4D spacetime Light cone Twins paradox

[eon_rider]

I'm an "on-my-own-free-time" arm-chair student of physics. Lol.
So if this question is way off the mark my apologies.
Feel free to let me know where I’m off base.

Anyway...

For me, a great visual example of the twin paradox was found at this site:
http://www.phy.syr.edu/courses/modules/LIGHTCONE/LightClock/default.html
(Thanks Janus it was your link posted on another thread I used to get there)

So, the second to last AVI or RAM at the bottom of the page shows the twins or (A) and (B)s full journey within their own world lines traveling through 4d-spacetime.

The twin (A) stays in his inertial frame and the twin (B) travels away from the first twin and then turns around.
It was said that at the point of the turn it is revealed who has the preferred reference frame.

So the paradox is resolved when the twin (B) turns around and returns to meet twin (A)

Twin (B)s change of direction shows that he can no longer be considered to be in the preferred reference frame. He will age more slowly also the faster he goes.

But my question is: When the twin (B) turns to return to twin (A) why can't it be said instead that twin (A) was the one doing the turning WRT to twin (B).
Or a third possibility could be that they both share half the turn, if one can put it that way.

So for the first half the trip twin (B)s clock runs slowly compared to twin(A) and for the second half of the trip twin (A)s clock runs slowly compared to twin (B)s. If they are doing a circular orbit around each other then perhaps the time dilation of both twins would just cancel out in real time.

Sorry for the convoluted example but there you go. That’s my question. Is this possible? At the end of the trip both twins aged at the same rate?

One can’t prove who was turning around whom and no one was discovered to be in the preferred inertial reference frame for the entire duration of the trip.

Both reference frames during all turns or orbits equally cancel out each others time dilation. But time dilation still occurs as per usual.

Sigh..

Hope that made some sense.

Best

Eon.

 

[Jimmy Snyder]

But my question is: When the twin (B) turns to return to twin (A) why can't it be said instead that twin (A) was the one doing the turning WRT to twin (B).

The answer to this is in your own post a couple of lines up.
Twin (B)s change of direction shows that he can no longer be considered to be in the preferred reference frame.
In other words, it is twin (B) and twin (B) only that gets mushed up into the nose cone of his rocket when it slows down to turn around. That's how he knows he's the one that will be younger when the two meet up.

Or a third possibility could be that they both share half the turn, if one can put it that way.

In that case, they would both get mushed up and indeed be the same age when they meet. There is still no paradox because as in the first case, the problem goes away when you account for all the frames being used. At the base of the twins seeming-paradox is the pretense that you are using two frames of reference when you are using three. In this case you are using five. All of the paradox goes away when you account for what happens as you jump from one frame to another.

 

[eon_rider]

Thanks for the reply

Thanks for the reply.
What you say makes sense.
But I'm still not sure on some of these points.

The answer to this is in your own post a couple of lines up.
Twin (B)s change of direction shows that he can no longer be considered to be in the preferred reference frame.

Lets say twin B is unaware of his flight path. So too twin A.

it is twin (B) and twin (B) only that gets mushed up into the nose cone of his rocket when it slows down to turn around. That's how he knows he's the one that will be younger when the two meet up..

So the knowledge of being mushed up makes him age?
Also...
Let's say the acceleration or change in direction were slow enough or the orbit very large (computer controlled) as to make the change in G's unnoticable then why would anyone need to get mushed up or be aware that they were turning. It can be set up so that the twins are unaware of the flight path and unable to tell.

The data for both twins trips could be compiled at the end of the journey.Do they suddenly age as soon as the data is compiled?

In this case you are using five.

Five? I believe you but I'm only counting 2.

Thanks again,

Eon.

PS... Please don't hit me with advanced maths of the lorentz tranformation as I'm sure I can plug in the numbers with a little help, but I'm trying to think through this conceptually. BTW I say that with humor..LOL...it's all good.

 

[eon_rider]

Quote:
eon_rider wrote:
"Or a third possibility could be that they both share half the turn, if one can put it that way."

J.S. Wrote: "In that case, they would both get mushed up and indeed be the same age when they meet"

Ok. I get this part.

Many Thanks

E

 

[Jimmy Snyder]

Lets say twin B is unaware of his flight path. So too twin A.

Again, that cramped feeling up there in the nose cone will tell him that something is going on. Twin A never experiences that feeling.

So the knowledge of being mushed up makes him age?

Well, that's sort of what I said, but it's not true. Here's the real skinny. When the rocket accelerates, the frame changes. It is the frame change that causes the difference in ages. Being all mushed and stuff tells you who is experiencing the frame change and who is not.

Let's say the acceleration or change in direction were slow enough or the orbit very large (computer controlled) as to make the change in G's unnoticable then why would anyone need to get mushed up or be aware that they were turning. It can be set up so that the twins are unaware of the flight path and unable to tell.

Yes it can. But again, it is not the awareness that is the issue, it is the frame change. Traditionally, the twins seeming-paradox is presented with no acceleration for most of the journey so that special relativity can be invoked. If you accelerate slowly, you lose this angle. You are accelerating a great deal of the time and now you need general relativity.

The data for both twins trips could be compiled at the end of the journey.Do they suddenly age as soon as the data is compiled?

No, but there is a lot of insight in this question. If we assume that the traveling twin is not accelerating on the outward leg, accelerates instantaneously to turn 180 degrees and then no more acceleration for the return journey, then the traveling twin will 'see' the stationary twin age suddenly at the moment of turning. Actually see is a poor word for it. As you put it, they will need to wait until the data is compiled. But although they must wait, when they do finally see the data, it will show that the aging took place at the turn in the journey.

Five? I believe you but I'm only counting 2.

1. A's outward leg
2. A's inward leg.
3. B's outward leg.
4. B's inward leg.
5. The frame of the meeting point.

It is common to think of this last frame as not moving in space, but in time only, although that is a frame dependent concept. Interestingly, traveling with this 'non-moving' frame takes the longest proper time to traverse the two meeting events. That's how it is that the stationary twin ages more than the traveling one.

 

[eon_rider]

Again, that cramped feeling up there in the nose cone will tell him that something is going on. Twin A never experiences that feeling.

Roger that.

Here's the real skinny. When the rocket accelerates, the frame changes. It is the frame change that causes the difference in ages.

Cool. Well said.

Yes it can. But again, it is not the awareness that is the issue, it is the frame change. Traditionally, the twins seeming-paradox is presented with no acceleration for most of the journey so that special relativity can be invoked. If you accelerate slowly, you lose this angle. You are accelerating a great deal of the time and now you need general relativity.

Well said. Frame change and G.R. involved here.
Got it. Cool.

No, but there is a lot of insight in this question. If we assume that the traveling twin is not accelerating on the outward leg, accelerates instantaneously to turn 180 degrees and then no more acceleration for the return journey, then the traveling twin will 'see' the stationary twin age suddenly at the moment of turning. Actually see is a poor word for it. As you put it, they will need to wait until the data is compiled. But although they must wait, when they do finally see the data, it will show that the aging took place at the turn in the journey.

Again. Really understandable and no maths! thanks.
The insight was yours. I just had a hunch.
The rapid aging is strange but interesting.

1. A's outward leg
2. A's inward leg.
3. B's outward leg.
4. B's inward leg.
5. The frame of the meeting point.
It is common to think of this last frame as not moving in space, but in time only, although that is a frame dependent concept. Interestingly, traveling with this 'non-moving' frame takes the longest proper time to traverse the two meeting events. That's how it is that the stationary twin ages more than the traveling one.

Sure enough that's IS 5. Very Clear.
Appreciate your expertise and your time.
Awsome.

all the best,

Eon.

 

[jtbell]

Traditionally, the twins seeming-paradox is presented with no acceleration for most of the journey so that special relativity can be invoked. If you accelerate slowly, you lose this angle. You are accelerating a great deal of the time and now you need general relativity.

No, you do not. This is a common misconception. Special relativity can handle acceleration by integrating over a sequence of "instantaneously co-moving inertial reference frames," one for each velocity the accelerating observer "passes through."
What special relativity can't handle is situations where the effects of gravity are significant, that is, where you have to deal with curved spacetime.

 

[Jimmy Snyder]

Special relativity can handle acceleration by integrating over a sequence of "instantaneously co-moving inertial reference frames," one for each velocity the accelerating observer "passes through."

eon_rider, jtbell may be right about this. An easier way to see that there is a problem with slowly accelerating is that in the end you still need to turn the frame around 180 degrees. Although it does matter whether you do that quickly or slowly, the mere fact that you did it is enough to void your twin paradox warranty.

I'm not 100% sure that jtbell is right though. After all, acceleration is equivalent to gravity. How much acceleration do you need to curve space? I would have guessed any at all. A while back I was told that the concept of CMRF was a stop-gap measure to approximate GR in the years between 1905 and 1915 when there was no GR. I was told that it is accurate enough and convenient enough for some calculations, but theoretically eclipsed after GR was introduced. Am I misinformed?

 

[JesseM]

No, you do not. This is a common misconception. Special relativity can handle acceleration by integrating over a sequence of "instantaneously co-moving inertial reference frames," one for each velocity the accelerating observer "passes through."

You'd have to be careful about what you integrate, though--you couldn't just use the same quantities you'd use if you were sticking to a single inertial frame. For example, if you find the function for the velocity of another clock v(t) as a function of its velocity in your instantaneous co-moving inertial frame at proper time t on your own clock, you cannot then just integrate $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$ to find the total time elapsed on that clock between two proper times $$t_0$$ and $$t_1$$ on your clock (say, between the time you departed from the other clock and the time you reunited with it), because that would fail to take into account the way your definition of simultaneity is changing as your instantantaneous co-moving inertial frame is changing.

 

[JesseM]

I'm not 100% sure that jtbell is right though. After all, acceleration is equivalent to gravity. How much acceleration do you need to curve space? I would have guessed any at all.

The effects experienced by an observer in a small region undergoing acceleration are equivalent to the effect the observer would experience in a gravitational field, but that doesn't mean that acceleration itself causes spacetime to be curved, you can talk about accelerating objects in flat spacetime (in practice I guess the energy used to accelerate an object would have to contribute very slightly to the curvature of spacetime, but I think that for ordinary human-scale objects undergoing human-scale accelerations the effect would be tiny, too small to measure).

I think the equivalence principle does mean that if you have an object experiencing constant acceleration in flat spacetime, you can come up with a new coordinate system where the object is at rest and the G-forces experienced by the object are now explained in terms of a uniform gravitational field--that's what this page on the GR analysis of the twin paradox seems to be saying. A question for the GR experts here--I think I remember from a previous discussion someone telling me that in GR the curvature of spacetime does not depend on your coordinate system, so can a "uniform gravitational field" exist in flat spacetime? If the curvature of space alone can vary depending on your choice of coordinate system, does a uniform gravitational field in flat spacetime mean that space is curved even though spacetime is not?

 

[Jimmy Snyder]

The effects experienced by an observer in a small region undergoing acceleration are equivalent to the effect the observer would experience in a gravitational field, but that doesn't mean that acceleration itself causes spacetime to be curved.

It sounds fishy to me. You seem to be saying that acceleration is equivalent to a special kind of gravity that doesn't curve spacetime.

Here is my take on things. I don't think that use of the MCRF is strictly allowable in SR, you just get away with it because the calculations work. In that respect it is like using Newton's equations to calculate rocket trajectories. It is adequate for calculations, but theoretically incorrect. In other words, there are three ways of handling the twins:

1. SR
2. SR + MCRF
3. GR

In this scheme, SR by itself is not adequate to explain the 'slowly accelerated' version of the twins seeming-paradox. However, I agree with jtbell that for computational purposes, GR is not necessary either. Instead SR + MCRF can be used for the resolution of the seeming-paradox. But only because it gets the right numbers. A full, theoretically correct analysis requires GR. These are the opinions of a non-physicist.

 

[JesseM]

It sounds fishy to me. You seem to be saying that acceleration is equivalent to a special kind of gravity that doesn't curve spacetime.

Well, the equivalence principle only deals with local regions, and locally spacetime is always near-flat...if you're in free fall, you experience all the same phenomena in your local region as you would if you were moving inertially in flat spacetime. The feeling of being pulled by gravity is because some non-gravitational force (like the electromagnetic force from the ground pushing up on your feet) is preventing you from taking the geodesic that you would in free-fall, and you feel the same thing as you would if a non-gravitational force was accelerating you in empty space (which also prevents you from following a geodesic).

Here is my take on things. I don't think that use of the MCRF is strictly allowable in SR

Sorry, what does MCRF stand for?

you just get away with it because the calculations work. In that respect it is like using Newton's equations to calculate rocket trajectories. It is adequate for calculations, but theoretically incorrect.

It's theoretically incorrect because the predictions of Newtonian mechanics deviate slightly from those of relativity, but not much if the velocities are small compared to light. On the other hand, GR should give exactly the same predictions as SR in flat spacetime. And again, it's certainly possible to have acceleration in flat spacetime...although as I said it would take some energy to accelerate a massive object and that energy would curve spacetime slightly, you can always get around this by talking about a "test particle" of arbitrarily small mass. Acceleration itself does not curve spacetime, only matter/energy does.

 

[eon_rider]

Just a thought.

Is there a tension between the non-locality of Q.M and the private nature of inertial frames and world lines in S.R?
In other words. Let’s say the twins make their flight and return.
But the trip is done in a double blind way where the twins don’t know their flight path nor does the scientist. (at the NASA of the future let's pretend)

Then the data is compiled. At the moment of compiling the data (doing the math) could some kind of non-binary wave function collapse and suddenly one twin ages the correct amount according to S.R.

LOL. I realize this is a complete fishing expedition and I’m just throwing out 3
ideas or more as a non-expert but its interesting to me.
Those concepts being:

1) non-locality in Q.M. – Copenhagen interpretation or some other well respected one.

2) collapse of a wave function (state of existence?) (is the cat alive or not) and can this be applied to does one twin age or not only at the moment of observing the data and compiling it.

3) The rules of S.R stating that each world line is inertial and private?

I’m so very much reaching here so I don’t mind if you all laugh.

But can Q.M issues run into S.R. issues when dealing with the twin paradox?
Eon.

 

[my_wan]

I’m so very much reaching here so I don’t mind if you all laugh.

But can Q.M issues run into S.R. issues when dealing with the twin paradox?

No. The nature of QM plays no role in relativity whatsoever. The intervals that determine relative aging rates is completely defined and measurable at all time in relativity. There are no collapsing wavefunctions and such. Changes in relative aging rates are completely determined by who accelerates, a little over a long time or a lot over a short time doesn't matter. Who accelerates can be agreed upon by all observers in all frames of reference and has nothing to do with whether anyone observed any part of it or not. Once you understand what's involved it's not even a real paradox. The apparent contradiction comes from the fact that you both appear to aging faster than the other until one of you accelerates to other. This is because you both disagree on when now is.

 

[my_wan]

I think I remember from a previous discussion someone telling me that in GR the curvature of spacetime does not depend on your coordinate system, so can a "uniform gravitational field" exist in flat spacetime? If the curvature of space alone can vary depending on your choice of coordinate system, does a uniform gravitational field in flat spacetime mean that space is curved even though spacetime is not?

I have some of the same issues in dealing with GR so I posted a question https://www.physicsforums.com/showthread.php?t=96927" in hopes of fully articulating the situation.

 

[eon_rider]

No. The nature of QM plays no role in relativity whatsoever. The intervals that determine relative aging rates is completely defined and measurable at all time in relativity. There are no collapsing wavefunctions and such. Changes in relative aging rates are completely determined by who accelerates, a little over a long time or a lot over a short time doesn't matter. Who accelerates can be agreed upon by all observers in all frames of reference and has nothing to do with whether anyone observed any part of it or not. Once you understand what's involved it's not even a real paradox. The apparent contradiction comes from the fact that you both appear to aging faster than the other until one of you accelerates to other. This is because you both disagree on when now is.

Well said my_wan. Cool.

I should have googled it first.

When I did articles and papers from universities and arXiv.org
came up on the subject of Q.M and S.R and the peaceful co-existence between the two for the most part, but not always.
But as you say for the most part there is no issue. I believe.

Thanks for the reply.

best

Eon.

 

[Jimmy Snyder]

what does MCRF stand for?

Momentarily Comoving Reference Frame. The postulates of SR constrain the theory to inertial reference frames. As I understand it, the use of MCRF grew up in the years between the introduction of SR in 1905 and that of GR in 1915 as a way of dealing with non-inertial frames. It is still used now because it gets the right answer in restricted cases. It takes advantage of the fact that slightly non-inertial frames are almost inertial. You integrate across inertial reference frames that are momentarily comoving with the non-inertial frame. It is computationally convenient and accurate (definition of slightly non-inertial: if MCRF gets the wrong answer then the non wasn't slight) but has no theoretical underpinings. SR doesn't allow you to work with non-inertial frames, and MCRF doesn't correctly account for all of the GR effects of spacetime curvature.

Edited to removed references to acceleration.

 

[Jimmy Snyder]

Well, the equivalence principle only deals with local regions, and locally spacetime is always near-flat.

We don't seem to disagree. Near-flat is far enough away from flat for SR to fail. The postulates of SR state that they are limited to inertial frames. Thus MCRF.

I don't fully understand your distinction between acceleration curving spacetime and energy curving spacetime. Can there be acceleration without energy? Using an admitedly Newtonian point of view: Acceleration is proportional to force and an accelerating particle cannot be stationary for any period of time. Multiply the force times the distance moved and you get energy.

Below is a famous picture of acceleration causing curvature. I have been told that there is a flaw in this picture, but I don't know what the flaw is, so I present it with the hopes that someone can explain it to me.

Consider a disk spinning about its center like a roulette wheel. Consider a portion of the circumference small enough to be nearly linear. As the length of this small piece is parallel to its motion, a stationary observer at a point on the axis of rotation will observe a foreshortening. Now consider the radius connecting the center of the disk to the center of the small piece of the circumerence. This distance is perpendicular to the motion and so is not foreshortened. As the piece of circumference was arbitrary, the entire circumference is foreshortened. So the observer notes that one of the properties of flat space $c = 2\pi{r}$ does not hold. The cause is explained entirely by the acceleration of the disk without any mention of forces or energy.

 

[masudr]

SR is about looking at the world from an inertial reference frame. That's easy enough. So I can sit still in an inertial reference frame, and watch the famous twins, both going off in opposite directions, and then one turning around and going back to the other. I sit in my inertial armchair and watch all this occur. There has been no need to do any physics in a non-inertial reference frame.

The proper time elapsed by a particle in a path is given by:
$$\Delta\tau = \int^{\lambda=b}_{\lambda=a} \sqrt{-\eta_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}} d\lambda$$
where $x^\mu(\lambda)$ is a parametrization of the worldline of that particle (I'm quoting more or less straight from Carroll). Furthermore, $\eta_{\mu\nu} = \mbox{diag}(-1, 1, 1, 1)$.

No GR involved whatsoever.

 

[jcsd]

Momentarily Comoving Reference Frame. The postulates of SR constrain the theory to inertial reference frames. As I understand it, the use of MCRF grew up in the years between the introduction of SR in 1905 and that of GR in 1915 as a way of dealing with non-inertial frames. It is still used now because it gets the right answer in restricted cases. It takes advantage of the fact that slightly non-inertial frames are almost inertial. You integrate across inertial reference frames that are momentarily comoving with the non-inertial frame. It is computationally convenient and accurate (definition of slightly non-inertial: if MCRF gets the wrong answer then the non wasn't slight) but has no theoretical underpinings. SR doesn't allow you to work with non-inertial frames, and MCRF doesn't correctly account for all of the GR effects of spacetime curvature.
Edited to removed references to acceleration.

When does inetgrating over the MCIFs get a wrong answer? The answers certainly aren't any different from GR sans curvature.

The postulates of SR apply to inertial frmaes, but there's ntohing to stop us constructing non-inertial frames. Plus once you accept that SR naturally leads to the description of spacetime in terms of a Minkowski manifold, the only difference between inertial and non-inertial farmes is that inertial frames have constant basis fields.

 

[jcsd]

We don't seem to disagree. Near-flat is far enough away from flat for SR to fail. The postulates of SR state that they are limited to inertial frames. Thus MCRF.

SR always applies locally in GR.

I don't fully understand your distinction between acceleration curving spacetime and energy curving spacetime. Can there be acceleration without energy? Using an admitedly Newtonian point of view: Acceleration is proportional to force and an accelerating particle cannot be stationary for any period of time. Multiply the force times the distance moved and you get energy.

Acceleration doesn't curve spacetime as such. You can still have accelarting observers against a Minkowskian background.

Below is a famous picture of acceleration causing curvature. I have been told that there is a flaw in this picture, but I don't know what the flaw is, so I present it with the hopes that someone can explain it to me.
Consider a disk spinning about its center like a roulette wheel. Consider a portion of the circumference small enough to be nearly linear. As the length of this small piece is parallel to its motion, a stationary observer at a point on the axis of rotation will observe a foreshortening. Now consider the radius connecting the center of the disk to the center of the small piece of the circumerence. This distance is perpendicular to the motion and so is not foreshortened. As the piece of circumference was arbitrary, the entire circumference is foreshortened. So the observer notes that one of the properties of flat space $c = 2\pi{r}$ does not hold. The cause is explained entirely by the acceleration of the disk without any mention of forces or energy.

read this: http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

i.e. For SR the basic conclusion should be that such discs do not exist in SR.

 

[Jimmy Snyder]

When does inetgrating over the MCIFs get a wrong answer?

Unfortunately, a lot of what I say is just parroting what I read. I have no deep understanding of it. I'll guess that using the MCRF never gives the same answer as the full GR treatment unless there is no curvature. The reason I make this guess is that if it were not true, then there would be no need for GR. The amount of difference between the MCRF calculated result and the GR result is probably related to the amount of spacetime curvature. When that curvature is small, the difference is below the threshold of what can be measured. When the curvature is larger than some threshold value, the difference becomes measureable.

So, to answer your question, if GR gives a value that conforms with measurement, and using MCRF gives a measurably different result, that is when MCRF gets a wrong answer.

I am currently reading "A First Course in GR" by Schutz. I note that in many equations terms that are quadratic or higher in small quantities are ignored. I realize that in some cases this is justified because the small quantity is a differential which will tend to zero, however not all small quantities treated this way are differentials. This is of course the only practical way to get pretty compact equations. However, nature is not concerned with this particular aspect of beauty and practicality and retains those terms. Are you comparing the results of using MCRF to the results of using this smoothed out version of GR? It reminds me that Newton's equations of motion are correct. When there is no motion that is.

 

[Jimmy Snyder]

When three people tell you you're drunk, it's time to lie down, and so I will. But not without a few questions first.

1. If I'm the accelerating twin, but think I'm in a gravity field, do I observe curvature?

2. If no, how do I explain why I think I'm in a gravity field?

3. If yes, then is curvature a frame dependent concept? Does my twin observe flatness where I observe curvature?

4. If no, that is if I and all other observers observe curvature, then how can SR give good results?

 

[JesseM]

3. If yes, then is curvature a frame dependent concept? Does my twin observe flatness where I observe curvature?

Yes, I'm wondering about this too...like I said earlier, I had gotten the impression from a discussion on an earlier thread that spacetime curvature was frame-independent but spatial curvature alone was frame-dependent, is this correct? If so, when they say in the GR analysis of the twin paradox that you can use a coordinate system where the traveling twin is at rest the whole time but feels the effects of a uniform gravitational field at the moment that an inertial frame would say she accelerates, would it still be true that spacetime is flat in this coordinate system, and does the uniform gravitational field imply space becomes curved in this coordinate system?

 

[masudr]

Gravity is locally equivalent to an accelerating frame. You can always locally choose a frame such that there is no acceleration. However, you cannot make all the second-order derivatives disappear - this is what the curvature tensor is.

The curvature tensor is certainly not frame-dependent it transforms just like a (3,1) tensor should transform. The point is that if either twin parallel transported a vector around an infinitesimally small loop, then it would not be rotated by any amount. However, if an observer is in a true gravitational field then the curvature tensor would not be 0.

 

[Jimmy Snyder]

Gravity is locally equivalent to an accelerating frame.

Finally, the light goes on in my dim mind. Twin B knows very well that he is not in a gravity field. Just before he was reduced to a protoplasmic pulp by the g's he pulled, he dropped a feather and a hammer. Naturally, they moved along geodesics, but the distance between them didn't change i.e. he failed to see any curvature. I'm sorry it took me so long to realize this.

The lesson I have learned is this. The equavalence principle is only true locally. Acceleration does not cause curvature, only massenergy does. Thank you.

Edit: This isn't right either is it? Sigh. I'll get it someday. The hammer and feather attract each other and so the distance between them does change. But no so much as we would expect if there were gravity causing the twin's acceleration.

 

[JesseM]

The curvature tensor is certainly not frame-dependent it transforms just like a (3,1) tensor should transform.

But you're talking about a tensor that describes spacetime curvature, right? If you pick a global coordinate system whose time parameter can be used to foliate 4D spacetime into a series of curved 3D surfaces, the curvature of space depends on your coordinate system, right? My understanding here was based on an earlier discussion on this thread, where on p. 5 I asked whether a universe totally devoid of matter and energy would be spatially flat or have negative spatial curvature, and in post #64 pervect said either one could be true depending on your choice of coordinate system, referencing this page which discusses the problem.

If this is correct, what I'm wondering is whether, in the GR analysis of the twin paradox discussed on John Baez's page above, the "uniform gravitational field" experienced by the traveling twin in a coordinate system where she is at rest throughout the trip would involve curved space in her coordinate system, even if spacetime is still flat in this system.

 

[pervect]

If this is correct, what I'm wondering is whether, in the GR analysis of the twin paradox discussed on John Baez's page above, the "uniform gravitational field" experienced by the traveling twin in a coordinate system where she is at rest throughout the trip would involve curved space in her coordinate system, even if spacetime is still flat in this system.

The Riemann curvature for the following metric

ds^2 = (1+gz)^2 dt^2 - dx^2 - dy^2 - dz^2

which represents a "uniform gravitational field", aka an accelerated observer's coordinate system, aka the Rindler metric, is zero.

 

[JesseM]

The Riemann curvature for the following metric
ds^2 = (1+gz)^2 dt^2 - dx^2 - dy^2 - dz^2
which represents a "uniform gravitational field", aka an accelerated observer's coordinate system, aka the Rindler metric, is zero.

Does Riemann curvature represent spacetime curvature or just spatial curvature? Remember, I know next to nothing about the mathematics of GR...

 

[Jimmy Snyder]

read this: http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html
i.e. For SR the basic conclusion should be that such discs do not exist in SR.

Thanks for this link. I didn't get a simple answer to the question: Does the disk cause spacetime to curve? Currently I don't think it does, but things are changing fast for me.

In the book "The Elegant Universe" (No, Professor Greene, string theory is elegant, the universe is a mess*), Brian Greene describes the relationship between the disk and curvature in the section entitled "Acceleration and the Warping of Space and Time". He never says that the argument is valid, he only says that it is the argument that Einstein used to justify the idea that Gravity causes curvature, similar to my own image (which I got from reading stuff like this)

Gravity = Acceleration = curvature.

But he doesn't say that it isn't valid either. I have been careful to say curvature and not curved space or curved spacetime because Professor Greene is rather ambiguous on this detail. At points in his narative he definitely indicates that he deals separately with curved space and curved time. But he also mentions Minkowski's quote about the decline of the independent ideas of space and time.

* I ripped this off wholesale from Sean Carroll. If you have never heard Professor Carroll speak, drop everything and go listen.

 

[jcsd]

Unfortunately, a lot of what I say is just parroting what I read. I have no deep understanding of it. I'll guess that using the MCRF never gives the same answer as the full GR treatment unless there is no curvature. The reason I make this guess is that if it were not true, then there would be no need for GR. The amount of difference between the MCRF calculated result and the GR result is probably related to the amount of spacetime curvature. When that curvature is small, the difference is below the threshold of what can be measured. When the curvature is larger than some threshold value, the difference becomes measureable.
So, to answer your question, if GR gives a value that conforms with measurement, and using MCRF gives a measurably different result, that is when MCRF gets a wrong answer.

There are some situations which cannot be decribed by SR, this is when there is curvature but as I said accelartaion in itself does not imply curvature. The use of the MCIFs (I prefer momentarily comoving inertial frame) gives the correct results (i.e. the ones that conform with GR) within it's limit of appplicabilty. There are of course difficulties (I use the word loosely as I'm ceratinly not implying that there are flaws) that you encounter in non-ienrtial frames in SR that you don't encounter in non-inertial frames (if you read the link in my last post one of these is very briefly mentoned - the difficulty of defining simulatneity), but these are the same general difficulties that occur in GR.

In GR you can always apply SR locally due to the tangent space to an event in spacetime, this means that when considering a vanishingly small region of spacetime GR reduces to SR.

I am currently reading "A First Course in GR" by Schutz. I note that in many equations terms that are quadratic or higher in small quantities are ignored. I realize that in some cases this is justified because the small quantity is a differential which will tend to zero, however not all small quantities treated this way are differentials. This is of course the only practical way to get pretty compact equations. However, nature is not concerned with this particular aspect of beauty and practicality and retains those terms. Are you comparing the results of using MCRF to the results of using this smoothed out version of GR? It reminds me that Newton's equations of motion are correct. When there is no motion that is.

GR reduces to SR in flat spacetime.