mind? Or do you mean B fires rockets and feels acceleration?

(There are also issues regarding how B's "time dilation" is defined if he is not static relative to the plate.)

Quote Quote by arindamsinha View Post
This acceleration is what I am saying could also be considered as a gravity by the equivalence principle, even if of a non-tidal nature.
If B free-falls towards the plate, then of course his "acceleration" could be considered as gravity; it *is* due to gravity.

If B fires rockets, yes, he could do so in such a way that he feels the same acceleration as A (who must feel acceleration in order to be held static at his distance from the plate). And yes, locally, by the EP, B would not be able to tell the difference between his state and A's state. But if the "gravity" is in fact "non-tidal", then this scenario doesn't really involve GR at all, because B is in a flat spacetime.

(Btw, this brings up a key reason why I'm skeptical that this spacetime would actually be flat in the vacuum region outside the plate. If it is, A must also be in flat spacetime, but if so, he should not need to feel acceleration in order to stay at rest with respect to B, if B is floating freely very far away from the plate.)

Quote Quote by arindamsinha View Post
[*]SR, from its premises, deals with velocity (time dilation and other corresponding phenomena) only
This is not correct; acceleration is easily derivable from SR's premises. It's just the derivative of 4-velocity with respect to proper time. There's nothing extra that needs to be added to make that well-defined.

Quote Quote by arindamsinha View Post
[*]By the equivalence principle, we can consider any such acceleration as being equivalent to gravity.
Locally, yes. See further comments below.

Quote Quote by arindamsinha View Post
Whether such acceleration is tidal or not does not come in the way, because even gravity can be non-tidal
For the reasons I gave above, I'm not sure you can actually have "non-tidal gravity" like this over a finite region. I would want to see a solution of the EFE that actually showed this.
tom.stoer
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Dec20-12, 12:42 AM
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Quote Quote by arindamsinha View Post
So, to summarize my logic on this thread, this is what I am saying:
  • SR, from its premises, deals with velocity (time dilation and other corresponding phenomena) only
  • In certain situations, acceleration is introduced in SR framework, either to resolve paradoxes or for many other purposes
  • By the equivalence principle, we can consider any such acceleration as being equivalent to gravity. Whether such acceleration is tidal or not does not come in the way, because even gravity can be non-tidal
  • In doing so, the distinction between SR and GR is blurred, and we are really dealing with the more general GR theory (which encompasses the velocity aspect of SR anyway)
What is wrong with this logic?
Acceleration is not "introduced" in SR, it's an integral element; it has no different status than velocity (not w.r.t. reference frames).

The first problematic statement is "we can consider any such acceleration as being equivalent to gravity"; as I said a couple of times we must be very careful b/c their equivalence holds regarding local observations only. That means that local effects (observations) following from acceleration are locally equivalent to gravity, but gravity is not equivalent to acceleration in general b/c this would ignore effects which are not observable locally (like tidal effects).

The second problematic (wrong) conclusion is that it's logically wrong to infer a general conclusion (the statement regarading complete equivalence of gravity and accelerarion) from a very special premise (your "physical example", non-tidal gravity, ...). To summarize what's wrong with your logic: All owls are animals; but not all animals are owls.

In addition it's problematic (we haven't touched this issue so far!) that you seem to narrow your perspective to motion of test particles. That excludes dynamics of spacetime, i.e. GR as a field theory. There are phenomena (like gravitational waves which are measurable in principle) which do not follow from geodesic motion of test particles but require the Einstein field equations; narrowing gravity to local acceleration means that you have to exclude these effects, your that you may describe the observable effects on test particles but w/o explaining the underlying dynamics. To summarize what's wrong with this perspective: it's like being satisfied that the tv displays a picture (based on motion of electrons) but ignoring how it is broadcasted (based on electromagnetic waves).
haushofer
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Dec20-12, 04:13 AM
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Quote Quote by arindamsinha View Post
[*]SR, from its premises, deals with velocity (time dilation and other corresponding phenomena) only.[*]In certain situations, acceleration is introduced in SR framework, either to resolve paradoxes or for many other purposes.
[/B]
This is a very strange statement. In SR you have a class of observers which is equivalent: the inertial observers. However, this is also true in Newtonian physics. Conceptually there is absolutely no difference, only the group of symmetries is different (Galilei v.s. Poincaré). That's why you have time dilatation in SR, whereas in classical mechanics time is absolute.

I can replace "SR" with "classical mechanics" in your statements. How would you feel about it then?

And the analogy goes further. Introducing gravity into the Newtonian framework extends the Galilei transformations to include arbitrary accelerations. Introducing gravity into the SR framework extends the Poincaré transformations to general coordinate transformations. The equivalence principle also holds in Newtonian gravity, but it is more restricted than in GR: the class of observers which is equivalent is smaller compared to GR. To push the analogy to the extremes: I can reformulate Newtonian gravity to be spacetime curvature (Newton-Cartan).

You seem to have a funny notion of how acceleration is treated in mechanics.
sshai45
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Dec20-12, 04:57 AM
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So does this mean that we can use SR alone to answer the question of what an accelerating observer sees, and we don't need GR in that case, unless gravity is also involved? Because some people here were saying you need GR despite it not involving gravity.
haushofer
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Quote Quote by sshai45 View Post
So does this mean that we can use SR alone to answer the question of what an accelerating observer sees, and we don't need GR in that case, unless gravity is also involved? Because some people here were saying you need GR despite it not involving gravity.
I've not read all the posts, but to me that would be exactly the same as saying that in order to describe what accelerating observers in classical mechanics observe we need to invoke the Newtonian potential.

Locally we can do that due to the equivalence principle. We're not obliged to.

In that sense there is no difference between classical mechanics involving Newtonian gravity and General Relativity, I would say. But perhaps I'm missing something fundamental here.
DaleSpam
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Quote Quote by sshai45 View Post
So does this mean that we can use SR alone to answer the question of what an accelerating observer sees, and we don't need GR in that case, unless gravity is also involved?
Yes. That is correct.
DaleSpam
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Quote Quote by arindamsinha View Post
This thread is not about whether there is a difference between gravity and acceleration.
Then you need to write more carefully. I also thought that was exactly the topic of the thread.

Quote Quote by arindamsinha View Post
I am taking a non-tidal gravity example just to eliminate this type of complication. In the gravity field of an infinite plate, the gravitational acceleration and potential do not vary with distance from the plate. Therefore the ball of dust particles will not stretch as it accelerates.

Therefore, it should be possible to treat any acceleration as equivalent gravity, ignoring whether such gravity is tidal or not.
Non-tidal gravity is equivalent to acceleration, and can be treated with SR alone. However, I would object to your last phrase. You cannot ignore if gravity is tidal or not, it must be non-tidal. I.e. the spacetime curvature must be 0 for SR to apply.

Quote Quote by arindamsinha View Post
Excellent. This is the point I was trying to make. Whether gravity is tidal or not doesn't matter in the equivalence principle.
Again, no, it does matter. It must be non-tidal for the equivalence principle to hold.

Quote Quote by arindamsinha View Post
The gravitational acceleration and potential caused by an infinite plate does not vary in magnitude with distance from the plate, and is therefore not tidal. As per my understanding, this means there is no spacetime curvature caused by such gravity. (You yourself stated in the previous post that spacetime curvature = *tidal* gravity.)
There is tidal gravity (spacetime curvature) within the plate.

Quote Quote by arindamsinha View Post
So, to summarize my logic on this thread, this is what I am saying:
  • SR, from its premises, deals with velocity (time dilation and other corresponding phenomena) only
  • In certain situations, acceleration is introduced in SR framework, either to resolve paradoxes or for many other purposes
  • By the equivalence principle, we can consider any such acceleration as being equivalent to gravity. Whether such acceleration is tidal or not does not come in the way, because even gravity can be non-tidal
  • In doing so, the distinction between SR and GR is blurred, and we are really dealing with the more general GR theory (which encompasses the velocity aspect of SR anyway)
What is wrong with this logic?
Point 1 is wrong. SR can also deal with acceleration, otherwise the equivalence principle wouldn't even make sense.

Point 2 is correct.

Point 3 is incorrect, it must be non-tidal (flat spacetime) to use SR

Point 4 doesn't follow from the above. The distinction between GR and SR is the EFE and the resulting tidal gravity. If you are not using tidal gravity then you do not need the EFE and can simply assume a flat spacetime and use SR. If you use the EFE then you are doing GR. Of course, even when you are using the EFE the distinction is somewhat blurry as you can always consider a small enough region where tidal effects are negligible (as you propose above) and use SR locally within that region. So I would agree that the distinction can be blurry, but not for bulleted reasons.
Naty1
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Dec20-12, 07:08 AM
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arindamsinha:

I would simply say you are attempting to extend the equivalence principle beyond it's intent.

If you haven't seen it, you might find the discussions of various 'equivalence principles' in Wikipedia of interest: http://en.wikipedia.org/wiki/Equivalence_principle

But even here there seems to be much to object to, for example here in the first section of detailed discussion. I don't see any mention of 'local' equivalence here, so one reading this might easily draw incorrect inferences:

The equivalence principle was properly introduced by Albert Einstein in 1907, when he observed that the acceleration of bodies towards the center of the Earth .... is equivalent to the acceleration of an inertially moving body that would be observed on a rocket in free space being accelerated at a rate of 1g. Einstein stated it thus:
we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system.
—Einstein, 1907
stevendaryl
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Dec20-12, 08:34 AM
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Quote Quote by arindamsinha View Post
So, to summarize my logic on this thread, this is what I am saying:
  • SR, from its premises, deals with velocity (time dilation and other corresponding phenomena) only
  • In certain situations, acceleration is introduced in SR framework, either to resolve paradoxes or for many other purposes
  • By the equivalence principle, we can consider any such acceleration as being equivalent to gravity. Whether such acceleration is tidal or not does not come in the way, because even gravity can be non-tidal
  • In doing so, the distinction between SR and GR is blurred, and we are really dealing with the more general GR theory (which encompasses the velocity aspect of SR anyway)
What is wrong with this logic?
Number 1 is completely wrong. SR can deal with acceleration just as well as Newtonian physics can. Nobody would ever say that Newtonian physics only deals with unaccelerated motion--that would make it pretty useless, since the fundamental equation of Newtonian physics is: [itex]F = M A[/itex]. SR is a replacement for Newtonian physics, a modification of Newtonian physics. It wouldn't be a very good replacement if it didn't deal with accelerations. SR has it's own version of [itex]F=M A[/itex]: In inertial cartesian coordinates,

[itex]m \dfrac{d^2}{d\tau^2} X^\mu = F^\mu[/itex]

where [itex]X^\mu[/itex] is the coordinates of the object, [itex]F^\mu[/itex] is the 4-force, and [itex]\tau[/itex] is the proper time.

SR works perfectly well to describe acceleration, just like Newtonian physics does.

Now, what isn't straight-forward in SR is the use of an accelerated coordinate system. That is different from accelerated objects. You can describe accelerating objects using inertial coordinates. But what you can't do is to treat the accelerating object as if it were at rest using inertial coordinates. If you want to describe an accelerated object as if it were at rest, you need noninertial coordinates. SR can easily be extended to handle noninertial, curvilinear coordinates, just like Newtonian physics can. You can do Newtonian physics in accelerated or curvilinear coordinates, but if you do, there are additional terms in the equations of motion that are technically what's called connection coefficients, but which people often call "fictitious forces". Inertial "g" forces, centrifugal force, the Coriolis force, these are all not forces at all, but are just additional terms that appear in the equations of motion when you use noninertial, curvilinear coordinates. These additional terms are present in both SR and in Newtonian physics. SR no more requires GR to handle noninertial coordinates than Newtonian physics requires Newtonian gravity to handle noninertial coordinates. All it takes is calculus, knowing how equations of motion change form when you change coordinate systems.

So of course SR can handle accelerations! Of course SR can handle noninertial, curvilinear coordinates! The effects of acceleration on clocks is a prediction of SR alone, using noninertial coordinates. There is absolutely no need for GR to handle accelerated rockets!

The point of the equivalence principle is not to help SR deal with accelerations. SR deals with accelerations perfectly well without the equivalence principle, and the EP does nothing to change how SR deals with accelerations. This was a misconception in the early days of GR, but it's a fallacy. The so-called "GR" resolution to the twin paradox isn't GR at all---it's simply SR in noninertial coordinates.

What you can't do with SR alone is to deal with the effects of gravitational attraction between massive objects. I'm specifically using that phrase, because of course SR can deal with the pseudo-gravitational forces that result from using a noninertial coordinate system. But SR can't deal with "gravitational attraction", because there was no theory of gravity that is compatible with SR.

The equivalence principle says that for a test body (one that is small enough that it doesn't affect other objects very much), the effects of gravitational attraction in a small region of spacetime is the same as the affects of "fictitious forces" encountered in SR or Newtonian physics when you use noninertial coordinate systems.

So the point of the EP is not that it allows you to handle acceleration---you don't need the EP for that. The point is that it allows you to handle the effects of gravitational attraction, which you COULDN'T do without the EP.
nitsuj
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Dec20-12, 09:12 AM
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The quasi-force felt from gravity and proper acceleration is the same; that's the foundation that leads to the equivalence principle...the foundation in a chain of reasoning.

With any description I've read about the EP it is always very clearly stated this is subject to comparative issues with respect to accuracy (local) of measurement....big deal. The point of the EP is clear as day. To raise the point that it doesn't hold strictly is a weak point.

That doesn't dissolve the the point of the EP. Yes the nature of the curvature is is different, the effect is equivalent, and yes in spite of magnitudes not being equivalent. (that I don't now, i don't know math but Peter said they are not equal even if acceleration is of the same magnitude; the magnitude of the curvature would be different)

arindamsinha, with respect to the break in symmetry in the twin paradox.

Me high up in the sky with a clock, you on Earth with a clock we will both observe that your clock is running slower; it is not a symmetrical scenario. That is coordinate acceleration, equivalent to the proper acceleration in the twin paradox. In other words proper acceleration is not symmetrical. The situation does become symmetrical once the motion is again inertial.

During the acceleration on the return trip the traveling observer will see the at home observers time advance very very quickly, well beyond their own time. Once the acceleration has stopped and motion is again inertial (symmetrical) the traveling twin will again observe the at home observer's clock as ticking slowly, however the "accumulation" of time by the at home twin as observed by the traveling twin is so great that even upon arrival the at home twins "accumulation" of proper time is "ahead" of the traveling twins "accumulation" of proper time.

In other words after the acceleration the slowly tick clock of the at home twin doesn't tick slow enough to close the "gap" between the proper times of the twins.
stevendaryl
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Dec20-12, 10:14 AM
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Quote Quote by nitsuj View Post
During the acceleration on the return trip the traveling observer will see the at home observers time advance very very quickly, well beyond their own time. Once the acceleration has stopped and motion is again inertial (symmetrical) the traveling twin will again observe the at home observer's clock as ticking slowly, however the "accumulation" of time by the at home twin as observed by the traveling twin is so great that even upon arrival the at home twins "accumulation" of proper time is "ahead" of the traveling twins "accumulation" of proper time.
There was a discussion of "acceleration-dependent time dilation" in some other topic a while ago. It's superficially puzzling. You can explain the differential rates of clocks in the front and rear of an accelerating clock in terms of the relativity of simultaneity: Even if the two clocks are synchronized in the original rest frame, the clock in front will be ahead in the co-moving frame. But after the rocket STOPS accelerating, as you say, there is still a difference between the times on the two clocks that never goes away. (and this difference is visible even in the original rest frame) Where did that come from?

The answer is length contraction: In the original rest frame, the rocket's length decreases. The front of a rocket travels slightly less than the rear of the rocket. So it's average speed is slightly smaller. So it has slightly less time dilation.
tom.stoer
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Dec20-12, 11:24 AM
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I think my post #31 fully explains time dilation including non-inertial motion
PeterDonis
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Quote Quote by nitsuj View Post
The quasi-force felt from gravity and proper acceleration is the same; that's the foundation that leads to the equivalence principle
I think you are trying to say something correct here, but your language is sloppy. The EP does *not* say that free fall is equivalent to proper acceleration; an object that is only subject to "the quasi-force felt from gravity" is in free fall, so such an object is *not* equivalent to an object feeling proper acceleration. An object that is held at rest in a gravitational field does feel proper acceleration, and can be considered equivalent (locally) to an object in an accelerating rocket, but that acceleration is not "from gravity"; it's from whatever is holding the object at rest in the field by pushing on it (a rocket, the surface of the Earth, etc.).

Quote Quote by nitsuj View Post
Yes the nature of the curvature is is different, the effect is equivalent, and yes in spite of magnitudes not being equivalent. (that I don't now, i don't know math but Peter said they are not equal even if acceleration is of the same magnitude; the magnitude of the curvature would be different)
Again, I think you're trying to say something correct here, but your language is sloppy. (Also, I'm not sure you're paraphrasing what I said correctly.) The EP does *not* say that path curvature of a worldline (which is felt as proper acceleration by an object traveling along that worldline) is equivalent to spacetime curvature (which is observed as tidal gravity). They are two different things, and they are independent; one can have a worldline with path curvature in flat spacetime (no tidal gravity), and one can observe tidal gravity purely from the motion of free-falling objects (no path curvature).

Quote Quote by nitsuj View Post
Me high up in the sky with a clock, you on Earth with a clock [U]we will both observe that your clock is running slower
If you are both at rest with respect to each other and the Earth, yes. That means you must be stationary high up in the sky, either on top of a tall tower or using a rocket to hover.

Quote Quote by nitsuj View Post
That is coordinate acceleration, equivalent to the proper acceleration in the twin paradox.
This is wrong. Coordinate acceleration is *not* equivalent to proper acceleration. The two of you sitting at rest relative to each other and the Earth are feeling proper acceleration, but in a frame at rest with respect to you and the Earth, you have zero coordinate acceleration. If you, high up in the sky, drop a rock and watch it free fall, the rock has zero proper acceleration, but in a frame at rest with respect to you and the Earth, the rock does have coordinate acceleration.

Quote Quote by nitsuj View Post
In other words proper acceleration is not symmetrical.
Yes, if only one of the observers is feeling it. In other words, proper acceleration is different from free fall.

Quote Quote by nitsuj View Post
During the acceleration on the return trip the traveling observer will see the at home observers time advance very very quickly, well beyond their own time. Once the acceleration has stopped and motion is again inertial (symmetrical) the traveling twin will again observe the at home observer's clock as ticking slowly
Once more, I think you are trying to say something correct, but your language is sloppy. The traveling observer will actually *see* (as in, with a telescope) the home observer's clock running fast as soon as he is moving back homeward, even while he is moving inertially. (In fact, by idealizing the acceleration to be very short, its effect on what the traveling observer sees with his telescope becomes negligible.)

The traveling observer *calculates* that the home observer's clock is running slow compared to his, after correcting for the relativistic Doppler effect and light travel time delay, while he is moving inertially. While he is accelerating, *if* he adopts the view that he is at rest in a gravitational field, he can *calculate*, in the non-inertial frame in which he is at rest, that he is subject to a large "gravitational time dilation" relative to the stay-at-home twin, who is at a much higher altitude in the field than he is. But he does not see this directly; it's only a calculation, and it is frame-dependent.


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