Resolve to the Twins Paradox – Frame of Origin

[James S Saint]

I have read many arguments concerning the issue of the famed “Twin Clock Paradox”. Very many arguments go through considerably complex explanations so as to defend Special Relativity or dispute it. But the explanation is actually much simpler.

In every proposed twins paradox, there is a Frame of Origin. The frame of origin is where the twins were at rest with respect to each other and to where they return. In the original scenario, that was “the Earth” and is referred to as “the inertial frame”.

It is called “the inertial frame” because it is the one frame declared “inertial; not movable”. But the issue comes to mind as to how either frame can be considered the one that isn’t moving. SR declares that such a notion is irrelevant. But if it is ignored, an apparent paradox arises.

So as to clear up the picture a bit, let’s presume that neither twin stands perfectly still, but rather one twin (Twin A) just very minutely drifts away from the Frame of Origin while the other twin (Twin B) takes off in his ultra fast rocket. This is merely to provide distinction and reveal that there are always 3 frames involved.

And further, so as to get away from the complexities of gravitation issues, let's put them out in space originally standing on a small floating space station far away from any gravity field.

The time dilation factor is always applied to whichever one leaves the Frame of Origin.

Twin B gets in his rocket and rockets away. Twin A gets into his rocket and sticks ores out his windows and begins to row hoping to catch a few gas particles drifting by from Twin B’s exhaust.

If we apply the time dilation to the one doing all of the serious moving, Twin B’s clock will run slower and he will age less. We cannot simply reverse the reasoning and say that it is all relative because Twin A is not moving away from the Frame of Origin.

That is pretty much the end of the story, but…

Some scenarios remove any sign of the Frame of Origin so as to make their point clear. The problem is that in every case, there must always be a Frame of Origin where their clocks were staying in sync whether anything is standing as a sign of it or not. If two items are moving away from each other, there must be a frame indicating the prior situation before they began moving. The issue becomes how to find out where that frame would be.

If the ships merely see nothing but each other separating, they might not know which is accelerating from the Frame of Origin, but that is only a matter of their awareness. If they later find that one has greater time dilation than the other, it can be deduced as to which was accelerating more from the Frame of Origin.

If they see no difference in their clocks, then they can deduce that they were both leaving the Frame of Origin merely in different directions.

Without determining the Frame of Origin, the Twins cannot know which of them will age more until they meet again.
Make sense?

 

[JesseM]

No--the twin paradox still works fine if you don't assume the two observers actually started out at rest relative to each other, you could imagine one astronaut who's spent his whole life on one spaceship and another who's spent his whole life on a different spaceship, with the two ships having been in motion relative to another since the astronauts were born. Then if the two ships pass right next to each other at some moment in time, they can quickly compare ages at the moment they are right next to each other, then both ships travel on inertially at the same relative speed. Then after they have moved apart for a while, either one can fire its rockets in the direction of the other ship, so that the distance between them decreases until they pass one another again and compare their ages again. Then whichever ship fired its rockets, the astronaut on that ship will have aged less in total than the ship that traveled inertially between the two meetings.

 

[Sakha]

Then whichever ship fired its rockets, the astronaut on that ship will have aged less in total than the ship that traveled inertially between the two meetings.

So its the acceleration that 'decides' who ages slower?
Is there any formula involving acceleration that gives a math solution to the paradox?

 

[James S Saint]

No--the twin paradox still works fine if you don't assume the two observers actually started out at rest relative to each other, you could imagine one astronaut who's spent his whole life on one spaceship and another who's spent his whole life on a different spaceship, with the two ships having been in motion relative to another since the astronauts were born. Then if the two ships pass right next to each other at some moment in time, they can quickly compare ages at the moment they are right next to each other, then both ships travel on inertially at the same relative speed. Then after they have moved apart for a while, either one can fire its rockets in the direction of the other ship, so that the distance between them decreases until they pass one another again and compare their ages again. Then whichever ship fired its rockets, the astronaut on that ship will have aged less in total than the ship that traveled inertially between the two meetings.

Well other than being more complicated, I don't see how that is any different than what I said. But consider this;

Twin B takes off for a few years and returns. As he passes by at constant velocity (no rockets), Twin A accelerates to catch up to him. Which has aged more?

Obviously (I hope) Twin B will still be the more aged. Yet it was Twin A that fired his rockets.

But on the other hand, upon returning, Twin B could have fired his rockets so as to slow and meet Twin A. Still Twin B is the more aged.

Who fired their rockets after leaving the Frame of Origin, doesn't resolve who is going to age more.

 

[JesseM]

So its the acceleration that 'decides' who ages slower?
Is there any formula involving acceleration that gives a math solution to the paradox?

The fact that one twin accelerated between meetings and the other didn't means that the twin that accelerated is guaranteed to have aged less, regardless of the actual magnitude and duration of the acceleration. This is closely analogous to the fact in ordinary 2D Euclidean geometry that if you have a straight line between two points (constant slope in any Cartesian coordinate system) and a non-straight path between the same two points (change in slope somewhere along the path, analogous to change in velocity) then the non-straight path is always guaranteed to have a greater total length, not because it accumulated all the extra length on the section of the path with a changing slope (the non-straight path might consist of two straight segments with different slopes joined by a very short curved section, like a bendy straw), but rather because in Euclidean geometry a straight line is the shortest distance between points. I fleshed out the mathematical details of this analogy in post #8 here if you're interested.

 

[James S Saint]

So its the acceleration that 'decides' who ages slower?
Is there any formula involving acceleration that gives a math solution to the paradox?

No. It is not the acceleration deciding which by calculation (not the amount of acceleration, but merely which accelerated away). The acceleration merely identifies who moved from the Frame of Origin.

You could assume instantaneous acceleration without affecting the results. The question is only who left the Frame of Origin where the clocks were in sync. If they both left, both will age with comparative dilations.

 

[JesseM]

Well other than being more complicated, I don't see how that is any different than what I said.

It's different because the two twins don't have a common "frame of origin", they started in different frames. I thought your argument was saying it was important to identify the "frame of origin"? Which astronaut's frame is the "frame of origin" in my scenario where they just pass each other at constant velocity, having lived their whole lives on ships moving inertially with different rest frames?

Twin B takes off for a few years and returns. As he passes by at constant velocity (no rockets), Twin A accelerates to catch up to him. Which has aged more?

Does "takes off for a few years" mean A and B were moving apart for a while, so the distance between them was increasing? If so, how can Twin B "return" to "pass by" A unless he fires his rockets so that he is traveling back towards A for a while?

Obviously (I hope) Twin B will still be the more aged.

Not if Twin B has to fire his rockets at the midpoint of the journey while A only fires his at the end, as suggested above. If that's not what you're imagining, please explain the scenario in more detail.

Yet it was Twin A that fired his rockets.

If what I'm saying above is right, they both fired their rockets, at different times. The fact that Twin B had to fire them at the point where the distance between the twins was greatest, while Twin A only fired them at the very end when they were already very close, helps explain why twin B has aged less. Consider the geometric analogy above I mentioned in my post to Sakha. Suppose you have two points A and B on a 2D plane, and one path between them is mostly straight (and almost parallel to a perfect straight-line path between A and B) and only has a bend right before it meets B. But the other path takes off in a straight line at a totally different angle from the perfect straight-line path, then has a bend in the middle followed by another straight section at a different angle which goes to B (like a V shape with a slightly rounded bottom and the two tips of the V ending at A and B). These paths both contain sections of changing slope (analogous to changing velocity), but the second path with a bend in the middle is obviously going to be longer than the first path that is much closer to the straight-line path and only has a bend at the very end.

Who fired their rockets after leaving the Frame of Origin, doesn't resolve who is going to age more.

In a scenario where one of them moves totally inertially between their two meetings while the other fires his rockets at some point between the two meetings, yes it doesn, the one that accelerated will always have aged less.

 

[Sakha]

So I deduce acceleration is not relative, i.e only one of them feels the force.
This takes me back to my high school physics teacher - "That tree got in my way, not my fault!" (Relatives velocities week), so with SR its really his fault, he 'accelerated' towards the tree and not viceversa.

 

[JesseM]

So I deduce acceleration is not relative, i.e only one of them feels the force.

Right, in SR you know if you're accelerating since you can feel G-forces, as measured by an accelerometer.

 

[Sakha]

Ok, you cleared a lot of confusing thought from my head. Now I'll sleep very well.
I don't know why I assumed that acceleration could be relative.

Nighties. It's your fault if I dream with spaceships and elder astronauts!

 

[James S Saint]

Jesse, in your version, they have to remember who has fired rockets and how much. In mine, they have to remember who left the Frame of Origin wherein their clocks were in sync. If they have no memories, they are pretty much screwed until they meet gain. Knowing who fired rockets last doesn't really help.

 

[Dale]

Without determining the Frame of Origin, the Twins cannot know which of them will age more until they meet again.

No, the age (proper time) is a Lorentz invariant, so it can be calculated in any inertial frame and all inertial frames will agree on its value. It is certainly not restricted to the frame of origin.

 

[JesseM]

Jesse, in your version, they have to remember who has fired rockets and how much. In mine, they have to remember who left the Frame of Origin wherein their clocks were in sync. If they have no memories, they are pretty much screwed until they meet gain. Knowing who fired rockets last doesn't really help.

Well, you didn't address the question of which is supposed to be the "frame of origin" in my scenario where they pass each other inertially--are you saying that your approach just can't tell us who'll be older in this scenario?

And consider another scenario: suppose we two twins that grew up on a small space station, then at some point one twin gets on a shuttle that's docked to the station and accelerates to build up a large velocity relative to the station, then coasts away inertially for a while. Would you say that the station's frame is the "frame of origin" in this case? If so, what do you predict would happen if, after the two twins have been moving apart inertially for a while, the station fires its own rockets to accelerate in the direction of the shuttle, reaching a high enough velocity that it is eventually able to catch up with the shuttle? Would you say that since the shuttle was the one that initially left the frame of origin, the twin on the shuttle should be younger when the two twins reunite?

 

[James S Saint]

No, the age (proper time) is a Lorentz invariant, so it can be calculated in any inertial frame and all inertial frames will agree on its value. It is certainly not restricted to the frame of origin.

No matter what frame you are in, you have to know in what frame their clocks would be in sync else you can't even use Lorentz. That is the "Frame of Origin".

Well, you didn't address the question of which is supposed to be the "frame of origin" in my scenario where they pass each other inertially--are you saying that your approach just can't tell us who'll be older in this scenario?

If they do not know at what frame their clocks would be in sync, they cannot predict who has aged more other than to just look and see.

And consider another scenario: suppose we two twins that grew up on a small space station, then at some point one twin gets on a shuttle that's docked to the station and accelerates to build up a large velocity relative to the station, then coasts away inertially for a while. Would you say that the station's frame is the "frame of origin" in this case? If so, what do you predict would happen if, after the two twins have been moving apart inertially for a while, the station fires its own rockets to accelerate in the direction of the shuttle, reaching a high enough velocity that it is eventually able to catch up with the shuttle? Would you say that since the shuttle was the one that initially left the frame of origin, the twin on the shuttle should be younger when the two twins reunite?

The station in that is not the Frame of Origin. The station was "at the frame of origin" until it accelerated. And the end result would be that they simply aged the same because they both moved away and established a knew frame of sync when they met up again.

 

[Dale]

No matter what frame you are in, you have to know in what frame their clocks would be in sync else you can't even use Lorentz. That is the "Frame of Origin".

The clocks never need to be in sync. Age (proper time) is a frame invariant quantity that is not dependent on any frame variant synchronization convention. In fact, the proper time is even invariant under transformations to non-inertial coordinate systems or in curved spacetime, both of which may have arbitrary synchronization conventions.

The frame of origin is certainly a very convenient frame to use, but it is not at all necessary.

 

[JesseM]

If they do not know at what frame their clocks would be in sync, they cannot predict who has aged more other than to just look and see.

Yes they can, they can just use SR to predict how much each one ages. If you know each twin's velocity as a function of time v(t) in the coordinates of any inertial frame, you can predict how much each ages between the time coordinate t₀ when they depart from one another and the coordinate time t₁ when they reunite, by doing the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2 } \, dt$$. This will give the same answer regardless of what frame you use. Also, if we assume the acceleration times are extremely brief, then the elapsed time can be approximated by just knowing the time in the inertial frame of each constant-velocity phase of the trip: if one twin travels at speed v₁ for time t₁ for the first part, then quickly turns around and travels at speed v₂ for time t₂ for the second part until reuniting with the other twin, then that twin's total aging will be $$t_1 * \sqrt{1 - v_1^2 /c^2 } + t_2 * \sqrt{1 - v_2^2 /c^2 }$$. Again, you can use any frame you want to define the coordinate times of different phases of the trip and the coordinate velocities in each phase, you'll always get the same answer for the total aging. I gave a numerical illustration in post #63 of this thread:

Let's call the inertial (Earth-bound) twin "Terence" and the traveling twin "Stella", following the Twin Paradox FAQ. First let's look at the numbers in Terence's rest frame. Suppose that in this frame, Stella travels away from Terence inertially at 0.6c for 10 years, at which point she is at a distance of 0.6*10 = 6 light-years from Earth in this frame, then she turns around (i.e. she accelerates, a non-inertial motion which will cause her to experience G-forces that show objectively that she wasn't moving inertially) and heads back towards Terence at 0.6c, finally reuniting with Terence after 20 years have passed since her departure in this frame. Since Terence is at rest in this frame, he has aged 20 years. But since Stella was moving at 0.6c in this frame, the time dilation formula tells us her aging was slowed down by a factor of $$\sqrt{1 - 0.6^2}$$ = 0.8, so she only aged 0.8*10 = 8 years during the outbound leg of her trip, and another 0.8*10 = during the inbound leg, so she has only aged 16 years between leaving Earth and returning.

Now let's analyze the same situation in a different inertial frame--namely, the frame where Stella was at rest during the outbound leg of her trip (she can't also be at rest during the inbound leg in this frame, since this is an inertial frame while Stella accelerated between the two legs of the trip). In this frame, Terence on Earth is initially moving away from Stella at 0.6c while she remains at rest. In Terence's frame, remember that Stella accelerated when she was 6 light-years away from Earth, so we can imagine she turns around when she reaches the far end of a measuring-rod at rest in Terence's frame and 6 light-years long in that frame, with Terence sitting on the near end; in the frame we're dealing with now, the measuring-rod will therefore be moving along with Terence at 0.6c, so it'll be shrunk via length contraction to a length of only 0.8*6 = 4.8 light-years. So, Stella accelerates when the distance between her and Terence is 4.8 light-years in this frame, and since Terence as moving away from her at 0.6c in this frame, they will be 4.8 light-years apart after 4.8/0.6 = 8 years have passed. During these 8 years, it is Terence's aging that is slowed down by a factor of 0.8, so while Stella ages 8 years during this leg, Terence only ages 0.8*8 = 6.4 years. Then Stella accelerates to catch up with Terence, while Terence continues to move inertially at 0.6c. Using the relativistic velocity addition formula, if Stella was moving at 0.6c in Terence's frame and Terence is moving at 0.6c in the same direction in this frame, then in this frame Stella must be moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 0.88235c during the inbound leg. And since Terence is still moving at 0.6c in the same direction, the distance between Stella and Terence will be closing at a "closing speed" of 0.88235c - 0.6c = 0.28235c. Since the distance was initially 4.8 light years at the moment Stella accelerated, in this frame it will take 4.8/0.28235 = 17 years for Stella to catch up with Terence on Earth. During this time Terence has aged another 0.8*17 = 13.6 years, so if you add that to the 6.4 years he had aged during the outbound leg, this frame predicts he has aged 20 years between Stella leaving and Stella returning, same as in Terence's frame. And since Stella is traveling at 0.88235c her aging is slowed by a factor of $$\sqrt{1 - 0.88235^2}$$ = 0.4706, so during those 17 years in this frame she only ages 0.4706*17 = 8 years during the inbound leg. If you add that to the 8 years she aged during the outbound leg, you find that this frame predicts she has aged 16 years between departing and returning, which again is the same as what was predicted in Terence's frame.

The station in that is not the Frame of Origin. The station was "at the frame of origin" until it accelerated. And the end result would be that they simply aged the same because they both moved away and established a knew frame of sync when they met up again.

Well, your prediction that they aged the same is incorrect. Even if the magnitude and duration of their accelerations relative to the inertial frame where the station was originally at rest was exactly the same, in this scenario the twin on the station would have aged less when they reunited. I can give a numerical derivation if you want, but the reason is not hard to understand intuitively in terms of the geometric analogy, where a path that's mostly straight with a small curve near one endpoint will be shorter than a path with two straight segments and a curve in the middle:

Consider the geometric analogy above I mentioned in my post to Sakha. Suppose you have two points A and B on a 2D plane, and one path between them is mostly straight (and almost parallel to a perfect straight-line path between A and B) and only has a bend right before it meets B. But the other path takes off in a straight line at a totally different angle from the perfect straight-line path, then has a bend in the middle followed by another straight section at a different angle which goes to B (like a V shape with a slightly rounded bottom and the two tips of the V ending at A and B). These paths both contain sections of changing slope (analogous to changing velocity), but the second path with a bend in the middle is obviously going to be longer than the first path that is much closer to the straight-line path and only has a bend at the very end.

 

[James S Saint]

The clocks never need to be in sync. Age (proper time) is a frame invariant quantity that is not dependent on any frame variant synchronization convention. In fact, the proper time is even invariant under transformations to non-inertial coordinate systems or in curved spacetime, both of which may have arbitrary synchronization conventions.

The frame of origin is certainly a very convenient frame to use, but it is not at all necessary.

This doesn't seem to approach the paradox question. "Proper time" has to get determined. This issue is about who gets to make that determination between two separating bodies.

Yes they can, they can just use SR to predict how much each one ages. If you know each twin's velocity as a function of time v(t) in the coordinates of any inertial frame, you can predict how much each ages between the time coordinate t0 when they depart from one another and the coordinate time t1 when they reunite, by doing the integral

You can't know that t0 without knowing the Frame of Origin. From "any" arbitrarily chosen frame, you can determine a change in their aging, but you can't know which frame to choose. If you choose a frame that is inertial to Twin B, then you will see yourself as the one moving away and not aging.

The Frame of Origin gives "proper time" to your choice.

Well, your prediction that they aged the same is incorrect. Even if the magnitude and duration of their accelerations relative to the inertial frame where the station was originally at rest was exactly the same, in this scenario the twin on the station would have aged less when they reunited. I can give a numerical derivation if you want, but the reason is not hard to understand intuitively in terms of the geometric analogy, where a path that's mostly straight with a small curve near one endpoint will be shorter than a path with two straight segments and a curve in the middle:

No actually, I didn't give it much thought and considered the idea that you would come back with critique. After a little thought, I can tell that because one left earlier and stayed at high velocity longer, he would have aged a little less until the other over accelerated to catch up. At that point, it is an issue of the exact details.

And I am a little curious now exactly how that would pan out. I'm not certain the resolution would be consistent.

If Twin B shot off at c then stopped and later Twin A did the same, they would be the same age due to the symmetry of their situation.

If Twin B shot off at c and didn't stop, then obviously Twin A would never catch him and they would not age any more, leaving Twin A still older.

Anywhere between those gets a little vague, but implies that Twin A would be merely less older than he would have been if he had stayed on the station. So you are very probably right even without the math.

 

[Mike_Fontenot]

So its the acceleration that 'decides' who ages slower?
Is there any formula involving acceleration that gives a math solution to the paradox?

The equation you're asking about DOES exist. Here's how I described it in a previous post:
________________________________________________________

In the simplest version of the traveling twin example (with a single instantaneous speed change at the midpoint), it IS possible to INFER the change in the home twin's age (according to the traveler), during the instantaneous turnaround: you know that both twins obviously must agree about their respective ages when they are reunited, and you know how much the traveler says his twin ages during the two inertial portions of his trip. So the home twin's ageing during the turnaround (according to the traveler) must be just enough to make the totals agree at the reunion.

But in only slightly more complicated examples (e.g., with multiple instantaneous velocity changes), it becomes important to KNOW how to directly calculate the amount of the home twin's ageing during each of the velocity changes. And in the case of FINITE accelerations, being able to directly calculate the home twin's ageing during each of the traveler's segments of finite acceleration is indispensable.

It turns out to be easy to do that. For the cases of instantaneous velocity changes, the required calculations are almost trivial to carry out. For piecewise-constant accelerations, the calculations are a bit more complex, but they can still be done, if necessary, with a good calculator. And with a computer program, they are very easy.

Both of the above types of problems can be handled with a simple equation that I derived many years ago, which I call "the CADO equation". The CADO equation follows directly from the Lorentz equations ... the CADO equation really just automates what you can deduce from the geometry of the Minkowski diagram.

I also, many years ago, implemented the CADO equation in a computer program I call "the CADO program".

It's also possible to use the CADO equation to do completely general acceleration profiles, but you (usually) can't do it in a closed-form way...it requires some numerical integrations. Fortunately, piecewise-constant accelerations are usually all you really need to be able to handle.

Here is a description of the CADO equation that I've posted previously, in other threads:
__________________________________________________ __________


Years ago, I derived a simple equation that relates the current ages of the twins, ACCORDING TO EACH TWIN. Over the years, I have found it to be very useful. To save writing, I write "the current age of a distant object", where the "distant object" is the stay-at-home twin, as the "CADO". The CADO has a value for each age t of the traveling twin, written CADO(t). The traveler and the stay-at-home twin come to DIFFERENT conclusions about CADO(t), at any given age t of the traveler. Denote the traveler's conclusion as CADO_T(t), and the stay-at-home twin's conclusion as CADO_H(t). (And in both cases, remember that CADO(t) is the age of the home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

(Although the dependence is not shown explicitly in the above equation, the quantities L and v are themselves functions of t, the age of the traveler).

The factor (c*c) has value 1 for these units, and is needed only to make the dimensionality correct. For simplicity, you can generally just ignore the c*c factor when using the equation.

The equation explicitly shows how the home twin's age will change abruptly (according to the traveler, not the home twin), whenever the relative speed changes abruptly.

For example, suppose the home twin believes that she is 40 when the traveler is 20, immediately before he turns around. So CADO_H(20-) = 40. (Denote his age immediately before the turnaround as t = 20-, and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home twin), and that their relative speed is +0.9 ly/y (i.e., 0.9c), when the traveler's age is 20-. Then the traveler will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 13

years old immediately before his turnaround. Immediately after his turnaround (assumed here to occur in zero time), their relative speed is -0.9 ly/y. The home twin concludes that their distance apart doesn't change during the turnaround: it's still 30 ly. And the home twin concludes that neither of them ages during the turnaround, so that CADO_H(20+) is still 40.

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years during his instantaneous turnaround.

The equation works for arbitrary accelerations, not just the idealized instantaneous speed change assumed above. I've got an example with +-1g accelerations on my web page:

http://home.comcast.net/~mlfasf

The derivation of the equation is given in my paper

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot

 

[Dale]

This doesn't seem to approach the paradox question. "Proper time" has to get determined. This issue is about who gets to make that determination between two separating bodies.

Anyone in any spacetime using any reference frame and any synchronization convention may make that determination. That is the whole point of using frame-invariant quantities.

 

[James S Saint]

Anyone in any spacetime using any reference frame and any synchronization convention may make that determination. That is the whole point of using frame-invariant quantities.

The point to the paradox confusion is that they do not have a "synchronization convention" so as to make the decision. The Frame of Origin gives that convention for them so as to resolve the question.

 

[JesseM]

You can't know that t0 without knowing the Frame of Origin. From "any" arbitrarily chosen frame, you can determine a change in their aging, but you can't know which frame to choose. If you choose a frame that is inertial to Twin B, then you will see yourself as the one moving away and not aging.

No, it doesn't matter what frame you choose. You can choose any inertial frame, and all frames will get the same answer for the elapsed total aging when they calculate $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2} dt$$. Also, note that if we know the time t₀ when the two twins departed from one another in some choice of frame, and also the position x₀ in the same frame, then to find the time t'₀ when the twins departed in a different frame moving at speed v relative to the first one, you can just use the standard Lorentz transformation equation:

$$t'_0 = \frac{1}{\sqrt{1 - v^2/c^2}} (t_0 - vx_0 /c^2)$$

Incidentally, I don't know what you mean by "a frame that is inertial to twin B". If twin B accelerates at some point in his journey, that's an objective fact that all inertial frames agree on (he will feel G-forces when he accelerates), and there is no inertial frame where twin B is at rest throughout the journey (though you can find an inertial frame where twin B was at rest during the coasting phase prior to accelerating to turn around, and another one where twin B was at rest during the coasting phase after accelerating to turn around)

The Frame of Origin gives "proper time" to your choice.

Proper time is a technical term, the proper time along a given path through spacetime (i.e. a given world line) refers to the time that would be read by a clock following that path. It's a frame-independent quantity, meaning all frames will agree in their predictions about the proper time elapsed between two points on a given path through spacetime.

No actually, I didn't give it much thought and considered the idea that you would come back with critique. After a little thought, I can tell that because one left earlier and stayed at high velocity longer, he would have aged a little less until the other over accelerated to catch up.

There is no objective frame-independent choice about which one "aged less" until they actually meet up and compare ages at a single point in spacetime. When you have two twins that are far apart, different frames can disagree on which has aged less since they departed one another. This has to do with the relativity of simultaneity which says that different frames disagree on whether two events at different spatial locations happened simultaneously; so for example if twin B departed from twin A when both were aged 30, and twin B quickly accelerated to a speed of 0.6c relative to twin A and then continued coasting from there, then in the rest frame of twin A the event of twin A turning 40 is simultaneous with the event of twin B turning 38 (so twin B has aged less in this frame), but in the frame where twin B is at rest during his coasting phase, the event of twin B turning 38 is simultaneous with the event of twin A being an age of 36.4 (so twin A has aged less in this frame). Of course, if either twin A or twin B later accelerates so eventually the two twins are able to meet at a common location and compare ages, both frames will agree on their ages when they meet, and thus both frames agree on who ages less in total despite disagreeing on who was aging slower during the initial coasting phase. Look at my post #63 from another thread which I quoted in post #16 of this one for a numerical example of how this works out.

And I am a little curious now exactly how that would pan out. I'm not certain the resolution would be consistent. If Twin B shot off at c

It's not actually possible for an object with nonzero rest mass to move at c, but they can move at some high fraction like 0.8c.

then stopped and later Twin A did the same, they would be the same age due to the symmetry of their situation.

Well, if in the original rest frame twin B first moved at 0.8c for 10 years and then was at rest for 10 years, while twin A first was at rest for 10 years and then moved at 0.8c for 10 years, then they would be the same age when they met due to symmetry. But that wasn't the situation I was talking about. I was imagining twin B briefly accelerated away and then continued on at some constant velocity like 0.8c, never accelerating again, while twin A remained at rest for a while and then accelerated to an even higher velocity than 0.8c so he could eventually catch up with twin B. In this case twin A would have aged less than twin B when they met.

 

[Dale]

The point to the paradox confusion is that they do not have a "synchronization convention" so as to make the decision. The Frame of Origin gives that convention for them so as to resolve the question.

So does any other frame.

In any arbitrary spacetime pick some reference frame x with metric g and some worldline P then the proper time is given by:

$$\tau = \int_P \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu}}$$

All reference frames will agree on this invariant number, not just the frame of origin. This is why invariant quantities are so valuable. You don't need a special Master Frame Of Correct Calculations, you can use any convenient one. You seem to miss the whole point of relativity.

 

[James S Saint]

All reference frames will agree on this invariant number, not just the frame of origin. This is why invariant quantities are so valuable. You don't need a special Master Frame Of Correct Calculations, you can use any convenient one. You seem to miss the whole point of relativity.

You seem to be missing the cause of the confusion. IF they know the "g" value from any particular frame, they can use that. But how would they ever know that or agree on it?

They cannot merely say, "well, I have acceleration g1 and you have acceleration g2 so now let's do the math." Arbitrarily chosen g values will not yield the same result. They have to know a g value from a particular frame. Again, the Frame of Origin gives them a frame to agree upon where they knew they were in sync or they can measure their g value, not merely pick one out of thin air.

 

[Dale]

Careful, g is the metric tensor, it is not an acceleration. Are you familiar with tensors in general and the metric tensor in specific?

 

[JesseM]

You seem to be missing the cause of the confusion. IF they know the "g" value from any particular frame, they can use that. But how would they ever know that or agree on it?

They cannot merely say, "well, I have acceleration g1 and you have acceleration g2 so now let's do the math." Arbitrarily chosen g values will not yield the same result. They have to know a g value from a particular frame. Again, the Frame of Origin gives them a frame to agree upon where they knew they were in sync or they can measure their g value, not merely pick one out of thin air.

If by "g" you mean the acceleration (although as DaleSpam said this is not what he meant), both acceleration and velocity in a particular frame are just time-derivatives of the function giving position as a function of time in that frame, x(t). And if you know x(t) in one inertial frame, it's a trivial matter to find x'(t') of the same object in the x', t' coordinates of any other inertial frame moving at speed v relative to the first. Just use the Lorentz transformation:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c²)
where gamma = 1/sqrt(1 - v²/c²)

which can be solved for x and t to get the reversed transformation:

x = gamma*(x' + vt')
t = gamma*(t' + vx'/c²)

For example, if in the first frame the object has x = 0.25t² (in units of light-seconds and seconds so c=1), which implies a constant acceleration of a=0.5 in that frame, then just substitute the above reversed Lorentz transformation equations into this to get:

gamma*(x' + vt') = 0.25*gamma²*(t' + vx'/c²)²

You can solve this for x' as a function of t' with a little algebra, and once you know x'(t') just take derivatives with respect to t' to find the velocity and acceleration as a function of time in the x',t' frame.

 

[James S Saint]

If by "g" you mean the acceleration (although as DaleSpam said this is not what he meant), both acceleration and velocity in a particular frame are just time-derivatives of the function giving position as a function of time in that frame, x(t). And if you know x(t) in one inertial frame,

But you can't know x(t) until you decide on an inertial frame. How do you decide if t = 10 or 1000, not to mention x. Each Twin would still argue.

..and g referring to "gravity" is irrelevant to the conversation so I took him to mean merely acceleration, else why bring it up at all. To know any g that would help would require knowing some common g frame thus not resolve anything, especially if no such g frame was handy.

 

[JesseM]

But you can't know x(t) until you decide on an inertial frame.

Of course. The point is that if you know it in one inertial frame, it's trivial to translate it to any other inertial frame. Thus there is nothing to compel you to use a particular frame, like the frame where both are originally at rest (if they are originally at rest relative to one another as opposed to just passing each other inertially as in my earlier example).

How do you decide if t = 10 or 1000, not to mention x. Each Twin would still argue.

Argue about what? Even if each twin uses a different inertial frame to do their calculations, they will both make the same prediction about how much each twin will have aged when they reunite. Of course the two frames will differ on frame-dependent questions like coordinate velocity and acceleration, but as long as they both understand that these quantities are frame-dependent and that all frames are equally valid, there's no need for them to argue about their differing values for these quantities either.

..and g referring to "gravity"

No, the metric tensor g is not the same as "gravity", rather it describes the curvature of spacetime, including the possibility that spacetime is uncurved as in special relativity. It's also what you use to calculate the coordinate-invariant "proper time" along any world line in any coordinate system (and in differential geometry, defining the curvature of a surface is the same as defining some coordinate-invariant notion of "distance" along arbitrary paths on the surface), for example in an inertial coordinate system in flat spacetime it can be written as ds² = dt² - (1/c²)(dx² + dy² + dz²).

 

[Dale]

..and g referring to "gravity" is irrelevant to the conversation so I took him to mean merely acceleration, else why bring it up at all. To know any g that would help would require knowing some common g frame thus not resolve anything, especially if no such g frame was handy.

For the third time, g is the metric tensor, it is not gravity and it is not acceleration. A coordinate system is just a set of four numbers used to label events in spacetime, it has no physical significance in itself. The metric tensor, g, is the thing which connects a coordinate system to physics. It quantifies the relationship between the coordinate labels and physical lengths, durations, and angles, as well as the causal structure of spacetime.

When you say "inertial frame" what you are saying is simply that you have a coordinate system with a particular metric tensor, specifically:

$$g=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -\frac{1}{c^2} & 0 & 0 \\ 0 & 0 & -\frac{1}{c^2} & 0 \\ 0 & 0 & 0 & -\frac{1}{c^2} \end{array} \right)$$

While that is nice and convenient, it is by no means necessary and any coordinate system with any metric may be used equally as well, the results are guaranteed to be the same.

 

[Mentz114]

James,
you're a hard man to convince. I'll try an analogy, which I hope you don't think is an oversimplification.

Road vehicles have odometers which measure what distance the vehicle travels. They do this however much the car might accelerate/decelarate/move at a constant speed. And everyone agrees on these readings.

Clocks are like odometers that measure proper length ( which is c times proper time). They probably have wheels in contact with spacetime . If you have two identical clocks and you're right by them, they are measuring time in your frame. But if one clock is taken on a round trip, it will measure the proper time of its journey, and when it arrives back, it will show less time than the clock that stayed with you. And, crucially, all inertial observers (from whose frames you and the traveling clock may be moving) will agree on the clock times.

So there's no particular frame required, it's all down to the proper time.

Consider two travellers, one we give the worldline x = 0, which means we're at rest in this frame, and we give the traveller the worldline x_T= 0.001*(t²-1)⁴. This is a round trip because for the traveller x_T=0 at t = -1 and t = 1. The maximum distance ( the turning point) is at t = 0 when x_T=0.001. I'm using units where light travels 1 length unit in 1 sec, so c = 1.

This is all we need to know to calculate the times on both clocks when the traveller returns. Furthermore, the calculation will give the same answer if done from another inertial frame.