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Acceleration in SR makes it equivalent to GR? 
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#37
Dec1912, 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. I am taking a nontidal 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. 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, highdensity, flat discshaped 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 galaxysize disc, and somewhere close to its centre/axis. 'A' can be considered to be in nontidal 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, edgewise 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 nontidal nature. So, to summarize my logic on this thread, this is what I am saying:



#38
Dec2012, 12:14 AM

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


#39
Dec2012, 12:42 AM

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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", nontidal 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). 


#40
Dec2012, 04:13 AM

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P: 901

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 (NewtonCartan). You seem to have a funny notion of how acceleration is treated in mechanics. 


#41
Dec2012, 04:57 AM

P: 64

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.



#42
Dec2012, 05:49 AM

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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. 


#43
Dec2012, 06:15 AM

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#44
Dec2012, 06:46 AM

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Point 2 is correct. Point 3 is incorrect, it must be nontidal (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. 


#45
Dec2012, 07:08 AM

P: 5,632

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: 


#46
Dec2012, 08:34 AM

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[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 4force, and [itex]\tau[/itex] is the proper time. SR works perfectly well to describe acceleration, just like Newtonian physics does. Now, what isn't straightforward 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 socalled "GR" resolution to the twin paradox isn't GR at allit'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 pseudogravitational 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 accelerationyou 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. 


#47
Dec2012, 09:12 AM

P: 1,098

The quasiforce 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. 


#48
Dec2012, 10:14 AM

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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. 


#49
Dec2012, 11:24 AM

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I think my post #31 fully explains time dilation including noninertial motion



#50
Dec2012, 11:53 AM

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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 noninertial frame in which he is at rest, that he is subject to a large "gravitational time dilation" relative to the stayathome 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 framedependent. 


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