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.

 

[Naty1]

Does putting the concept of acceleration in SR make it equivalent to GR?... an acceleration IS gravity by the equivalence principle, so what's the difference?

After reading all the posts, I come to the conclusion that while the answer is obvious [‘no’, Einstein was not fooled into spending ten years developing GR for naught.]
explaining just why acceleration is similar to gravity, not identical, to someone who takes literally they are the same, is not so easy.

Here are some related ideas indicating why gravity and acceleration are NOT identical.

First a general perspective…. a quote I kept from a discussion in these forums...from a highly regard classic textbook [Misner, Thorne, Wheeler]:

...nowhere has a precise definition of the term “gravitational field” been given --- nor will one be given. Many different mathematical entities are associated with gravitation; the metric, the Riemann curvature tensor, the curvature scalar … Each of these plays an important role in gravitation theory, and none is so much more central than the others that it deserves the name “gravitational field.”

So one answer is that none of these entities exist in SR…so it can’t entirely manifest gravity.

I especially like both PeterDonis and Dalespam’s posts which seem like two sides of the same coin: Peter:

This [gravity acceleration similar] 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.

Dalespam:

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.

And Dalespam’s post #32 makes these ideas explicit via example:

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?

And that idea got me thinking about a source of gravity in SR that explains or replicates those multiple sources of gravity in GR. [there is none]. In each we have gravitational and inertial acceleration, locally, but that’s about as far as I can see the ‘equivalence’.

So a way of approaching this difference between acceleration in SR and gravity in GR seems to me : E =mc² plus special relativity does NOT equal GR. Wikipedia:

According to the Einstein field equation, the gravitational field is locally coupled not only to the distribution of non-gravitational mass-energy, but also to the distribution of momentum and stress (e.g. pressure, viscous stresses in a perfect fluid). So there are components, sources, of gravity in the Stress Energy Tensor that are not available from SR, even with mass energy equivalence.

As if this were not enough to distinguish acceleration in SR with gravity in GR also consider that in GR knowing all about the sources (the stress-energy tensor ) isn't enough to tell you all about the curvature. When we are in truly empty space, there's no Ricci curvature, so a ball of coffee grounds doesn't change volume. But there can be Weyl curvature due to gravitational waves, tidal forces, and the like. So the shape can change: Gravitational waves and tidal forces tend to stretch things out in one direction while squashing them in the other. You can't find this in SR.

Since curvature of spacetime is gravity, another example of how SR and GR are similar locally, not globally:

There is no cosmological expansion in SR: No cosmic acceleration!
Spacetime in GR is curved. If you split spacetime into "space" (3D) and "time" [so as to insure that the "time" direction at every "spatial location" points to the future and is the potential worldline of an observer] SR and GR are vastly different regarding acceleration:
If you do this split in empty flat spacetime, two observers at "rest" in "space" don't find that the distance between them increases with "time". But when you do it in the matter-containing curved spacetime used to model our universe, two observers at "rest" in "space" do find that the "spatial distance" between them increases with "time".
[atyy posted the above idea from Wikipedia in another thread]

Also, another difference between gravity and acceleration is discussed by Roger Penrose:

Penrose's book Shadows of the Mind, Section 4.4, where he talks about causality and light-cone tilting, something that becomes very evident in highly “curved” space-times.

“The reason for this is that gravity actually influences the causal relationships between space-time events, and it is the only physical quantity that has this effect. Another way of phrasing this is that gravity has the unique capacity to 'tilt' light cones. No physical field other than gravity can tilt light cones, nor can any collection whatever of non-gravitational physical influences”…so gravity is something that is simply different from all other known forces and physical influences….”

Also:
Stevendaryl: I liked your historical story early in this thread…..here is an additional piece you might include as a one liner:

http://en.wikipedia.org/wiki/Spacetime#Mathematical_concept


“While spacetime can be viewed as a consequence of Albert Einstein's 1905 theory of special relativity, it was first explicitly proposed mathematically by one of his teachers, the mathematician Hermann Minkowski, in a 1908 essay[7] building on and extending Einstein's work. His concept of Minkowski space is the earliest treatment of space and time as two aspects of a unified whole, the essence ofspecial relativity. (For an English translation of Minkowski's article, see Lorentz et al. 1952.) The 1926 thirteenth edition of the Encyclopædia Britannica included an article by Einstein titled "Space–Time".[8]) The idea of Minkowski space led to special relativity being viewed in a more geometrical way.”

 

[arindamsinha]

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.

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.

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.

...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.

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.)

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]

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?

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".

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.

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.

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.

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.

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.

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.)

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.)

[*]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.

[*]By the equivalence principle, we can consider any such acceleration as being equivalent to gravity.

Locally, yes. See further comments below.

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]

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]

[*]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]

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]

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.

 

[Dale]

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.

 

[Dale]

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.

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.

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.

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.

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]

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]

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: F = M A. 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 F=M A: In inertial cartesian coordinates,

m \dfrac{d^2}{d\tau^2} X^\mu = F^\mu

where X^\mu is the coordinates of the object, F^\mu is the 4-force, and \tau 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]

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]

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]

I think my post #31 fully explains time dilation including non-inertial motion

 

[PeterDonis]

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.).

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).

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

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.

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.

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.

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.