Was Einstein lucky when not considering twin paradox as paradox?

[Fantasist]

It's possible that the confusion in that paper is an accurate reflection of the confusion of physicists (including Einstein himself) in the early days of relativity. But just because people were confused about it in the past doesn't mean that we need to confuse ourselves in the same way.

The OP was asking about the perception of the twin paradox in the early days (starting with Einstein's paper). If discussing this issue still causes confusion today, it shows that probably not everything is as clear-cut here as it is sometimes portrayed. Only a continued discussion of confusing issues can lastly lead to full clarification, not their suppression.

 

[stevendaryl]

The OP was asking about the perception of the twin paradox in the early days (starting with Einstein's paper). If discussing this issue still causes confusion today, it shows that probably not everything is as clear-cut here as it is sometimes portrayed. Only a continued discussion of confusing issues can lastly lead to full clarification, not their suppression.

I'm not talking about suppression. I'm talking about intentionally introducing misconceptions, and then trying to clear them up. I don't see that that's helpful.

 

[Fantasist]

In the English translation On the Electrodynamics of Moving Bodies at the top of page 11, the final paragraph of §4.

"Thence we conclude that a balance-clock at the equator must go more slowly, by a very
small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."​

(I assume you realize that circular motion, around the equator in this case, is non-inertial motion.)

You possibly misunderstood this paragraph. Einstein's argument is as follows here:

1) a circle can be approximated by a polygon (by making the sides infinitesimally small)
2) a polygon is a piece-wise linear curve, so we have a piece-wise constant velocity vector that only changes in direction (and thus leaves the time dilation factor constant)
3) so for a circle of circumference C, the time dilation is the same as for a straight line of length C (assuming the speed v is the same).

So rather on the contrary, for a circular orbit there is a time dilation as it can be approximated by a (piece-wise) inertial motion (for which Einstein knows how to derive the time dilation).

 

[stevendaryl]

if the reference frame is linearly accelerating then one wouldn't need to do anything with Newton's laws and they would take their simplest form anyway. objects that move under the influence of the same force as the Reference frame would appear to stand still or move uniformly and objects that would appear to accelerate would be the once that move with different accelerations than the reference frame. anyways, we would have just a shift in perception of what's accelerating and what's not and not a change in laws.

I don't think that's correct. If you do Newtonian physics using an accelerated coordinate system, then Newton's laws don't hold.

In an inertial coordinate system, you have:

m \frac{d^2 x^j}{dt^2} = F^j (F = ma)

Now, switch to a new coordinate system x'^j = x^j + A^j t^2

In this new coordinate system, you have:

m \frac{d^2 x'^j}{dt^2} - m A^j = F^j

This does not have the same form. You could try to restore it to the same form by moving the constant acceleration m A^j to the other side:

m \frac{d^2 x'^j}{dt^2} = F'^j = m A^j + F^j

so you have a new "ficititious force" m A^j. But this new force DOESN'T obey Newton's laws. In particular, it doesn't obey the third law, "For every action, there is an equal and opposite reaction". If that's interpreted to mean that whenever there is a force on one object, that object exerts an equal and opposite force, then that's false for fictitious forces. The mass m has a force mA^j exerted on it, but it doesn't exert an equal and opposite force on anything. Momentum is not conserved in this new coordinate system.

 

[stevendaryl]

You possibly misunderstood this paragraph. Einstein's argument is as follows here:

1) a circle can be approximated by a polygon (by making the sides infinitesimally small)
2) a polygon is a piece-wise linear curve, so we have a piece-wise constant velocity vector that only changes in direction (and thus leaves the time dilation factor constant)
3) so for a circle of circumference C, the time dilation is the same as for a straight line of length C.

So rather on the contrary, for a circular orbit there is a time dilation as it can be approximated by a (piece-wise) inertial motion (for which Einstein knows how to derive the time dilation).

What you're saying is always true. You can always approximate the time dilation for any noninertial motion by breaking it up into small segments, and approximate those segments by constant-velocity segments.

That prescription gives rise to the following formula for computing proper time for noninertial motion:

The proper time for taking a path x(t) from time t=t_1 to t=t_2 is

\int_{t_1}^{t_2} \sqrt{1-\frac{(\frac{dx}{dt})^2}{c^2}} dt

 

[Dale]

Inertial motion is what Special Relativity is based on

This is not quite correct. Special relativity is based on inertial frames, not inertial motion. There is an important difference between the two. Even in the first paper, Einstein's "on the electrodynamics of moving bodies", it was clear how to correctly treat non inertial motion (see section 4). To the OP's question it wasn't luck and it was not unaddressed by Einstein.

EDIT: I see stevendaryl made the same point first!

 

[Dale]

I already added this in my last example as you can see, but this doesn't help.

I didn't see the "last example" you are referring to, but it does completely resolve the issue. The accelerating twin is the one which interacts with the external field or the other particle.

 

[Fantasist]

What you're saying is always true. You can always approximate the time dilation for any noninertial motion by breaking it up into small segments, and approximate those segments by constant-velocity segments.

Still, it is only a geometrical argument. Each of the twins could describe the motion of the other this way. So how do we single out one of the twins on this basis?

 

[adoion]

I don't think that's correct. If you do Newtonian physics using an accelerated coordinate system, then Newton's laws don't hold.

In an inertial coordinate system, you have:

m \frac{d^2 x^j}{dt^2} = F^j (F = ma)

Now, switch to a new coordinate system x'^j = x^j + A^j t^2

In this new coordinate system, you have:

m \frac{d^2 x'^j}{dt^2} - m A^j = F^j

This does not have the same form. You could try to restore it to the same form by moving the constant acceleration m A^j to the other side:

m \frac{d^2 x'^j}{dt^2} = F'^j = m A^j + F^j

so you have a new "ficititious force" m A^j. But this new force DOESN'T obey Newton's laws. In particular, it doesn't obey the third law, "For every action, there is an equal and opposite reaction". If that's interpreted to mean that whenever there is a force on one object, that object exerts an equal and opposite force, then that's false for fictitious forces. The mass m has a force mA^j exerted on it, but it doesn't exert an equal and opposite force on anything. Momentum is not conserved in this new coordinate system.

as I mentioned the new coordinate system would be just shifted compared to the old one, your fictious force $mA^j$ would only be present from the old coordinate systems point of view.
but if you fix your new reference frame is accelerating compared to the old one then from the new RF point of view (witch you suppose to be stationary) the old RF is accelerating and you would have fictious forces if you look at the old one from the new RF.
But the laws you can state the same as in the old one.

while when the RF is rotating you would clearly have laws involving rotational forces right away, and if you step out of it you would have Newton's ordinary laws. you would not be able to compensate and have only linear acceleration in the rotating RF-laws.

so clearly there is a sense of linear symmetry in our universe but not symmetry between rotational and linear reference frames, this is a consequence of the way the universe is setup and how things are moving.

 

[Dale]

I'm saying that the case of a noninertial clock is a deduction from Einstein's paper.

Furthermore, it is a deduction which he explicitly makes.

 

[Dale]

but how does one make sure that a force is or is not fictional?

One uses an accelerometer.

if the reference frame is linearly accelerating then one wouldn't need to do anything with Newton's laws and they would take their simplest form

This is not correct. In a linearly accelerating reference frame there is a fictitious force. Accelerometers at rest in linearly accelerating reference frames register the acceleration.

both of the twins must make their measurements from their point of view and since in both cases the laws of physics take their simplest form, both of them are correct in assuming that their reference frame is inertial and that the other one is accelerating.

This is simply false. An accelerometer aboard the traveling twin's ship registers a sharp spike halfway through the journey. He knows that his rest frame is not inertial and there are no "laws of physics take their simplest form" which would explain that accelerometer reading.

 

[Nugatory]

But how does one make sure that a force is or is not fictional? one has to find the source of the force or one has to find other reference frames in which those forces disappear

Yes, that is pretty much it. Of course finding such a reference frame is a purely mathematical exercise - I just need to find a coordinate transformation that makes the coordinate accelerations go away.

If we observe a force that accelerates all objects equally regardless of their mass (as does centrifugal and coriolis force) that's a fairly solid hint that we're dealing with a fictional force and that such a coordinate transformation exists. Indeed, the only non-fictional force with that property is Newtonian gravity - and GR eliminates that special case by providing a mathematical framework in which it is also a fictional force.

 

[stevendaryl]

as I mentioned the new coordinate system would be just shifted compared to the old one, your fictious force $mA^j$ would only be present from the old coordinate systems point of view.

I know that's what you said, but it's not true. That term is in the equations, either as a "force" term (on the F side of F=ma) or as an acceleration term (on the ma side of F=ma). Neither choice leaves Newton's laws unchanged. Either you have to modify the notion of "acceleration" to include terms due to noninertial, non-Cartesian coordinates, or you modify the third law, and allow for forces without a "reaction" counterpart.

but if you fix your new reference frame is accelerating compared to the old one then from the new RF point of view (witch you suppose to be stationary) the old RF is accelerating and you would have fictious forces if you look at the old one from the new RF.

I'm sorry, but that's just not true. In terms of x^j, you have:

m \frac{d^2 x^j}{dt^2} = F^j (no fictitious forces)

In terms of x'^j, you have:
m \frac{d^2 x^j}{dt^2} = F^j + m A^j (fictitious forces are present)

Some frames have fictitious forces, and some frames do not. The ones that do not are the inertial frames.

 

[Dale]

as I mentioned the new coordinate system would be just shifted compared to the old one, your fictious force $mA^j$ would only be present from the old coordinate systems point of view.

No, this is not true. The fictitious force is intrinsic to the coordinate system itself and not required to be in reference to any other coordinate system's point of view. That fictitious force must be included to match the observations of motion within that frame alone.

but if you fix your new reference frame is accelerating compared to the old one then from the new RF point of view (witch you suppose to be stationary) the old RF is accelerating and you would have fictious forces if you look at the old one from the new RF.

This is also not true. The lack of fictitious force in the inertial reference frame is inherent to the frame itself and does not require any reference to any other frame. The fictitious force must be absent to match the observations of motion within that frame alone. If you include it you would get the wrong motions.

EDIT: or in other words: "What stevendaryl said".

 

[stevendaryl]

Still, it is only a geometrical argument. Each of the twins could describe the motion of the other this way. So how do we single out one of the twins on this basis?

You don't. You approximate each twin's path by a bunch of little constant-velocity paths, and you compute the proper time for each path using the formula: \delta \tau = \sqrt{\delta t^2 - \frac{1}{c^2} \delta x^2} You add up \delta \tau for each segment, and that gives you how much each twin ages along his path. There is no singling out of one twin over the other.

There is a choice that must be made, which is to pick an inertial coordinate system for measuring \delta x and \delta t. But every inertial coordinate system will give the same value for \delta \tau. It's exactly like computing the length of a line segment in Euclidean geometry. The length is given by: \delta L= \sqrt{\delta x^2 + \delta y^2}. You can choose any Cartesian coordinate system you like to measure \delta x and \delta y, and you will get the same answer for \delta L.

 

[PeterDonis]

if you have only 2 point particles an nothing else, how do you determent witch one is accelerating?

Easy: attach an accelerometer to each particle. The one whose accelerometer reads nonzero for some portion of the trip is the one who accelerated.

DaleSpam and stevendaryl have correctly pointed out that, strictly speaking, for one of the particles to accelerate in the above sense (i.e., for its accelerometer to read nonzero at some point), there must be other "stuff" present in the scenario. The "point particle" whose accelerometer reads nonzero has to exchange momentum with something (for example, a rocket exhaust). But you don't have to know any of the details of how that happens to know which particle accelerated from the accelerometer readings.

 

[harrylin]

It may be merely a student exercise today, but exactly what we are discussing here was a serious issue for Einstein already before he published his GR, which was discussed by leading scientists at the time. I quote from the Wikipedia article
Starting with Paul Langevin in 1911, there have been various explanations of this paradox. [..]

I gave you the link to his paper; you can check for yourself that, contrary to Wikipedia's suggestion*, it was not considered to be an existing paradox. Instead it was an original, non-paradoxical example of predictions based on what Einstein later named "special relativity".

*About Wikipedia, I'm not sure that the person who wrote that intended to make your claim, it may be just poor phrasing. And you can use the back-in-time feature of Wikipedia to find different opinions. ;)

 

[harrylin]

[..] In retrospect, General Relativity was not needed to describe things from the point of view of an accelerated coordinate system. That description is derivable from SR alone. And that description has terms that are "gravity-like", but all within SR. GR is only needed if you want to describe real gravity, due to the presence of massive objects.

"In retrospect"? Why do you think that this was not understood at that time? It had been straightforward in classical mechanics to describe things from the point of view of an accelerated coordinate system, so it seems unlikely that this wasn't understood from the start in SR. (sorry for going slightly off topic, but it fits rather well in the discussion here).

 

[stevendaryl]

"In retrospect"? Why do you think that this was not understood at that time? It had been straightforward in classical mechanics to describe things from the point of view of an accelerated coordinate system, so it seems unlikely that this wasn't understood from the start in SR. (sorry for going slightly off topic, but it fits rather well in the discussion here).

You're certainly right, that noninertial frames came up in Newtonian mechanics. So where did the idea that GR was necessary to handle an accelerated reference frame come from?

I think that part of it is the insistence on relativity. Although Newtonian mechanics also satisfied a principle of (Galilean) relativity, I don't think that it played that much role in the teaching and application of the subject. Nobody bothered (as far as I know) to try to write Newtonian mechanics in a way that treated all coordinate systems equally. I don't think that the latter was developed until after GR (the Newton-Cartan formulation of Newtonian physics).

 

[adoion]

Easy: attach an accelerometer to each particle. The one whose accelerometer reads nonzero for some portion of the trip is the one who accelerated.

DaleSpam and stevendaryl have correctly pointed out that, strictly speaking, for one of the particles to accelerate in the above sense (i.e., for its accelerometer to read nonzero at some point), there must be other "stuff" present in the scenario. The "point particle" whose accelerometer reads nonzero has to exchange momentum with something (for example, a rocket exhaust). But you don't have to know any of the details of how that happens to know which particle accelerated from the accelerometer readings.

you would have to calibrate the accelerometers differently in order for both of them to be zero if they accelerate differently, because if you assume that your accelerating RF is actually stationary then you calibrate your accelerometer to read zero in your reference frame, don't you?

and there is another thing, if you change acceleration in a linear fashion then your laws are not the simplest anymore and maybe its the way to think about the twin paradox.

 

[adoion]

also an accelerometer doesn't measure acceleration in free fall only when you stand on the Earth's suface does it do so, you can see that from the link to Wikipedia http://en.wikipedia.org/wiki/Accelerometer

 

[Dale]

you would have to calibrate the accelerometers differently in order for both of them to be zero if they accelerate differently, because if you assume that your accelerating RF is actually stationary then you calibrate your accelerometer to read zero in your reference frame, don't you?

Although calibration is necessary for real devices, for the purpose of thought experiments it is generally a detail which is glossed over. We simply assume ideal measuring devices such as rods, clocks, and accelerometers.

In principle, it is not a bad assumption. If you miscalibrate the accelerometer then you will detect violations of the conservation of momentum which are not accounted for. You will not be able to get experiments to match the known laws of physics. So a miscalibrated accelerometer will be something which is experimentally detectable in the end.

also an accelerometer doesn't measure acceleration in free fall only when you stand on the Earth's suface does it do so, you can see that from the link to Wikipedia http://en.wikipedia.org/wiki/Accelerometer

Yes. This is why in relativity free-fall frames are inertial and frame attached to the Earth's surface is non-inertial.

 

[adoion]

Although calibration is necessary for real devices, for the purpose of thought experiments it is generally a detail which is glossed over. We simply assume ideal measuring devices. In principle, it is not a bad assumption. If you miscalibrate the accelerometer then you will detect violations of the conservation of momentum which are not accounted for. You will not be able to get experiments to match the known laws of physics. So a miscalibrated accelerometer will be something which is experimentally detectable in the end.

what I mean is that if you for example take the equivalence principle, take an elevator witch you can think of as the rocket of the traveling twin.

if the elevator is accelerating with uniform acceleration it acts just like if it were in free fall.
so even in an accelerated RF (elevator) one would not measure acceleration even with a "accelerometer", the thing is that the rocket of the traveling twin travels with different accelerations not only with different speeds, so that that might break the symmetry somehow.
im not yet sure how but I think its the right way to go.

 

[stevendaryl]

what I mean is that if you for example take the equivalence principle, take an elevator witch you can think of as the rocket of the traveling twin.

if the elevator is accelerating with uniform acceleration it acts just like if it were in free fall.

That's not true. If you drop a ball inside an elevator that is accelerating upward with uniform acceleration, then the ball will drop to the floor. If the elevator is accelerating downward, then the ball will rise to the ceiling. But if the elevator is in freefall, then the ball will just drift where you dropped it.

An elevator in freefall is completely different from an elevator with uniform acceleration.

The equivalences are: An elevator in freefall near a planet is equivalent to an inertial elevator in empty space. An elevator at rest on a planet is equivalent to an elevator accelerating upward in empty space.

An accelerometer cannot distinguish between freefall and inertial motion. So, from the point of view of General Relativity, they are both inertial. An accelerometer cannot distinguish between an accelerating elevator and an elevator at rest in a gravitational field, so from the point of view of GR, they are both noninertial.

[QUOTEso even in an accelerated RF (elevator) one would not measure acceleration even with a "accelerometer"[/QUOTE]

That's not true.

As DaleSpam said, freefall is considered inertial motion from the point of view of General Relativity, precisely because an accelerometer would show no acceleration.

 

[adoion]

As DaleSpam said, freefall is considered inertial motion from the point of view of General Relativity, precisely because an accelerometer would show no acceleration.

I forgot to mention:
if the elevator (rocket) accelerates under the influence of an gravitational field (at the turnaround) then what would we have ?

an inertial frame that accelerates ??

even from the point of view of special relativity, this has to be a IRF since the laws of stay unchanged in their simplest form.

 

[stevendaryl]

I forgot to mention:
if the elevator (rocket) accelerates under the influence of an gravitational field (at the turnaround) then what would we have ?

an inertial frame that accelerates ??

Yes. There are two different notions of "acceleration": coordinate acceleration, and proper acceleration. Coordinate acceleration depends on your coordinate system. For example, the path: (x=0, y=vt) has zero coordinate acceleration in Cartesian coordinates. But if you switch to polar coordinates: r = \sqrt{x^2 + y^2}, \theta = tan^{-1}(\frac{y}{x}), then the coordinate acceleration in terms of r and \theta is nonzero. Proper acceleration is what is measured by an accelerometer, and it is independent of what coordinate system you use.

even from the point of view of special relativity, this has to be a IRF since the laws of stay unchanged in their simplest form.

 

[adoion]

Yes. There are two different notions of "acceleration": coordinate acceleration, and proper acceleration. Coordinate acceleration depends on your coordinate system. For example, the path: (x=0,y=vt) (x=0, y=vt) has zero coordinate acceleration in Cartesian coordinates. But if you switch to polar coordinates: r=x 2 +y 2 − − − − − − √ r = \sqrt{x^2 + y^2}, θ=tan −1 (yx ) \theta = tan^{-1}(\frac{y}{x}), then the coordinate acceleration in terms of r r and θ \theta is nonzero. Proper acceleration is what is measured by an accelerometer, and it is independent of what coordinate system you use.

an accelerometer would measure zero in the above case since the rocket would be in free fall.

 

[stevendaryl]

an accelerometer would measure zero in the above case since the rocket would be in free fall.

Right, an accelerometer measures proper acceleration, not coordinate acceleration.

 

[Dale]

an accelerometer would measure zero in the above case since the rocket would be in free fall.

Yes. And therefore the rockets frame is inertial.

 

[PeterDonis]

an accelerometer doesn't measure acceleration in free fall

It measures zero acceleration in free fall. Zero is a perfectly good measurement result.

 

[harrylin]

You're certainly right, that noninertial frames came up in Newtonian mechanics. So where did the idea that GR was necessary to handle an accelerated reference frame come from?

I think that part of it is the insistence on relativity. [...] Nobody bothered (as far as I know) to try to write Newtonian mechanics in a way that treated all coordinate systems equally. I don't think that the latter was developed until after GR (the Newton-Cartan formulation of Newtonian physics).

I had -and still have- an issue with your "retrospective" because I was not thinking about Newtonian mechanics but about special relativistic mechanics with classical "know-how". As a matter of fact, now that I think of it: Einstein even developed GR based on the understanding that SR can handle accelerated frames! But I think that you are right that many people for some time afterward lacked that understanding, if that is what you meant. It's a mystery to me how this original understanding which was rather well explained in papers could have been lost or mixed up for a while.

 

[harrylin]

I forgot to mention:
if the elevator (rocket) accelerates under the influence of an gravitational field (at the turnaround) then what would we have ? [...]
even from the point of view of special relativity, this has to be a IRF since the laws of stay unchanged in their simplest form.

It is not a universal IRF. In case you forgot: you can infer my answer (and even infer Einstein's answer), from my earlier reply here:
https://www.physicsforums.com/threa...paradox-as-paradox.780185/page-2#post-4907595

 

[stevendaryl]

I had -and still have- an issue with your "retrospective" because I was not thinking about Newtonian mechanics but about special relativistic mechanics with classical "know-how". As a matter of fact, now that I think of it: Einstein even developed GR based on the understanding that SR can handle accelerated frames! But I think that you are right that many people for some time afterward lacked that understanding, if that is what you meant. It's a mystery to me how this original understanding which was rather well explained in papers could have been lost or mixed up for a while.

I think it was just a matter of figuring out the right pedagogy. Technically, there were no problems in applying SR to noninertial coordinate systems. The issue was how to "frame" what you were doing. Einstein's paper, which does invoke GR to explain the paradox, is an example of bad pedagogy. There is nothing "GR" about it, except for the fact that Einstein maybe was a little unclear about the distinction between the use of noninertial coordinates and gravity. Gravity requires noninertial coordinates, but not the other way around.

 

[PhoebeLasa]

[...]
Einstein's paper, which does invoke GR to explain the paradox, is an example of bad pedagogy. There is nothing "GR" about it, except for the fact that Einstein maybe was a little unclear about the distinction between the use of noninertial coordinates and gravity.

I think there was a specific reason Einstein used GR to resolve the twin paradox. He wanted to construct an analogous scenario (via the equivalence principle) in which the "rocket-twin" could say that he was absolutely stationary and unaccelerated during the whole time that the twins were separated. When he fired his rocket engine, he was doing it strictly to counteract the spatially-uniform gravitational field that is somehow momentarily switched on, so that the rocket-twin would remain stationary and unaccelerated. That momentarily switched-on gravitational field causes the "home twin" (the twin who has no rocket) to accelerate, reverse course, and move toward the "traveler". The resulting conclusion using this GR scenario is that the rocket-twin will say that the "home-twin" suddenly gets much older while that gravitational field is switched on.

The exact same result (regarding the rocket-twin's conclusion about the home-twin suddenly getting much older during the turnaround) is obtained without recourse to GR (and without any gravitational fields), purely from SR, using a non-inertial reference frame for the rocket-twin which is formed by piecing together multiple inertial frames that are each momentarily co-moving with the rocket-twin at different instants of his life. The rocket-twin is always at the spatial origin of his non-inertial reference frame, but he never contends that he doesn't accelerate. He knows that he accelerates, and reverses course, when he turns on his rocket. And he knows that it is the home-twin who is unaccelerated for the whole trip.

There is a difference between being "always absolutely at rest" (Einstein's GR scenario for the rocket-twin) versus "being always at the spatial origin of your own personal reference frame, but accelerating at will using your rocket engine" (the SR scenario). But what the rocket-twin says about the home-twin suddenly getting much older during the turnaround is exactly the same for both scenarios (even though it's a different twin doing the turnaround in the two cases).

 

[Dale]

using a non-inertial reference frame for the rocket-twin which is formed by piecing together multiple inertial frames that are each momentarily co-moving with the rocket-twin at different instants of his life.

This is not the only method of forming a non inertial coordinate systems, and as mentioned before it has its own problems.

 

[PhoebeLasa]

This is not the only method of forming a non inertial coordinate systems, and as mentioned before it has its own problems.

But the momentary co-moving inertial frames method is the only (SR) method that exactly agrees with the often-cited standard GR method ... both give the result that the rocket-twin says that the home-twin suddenly gets much older during the turnaround. Alternative SR methods that have been proposed don't agree with the standard GR method.

 

[PeterDonis]

Alternative SR methods that have been proposed don't agree with the standard GR method.

Really? Which ones?

 

[pervect]

But the momentary co-moving inertial frames method is the only (SR) method that exactly agrees with the often-cited standard GR method ... both give the result that the rocket-twin says that the home-twin suddenly gets much older during the turnaround. Alternative SR methods that have been proposed don't agree with the standard GR method.

I'm not quite sure what your point is? The momentarily co-moving frames method is popular (deservedly so), but it's not the only method. You'll find some disscusion of Dolby & Gull's "radar time" on Physics Forums and the literature, for instance. See for instance http://arxiv.org/abs/gr-qc/0104077 "On Radar Time and the Twin `Paradox".

In the present paper we recall the definition of ‘radar time’ (and related ‘radar distance’) and emphasise that this definition applies not just to inertial observers, but to any observer in any spacetime. We then use radar time to derive the hypersurfaces of simultaneity for a class of traveling twins, from the ‘Immediate Turn-around’ case, through the ‘Gradual Turn-around’ case, to the ‘Uniformly Accelerating’ case. (The
‘Immediate Turn-around’ and ‘Uniformly Accelerating’ cases are also discussed in Pauri et al.

We show that in all cases this definition assigns a unique time to any event with which Barbara can send and
receive signals,

Editorial note. It isn't obvious, but Barabara can NOT send and receive signals from all space-time events! To give a specific example, if Barbara accelerates at 1g, and Obe stays behind. If Babara leaves in the year 3000 as measured by Obe's calendear, Barbara will never receive a signal sent by Obe in year 3001 or later.

and that this assignment is independent of any choice of coordinates. We then demonstrate that brief periods of acceleration have negligible effect on the radar time assigned to distant events, in contrast with the sensitive dependence of the hypersurfaces implied by Figures 1 and 2. By viewing the situation in different coordinates we further demonstrate the coordinate independence of radar time,
and note that there is no observational difference between the interpretations in which the differential aging is ‘due to Barbara’s acceleration’ or ‘due to the gravitational field that Barbara sees because of this acceleration’.

So to summarize, while the momentarily co-moving frame method is popular (and deservedly so, though I didn't get into it's nice quantities), in some circumstances other methods such as Dolby & Gull's "radar simultaneity" might be better. In the abstract framework of things, the point is that simultaneity is relative, and different simultaneity conventions have different strengths and weaknesses.

Additionally, it's important to note that accelerating observers cannot receive signals from all of space-time, and this in many circumstances effectively prevents an accelerating observer from defining the notion of "at the same time" to certain events behind them, including events that happen at their point of departure after "a long enough time", due to the fact that the accelerating observer can't receive signals from these events as long as they keep accelerating.

Doby and Gull's method isn't an exception to this - while it has some good qualities, it can't handle the situation where Barabara doesn't receive signals from Obe, this is pointed out in the paper but not emphasized.

 

[harrylin]

I think it was just a matter of figuring out the right pedagogy. Technically, there were no problems in applying SR to noninertial coordinate systems. The issue was how to "frame" what you were doing. Einstein's paper, which does invoke GR to explain the paradox, is an example of bad pedagogy. There is nothing "GR" about it, except for the fact that Einstein maybe was a little unclear about the distinction between the use of noninertial coordinates and gravity. [..]

That can't be right. Einstein explains in that very same paper why the twin calculation can hardly be considered paradoxical in SR - at least, it surely wasn't paradoxical for people who correctly understood SR at the time. And Einstein understood rather well how to deal with accelerating frames, as -once more- his "induced gravitational fields" were in fact derived from his calculations with accelerating frames. Working backwards, he did not make any calculation error concerning accelerating frames as far as I can tell, and also according to the Physics FAQ: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html

Perhaps many people who did not correctly understand SR, got the wrong impression that GR had to be used for accelerated objects frames and even accelerated objects, because Einstein argued that GR could be used like that?

 

[stevendaryl]

That can't be right. Einstein explains in that very same paper why the twin calculation can hardly be considered paradoxical in SR - at least, it surely wasn't paradoxical for people who correctly understood SR at the time. And Einstein understood rather well how to deal with accelerating frames, as -once more- his "induced gravitational fields" were in fact derived from his calculations with accelerating frames

My point, as I said, is that the calculation has nothing really to do with GR, except in the sense that SR is a limiting case of GR, so any SR calculation is automatically a GR calculation.

Perhaps many people who did not correctly understand SR, got the wrong impression that GR had to be used for accelerated objects frames and even accelerated objects, because Einstein argued that GR could be used like that?

As I said, he's not using GR at all in that calculation. He's using SR in noninertial coordinates.

 

[JesseM]

There's a widespread misconception that you need general relativity in situations involving acceleration, but it's just not true; special relativity handles acceleration just fine. You can google for "Rindler coordinates" for one example, and you'll find another example (a clock experiencing uniform circular motion due to the Earth's rotation) in Einstein's original 1905 paper to which ghwellsjr gave you a link above.

You can also forego accelerating coordinate systems, and just analyze the time elapsed on an accelerating clock using the coordinates of some inertial frame in which you know the clock's coordinate position and velocity as a function of coordinate time. The trick is to approximate a smoothly-varying path by a polygonal path made up of a bunch of short inertial segments lasting a coordinate time \Delta t, that way the time elapsed on the clock on each segment will be \sqrt{1 - v^2/c^2} \Delta t, and then you can just add up the clock times on all the segments (using the appropriate v for each segment, which can vary from one to another) to get the total time elapsed on the polygonal path. Then you let the time of each segment become infinitesimal, so the sum becomes an integral and the total time elapsed on a clock with velocity as a function of time v(t) is just \int \sqrt{1 - v(t)^2/c^2} dt.

Einstein doesn't go into detail, but he does allude to this method at the end of section 4 of the 1905 paper, when he writes:

It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be \frac{1}{2} tv^2/c^2 second slow.

(note that the factor he gives is the result of a first-order approximation to the fully accurate formula t\sqrt{1 - v^2/c^2})

 

[harrylin]

My point, as I said, is that the calculation has nothing really to do with GR [..] He's using SR in noninertial coordinates.

I see that you disagree with his argument; and for reasons different from yours, so do I. :)
In fact, I won't discuss disagreements you may have with Einstein about his theory, as you here agree with the point that I tried to make (sorry if that was not clear!): Einstein and contemporaries understood perfectly well how to handle accelerations with SR - even accelerating frames. Furthermore, he did not pretend that GR is required to handle acceleration. Nevertheless, the misconception about acceleration did come about. Thus my suggestion remains that perhaps many people who did not correctly understand SR, got the wrong impression that GR had to be used for accelerated objects frames and even accelerated objects, because Einstein argued that GR could be used like that.

 

[stevendaryl]

I see that you disagree with his argument.

No, I don't disagree with his argument. I disagree that his argument involves General Relativity. Going from SR to GR consists of two steps:

1. Allowing the metric tensor g_{\mu \nu} to be dynamic, instead of constant everywhere.
2. Describing how the metric tensor is affected by mass/energy/momentum (the field equations--these describe how spacetime curvature is affected by stress-energy, but in the most basic version of GR, the curvature tensor is a function of the metric only).

Those are the only two additions to SR to produce GR, as far as I know. (There is perhaps a little ambiguity in #1, because it's not always straight-forward how to generalize the physics of flat spacetime to curved spacetime.) Neither is relevant in the twin paradox.

 

[harrylin]

No, I don't disagree with his argument. I disagree that his argument involves General Relativity. [..].

Once more (and with this last repetition I end my discussion about this side topic): that is your main disagreement with Einstein. He admitted that " it is certainly correct that from the point of view of the general theory of relativity we can just as well use coordinate system K' as coordinate system K " , and the rest of that paper is his argument in defence of that position against critics.

 

[stevendaryl]

He admitted that " it is certainly correct that from the point of view of the general theory of relativity we can just as well use coordinate system K' as coordinate system K " , and the rest of that paper is his argument in defence of that position against critics.

I agree that it is correct from the point of view of GR, but it's ALSO correct from the point of view of SR.

 

[harrylin]

I agree that it is correct from the point of view of GR, but it's ALSO correct from the point of view of SR.

Einstein clarified that if one would make the mistake to consider the accelerating system K' as a valid "rest" system from the point of view of SR, in which only Galilean reference systems such as K are equivalent, one would create a twin paradox in SR. That has now been explained on this forum many times, even in this thread. The point from which you distracted is the fact that Einstein and his contemporaries knew very well how to handle acceleration in SR.

 

[stevendaryl]

Einstein clarified that if one would make the mistake to consider K' as a valid "rest" system from the point of view of SR, in which only Galilean reference systems such as K are equivalent, one would create a twin paradox in SR.

Yes, the equations do not have the same form in a non-inertial coordinate system. But that's a fact about SR. The derivation that Einstein gave is an SR derivation. Of course, it's valid in GR, as well, but there is nothing specifically GR about it.

His derivation is an SR derivation, because he did not make use of any of the differences between SR and GR.

 

[stevendaryl]

Yes, the equations do not have the same form in a non-inertial coordinate system. But that's a fact about SR. The derivation that Einstein gave is an SR derivation. Of course, it's valid in GR, as well, but there is nothing specifically GR about it.

His derivation is an SR derivation, because he did not make use of any of the differences between SR and GR.

The equations of SR have the same form in any inertial frame. But to figure out the form in a noninertial frame, all it takes it calculus. I consider that still SR to treat the accelerated frame as "at rest"--that's just a coordinate transformation. The fact that when using noninertial coordinates, the metric tensor becomes position-dependent just falls right of the coordinate transformation. So position-dependent time dilation in noninertial coordinates is inherent in SR. There is nothing "General Relativistic" about it. To call the position-dependence of the metric a "gravitational field" is just picturesque language. There is still no additional assumptions involved, as far as I can tell, beyond those of SR. So it's a mistake to call it a "GR" resolution, because there is nothing in it that isn't already implicit in SR.

I suppose that GR is involved when you say that the fake gravitational field that results from noninertial coordinates is no less real than the gravitational field due to planets, but since the equations don't depend on how "real" the gravitational field is, I just don't see how the derivation could be considered a GR derivation.

 

[harrylin]

Yes, the equations do not have the same form in a non-inertial coordinate system. But that's a fact about SR. The derivation that Einstein gave is an SR derivation. Of course, it's valid in GR, as well, but there is nothing specifically GR about it.

His derivation is an SR derivation, because he did not make use of any of the differences between SR and GR.

K' is invalid as "rest frame" in SR. Despite our differing disagreements with Einstein, we agreed a long time ago on the point that I made, which is that Einstein and others understood acceleration in SR. It was not in "retrospect" that General Relativity was not needed to calculate the twin problem. Einstein never suggested that GR would be needed for the calculation. However, from my discussion with you I now slightly change my hypothesis about how that misunderstanding may have come about. For it now seems plausible to me that many people may have misunderstood Einstein's arguments in his papers from 1916-1918 that GR could be used for accelerated frames and even accelerated objects, so that they misconstrued that according to Einstein GR had to be used. And that's all that I will hypothesize about that. :)

 

[stevendaryl]

K' is invalid as "rest frame" in SR.

SR as a theory of physics is not about rest frames. That's a way to talk about SR, and a way to derive the Lorentz transformations, but as a theory of physics, it makes claims about the behavior of clocks and rods and light signals and so forth. Those claims can be expressed in any coordinates you like. The fact that they were originally derived for inertial reference frames is irrelevant, except for historical interest.

So K' is not an inertial reference frame. That's certainly true. What does that fact have to do with whether using K' to describe the twin paradox involves GR or not? If I have a description of what goes on, according to frame K, and I know the coordinate transformation connect coordinates of frame K to coordinates of frame K', then I know how to describe things in frame K'. I don't need to know whether K' is a "valid rest frame". If by "valid rest frame" you mean "inertial rest frame", then It's not, and it doesn't matter.

There is nothing about using K' that requires going beyond SR.