Acceleration in SR makes it equivalent to GR?

[arindamsinha]

Does putting the concept of acceleration in SR make it equivalent to GR?

I see a lot of sources (including many posts in this forum) which seem to say that SR can handle acceleration fine, but not gravity. However, an acceleration IS gravity by the equivalence principle, so what's the difference? (That historicallly EP was developed after SR does not matter, I think).

In this context, I have seen a lot of explanations saying that the Twin Paradox can be resolved within SR framework and does not require GR. The main differentiator though is that one of the twins preferentially 'feels acceleration'. Isn't it ultimately then putting the explanation in GR territory without openly acknowledging it?

 

[K^2]

SR can handle acceleration of world lines. It cannot handle acceleration of frames of reference. If you consider a space station from which a spaceship accelerates away, we have no problem describing this with SR from perspective of space station. The station does not accelerate, so the frame is inertial, and there is no problem. But if you try to describe the same situation from perspective of the accelerating ship, you are now in accelerated frame of reference, and you need GR to address it properly.

Gravity is caused by acceleration of the frame of reference. While standing on the ground, according to GR, the pull you are experiencing is due to you and ground accelerating upwards at 9.8m/s². So if you want to describe motion of object relative to ground, you are working in an accelerated reference frame.

 

[Bill_K]

In both Special and General Relativity there is simply no such thing as an accelerated reference frame in the ordinary sense. Any set of accelerated coordinates such as linear acceleration or rotation eventually exceeds the speed of light at large distances. Acceleration must be described as a collection of accelerating world lines, or equivalently as a collection of Lorentz frames, one at each point. This is not a problem, just a different set of concepts. If treated correctly, acceleration is well within the province of Special Relativity.

 

[K^2]

Any set of accelerated coordinates such as linear acceleration or rotation eventually exceeds the speed of light at large distances.

Or you can just deal with the fact that some 4-velocities will have imaginary components.

 

[stevendaryl]

Does putting the concept of acceleration in SR make it equivalent to GR?

I see a lot of sources (including many posts in this forum) which seem to say that SR can handle acceleration fine, but not gravity. However, an acceleration IS gravity by the equivalence principle, so what's the difference? (That historicallly EP was developed after SR does not matter, I think).

In this context, I have seen a lot of explanations saying that the Twin Paradox can be resolved within SR framework and does not require GR. The main differentiator though is that one of the twins preferentially 'feels acceleration'. Isn't it ultimately then putting the explanation in GR territory without openly acknowledging it?

You can think of the logical progression from SR to GR this way (this is not the way Einstein thought of it, but in hindsight it is a way to think about it):

1. Describe SR in inertial, cartesian coordinates.
2. Describe SR in accelerated, curvilinear coordinates. The physical content of SR is not changed in the transition from 1. to 2. If you know how SR works in inertial, cartesian coordinates, then you can derive how it works in accelerated, curvilinear coordinates by just using calculus. There is no new physics. However, using curvilinear coordinates allow us to describe how things look from the point of view of an accelerated observer. What we find when we do this is:
   • If you drop an object, it appears to fall away as if there were a universal force (acting the same way on all objects) were acting on it.
   • There are apparent velocity-dependent forces, such as the Coriolis force, Centrifugal force.
   • Clock rates appear to be position-dependent.
   Once again, these effects are not new physical predictions. They are simply the predictions of ordinary SR described in different coordinates.
3. Note that SR in curvilinear coordinates has a lot of the same features as Newtonian gravity:
   • There is an apparent force affecting all objects.
   • This force is universal, leading to the same (coordinate) acceleration for all objects, regardless of their mass or composition.
4. Descrbe SR in curved spacetime. Once you have figured out how to do SR using curvilinear coordinates, it's not a huge leap to asking how to do SR in curved spacetime. Here an analogy with 2D Euclidean geometry might help. Most of Euclid's geometry is about flat, 2D spaces. So knowing Euclidean geometry, how can we analyze things on a curved surface such as the surface of the Earth? The key insight is this: any small enough region on a curved surface will look approximately flat. (This actually isn't true if the surface is fractal, but let's ignore that complication, since I don't know how to deal with it.) So a recipe for analyzing geometry on the surface of the Earth is:
   • Partition the surface into lots of overlapping small regions.
   • Use ordinary flat 2D geometry within a region.
   • Use curvilinear coordinates to interpolate between neighboring regions.
   The same recipe can be used to do SR in curved spacetime: Partition spacetime into lots of little regions of spacetime. Apply SR within each region. Use curvilinear coordinates to interpolate between neighboring regions.
   
   This development actually goes beyond SR, but not a whole lot beyond SR. If you know how some physical theory, such as Maxwell's equations, works in SR, then you can come up with a pretty good guess as to how it works in curved spacetime. It's not uniquely determined what the theory ought to be like in curved spacetime, because there might be effects that couple to the curvature tensor that you can't guess just from knowing the flat spacetime version of the theory. But you can get a good idea of how things work in regions where the curvature is negligible.
5. Speculate that gravity is nothing but a manifestation of curved spacetime. As noted in Development 2., SR in curvilinear coordinates has some of the same features as gravity. But it's "fake" gravity, because it can always be eliminated through choosing inertial cartesian coordinates. On the other hand, doing SR in curved spacetime requires the use of curvilinear coordinates when considering a large enough region of spacetime. So the "fake" gravitational effects cannot be eliminated completely. So one might guess that "real" gravity is just a manifestation of noninertial, curvilinear coordinates that are forced on us because spacetime is curved. This guess is the Equivalence Principle.
6. Describe the dynamics of the curvature tensor. Development 3. is only half-way to General Relativity. It describes physics in a fixed curved "background" spacetime. It tells you how spacetime curvature affects physics (or at least, it gives you an approximate idea), but it doesn't tell you how physics affects spacetime curvature. Once you have an equation describing how matter and energy affects the curvature tensor, then you've got GR.

This is not the historical order of things: Einstein guessed the equivalence between gravity and noninertial coordinates (Development 4) before systematically developing the theory of curved spacetime (Development 3). Developments 3 and 5 were never done separately--Einstein didn't explore SR in curved spacetime before trying to come up with equations describing the curvature.

To do physics on board an accelerating rocket requires only developments 1 and 2. To do physics near the surface of the Earth requires the Equivalence Principle (development 4).

 

[nitsuj]

In this context, I have seen a lot of explanations saying that the Twin Paradox can be resolved within SR framework and does not require GR. The main differentiator though is that one of the twins preferentially 'feels acceleration'. Isn't it ultimately then putting the explanation in GR territory without openly acknowledging it?

I would guess the issue of acceleration is used to illustrate the "break in symmetry".

The details of how acceleration breaks the symmetry can be as simple as this twin felt acceleration.

 

[PeterDonis]

However, an acceleration IS gravity by the equivalence principle, so what's the difference?

This is only true locally. Once you go beyond a small local patch of spacetime, there *is* a difference between acceleration and gravity: gravity requires spacetime curvature. SR can only deal with flat spacetime; to deal with curved spacetime requires GR. That's the difference.

In this context, I have seen a lot of explanations saying that the Twin Paradox can be resolved within SR framework and does not require GR.

The standard twin paradox scenario is set in flat spacetime, so yes, SR can resolve it just fine. There are alternate versions set in curved spacetime which require GR to resolve.

The main differentiator though is that one of the twins preferentially 'feels acceleration'. Isn't it ultimately then putting the explanation in GR territory without openly acknowledging it?

No; observers can feel acceleration in flat spacetime. They just need to turn on their rocket engines.

 

[Dale]

However, an acceleration IS gravity by the equivalence principle, so what's the difference?

The difference is tidal gravity. If you have no tidal gravity then you have a flat spacetime and can use SR. If there is tidal gravity then spacetime is curved and you need GR and the EFE.

 

[nitsuj]

In books I have read that mention the "Weak / Strong Equivalence Principle" mention with a "big BUT" this is only true locally, just like Peter Donis already mentioned.

As you've seen it is a great segue into similar GR principals...locally


Curved spacetime in SR = Ship is inertial fires photon and accelerates while photon is en route. Not for real, but the picture is great!

 

[PeterDonis]

Curved spacetime in SR = Ship is inertial fires photon and accelerates while photon is en route.

This isn't curved spacetime. The worldline of the ship is curved--it has nonzero proper acceleration, for at least a portion of it--but spacetime itself is still flat.

 

[nitsuj]

This isn't curved spacetime. The worldline of the ship is curved--it has nonzero proper acceleration, for at least a portion of it--but spacetime itself is still flat.

Never said it was...
"Not for real but the picture is great!" and it is...weak equivalence.

Dalespam already improved the distinction with mentioning tidal forces/

 

[haushofer]

Hi,

check out these notes by 't Hooft,

http://www.staff.science.uu.nl/~hooft101/lectures/blackholes/BH_lecturenotes.pdf

chapter 3. I think that answers your question :)

 

[PeterDonis]

Never said it was...

Well, you did say "curved spacetime in SR". What was that supposed to mean? If you were drawing an analogy between spacetime curvature in *GR* (not SR) and path curvature of an accelerated worldline, I don't think that analogy is valid.

 

[nitsuj]

Well, you did say "curved spacetime in SR". What was that supposed to mean? If you were drawing an analogy between spacetime curvature in *GR* (not SR) and path curvature of an accelerated worldline, I don't think that analogy is valid.

oh comon Peter, pretty sure it's considered equivalent... at least in principle

The photon makes a nice curved path & visually is great picture of curvature. True it is not because of the spacetime, merely because the scenario uses proper acceleration to get the curved path.

 

[PeterDonis]

oh comon Peter, pretty sure it's considered equivalent... at least in principle

And I'm pretty sure it's not. Path curvature and spacetime curvature are different things. They are not equivalent, and pretending that they are can easily cause confusion. I've seen it happen on previous threads here on PF.

The photon makes a nice curved path & visually is great picture of curvature.

Of *path* curvature. *Not* of spacetime curvature. (And even path curvature in this case is problematic because the photon is traveling on a geodesic; it's the observer's path that is curved.)

True it is not because of the spacetime

Exactly. If you want to claim that the two cases are nevertheless somehow "equivalent", the burden of proof is on you to demonstrate the equivalence. Just waving your hands and saying you're "pretty sure" is not sufficient.

 

[nitsuj]

burden of proof! so formal

Peter, I'll have you know I didn't merely wave my hands; I also stomped my feet!

I'm going on proper acceleration "force" is same as gravitational "force", that the ship observer couldn't distinguish the difference between the curved path of the photon being because of curved space time or proper acceleration just by looking at the photon path.

 

[PeterDonis]

I'm going on proper acceleration "force" is same as gravitational "force", that the ship observer couldn't distinguish the difference between the curved path of the photon being because of curved space time or proper acceleration just by looking at the photon path.

And this is incorrect. By hypothesis, the photon path is a geodesic; that means it has zero proper acceleration. The proper acceleration of any path is an invariant, so any observer can in principle calculate it and get the same answer, including the ship observer.

You have just exhibited the confusion I was talking about.

 

[grav-universe]

Does putting the concept of acceleration in SR make it equivalent to GR?

I see a lot of sources (including many posts in this forum) which seem to say that SR can handle acceleration fine, but not gravity. However, an acceleration IS gravity by the equivalence principle, so what's the difference?

Acceleration in SR is equivalent to gravity in a constantly accelerating field for a single accelerating observer, for instance, right. The gravitational field around a mass, however, is not constant, but varies with distance from the center, so SR is only valid locally, neglecting the tidal gradient of the field, while GR relates the positions and accelerations between many static observers. That is to say, static observers at varying distances from the center of a gravitational field will measure different local accelerations, so for a body in freefall, we must not only apply SR or the equivalence principle for the local acceleration of a single static observer, but we must apply full GR in order to relate the different accelerations of a family of static observers at varying distances to each other in order to determine what each measures of the freefalling body at different times or at different places while falling through the field.

 

[PeterDonis]

Acceleration in SR is equivalent to gravity in a constantly accelerating field

More precisely, proper acceleration in free space is equivalent, locally, to being *at rest* in a static gravitational field of the appropriate strength to require the same proper acceleration to hold you at rest. So, for example, if you are inside your rocket ship and are feeling a 1 g acceleration, you can't tell, locally, whether that's due to your rocket engine firing in free space, or your rocket sitting at rest on the surface of the Earth. But as you point out, if you are allowed to make measurements over a finite region, you can tell the difference by the presence of tidal gravity in the second case, but not in the first.

 

[Naty1]

This is a very interesting discussion. I have tried to compare statements above with explanations I have in my notes...some successful, some I am still puzzling about.

Dalespams is one of the puzzles:

[However, an acceleration IS gravity by the equivalence principle, so what's the difference? The difference is tidal gravity. /QUOTE]

If this can be taken as , 'one of the differences' instead of 'the difference' I think I get it.


I have two comments and would be interested in reactions...one relates to time, one to space...

[If I had to fit these into the existing discussion, I'd Have to stuff them in as details in stevendaryl's point 6.]

[This are from somewhere in Wikipedia]:

Experimental data shows that time as measured by clocks in a gravitational field—proper time, to give the technical term—does not follow the rules of special relativity. In the language of spacetime geometry, it is not measured by the Minkowski metric. ...The metric tensor of GR ...is not the Minkowski metric of special relativity,

[This is separate from the 'curvature tensor', right??]

In Riemannian geometry, the scalar curvature (or Ricci scalar) is the simplest curvature invariant of a Riemannian manifold. ...In two dimensions, the scalar curvature is twice the Gaussian curvature, and completely characterizes the curvature of a surface…..When the scalar curvature is positive at a point, the volume of a small ball about the point has smaller volume than a ball of the same radius in Euclidean space

and I am reminded of Einstein's predictions confirmed by Eddington's observations during an eclipse...

 

[Dale]

If this can be taken as , 'one of the differences' instead of 'the difference' I think I get it.

I cannot think of another difference, but I would certainly feel comfortable with a weaker statement, like the one you suggest.

 

[nitsuj]

And I'm pretty sure it's not. Path curvature and spacetime curvature are different things. They are not equivalent, and pretending that they are can easily cause confusion. I've seen it happen on previous threads here on PF.



Of *path* curvature. *Not* of spacetime curvature. (And even path curvature in this case is problematic because the photon is traveling on a geodesic; it's the observer's path that is curved.)



Exactly. If you want to claim that the two cases are nevertheless somehow "equivalent", the burden of proof is on you to demonstrate the equivalence. Just waving your hands and saying you're "pretty sure" is not sufficient.

Wow, just finished reading about Einsteins insight regarding the equivalence principle & this is exactly how it was proved. Eddington checked for curvature caused by the sun.

Einstein thought of the scenario regarding the inertial ship, firing a photon, and accelerating at a right angle to the direction of the photon and that resulting in a curved path...and the it would be EQUIVALENT to the geometric curvature caused by gravity.

It sucks, having taken your spat that this is not true as being an informed response.

It doesn't even remotely "jive" with the history of the development & subsequent proof of the equivalence principle. That is, the curvature of the photon caused by acceleration is equivalent to the curvature to gravity, just like I said but that you said was incorrect.

So, I guess the burden of proof is on you then?

 

[nitsuj]

More precisely, proper acceleration in free space is equivalent, locally, to being *at rest* in a static gravitational field of the appropriate strength to require the same proper acceleration to hold you at rest. So, for example, if you are inside your rocket ship and are feeling a 1 g acceleration, you can't tell, locally, whether that's due to your rocket engine firing in free space, or your rocket sitting at rest on the surface of the Earth. But as you point out, if you are allowed to make measurements over a finite region, you can tell the difference by the presence of tidal gravity in the second case, but not in the first.

So what again was your dispute with what I had said?

 

[PeterDonis]

Wow, just finished reading about Einsteins insight regarding the equivalence principle & this is exactly how it was proved.

Reference, please? I suspect you are misinterpreting, or at least mis-stating, whatever it is you are reading.

Einstein thought of the scenario regarding the inertial ship, firing a photon, and accelerating at a right angle to the direction of the photon and that resulting in a curved path...and the it would be EQUIVALENT to the geometric curvature caused by gravity.

You are (or at least appear to be) mis-stating what Einstein claimed was equivalent to what. He claimed that being in a rocket ship under acceleration was equivalent to sitting at rest in a gravitational field. That meant that, since a photon's path looks curved to the observer in the rocket ship, it should also look curved to the observer at rest in the gravitational field.

However, that is not the same as claiming that the path curvature of the accelerated observer's worldline is equivalent to the spacetime curvature that makes the photon's path *look* curved. The photon's path is *not* curved in any physical sense; it's straight. So it has no curvature to be equivalent to any other kind of curvature. There is nothing in flat spacetime that is "equivalent" to spacetime curvature.

It's also worth noting that the "equivalence" reasoning given above, for why the path of a photon passing close to a massive object like the Sun should look "curved", doesn't actually give the correct numerical answer. The correct answer for how "curved" the photon's path looks is *twice* the answer you get just from the "equivalence" reasoning.

It sucks, having taken your spat that this is not true as being an informed response.

It sucks, having someone make claims about something he read without actually giving me a chance to read the source to see whether he is interpreting it correctly.

So, I guess the burden of proof is on you then?

No, the burden is on you to provide an actual reference. Again, I suspect you are misinterpreting or at least mis-stating what you are reading, but I can't tell without being able to read it myself.

 

[PeterDonis]

So what again was your dispute with what I had said?

You appeared to be claiming that path curvature is equivalent to spacetime curvature. It isn't.

 

[arindamsinha]

Great discussion and inputs. However, I would like to pull out a few specific points which seem very relevant to my original question.

SR can handle acceleration of world lines. It cannot handle acceleration of frames of reference. If you consider a space station from which a spaceship accelerates away, we have no problem describing this with SR from perspective of space station. The station does not accelerate, so the frame is inertial, and there is no problem. But if you try to describe the same situation from perspective of the accelerating ship, you are now in accelerated frame of reference, and you need GR to address it properly.

..

That is tantamount to saying the Twin Paradox requires GR for explanation, when seen from the traveling twin's perspective. Not what I have seen anywhere else, as the resolution seems to have been within SR from both observers' perspective, inclusive of acceleration. However, very relevant to the question I asked, if this is true. Will wait to hear more.

You can think of the logical progression from SR to GR this way...

Very good historical summary. I will look through in detail later and see if it raises further questions.

I would guess the issue of acceleration is used to illustrate the "break in symmetry"...

You highlight one of the points I am trying to get answered - once the acceleration stops, why does the 'symmetry remain broken'?

The acceleration (assuming it is quick and sharp) is not causing much of the time dilation, velocity is. So what has the acceleration changed? Are we not going into GR territory when we attempt to answer this question?

This is only true locally. Once you go beyond a small local patch of spacetime, there *is* a difference between acceleration and gravity: gravity requires spacetime curvature...

While this is true around a large spherical mass, I cannot agree to this in general. Gravity of an infinite (or say very large) flat object causes no real spacetime curvature, but does create gravitational potential. The equivalence principle will still hold. (I can expand on this with an example, if needed).

No; observers can feel acceleration in flat spacetime. They just need to turn on their rocket engines.

... and the moment they do, spacetime is no longer flat between them and any inertial frames nearby. This may sound like heresy, but let's think about this. We can consider the accelerating rocket to be in a different gravity by the equivalence principle. This is the crux of my original question.

The difference is tidal gravity. If you have no tidal gravity then you have a flat spacetime and can use SR. If there is tidal gravity then spacetime is curved and you need GR and the EFE.

Again, I believe that the equivalence principle will still hold, even if the gravity were not tidal, and we would not be able to differentiate between an acceleration and gravity. My answer to PeterDonis's point above applies to this as well.

So when we are talking about acceleration and gravity equivalence, the tidal nature, though important in some cases, is not relevant to my original question.

 

[tom.stoer]

I think the problem is the following: what the equivalence principle says is that one cannot distinguish between the effects of a gravitational field and acceleration using purely local experiments and observations; but the equivalence principle does not say that acceleration and gravity are the same!

You cannot distinguish between rain and a sprinkler just by looking at the wet grass; but taking more than just one single observation into account, you can ;-)

 

[arindamsinha]

I think the problem is the following: what the equivalence principle says is that one cannot distinguish between the effects of a gravitational field and acceleration using purely local experiments and observations; but the equivalence principle does not say that acceleration and gravity are the same!

You cannot distinguish between rain and a sprinkler just by looking at the wet grass; but taking more than just one single observation into account, you can ;-)

Cannot agree on this. Look at my points on the previous post about the gravity of a very large flat object (or infinite plane). The above experimentation will not allow differentiation between that gravity and acceleration. The equivalence principle works irrespective, whether you can experimentally differentiate gravity and acceleration, or not. This is only sidestepping the point I am trying to have discussed in this topic.

 

[tom.stoer]

In general there are experiments where you CAN distinguish between gravity and acceleration. The acceleration, e.g. due to propulsion of a rocket, acts on the rocket as a whole; it causes a curved non-geodesic motion of this rocket in spacetime; this is totally different from gravity which may result in tidal forces, distorting the rocket.

It's only due to artificial field configurations, restricted observations, purely local experiments etc. that gravity and acceleration look identical.

Do you understand the difference between "looking the same locally" and "being identical in general"?

 

[A.T.]

No; observers can feel acceleration in flat spacetime. They just need to turn on their rocket engines.

... and the moment they do, spacetime is no longer flat between them and any inertial frames nearby.

Keep in mind that that space time can be "flat" (no intrinsic curvature), but still "distorted" so it produces gravity. A simplified example is the cone surface:

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif

From: http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

 

[tom.stoer]

That is tantamount to saying the Twin Paradox requires GR for explanation, when seen from the traveling twin's perspective. Not what I have seen anywhere else, as the resolution seems to have been within SR from both observers' perspective, inclusive of acceleration. However, very relevant to the question I asked, if this is true. Will wait to hear more.

GR is not required for resolution of the twin paradox even for accelerated or non-inertial motion (GR is only required for curvature effects i.e. non-flat spacetime)

The acceleration (assuming it is quick and sharp) is not causing much of the time dilation, velocity is. So what has the acceleration changed? Are we not going into GR territory when we attempt to answer this question?

Acceleration does in no way affect time dilation; the formula is

$$\tau = \int_0^t dt\,\sqrt{1-\vec{v}^2/c^2}$$

Here τ is the proper time a moving (possibly accelerated) clock, t is the coordinate time, i.e. the proper time of an inertial clock. As you can see only speed is relevant.

Gravity of an infinite (or say very large) flat object causes no real spacetime curvature, but does create gravitational potential. The equivalence principle will still hold. (I can expand on this with an example, if needed).

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

... and the moment they do, spacetime is no longer flat between them and any inertial frames nearby. This may sound like heresy, but let's think about this. We can consider the accelerating rocket to be in a different gravity by the equivalence principle. This is the crux of my original question.

I think I begin to understand you problem: accelerated motion in flat spacetime is still described by flat spacetime (i.e. a spacetime with vanishing curvature). Acceleration causes the metric to become noin-trivial, but it remains flat. A simple example a polar coordinates which are non-trivial co,mpared to cartesian coordinates, but which still describe the same flat space!

So accelerated observers in flat spacetime are accelerated observers in flat spacetime ;-)

So when we are talking about acceleration and gravity equivalence, the tidal nature, though important in some cases, is not relevant to my original question.

OK, you original question was

Does putting the concept of acceleration in SR make it equivalent to GR?

No. Acceleration of observers (clocks, ..) is part of SR already.

... seem to say that SR can handle acceleration fine, but not gravity.

Yes

However, an acceleration IS gravity by the equivalence principle, so what's the difference?

No! As I said, the equivalence principle only says that one cannot distinguish gravity and acceleration locally; that does not imply that they are identical globally.

Wikipedia says "The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception"

... saying that the Twin Paradox can be resolved within SR framework and does not require GR.

Yes, this is true

The main differentiator though is that one of the twins preferentially 'feels acceleration'.

I don't like this explanation. The situation is best described in terms of a simple formula: Two twins A and B are traveling along two arbitrary world lines C_A and C_B. Start point and end point are identical, the curves are arbitrary, except for the fact that v < c must hold. That means that in principle both twins could be accelerated w.r.t. Now we introduce an inertial frame (one twin can be located in the origin of this frame, but that's not required). The result does not depend on this frame, it can be formulated frame-independent, but for practical calculations and simple explanations a frame is required.

Then the two proper times τ_A and τ_B measured by the twins A and B along their world lines C_A and C_B are

$$\tau_{A,B} = \int_0^t dt\,\sqrt{1-\vec{v}_{A,B}^2/c^2}$$

If one twin himself defines the inertial frame then his v is zero and his proper time is equal to the coordinate time.

Formulated that way there is no paradox to be resolved, there's only a formula for the calculation of proper times.

 

[Dale]

Again, I believe that the equivalence principle will still hold, even if the gravity were not tidal, and we would not be able to differentiate between an acceleration and gravity. My answer to PeterDonis's point above applies to this as well.

So when we are talking about acceleration and gravity equivalence, the tidal nature, though important in some cases, is not relevant to my original question.

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?

 

[PeterDonis]

... and the moment they do, spacetime is no longer flat between them and any inertial frames nearby.

This is incorrect. Spacetime is still flat when the rocket engine turns on. The rocket's *worldline* becomes curved instead of straight, but that doesn't produce any curvature of spacetime.

This may sound like heresy, but let's think about this. We can consider the accelerating rocket to be in a different gravity by the equivalence principle.

"Different gravity" in this sense is *not* the same as "different curvature of spacetime". Spacetime curvature = *tidal* gravity. You can have "gravity" present in the sense of the equivalence principle, without having tidal gravity present.

Again, I believe that the equivalence principle will still hold, even if the gravity were not tidal

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.

 

[PeterDonis]

Gravity of an infinite (or say very large) flat object causes no real spacetime curvature

Are you sure? Can you exhibit the solution of the Einstein Field Equation for this case that has a zero Riemann tensor?

(I can expand on this with an example, if needed).

Yes, please do.

 

[PeterDonis]

It's also worth noting that the "equivalence" reasoning given above, for why the path of a photon passing close to a massive object like the Sun should look "curved", doesn't actually give the correct numerical answer. The correct answer for how "curved" the photon's path looks is *twice* the answer you get just from the "equivalence" reasoning.

Actually, on re-reading this, it isn't correct and I shouldn't have said it this way. The equivalence principle is local; it can tell you that you should expect a photon's path to look curved if measured inside a box at rest on the Earth's surface, because it would look curved inside a box in an accelerating rocket. But the EP can't tell you about the actual bending of light around a massive object like the Sun, because that requires global knowledge of the gravitational field, including how it changes direction from place to place. The EP doesn't cover that.

The calculation that gives an answer equal to half of the correct answer is a sort of "naive" Newtonian calculation that treats the photon as a "particle" that is "falling" in the spherical field of the Sun, and calculates its trajectory the same way one would for any other fast-moving particle on a hyperbolic orbit. But since this calculation uses the spherical field, not a local approximation, it isn't correctly described as an "equivalence principle" calculation.