Acceleration and the twin paradox

[m4r35n357]

we see that the traveling twin A (we’ll call her Alice)’s “path” through this ST diagram visually looks longer than Bob’s. However, the sides of the symmetrical stacked right triangles are subtracting not adding, as in the Euclidean case, and thus the magnitude of Alice’s resultant vector, or proper time/invariant interval, is actually shorter than Bob’s, not vice-versa. That is, for Alice’s journey out to her turnaround point 3 light years away, her time/age is (tau)^2=(5)^2-(3)^2= (tau)=4 years, whereas Bob’s is (tau)^2=(5)^2-(0)^2=(tau)=5 years.

Now I do have a little confusion here in the above example. That is, who’s proper time are we measuring or using here, Bob’s or Alice’s or both? (tau)=proper time is supposed to be invariant, right? But in the above example (tau) is 5 for Bob and 4 for Alice. Should I have re-arranged one of those equations?

Alice's dots are spaced further apart, too. But you have the right idea about pythagoras and subtracting, that is the essence of the relationship between time and space (the spacetime interval, to labour the point), notice we have a "3-5-4" triangle, not by accident! . Everyone has their own proper time, and you have worked out both Alice's and Bob's in this instance. I think you have it, just don't panic ;)

 

[PeterDonis]

it seems that, as Alice approaches Bob’s planet, the signals she has been sending will simply “stack up” on Bob during her final approach. Thus, as Alice is coming in the final stretch, Bob will see the entirety of Alice’s aging over her trip at an extremely accelerated rate. I guess I don’t see how, at the end of Alice’s trip, the balance of the sent radio signals (say there was one “marker” signal sent each year as in ghwellsjr example) don’t add up at the end.

Alice sees Bob's signals blueshifted for her entire journey--that is, she sees Bob's clock running faster than hers for the entire 4 years of her trip. So Bob's clock starts out 3 years behind hers (that is, the signal she receives from Bob when she starts her journey shows his clock 3 years behind her clock's reading when she starts); Bob's clock gains 4 years on Alice's during the journey (she sees 8 years' worth of Bob's signals in 4 years of her own time); so Bob's clock ends up reading 1 year ahead of Alice's when she reaches him.

Bob, however, only sees Alice's signals blueshifted for 2 years (5 years total time elapsed on Bob's clock, minus the 3 years of light travel time that it takes before Bob receives the signal Alice emits when she starts her journey). So even though he sees all of Alice's signals during the journey "stacked up" during that 2 years, it still isn't enough for her clock to "catch up" to his. He sees her clock reading 3 years behind his when Alice's "start of journey" signal reaches him; Alice's clock gains 2 years on his during the journey (he sees 4 years' worth of Alice's signals in 2 years of his own time); so when Alice reaches him, her clock is 1 year behind his.

 

[PhoebeLasa]

[...]
Here is a good reference explaining the mathematical requirements of a valid coordinate chart (chapter 2). As PAllen says one of the requirements is that it must be 1-to-1.

http://arxiv.org/abs/gr-qc/9712019

I'm not saying that Brian Greene apparently thinks that charts aren't required to be invertible.

I'm saying that Brian Greene apparently thinks that the accelerating traveler's perspective isn't required to be a chart.

 

[Dale]

I'm saying that Brian Greene apparently thinks that the accelerating traveler's perspective isn't required to be a chart.

Simultaneity requires a chart. Simultaneity means that two events share the same time coordinate. A time coordinate is part of a chart. So simultaneity necessarily implies a chart.

A "perspective" is not well-defined. But insofar as a "perspective" says anything about simultaneity then it must involve a chart.

 

[CKH]

what people want most from a "coordinate system of the traveling twin" is to be able to say: When the traveling twin is X years old, how old is the stay-at-home twin?

And the correct answer to this question is "mu"; as Nugatory pointed out, the question is ill-formed. IMO it's better to just face that up front, rather than trying to salvage people's pre-relativistic intuitions in some form. Understanding why the question is ill-formed is a key part of understanding relativity.

I think this is in fact what confused folks (like myself) what to know in order to remove the paradox. On the outbound leg, in the travelers frame, the stay at home clock runs slower, after the turnaround point in the travelers new frame, the stay at home clock still runs slower. These are both true statements from the Lorentz transform. Just as true as the calculation made above by the stay at home twin using the Lorentz transforms to predict that the returning twin will be younger.

So the paradox is how can we explain the fact that the travelers clock is behind the stay at home clock at the end of the trip from the travelers viewpoint, while on both legs of the journey, to the traveler, the stay at home clock ran slower. We can say what happens in the traveling twin's frame (which is not inertial at the turning point, but is on both legs). I believe the following is an answer.

At the turn around point, an abrupt change in the traveler's time coordinates occurs because the inertial frame of the traveler changes. When the traveler changes inertial frames in turning back, what was the current time on the stay at home clock in the outgoing frame changes abruptly to a later time in the new frame. The size of this jump depends on how far the traveler is from the stay at home twin and the travel velocity. This jump in the stay at home clock more than compensates for the slower running of the stay at home clock on the outbound and inbound legs of the journey as observed in the traveler's frame.

This is just a closer look at how the simultaneity relationship for the traveler changes. The change in time frames accounts for the apparently "missing" aging of the stay at home twin according to the traveling twin.

So if you want to ask in the travelers frame, how old is the stay at home twin, the answer is that the stay at home twin ages slower on the outgoing leg. At the turn around point, the stay at home twin suddenly ages by a large amount, on the return trip, the stay at home twin ages more slowly than the traveler.

If you make the turn around less abrupt, then for the traveler the stay at home twin will age very quickly during turn around but not so abruptly.

The situation is asymmetrical because the stay at home twin remains in the same inertial frame during the whole trip, while the travelers frame changes.

I don't know if that helps anyone else, but it helps me account for the "missing" aging of the stay at home twin. What this scenario looks like to the traveler when watching the stay at home clock with a telescope is different because the time delay of the observation is changing with distance.

 

[Dale]

So the paradox is how can we explain the fact that the travelers clock is behind the stay at home clock at the end of the trip from the travelers viewpoint, while on both legs of the journey, to the traveler, the stay at home clock ran slower.

What makes you believe that from the traveler's viewpoint the stay at home clock runs slower at all?

 

[PhoebeLasa]

what people want most from a "coordinate system of the traveling twin" is to be able to say: When the traveling twin is X years old, how old is the stay-at-home twin?
I think this is in fact what confused folks (like myself) what to know in order to remove the paradox. On the outbound leg, in the travelers frame, the stay at home clock runs slower, after the turnaround point in the travelers new frame, the stay at home clock still runs slower. These are both true statements from the Lorentz transform. Just as true as the calculation made above by the stay at home twin using the Lorentz transforms to predict that the returning twin will be younger.

So the paradox is how can we explain the fact that the travelers clock is behind the stay at home clock at the end of the trip from the travelers viewpoint, while on both legs of the journey, to the traveler, the stay at home clock ran slower. We can say what happens in the traveling twin's frame (which is not inertial at the turning point, but is on both legs). I believe the following is an answer.

At the turn around point, an abrupt change in the traveler's time coordinates occurs because the inertial frame of the traveler changes. When the traveler changes inertial frames in turning back, what was the current time on the stay at home clock in the outgoing frame changes abruptly to a later time in the new frame. The size of this jump depends on how far the traveler is from the stay at home twin and the travel velocity. This jump in the stay at home clock more than compensates for the slower running of the stay at home clock on the outbound and inbound legs of the journey as observed in the traveler's frame.

This is just a closer look at how the simultaneity relationship for the traveler changes. The change in time frames accounts for the apparently "missing" aging of the stay at home twin according to the traveling twin.

So if you want to ask in the travelers frame, how old is the stay at home twin, the answer is that the stay at home twin ages slower on the outgoing leg. At the turn around point, the stay at home twin suddenly ages by a large amount, on the return trip, the stay at home twin ages more slowly than the traveler.

If you make the turn around less abrupt, then for the traveler the stay at home twin will age very quickly during turn around but not so abruptly.

The situation is asymmetrical because the stay at home twin remains in the same inertial frame during the whole trip, while the travelers frame changes.

I don't know if that helps anyone else, but it helps me account for the "missing" aging of the stay at home twin. What this scenario looks like to the traveler when watching the stay at home clock with a telescope is different because the time delay of the observation is changing with distance.

Excellent post! I think you've nailed it.

 

[PeterDonis]

The situation is asymmetrical because the stay at home twin remains at rest in in the same inertial frame during the whole trip, while the travelers frame changes.

I added the bolded text to make it clear what being "in" an inertial frame, in the sense you are using the term, actually means, physically. I think this is an important distinction because the worldlines of both twins can be described in any inertial frame you like; when the traveling twin turns around, his worldline does not "disappear" from one inertial frame and "appear" in another. It is present in all inertial frames all the time. All that changes is the behavior of the traveling twin's spatial coordinates in different frames (before he turns around his spatial coordinates are constant in the outgoing frame; after he turns around his spatial coordinates are constant in the incoming frame).

 

[CKH]

What makes you believe that from the traveler's viewpoint the stay at home clock runs slower at all?

To be specific, I mean within the two legs of the trip, where the traveler is at rest in an inertial frame (a different frame in each leg), the Lorentz transform tells us that the home clock runs slower because the home clock is moving wrt to the traveler's rest frame.

To clarify more. At the turnaround (assume it is smooth and gradual turn), the traveler's rest frame is changing. As a result the "now" for the traveler is changing (his time axis is tilting) wrt to the stay at home clock. So the stay at home clock runs fast in the traveler's rest frame during turn around. When the turn around is completed and the traveler's rest frame is once again an inertial frame, the home clock once again runs slower than the traveler's.

 

[Dale]

To be specific, I mean within the two legs of the trip, where the traveler is at rest in an inertial frame

Yes, that is the correct way to say it. The traveler is at rest in an inertial frame. In the scientific literature that frame is known as the momentarily co-moving inertial frame, or MCIF.

the Lorentz transform tells us that the home clock runs slower because the home clock is moving wrt to the traveler's rest frame.

And this is incorrect because the traveler's rest frame is not an inertial frame. Therefore the Lorentz transform does not tell us anything about it at all. The correct way to say this would be to use the same language as you used above. "The Lorentz transform tells us that the home clock runs slower because the home clock is moving wrt to the inertial frame where the traveler is at rest."

Note the difference. "The inertial frame where the traveler is at rest" vs. "the traveler's rest frame". The Lorentz transform tells us about the former and the standard time dilation formula applies to the former because it is inertial. The Lorentz transform tells us nothing about the latter and the standard time dilation formula does not apply to the latter because it is non-inertial.

To clarify more. At the turnaround (assume it is smooth and gradual turn), the traveler's rest frame is changing.

More correctly, the traveler's MCIF is changing. The traveler's rest frame is not changing, it is simply non-inertial.

 

[Dale]

One of the big standard pitfalls that students fall into regarding the twins scenario is the following. They chop off pieces of several different reference frames and stitch them together haphazardly. Then, like Dr. Frankenstein, they are surprised when their creation fails to behave normally.

Pieces of a pair of frames stitched haphazardly together does not form a valid chart, and if you do the stitching carefully it still does not form an inertial frame. It should be obvious that a Frankenstein frame will not behave like an inertial frame.

 

[DiracPool]

I just thought of another way of looking at the problem that actually makes more sense to me than looking at it from a strictly time dilation perspective.Let’s take again my scenario and ghwellsjr spacetime (ST) diagram from my post #99. Now, another way to look at the asymmetry presented when Alice begins to accelerate toward Bob’s planet is that the space between her and Bob becomes asymmetrical, and that asymmetry is triggered by Alice breaking the symmetry of their mutual inertial frame. Hence, now all of a sudden, according to Alice, her and Bob’s planet are not separated by 3 light years, they are separated by 2.4 light years. That’s how she now sees it, the instant she leaves her planet and starts moving toward Bob’s (assuming instant acceleration to 0.6c). Again, Bob doesn’t experience this, Bob still sees the distance between his and Alice’s planet to be 3 light years, and in fact, will remain seeing 3 light years as the distance even after Alice has landed on his planet.

So basically what we have here is a situation whereby both Bob and Alice experience their own proper time passage “normally” relative to one another, and, in addition, both agree on Alice’s velocity of 0.6c. What they disagree on primarily is the distance between their 2 planets. According to Alice, she is traveling at 0.6c for 4 years and, thus, is traversing a distance of 2.4 light years. According to Bob, however, Alice is traveling at 0.6c and is traversing a distance of 3 light years, but this is taking her 5 years to do, not 4. The only way to reconcile this discrepancy, for Bob, is for him to see her clock running at 4/5ths the speed of his, and for Alice to see Bob’s running at 5/4ths the speed of her clock. Am I accurate in my above assessment? One question I have here, though, is who looks length contracted to whom during the journey? It would seem that, since Bob’s conception of space hasn’t changed, he should see Alice’s spaceship in the normal dimensions, only see her clock running slower. Alice, on the other hand should see Bob’s world as contracted since her conception of the space has contracted from 3 light years to 2.4 light years. This model, however, seems to disagree with the popular conception that Bob sees both Alice’s clock slow and spaceship contract. Which one is it?

Also, it’s intuitive to see how, once Alice lands on Bob’s planet, her conception of the space between hers and Bob’s planet will once again revert to 3 light years, as will her perspective on Bob’s length, he will no longer be contracted. What might be less intuitive, though, is the fact that, although the conception of space will revert to “normal” once Alice and Bob are once again in a common rest frame, Alice will retain her younger age forever, presumably. In other words, their length/sizes will renormalize to the point where Alice’s trip doesn’t leave a lasting trace on their relative size differences, but the relative differences in their ages will leave a lasting trace. Why is such an asymmetry permanent in the time domain but only transient in the space domain?

 

[PeterDonis]

This model, however, seems to disagree with the popular conception that Bob sees both Alice’s clock slow and spaceship contract. Which one is it?

As far as what Alice and Bob actually see, each one sees the other's clock running fast (not slow), because of Doppler blueshift (as has already been discussed in this thread). Bob calculates that Alice's clock is running slow compared to his during her journey, by taking the raw data he actually sees (the Doppler blueshifted signals he gets from Alice), and correcting it for light travel time and the distance Alice was from him (according to the inertial frame in which he is at rest) when she emitted each signal. Alice does a similar calculation for Bob's clock, using the inertial frame in which she is at rest. So time dilation is not something that is directly observed; it's calculated based on other observations.

Length contraction is similar to time dilation in this respect: it's not directly observed, it's calculated based on other observations. For example, if Alice emitted two light signals at each tick of her clock, one from the front of her ship and one from the rear, Bob could use the differing times at which he received the front and rear signals, along with their "time stamps" (what Alice's clock read when each one was emitted), and along with the other data he has, to calculate that Alice's ship is length contracted. But he won't directly observe it as contracted. If it's coming straight at him, he'll just directly observe the nose of the ship, which won't tell him anything about its length; if its direction of motion has a component perpendicular to his line of sight, he will directly observe the ship to be rotated (Google Penrose-Terrell rotation), not contracted. (And again, similar remarks apply to Alice observing Bob.)

 

[DiracPool]

As far as what Alice and Bob actually see, each one sees the other's clock running fast (not slow), because of Doppler blueshift (as has already been discussed in this thread). Bob calculates that Alice's clock is running slow compared to his during her journey, by taking the raw data he actually sees (the Doppler blueshifted signals he gets from Alice), and correcting it for light travel time and the distance Alice was from him (according to the inertial frame in which he is at rest) when she emitted each signal. Alice does a similar calculation for Bob's clock, using the inertial frame in which she is at rest. So time dilation is not something that is directly observed; it's calculated based on other observations.

Ah yes, an important distinction. Thank you for reminding me of that.

 

[CKH]

Yes, that is the correct way to say it. The traveler is at rest in an inertial frame. In the scientific literature that frame is known as the momentarily co-moving inertial frame, or MCIF.

And this is incorrect because the traveler's rest frame is not an inertial frame. Therefore the Lorentz transform does not tell us anything about it at all. The correct way to say this would be to use the same language as you used above. "The Lorentz transform tells us that the home clock runs slower because the home clock is moving wrt to the inertial frame where the traveler is at rest."

I don't disagree, but note that I was assuming the simplified version of the trip that was being discussed. The traveler immediately reaches his "cruising speed" on the outbound leg, he then turns (reverses) quickly and then cruises back on the inbound leg. This means he is at rest in a constant inertial frame which is moving wrt to the home twin during the cruising legs of the journey. The outbound cruising frame is different from the inbound one. The Lorentz transformation shows that the home clock ticks slower in both these legs of the traveler's journey.

Where things change (in this simplified scenario) is in the turn around. During the turn around, the traveler's rest frame is not inertial. However, at every instant of the turn around there is a comoving inertial frame in which the traveler is instantaneously at rest. So during this period of turn around the traveler's instantaneous inertial rest frame is continuously changing to a different inertial frame.

During that turn what is now for the traveler changes wrt the home clock. It changes because simultaneity with the home twin's world line is different in each momentary inertial rest frame of the traveler during his turn. [/QUOTE]

Note the difference. "The inertial frame where the traveler is at rest" vs. "the traveler's rest frame". The Lorentz transform tells us about the former and the standard time dilation formula applies to the former because it is inertial. The Lorentz transform tells us nothing about the latter and the standard time dilation formula does not apply to the latter because it is non-inertial.

More correctly, the traveler's MCIF is changing. The traveler's rest frame is not changing, it is simply non-inertial.

Yes the traveler's rest frame (over time) is not inertial, but at each instant it coincides with an inertial frame.

This is a cool thing in SR that allows you to analyze arbitrary motion with the Lorentz transforms. At every moment, an object is at rest in some inertial frame. The clock hypothesis tells us that acceleration (per se) does not affect clock rates. The only thing that affects clock rates is the current relative velocity of the two clocks. Those rates can always be computed with the Lorentz transform.

When an object accelerates, the particular inertial frame in which the object is at rest is continuously changing. Each of these inertial frames defines a line of simultaneity in spacetime, at the instant when the object is at rest in that frame. In the case of the traveler, the point at which this line intersects the home clock's time axis, gives the home clock time at that instant in the traveler's rest frame.

I hope I've got that right. You have to be very careful in describing frames or you mess up the story.

 

[CKH]

Length contraction is similar to time dilation in this respect: it's not directly observed, it's calculated based on other observations.

In principle, length contraction could be observed directly (but not from a single position). Imagine a photographic plate longer than the object you want to measure. As an object traverses the plate (parallel to and very near to the plate) a continuous string of flash bulbs are flashed simultaneously to illuminate the object and create an an "accurate" instantaneous image of the moving object on the plate.

You cannot capture this with a camera at some point along the path. The light travel time from each part of the object to the camera is different which distorts the picture.

This is another confusing factor when talking about these SR senarios. There is a notion of simultaneity in an inertial frame, however an observer at rest at some particular position in the frame does not see simultaneous events as such because (in general) the light from each event does not reach him simultaneously.

 

[CKH]

What might be less intuitive, though, is the fact that, although the conception of space will revert to “normal” once Alice and Bob are once again in a common rest frame, Alice will retain her younger age forever, presumably. In other words, their length/sizes will renormalize to the point where Alice’s trip doesn’t leave a lasting trace on their relative size differences, but the relative differences in their ages will leave a lasting trace. Why is such an asymmetry permanent in the time domain but only transient in the space domain?

The length change disappears when Alice reaches Bob and stops. The clock rate change also disappears when Alice reaches Bob and stops. The distinction is that during the trip the difference in clock rates accumulates to a difference time which does not disappear.

 

[harrylin]

[..] The only thing that affects clock rates is the current relative velocity of the two clocks. Those rates can always be computed with the Lorentz transform.[..]
I hope I've got that right. You have to be very careful in describing frames or you mess up the story.

Almost: "the current relative velocity of the two clocks" has issues, grammatically I think that it can only mean the velocity of one clock relative to the other clock. By leaving out referral to inertial reference systems, you reintroduce the twin paradox! ;)
Correcting your phrasing is a bit tricky if you want to maintain "affects" in your sentence; but it's similar to "kinetic energy" in classical mechanics, so you can get inspiration from there. :)

 

[DiracPool]

The length change disappears when Alice reaches Bob and stops. The clock rate change also disappears when Alice reaches Bob and stops. The distinction is that during the trip the difference in clock rates accumulates to a difference time which does not disappear.

That's just what I just said. My question is why is there such a distinction?

 

[harrylin]

That's just what I just said. My question is why is there such a distinction?

CKH used a keyword that you did not use (nor something similar to it): "accumulates". And he tried to clarify that there is not inherently a distinction between the "time domain" and the "space domain": clock frequency and its inverse, clock period ("time domain") as well as length ("space domain") do not accumulate. In contrast, clock time is commonly the summation of registered clock periods; as a result, clock time has memory - and it works just the same for age.

 

[Dale]

I don't disagree, but note that I was assuming the simplified version of the trip that was being discussed.

I was also assuming the simplified version. Even in the simplified version the traveler's rest frame is non-inertial.

This means he is at rest in a constant inertial frame which is moving wrt to the home twin during the cruising legs of the journey. The outbound cruising frame is different from the inbound one. The Lorentz transformation shows that the home clock ticks slower in both these legs of the traveler's journey.

More correctly, the Lorentz transformation shows that the home clock ticks slower in both the outbound MCIF and the inbound MCIF. The Lorentz transform is not a transformation between "legs of journeys" but between "inertial frames".

During the turn around, the traveler's rest frame is not inertial.

No, the traveler's rest frame is simply non-inertial. A reference frame is an entity which covers the entire spacetime, so it always includes the turnaround. Furthermore, since at this point a student has not been introduced to the mechanics of creating a non-inertial reference frame there is no guarantee that a non-inertial rest frame will ever coincide with any of the MCIFs.

However, at every instant of the turn around there is a comoving inertial frame in which the traveler is instantaneously at rest.

Yes, this is the correct way to say it.

During that turn what is now for the traveler changes wrt the home clock. It changes because simultaneity with the home twin's world line is different in each momentary inertial rest frame of the traveler during his turn.

There is no MCIF where this is true, and in a properly formed non-inertial frame it may also not be true. This is a "Frankenstein" assumption, not a correct conclusion from the math.

Yes the traveler's rest frame (over time) is not inertial, but at each instant it coincides with an inertial frame.

On the contrary, the traveler's rest frame NEVER coincides with an inertial frame. There is, at each event of any worldline, a MCIF. This frame only coincides with the traveler's rest frame if the traveler is inertial, which the traveling twin is not.

When an object accelerates, the particular inertial frame in which the object is at rest is continuously changing. Each of these inertial frames defines a line of simultaneity in spacetime, at the instant when the object is at rest in that frame. In the case of the traveler, the point at which this line intersects the home clock's time axis, gives the home clock time at that instant in the traveler's rest frame.

This approach to constructing a non-inertial reference frame has some well-known problems. However, even if you use this method to construct your non-inertial frame, you still cannot use the Lorentz transform to determine anything about its properties.

 

[CKH]

Almost: "the current relative velocity of the two clocks" has issues, grammatically I think that it can only mean the velocity of one clock relative to the other clock. By leaving out referral to inertial reference systems, you reintroduce the twin paradox! ;)
Correcting your phrasing is a bit tricky if you want to maintain "affects" in your sentence; but it's similar to "kinetic energy" in classical mechanics, so you can get inspiration from there. :)

You are a stickler, but absolutely right. I felt it wasn't precise when I phrased it that way, but was too lazy to spell it out. As soon as you start getting lazy like that, you begin to lose clarity and are at risk of misleading yourself as well as others. Peter also pointed out some looseness in my phrasing earlier. (I'm still on the learning curve in SR.)

It is common that threads go awry when someone uses loose terminology. It's often the source for protracted disagreements.

Thanks for elaborating and clarifying my response to DiracPool. I hope it helps.

 

[DiracPool]

CKH used a keyword that you did not use (nor something similar to it): "accumulates". And he tried to clarify that there is not inherently a distinction between the "time domain" and the "space domain": clock frequency and its inverse, clock period ("time domain") as well as length ("space domain") do not accumulate. In contrast, clock time is commonly the summation of registered clock periods; as a result, clock time has memory - and it works just the same for age.

You really inferred all of this from CKH's short passage...

The length change disappears when Alice reaches Bob and stops. The clock rate change also disappears when Alice reaches Bob and stops. The distinction is that during the trip the difference in clock rates accumulates to a difference time which does not disappear.

I don't see where "he tried to clarify that there is not inherently a distinction between the "time domain" and the "space domain"'. Am I missing something here? This is a new concept for me, clock time having "memory" and space not having memory? But on the other hand, you state that, "there is not inherently a distinction between the "time domain" and the "space domain"' I'm a bit confused. Perhaps I can get little bit more of an involved clarification of this?

 

[CKH]

I was also assuming the simplified version. Even in the simplified version the traveler's rest frame is non-inertial.

My point was that during the main portions of the trip (in this scenario) the traveler is in inertial motion. You can analyze the trip piecewise and I think it's helpful because it highlights how the paradox is resolved.

More correctly, the Lorentz transformation shows that the home clock ticks slower in both the outbound MCIF and the inbound MCIF. The Lorentz transform is not a transformation between "legs of journeys" but between "inertial frames".

Those portions of the traveler's trip are at rest in inertial frames.

No, the traveler's rest frame is simply non-inertial. A reference frame is an entity which covers the entire spacetime, so it always includes the turnaround.

Are you suggesting that the traveler's rest frame is non-inertial at all times, because it is non-inertial at some times?

Furthermore, since at this point a student has not been introduced to the mechanics of creating a non-inertial reference frame there is no guarantee that a non-inertial rest frame will ever coincide with any of the MCIFs.

I don't understand what you are getting at here. When we look at the traveler's motion over an infinitesimal period, his velocity is essential constant so during that period he moves at rest in an inertial frame as far as velocity is concerned. The traveler's acceleration remains, but as pointed out, that does not affect his clock rate and can be ignored.

There is no MCIF where this is true, and in a properly formed non-inertial frame it may also not be true. This is a "Frankenstein" assumption, not a correct conclusion from the math.

Why can we not analyze a physical situation in parts? The whole point of this scenario is to reveal where the incomplete description of the journey (considering only the cruising legs) goes awry.

On the contrary, the traveler's rest frame NEVER coincides with an inertial frame. There is, at each event of any worldline, a MCIF. This frame only coincides with the traveler's rest frame if the traveler is inertial, which the traveling twin is not.

The traveler's rest frame does coincide with an inertial frame during the cruising legs of the journey. During turnaround that's a fair statement if you consider that the traveler's frame is undergoing acceleration, and thereby cannot be considered inertial. But there is a resolution to that problem.

This approach to constructing a non-inertial reference frame has some well-known problems. However, even if you use this method to construct your non-inertial frame, you still cannot use the Lorentz transform to determine anything about its properties.

Why not? At each instant the traveler's velocity vector matches that of an inertial frame. The Lorentz transformation refers only to velocity. During a short period, the traveler's velocity is constant (uniform) to a first order approximation. It appears that the acceleration can be neglected at each moment because it does not significantly affect the velocity over such a short period. If that's not enough, the clock hypothesis tells us that the acceleration does not affect the infinitesimal analysis.

Isn't this how we extend SR to accelerated frames?

There must be some subtle point I'm missing, but it sounds like you are telling me that such an analysis of accelerated motion is invalid and the Lorentz transform cannot be applied. On the other hand, isn't it common knowledge that it can be applied through integration?

Einstein himself said somewhere(?) that we assume that the path of a moving object can be treated as a polygonal path to an approximation and that approximation becomes more accurate as the sides of the polygon become smaller. In saying this, he was implicitly saying that all higher derivatives of motion (after velocity) can be neglected according to SR. The velocity and all higher derivatives are undefined at the vertices of the polygon, but then no time passes at the vertices, so maybe that makes sense?

I'm willing to believe that Einstein's approach (as described) was not strictly mathematically correct. It would be nice if you could explain how this is done correctly. But I don't believe that the Lorentz transformation cannot be applied infinitesimally while velocity is changing (yet).

To go back to the super simple version where the traveler instantaneously reverses velocity at the turn around. His motion changes from rest in one inertial frame to rest in another inertial frame in that instant. In his new inertial frame the space of simultaneity has abruptly changed and the home clock jumps to a later time for the traveler in the new frame.

 

[PeterDonis]

Imagine a photographic plate longer than the object you want to measure. As an object traverses the plate (parallel to and very near to the plate) a continuous string of flash bulbs are flashed simultaneously to illuminate the object and create an an "accurate" instantaneous image of the moving object on the plate.

Simultaneously according to which observer?

 

[PhoebeLasa]

[...]
This approach to constructing a non-inertial reference frame has some well-known problems.
[...]

Some people believe that it has problems. Other people don't believe that. Apparently, Brian Greene is one of the people who don't believe that it has problems, because he used that approach in his book and in his NOVA series. And, apparently Taylor and Wheeler are two other people who didn't believe it has problems, because they used that approach in Example 49 (pp. 94-95) of their "Spacetime Physics" book.

 

[CKH]

Simultaneously according to which observer?

Simultaneous with clocks placed at the position of each flash bulb that have been synchronized according to the method in SR.

I think by convention when we say "observer" we mean sensors at a specific point in space that can measure only local values. However, you could also think about "extended observers" who are spatially distributed but make their observations synchronously at rest in the same inertial frame, as in this example of a photographic plate. This is a non-standard concept of "observer". I don't mean to confuse the correct definition of "observer", but there is more than one way to "observe" (measure).

Another way to observe (in an inertial rest frame) is to place one end of an optical fibre at each point where you want to make an observation. Gather all of the fibres so the other ends come to the vicinity of a normal local observer. Make the length of each fiber the same. Then you can watch the passing object without distortion (but with a fixed delay) and see the contraction.

 

[CKH]

Some people believe that it has problems. Other people don't believe that. Apparently, Brian Greene is one of the people who don't believe that it has problems, because he used that approach in his book and in his NOVA series. And, apparently Taylor and Wheeler are two other people who didn't believe it has problems, because they used that approach in Example 49 (pp. 94-95) of their "Spacetime Physics" book.

I'm waiting for details of Dale's objection. I'm also thinking about an analogy in mathematics when we determine the circumference of a circle by inscribing polygons. I argue: "As the sides of the polygon get smaller they are more nearly the same length as the corresponding portion of the circle, in the limit they are the same." Dale argues: "But the sides are straight while the circle is curved! In fact the curvature of the circle exists at every point and is constant. The comparison with the straight side of a polygon is wrong since the straight side has no curvature."

The proper resolution to this debate may be deeper mathematically, I'm not sure. It is an interesting point that is essential to proper use of integration. The circle argument above is not sufficiently formal. When we add up the sides of the polygon in the circle, we have to known that the remaining error in length of the circumference gets smaller as the sides get smaller (that the little errors add up, but the total error gets smaller as the sides get smaller).

 

[PAllen]

That's just what I just said. My question is why is there such a distinction?

Well, if there were a natural form of odometer, then length would have an analog of the twin differential aging. The difference in odometer reading would be stable, and have memory. However, we don't have any simple physical systems that behave like odometers in empty space. You can mathematically define one, with appropriate properties (I created a whole thread on this, with appropriate formulas). Unfortunately, there is no device that implements this mathematical odometer.

 

[PeterDonis]

Simultaneous with clocks placed at the position of each flash bulb that have been synchronized according to the method in SR.

Synchronized with respect to what object--what state of motion? Einstein clock synchronization is dependent on the state of motion; clocks synchronized with respect to the moving object are not synchronized with respect to the photographic plate, and vice versa.

 

[CKH]

Well, if there were a natural form of odometer, then length would have an analog of the twin differential aging. The difference in odometer reading would be stable, and have memory. However, we don't have any simple physical systems that behave like odometers in empty space. You can mathematically define one, with appropriate properties (I created a whole thread on this, with appropriate formulas). Unfortunately, there is no device that implements this mathematical odometer.

The symmetry is fixed!

 

[CKH]

Synchronized with respect to what object--what state of motion? Einstein clock synchronization is dependent on the state of motion; clocks synchronized with respect to the moving object are not synchronized with respect to the photographic plate, and vice versa.

OK. The original issue was about observing the length contraction which occurs in SR under specific conditions. In the experiment, there is a inertial frame S. Using instruments at rest in frame S, a measurement of the length of a moving object will be made. The moving object is at rest in another inertial frame S'. Let's suppose that x-axes of S and S' lie on the same line and that the object is on the x-axis and that S' (with the object) is moving in the x-direction in S.

Since S is an inertial frame, we can create synchronized clocks along the x-axis. Place a clock at rest at the origin of S then place additional clocks along the x-axis of S and synchronize them with the clock at the origin using the method prescribed in SR.

So now we can build up the remainder of the observation system. All instruments used are placed at rest in S. We use many clocks along the x-axis. Next to each clock we place a flashbulb (archaic term). We electrically attach each clock to it's flashbulb such that at a certain time the flashbulb will be fired by the clock. All clocks are synchronized and will flash their bulbs at the same certain time. Thus all bulbs will flash at the same time in frame S.

The flash bulbs are shielded such that the light shines only in the z direction (toward us in the diagram). We place a long photographic plate in front all the flash bulbs, such that there is a gap (in the z direction) for the moving object to pass between the photographic plate and line of flashbulbs. We know of a certain time when the moving object will be within that space so we set the "flash time" on each clock to that time.

Now we do the experiment. The moving object reaches some point at the "flash time". All the bulbs flash instantly at the same time capturing a shadow picture of the object on the photographic plate. In the picture, the length of the shadow is the contracted length of the object as viewed in frame S.

If you wished, you could make a much nicer picture using a cylindrical lens along the x-axis to focus light reflected from the object onto the photographic plate.

There is nothing novel about simultaneity in an inertial frame. An inertial frame can be assigned a four dimensional linear coordinate system in which one coordinate is time. All events with the same time coordinate are simultaneous. Thus it ought to be possible to explain this setup much more compactly. Like this.

In inertial frame S, flashbulbs arranged along the x-axis are triggered simultaneously. Shouldn't that be clear enough?

The ability to have such a coordinate system in an inertial frame is derived from Einstein's synchronization method. That is how we define a time axis (time coordinates) in the frame. The orientation of the spatial axes in the frame are arbitrary, but we choose a fixed set of spatial axes to measure things in the frame to avoid confusion.

We really should not have to go into such great detail as above. We can talk about a specific times and specific places in an inertial frame. Events located at the same time in the frame are simultaneous in the frame.

In the twin paradox (as described earlier), at the turn around the spatial and temporal axes of the instantaneous inertial frames of the traveler change, they tilt or rotate. This causes the hyperplane of current time in the travel's instantaneous frame to intersect the world line of the home clock at a different place, i.e. at different time in the home frame. This is visualized in Minkowski space.

I hope that's clearer now. If I'm using terminology incorrectly or mixed up, you'll let me know. It's tedious to always think in terms of a local observer and adjust for delays in receiving information when you can instead think about the changing inertial rest frames of the object in question. On the other hand if you want to know what such an observer actually sees, you can figure that out later.

 

[Dale]

Those portions of the traveler's trip are at rest in inertial frames.

Yes.

Are you suggesting that the traveler's rest frame is non-inertial at all times, because it is non-inertial at some times?

Yes, except that it doesn't even make sense to speak of a frame at a time. A frame, or more properly a coordinate chart, is mathematically a 4 dimensional entity. Specifically, it is a map from an open set in the spacetime to an open set in R4. So a chart is not something which exists at a time and changes over time, it is a single mathematical entity that covers all times in its domain.

I don't understand what you are getting at here. When we look at the traveler's motion over an infinitesimal period, his velocity is essential constant so during that period he moves at rest in an inertial frame as far as velocity is concerned.

Sure, you can always find a MCIF, but the point I was making is that there is no guarantee that a non-inertial rest frame ever matches any of the MCIFs.

Why can we not analyze a physical situation in parts?

You can, but you have to know what you are doing and you have to connect the parts correctly. Here is a good introduction: http://preposterousuniverse.com/grnotes/grnotes-two.pdf

Why not? At each instant the traveler's velocity vector matches that of an inertial frame. The Lorentz transformation refers only to velocity. During a short period, the traveler's velocity is constant (uniform) to a first order approximation. It appears that the acceleration can be neglected at each moment because it does not significantly affect the velocity over such a short period. If that's not enough, the clock hypothesis tells us that the acceleration does not affect the infinitesimal analysis.

Here you are discussing properties of the non-inertial observer's worldline. That is fine, but it doesn't tell us anything about its reference frame. The worldline is a 1D mathematical object, the reference frame is a 4D mathematical object.

Isn't this how we extend SR to accelerated frames?

No, see the lecture notes I posted.

There must be some subtle point I'm missing, but it sounds like you are telling me that such an analysis of accelerated motion is invalid and the Lorentz transform cannot be applied.

Yes, it cannot be applied. The Lorentz transform maps between inertial frames, which again are maps of 4D open sets in the manifold to open sets in R4.

On the other hand, isn't it common knowledge that it can be applied through integration?

I think that you must be referring to integration of the proper time along the worldline. That can certainly be done, but doing so does not give you a reference frame.

Einstein himself said somewhere(?) that we assume that the path of a moving object can be treated as a polygonal path to an approximation and that approximation becomes more accurate as the sides of the polygon become smaller.

The last paragraph of section 4 of his 1905 paper.

In saying this, he was implicitly saying that all higher derivatives of motion (after velocity) can be neglected according to SR. The velocity and all higher derivatives are undefined at the vertices of the polygon, but then no time passes at the vertices, so maybe that makes sense?

I'm willing to believe that Einstein's approach (as described) was not strictly mathematically correct. It would be nice if you could explain how this is done correctly. But I don't believe that the Lorentz transformation cannot be applied infinitesimally while velocity is changing (yet).

Einstein's approach is fine, but it is not an approach for building a non-inertial frame, it is an approach for calculating the proper time along a non-inertial worldline using a single inertial frame.

In his new inertial frame the space of simultaneity has abruptly changed and the home clock jumps to a later time for the traveler in the new frame.

This is not correct. In the new INERTIAL frame (4D chart covering all of spacetime) there is no time jump. In the previous INERTIAL frame there is also no time jump. There is no inertial frame where there is a time jump for either clock. Different inertial frames obviously disagree about the space of simultaneity, but there is no sense in which there is a jump in any inertial frame.

 

[Dale]

I'm waiting for details of Dale's objection. I'm also thinking about an analogy in mathematics when we determine the circumference of a circle by inscribing polygons. I argue: "As the sides of the polygon get smaller they are more nearly the same length as the corresponding portion of the circle, in the limit they are the same." Dale argues: "But the sides are straight while the circle is curved! In fact the curvature of the circle exists at every point and is constant. The comparison with the straight side of a polygon is wrong since the straight side has no curvature."

I like this analogy. My argument is that the circumference is a 1D measurement and the volume requires 3D, so you need something more or different than just a valid method of calculating the circumference.

 

[PeterDonis]

All instruments used are placed at rest in S.

Ok, that answers my question.

 

[harrylin]

You really inferred all of this from CKH's short passage...

Yes indeed - but that's maybe because I would have given roughly the same answer as he gave.

I don't see where "he tried to clarify that there is not inherently a distinction between the "time domain" and the "space domain"'. Am I missing something here? This is a new concept for me, clock time having "memory" and space not having memory? But on the other hand, you state that, "there is not inherently a distinction between the "time domain" and the "space domain"' I'm a bit confused. Perhaps I can get little bit more of an involved clarification of this?

He stated that the clock rate change also disappears. Clock rate [ Hz ] is simply the inverse of clock period [ s ]. A proper clock second is also in the time domain but does not accumulate time. :)
In contrast, clock time is the addition of all recorded clock seconds.

 

[Fantasist]

The situation is asymmetrical because the stay at home twin remains in the same inertial frame during the whole trip, while the travelers frame changes.

Why should that make a difference for the time dilation? Assume a further traveler, who, instead of turning around, travels straight on to a further destination twice the distance away. When he arrives there, should he not have experienced the same dilation as the traveler having returned to earth? But he was in an inertial frame all the time.

 

[PeterDonis]

Assume a further traveler, who, instead of turning around, travels straight on to a further destination twice the distance away. When he arrives there, should he not have experienced the same dilation as the traveler having returned to earth?

Yes, but unlike the case of the twin who returns, the "time dilation" here depends on adopting a particular simultaneity convention, that of the frame in which the Earth and the further traveler's destination are at rest (we assume they are both at rest relative to each other). The twin who returns ends up at the same spatial location as the stay-at-home twin, so no simultaneity convention is required to see that the stay-at-home twin is older; they can compare their clocks directly since they are co-located.

But he was in an inertial frame all the time.

Yes, and using that inertial frame's simultaneity convention, he can say that the Earth and his destination had much less elapsed time than he did during his journey (i.e., that they are time dilated, not him). This is possible because, as above, he is not spatially co-located with the same "stay-at-home" clock at the start and end of his journey; he starts out co-located with the Earth clock, and ends up co-located with the clock at his destination.

It is true that the difference in readings between the two "stay-at-home" clocks at the start and end of his journey (the Earth clock when he starts, and the destination clock when he arrives) will be much larger than the elapsed time on his own clock during the trip. But, using the simultaneity convention of the frame in which he travels, he will say that that's because the destination clock was way out of sync with the Earth clock at the start of his journey; it was reading a time much later than the Earth clock's time. The difference between that much later time and the time the destination clock reads when he arrives will be much smaller than his own elapsed time during the trip--hence, according to his traveling frame, the destination was time dilated, not him.

 

[stevendaryl]

Why should that make a difference for the time dilation? Assume a further traveler, who, instead of turning around, travels straight on to a further destination twice the distance away. When he arrives there, should he not have experienced the same dilation as the traveler having returned to earth? But he was in an inertial frame all the time.

In my opinion, it helps to work through the analogous calculations in Euclidean geometry.

Suppose you have a long straight road, R. You can use the road as the basis for a 2-dimensional coordinate system, x measures distances parallel to the road, and y measures distances perpendicular to the road.

Now, let a second road, R' cross the first at the point x=0, y=0. We could similarly set up a 2D coordinate system based on this road, with x' measuring distances parallel to R' and y' measuring distances perpendicular to R'. Then the relationship between these two coordinate systems is:

x' = \frac{1}{\sqrt{1+m^2}}(x + m y)
y' = \frac{1}{\sqrt{1+m^2}}(y - m x)

where m is the slope of the second road, relative to the first (slope is defined as tan(\theta) where \theta is the angle between the two roads).

Now let's consider points along the second road. Those are points where y'=0, since y' measures perpendicular distance away from the road. For those particular points, y = mx, and the relationship between x' and x becomes:

x' = \frac{1}{\sqrt{1+m^2}}(x + m y) =\sqrt{1+m^2} x

So, the value of x', which measures distance along the road, increases faster than x, by a factor of \sqrt{1+m^2}

Now, suppose that the second road is not straight. It goes for a certain distance at angle \theta (relative to the first road), then makes a turn so that it now is going at an angle of -\theta (relative to the first road) and then continues on until it intersects the first road again. Note that the slope is -m on the second leg of the journey.

Let the point where the roads intersect a second time be x=L. Now, we have two roads that both go from the point x=0, y=0 to the point x=L, y=0. One of the roads is straight, and one of the roads is bent.

We can calculate the length of the bent road using the relationship between x' and x, as follows:
L' = the change in x', but since we know the relationship between x' and x, we can write:

L' = \delta x' = \delta x \sqrt{1+m^2} = L \sqrt{1+m^2} (even though m changes, m^2 is constant).

So \delta x' > \delta x by a factor of \sqrt{1+m^2}. So the bent road is longer than the straight road.

At this point, you could say: It's a paradox! Slope is relative. If R' has slope m relative to R, then R has slope -m relative to R'. So you could use the exact same reasoning to compute \delta x in terms of \delta x', and come to the conclusion that \delta x > \delta x' by the same factor \sqrt{1+m^2}. In other words, from the point of view of road R, road R' is longer, and from the point of view of R', road R is longer!

Well, what's wrong with this reasoning is that R' is not a straight road. You can't use a bent road as the basis for a 2-dimensional coordinate system (or at least, not easily). So the point of view of road R' is not legitimate.

I hope you see that this situation is exactly analogous to the twin paradox. Instead of
x' = \sqrt{1+m^2} x,
in the twin paradox, we have
t' = \sqrt{1-\frac{v^2}{c^2}} t, where the factor \frac{v}{c} is a kind of "slope" comparing the two time axes.

 

[CKH]

Yes.

Yes, except that it doesn't even make sense to speak of a frame at a time. A frame, or more properly a coordinate chart, is mathematically a 4 dimensional entity. Specifically, it is a map from an open set in the spacetime to an open set in R4. So a chart is not something which exists at a time and changes over time, it is a single mathematical entity that covers all times in its domain.

When we talk about an inertial frame, we can assign that a 4-D coordinate system (why is this now called a "chart", perhaps more general somehow?). That coordinate system covers all of space and time (in SR). So an inertial frame is not something that changes with time, as you say.

The traveler experiences change over time as measured by the clock he carries with him. When he is moving inertially, he is at rest in some inertial frame and his clocks ticks at the same rate as clocks in that frame. When he moves with acceleration, at each instant of his time, he is at rest in a different inertial frame. Hyperplanes of simultaneity in each of these different frames intersect the worldline of the home clock at a different point. So, for the traveler, the simultaneous reading on the home clock changes because his motion becomes coincident with different inertial frames as his motion changes.

Sure, you can always find a MCIF, but the point I was making is that there is no guarantee that a non-inertial rest frame ever matches any of the MCIFs.

(Note when I've said the traveler's "instantaneous inertial frame" I mean the same thing as the traveler's "momentarily comoving inertial frame (MCIF)".)

This seems to be the central issue here. Can you explain what you mean in more detail, in SR where spacetime is flat?

This is a problem in GR I believe because the MCIF of an object only has useful coordinates in the vicinity of the object. Furthermore I think those local coordinates are linearly distorted by gravitational field in which the object exists.

I need understand what we mean by the "non-inertial rest frame", if we consider it as a 4-D coordinate system. We are trying to describe 4-D rest frame for the traveler (so over the whole trip). I can understand how a traveler in uniform motion has a 4-D rest frame with coordinate axes. When his motion is not inertial, how do you define his 4-D frame? Can you create a coordinate system for this frame over all spacetime?

However, to me it does makes sense to say there is an instantaneous chart at each point along on the traveler's world line.

You can, but you have to know what you are doing and you have to connect the parts correctly. Here is a good introduction: http://preposterousuniverse.com/grnotes/grnotes-two.pdf

Yes, the frames have to be connected using the traveler's clock, position and orientation. We he moves from one inertial frame to another over a short time Δt, we can conceptually synchronize the clocks in that new frame with his clock. Then, we can find (throughout spacetime) what events are simultaneous with the traveler at that time on the traveler's clock, including the simultaneous reading on the home clock.

I think we can also define his new position as the origin of the new inertial frame. Finally, assuming the traveler is not rotating, the orientations of his spatial axes are fixed.

In order to do all of the above, we need analyze the traveler's motion using a single inertial frame (a base frame) such as the home inertial rest frame. We track the motion of the traveler in that frame, we can define how the traveler's MCIF changes from moment to moment. Over a short period Δt in the home frame (our base frame), there are corresponding changes in position of the traveler Δx, Δy, Δz and a change in velocity Δv (due to acceleration). Using this information, we can describe the new MCIF at the new point in space time. We can calculate Δt' (the incremental change in the travers clock) using the initial velocity in interval Δt and the Lorentz transform.

Your reference looks like a nice introduction to general relativity which I want to study, but that is not an issue here since we have no gravitational fields in SR. It's already clear to me in GR you can only sensibly use the part on an MCIF which is in the immediate vicinity of an object, because spacetime is not flat globally but only approaches flatness locally.

So admittedly, this analysis of the twin paradox falls apart in GR in the presence of gravitation fields. In this case the question "at some moment on the traveler's clock, what is the simultaneous reading on the home clock", may have no meaningful answer. Perhaps this is why PeterDonis (I think) said the answer is "mu".

SR is a much simpler case and perhaps you can do things in SR that are not possible when a gravitation field is present and you must use GR. With that in mind, perhaps this whole analysis of mine is has limited usefulness. In that case, the twin paradox might better be answered by just analyzing the traveler's world line in the home frame. The basic question is how many times does the traveler's clock tick versus the home clock.

So let me try the twin paradox from that point of view. (This will bore most, but I want Dale or someone to help if I've got it wrong.)

There is a diagram in Minkowski spacetime in a earlier post that assumes the simplified case with instant acceleration of the traveler. It shows the world line of the home clock and that of the traveler in the rest frame of the home clock (the base frame). The world line of the home clock is a straight vertical line segment (between these events). The world line of the traveler is a dogleg ">" segment.

The wordline for an object is the path of the object in the Minkowski spacetime. Along a world line, a tangent ray (pointing upward toward positive time) is the positive time axis of the object in its rest frame at that point. In Minkowski space, clock ticks have equal length regardless of the direction of time. So the integrated length of a world line is the elapsed time along that world line (for a clock on that world line).

The shortest path from the event where the traveler leaves home and the event where he returns home is a straight line so such a worldline takes the least time. The dogleg path is longer in time on a clock following that path because the dogleg is longer.

OK, now here's where I'm losing it, the equation for proper time, i.e. the time on a clock for a world line.

I don't get why squared values of spatial distance are subtracted from squared values of time to get proper time, if what I said above is true in Minkowski space about path lengths. So I do not have a firm grip on Minkowski space.

Here you are discussing properties of the non-inertial observer's worldline. That is fine, but it doesn't tell us anything about its reference frame. The worldline is a 1D mathematical object, the reference frame is a 4D mathematical object.

See above for how we define a local inertial frame for the traveler at each point along traveler's worldline (which is plotted in the home rest frame). Valid for SR only.

The worldline as an description of the trip is adequate (in SR) to find these monetary internal frames (at least in the absence of rotation of the traveler).

No, see the lecture notes I posted.

Yes, it cannot be applied. The Lorentz transform maps between inertial frames, which again are maps of 4D open sets in the manifold to open sets in R4.

Those notes are about GR not SR. We have spatial flatness and uniformity in SR. I'm not sure how this is managed in GR where space isn't so simple. You can't do the integrations in the same way when gravity is involved and the Lorentz transforms may not be as easily applied.

Hill first, mountain later.

I think that you must be referring to integration of the proper time along the worldline. That can certainly be done, but doing so does not give you a reference frame.

The last paragraph of section 4 of his 1905 paper.

Einstein's approach is fine, but it is not an approach for building a non-inertial frame, it is an approach for calculating the proper time along a non-inertial worldline using a single inertial frame.

Because spacetime is flat here, you can also use MCIFs along the traveler's worldline and determine the simultaneous readings on the home clock for the traveler.

This is not correct. In the new INERTIAL frame (4D chart covering all of spacetime) there is no time jump. In the previous INERTIAL frame there is also no time jump. There is no inertial frame where there is a time jump for either clock. Different inertial frames obviously disagree about the space of simultaneity, but there is no sense in which there is a jump in any inertial frame.

A 4-D inertial frame doesn't do anything (it's fixed). As the traveler moves along his world line, he is momentarily at rest in different inertial frames. Thus over a short period of time, the traveler moves from one inertial rest frame to a different one. The traveler's hyperplane of simultaneity changes, so in the travel's simultaneous reading on the home clock changes over that period of time. So as you say, obviously his notion of simultaneity changes and thus the simultaneous reading on the home clock.

In the case where the traveler instantly reverses his velocity, the traveler "jumps" into a new inertial frame, causing the simultaneous reading on the home clock to "jump".

 

[Dale]

When we talk about an inertial frame, we can assign that a 4-D coordinate system (why is this now called a "chart", perhaps more general somehow?).

"Chart" is the technical term from Riemannian geometry. It is all spelled out in chapter 2 of the reference I provided ( https://www.physicsforums.com/threads/acceleration-and-the-twin-paradox.779110/page-7 ). Although the notes are from a course on general relativity, that specific chapter is only about Riemannian geometry, which is applicable to any branch of physics, including SR, QM, and even classical mechanics. If you want to learn about non-inertial frames in SR then you have to study the basics of charts and Riemannian manifolds.

That coordinate system covers all of space and time (in SR). So an inertial frame is not something that changes with time, as you say.

Yes.

The traveler experiences change over time as measured by the clock he carries with him. When he is moving inertially, he is at rest in some inertial frame and his clocks ticks at the same rate as clocks in that frame. When he moves with acceleration, at each instant of his time, he is at rest in a different inertial frame. Hyperplanes of simultaneity in each of these different frames intersect the worldline of the home clock at a different point.

All of this is correct, but ...

So, for the traveler, the simultaneous reading on the home clock changes because his motion becomes coincident with different inertial frames as his motion changes.

This doesn't necessarily follow. You can define a chart this way, but any other smooth one-to-one mapping which assigns a constant coordinate to the traveler is equally valid. Furthermore, this method of defining a chart can violate the requirement of being smooth and one-to-one, which invalidates it in those cases.

This seems to be the central issue here. Can you explain what you mean in more detail, in SR where spacetime is flat? ...

I need understand what we mean by the "non-inertial rest frame", if we consider it as a 4-D coordinate system. We are trying to describe 4-D rest frame for the traveler (so over the whole trip). I can understand how a traveler in uniform motion has a 4-D rest frame with coordinate axes. When his motion is not inertial, how do you define his 4-D frame? Can you create a coordinate system for this frame over all spacetime?

Any diffeomorphism from open sets in the manifold to open sets in R4 is a valid chart. See the reference above. The only requirement beyond being a valid chart is that in X's rest frame the spatial coordinates for X are constant. This leaves an immense amount of freedom for defining a non-inertial object's rest frame, and no standard convention.

Here is an example of an alternative method for constructing a non-inertial rest frame which is also commonly used: http://arxiv.org/abs/gr-qc/0104077

However, to me it does makes sense to say there is an instantaneous chart at each point along on the traveler's world line.

There is a MCIF. There is not an instantaneous 3D chart since that would not be a mapping from an open set of the manifold to an open set in R4.

Yes, the frames have to be connected using the traveler's clock, position and orientation.

More than that, you have to make sure that the resulting mapping is a diffeomorphism.

Your reference looks like a nice introduction to general relativity which I want to study, but that is not an issue here since we have no gravitational fields in SR.

The chapter I posted is a gentle introduction to Riemannian geometry, which is used for non-inertial frames in SR also. GR does not "own" Riemannian geometry. I agree that we should stick with only SR and not consider gravity. All of my above comments have been restricted to SR (and in fact, I have only been considering the instantaneous turn around scenario except when explicitly responding to a comment from someone else about the gradual turn around). So, there is no need to go on about GR vs SR, I am not using GR.

There is a diagram in Minkowski spacetime in a earlier post that assumes the simplified case with instant acceleration of the traveler. It shows the world line of the home clock and that of the traveler in the rest frame of the home clock (the base frame). The world line of the home clock is a straight vertical line segment (between these events). The world line of the traveler is a dogleg ">" segment.

The wordline for an object is the path of the object in the Minkowski spacetime. Along a world line, a tangent ray (pointing upward toward positive time) is the positive time axis of the object in its rest frame at that point. In Minkowski space, clock ticks have equal length regardless of the direction of time. So the integrated length of a world line is the elapsed time along that world line (for a clock on that world line).

Yes, that is all correct.

The shortest path from the event where the traveler leaves home and the event where he returns home is a straight line so such a worldline takes the least time. The dogleg path is longer in time on a clock following that path because the dogleg is longer.

Actually, a straight line is the longest timelike interval. Any timelike dogleg path is shorter. The "triangle inequality" is reversed for timelike intervals.

OK, now here's where I'm losing it, the equation for proper time, i.e. the time on a clock for a world line.

I don't get why squared values of spatial distance are subtracted from squared values of time to get proper time, if what I said above is true in Minkowski space about path lengths. So I do not have a firm grip on Minkowski space.

You may want to read chapter 1 of the reference I posted also: http://preposterousuniverse.com/grnotes/grnotes-one.pdf It gives a good introduction to Minkowski space in a way that prepares you for more in the future.

Hill first, mountain later.

That is fine, but then you should drop the topic of non-inertial frames altogether and concentrate on understanding Minkowski spacetime and four-vectors.

Because spacetime is flat here, you can also use MCIFs along the traveler's worldline and determine the simultaneous readings on the home clock for the traveler.

Not necessarily. Doing so can lead to mappings that are not diffeomorphisms, even in flat spacetime.

A 4-D inertial frame doesn't do anything (it's fixed). As the traveler moves along his world line, he is momentarily at rest in different inertial frames. Thus over a short period of time, the traveler moves from one inertial rest frame to a different one. The traveler's hyperplane of simultaneity changes, so in the travel's simultaneous reading on the home clock changes over that period of time. So as you say, obviously his notion of simultaneity changes and thus the simultaneous reading on the home clock.

In the case where the traveler instantly reverses his velocity, the traveler "jumps" into a new inertial frame, causing the simultaneous reading on the home clock to "jump".

Please read the material provided. Until then, you are just not prepared. You may also simply want to stick to inertial frames until you have a firm grasp on four-vectors, the spacetime interval, and so forth.

 

[harrylin]

My point was that during the main portions of the trip (in this scenario) the traveler is in inertial motion. You can analyze the trip piecewise and I think it's helpful because it highlights how the paradox is resolved.

Yes, that has been the standard treatment from the start - and your account was perfectly clear to me and several other participants. So, I hope that it was also understood by Diracpool!

However:

To go back to the super simple version where the traveler instantaneously reverses velocity at the turn around. His motion changes from rest in one inertial frame to rest in another inertial frame in that instant. In his new inertial frame the space of simultaneity has abruptly changed and the home clock jumps to a later time for the traveler in the new frame.

Surely you agree that simultaneity of an inertial frame is constant; thus, you likely meant something else than what you wrote (certainly no clock "jumps"!). But it's not clear to me what you meant with that, and it suggests a misunderstanding (although it was perhaps just figurative speaking). Also for Diracpool it may be useful to clarify this.

I waited to see if you would by yourself clarify this point, but this was not really the case, the physical meaning remained unclear:

[..] The traveler's hyperplane of simultaneity changes, so in the travel's simultaneous reading on the home clock changes over that period of time. [..]
In the case where the traveler instantly reverses his velocity, the traveler "jumps" into a new inertial frame, causing the simultaneous reading on the home clock to "jump".

I think that it was mentioned earlier in this thread that the traveler has a free choice of inertial frames. Real on-board clocks will not "jump", and consequently the home clock will also not jump to a later time if the traveler passively uses an extension of his on-board clocks to determine time at home. Instead, the traveler will first have to synchronize the on-board clocks to the new rest frame. Thus the simultaneous reading on the home clock is adjusted by the traveler, according to his/her choice of instruments and "maps".

 

[stevendaryl]

Surely you agree that simultaneity of an inertial frame is constant; thus, you likely meant something else than what you wrote (certainly no clock "jumps"!). But it's not clear to me what you meant with that, and it suggests a misunderstanding (although it was perhaps just figurative speaking). Also for Diracpool it may be useful to clarify this.

I think that the meaning is this: At any moment, the traveling twin has an associated momentarily comoving inertial frame. According to this inertial frame, the stay-at-home twin is a certain age. Then if you define "the stay-at-home twin's age current age" (according to the traveling twin) to be his age, according to the traveling twin's current comoving inertial frame, then the stay-at-home twin's current age jumps abruptly if the traveling twin accelerates suddenly. This jump can't be taken seriously as a physical change, because by this definition, the stay-at-home twin's current age can jump forward or backward, depending on how the traveling twin accelerates.

 

[harrylin]

[..]This jump can't be taken seriously as a physical change, because by this definition, the stay-at-home twin's current age can jump forward or backward, depending on how the traveling twin accelerates.

Yes, exactly that was my point. One has to be careful with fancy phrasings.

 

[jerromyjon]

The clock hypothesis states that the extent of acceleration doesn't influence the value of time dilation. In most of the former experiments mentioned above, the decaying particles were in an inertial frame, i.e. unaccelerated. However, in Bailey et al. (1977) the particles were subject to a transverse acceleration of up to ∼1018 g. Since the result was the same, it was shown that acceleration has no impact on time dilation.[27] In addition, Roos et al. (1980) measured the decay ofSigma baryons, which were subject to a longitudinal acceleration between 0.5 and 5.0 × 1015 g. Again, no deviation from ordinary time dilation was measured.[29]

 

[CKH]

"Chart" is the technical term from Riemannian geometry. It is all spelled out in chapter 2 of the reference I provided ( https://www.physicsforums.com/threads/acceleration-and-the-twin-paradox.779110/page-7 ). If you want to learn about non-inertial frames in SR then you have to study the basics of charts and Riemannian manifolds.

All of this is correct, but ...This doesn't necessarily follow. You can define a chart this way, but any other smooth one-to-one mapping which assigns a constant coordinate to the traveler is equally valid. Furthermore, this method of defining a chart can violate the requirement of being smooth and one-to-one, which invalidates it in those cases.

No it's not unique, but the relative time results are the same whatever free choices you make for frames, given that you synchronize the traveler's clock with the home clock at the departure event.

In the case I was elaborating, we freely pick some spatial orientation for the home frame and for simplicity choose the origin as the position of the home clock . We choose that the traveler's initial frame as identical to the home frame. Then we plot the traveler's path. The MCIFs of the traveler along the path then have no free choices. The spatial orientation remains constant for all MCIFs. The spatial origin is the traveler's position to maintain continuity with the departure frame. The time axes are tangent to the worldline, the current time in each MCIF is the traveler's proper time.

Anyhow, in this mapping all that matters to determine simultaneity are the time axes of the MCIFs (no choice there) and the departure clock synchronization of the home and traveler's clocks.

Furthermore, this method of defining a chart can violate the requirement of being smooth and one-to-one, which invalidates it in those cases.

Then let's not call it a chart if it has to satisfy such criteria.

Any diffeomorphism from open sets in the manifold to open sets in R4 is a valid chart. See the reference above. The only requirement beyond being a valid chart is that in X's rest frame the spatial coordinates for X are constant. This leaves an immense amount of freedom for defining a non-inertial object's rest frame, and no standard convention.

Just using the definitions provided in SR for inertial frames and simultaneity, I see no way of avoiding the fact the traveler's worldline has a tangent time axis at each point and the time on this axis at each point is the traveler's clock time. This time axis and proper time uniquely define a hyperplane containing all simultaneous events, including the corresponding event on the home clock's worldline.

I don't know why we need a diffeomorphism or a "valid chart" to do this analysis.

By constant spatial coordinates you mean constant orientation and constant position of the object?

Here is an example of an alternative method for constructing a non-inertial rest frame which is also commonly used: http://arxiv.org/abs/gr-qc/0104077

In that paper the first two examples use the same method I used. The author complains that there are events to which multiple values of time can be assigned by the traveler and that the home clock can jump backward and what not. Isn't this exactly what the definition of simultaneity and proper time in SR require?

Then the author defines another "method". For the new method invents a new definition for the time for remote events:

[QUOTE: From the paper] ]This formula simply says that an observer can assign a time to a distant event by sending a light signal to the event and back, and averaging the (proper) times of sending and receiving.[/QUOTE]

Sure you can make such a definition and it may have some very cool properties. It is a rather arbitrary choice, since after sending a light signal, the observer may have changed position and velocity in any number of ways before he receives the reflection.

This "method" has little or nothing to do with simultaneous readings by the traveler of the time on the home clock as defined by SR. I don't think we are free to change this relationship in SR.

There is a MCIF. There is not an instantaneous 3D chart since that would not be a mapping from an open set of the manifold to an open set in R4.

More than that, you have to make sure that the resulting mapping is a diffeomorphism.

I believe what you are talking about is defining a frame for the traveler (i.e. a 4-D coordinate system along the traveler's worldline). That is an interesting subject which I asked about in the last post because I could not see how to do it with uniqueness over all of space time. But, notice that it is not really required for this analysis.

These rules for a diffeomorphism place constraints on what parts of space time can be included in a frame constructed in this way. I can visualize a variety of "tubes" surrounding the worldline of the traveler that you could include in such a chart. But no one chart can cover all of spacetime in general without overlapping itself which violates the rules. For a given event, it may be possible to include that event in some chart, but it will not be a unique chart nor will the coordinates of the event be unique in all such charts. I can't see how to join such charts to cover all of spacetime. I suspect this chart notion is more useful in GR?

On the other hand, a global coordinate system for the travelers frame can be defined using the MCIFs on the world line. But in that case you have to accept that events do not a have unique coordinates in this frame. That's not the fault of the method, it simply a consequence of SR.

The chapter I posted is a gentle introduction to Riemannian geometry, which is used for non-inertial frames in SR also. GR does not "own" Riemannian geometry. I agree that we should stick with only SR and not consider gravity. All of my above comments have been restricted to SR (and in fact, I have only been considering the instantaneous turn around scenario except when explicitly responding to a comment from someone else about the gradual turn around). So, there is no need to go on about GR vs SR, I am not using GR.

OK thanks.

Actually, a straight line is the longest timelike interval. Any timelike dogleg path is shorter. The "triangle inequality" is reversed for timelike intervals.

Aha. I guess that's a result of the subtraction I mentioned. The very strange pythagorean theorem for spacetime. So, distance along a worldline is this quantity called "s" which is not proper time, but I see formulas for proper time that look exactly like proper length, expect that the sign of the squared sum is reversed so that one of these quantities is imaginary. Is time then an imaginary number in these formulas?

Thanks for the Minkowski reference, indeed I don't understand it yet.

 

[stevendaryl]

The very strange pythagorean theorem for spacetime. So, distance along a worldline is this quantity called "s" which is not proper time, but I see formulas for proper time that look exactly like proper length, expect that the sign of the squared sum is reversed so that one of these quantities is imaginary. Is time then an imaginary number in these formulas?

The modern way of doing things doesn't use imaginary values, it uses \delta \tau = \sqrt{\delta t^2 - \frac{1}{c^2}\delta x^2} for proper time (when c |\delta t| > |\delta x|) and uses \delta L = \sqrt{ \delta x^2 - c^2 \delta t^2} for proper distances (when c |\delta t| < |\delta x|).

I've asked before, and there seems to be no reason ever to compute spacetime interval along a path that is sometimes spacelike and sometimes timelike. That would give a complex number. But such a path can't represent the path of an object, so it's not clear what physical meaning could be associated with such a path.

 

[PeterDonis]

Then let's not call it a chart if it has to satisfy such criteria.

This may or may not be an issue, depending on what you're trying to do. See below.

I don't know why we need a diffeomorphism or a "valid chart" to do this analysis.

You don't if all you are trying to do is "solve" the twin paradox for a particular scenario. But people use charts for a lot of things that require the chart to satisfy the criteria DaleSpam gave.

The author complains that there are events to which multiple values of time can be assigned by the traveler and that the home clock can jump backward and what not. Isn't this exactly what the definition of simultaneity and proper time in SR require?

No.

Proper time for a particular observer is only meaningful for events on that observer's worldline. If the observer wants to assign a "time" to events not on his worldline, he has to pick a simultaneity convention, and there is no one unique way of doing that. Even the obvious way for an inertial observer, namely just using the coordinate time in a coordinate chart in which he is always at rest, is not the only possible way for that observer. Simultaneity is simply not a physically meaningful thing; it's a convention.

I don't think we are free to change this relationship in SR.

Sure we are. It may not make sense to do it for a particular problem, but there's nothing in the physics that forbids us from doing it. As above, simultaneity is not a physically meaningful thing; it's just a convention.

I believe what you are talking about is defining a frame for the traveler (i.e. a 4-D coordinate system along the traveler's worldline).

Technically, this is a "coordinate chart"; "frame" means something different. But "frame" is often used (not strictly correctly) to mean "coordinate chart".

I could not see how to do it with uniqueness over all of space time.

This depends on the spacetime. In flat spacetime, it is always possible to construct a valid chart (i.e., one meeting all of DaleSpam's requirements) that covers the entire spacetime, with any chosen timelike curve as its "time axis". The Dolby & Gull paper shows how to do this. Obviously, for timelike curves that are not geodesics, i.e., not inertial worldlines, the surfaces of constant time in such a chart will not match the notion of "simultaneity" that you have been using, since that notion does not lead to a valid chart on all of spacetime. That's fine because, once again, simultaneity is not a physically meaningful thing.

In curved spacetime, it may not be possible to construct a single valid chart that covers the entire spacetime, regardless of how we pick the "time axis". Even if it is possible, such a chart will probably be highly non-intuitive (for example, the Kruskal-Szekeres chart on Schwarzschild spacetime).

a global coordinate system for the travelers frame can be defined using the MCIFs on the world line. But in that case you have to accept that events do not a have unique coordinates in this frame.

Then this is not a "global coordinate system"; at least, not in any useful sense. How can a coordinate system be useful in a region of spacetime where it doesn't assign unique coordinates to events?

 

[CKH]

I think that the meaning is this: At any moment, the traveling twin has an associated momentarily comoving inertial frame. According to this inertial frame, the stay-at-home twin is a certain age. Then if you define "the stay-at-home twin's age current age" (according to the traveling twin) to be his age, according to the traveling twin's current comoving inertial frame, then the stay-at-home twin's current age jumps abruptly if the traveling twin accelerates suddenly. This jump can't be taken seriously as a physical change, because by this definition, the stay-at-home twin's current age can jump forward or backward, depending on how the traveling twin accelerates.

That is exactly what I was trying to do. The purpose was to find, for a given time on the traveler's clock, the corresponding simultaneous reading on the home clock as defined by SR in the traveler's frame .

As the traveler moves from the outgoing IRF to the incoming IRF (instantly), he keeps his clock reading and uses it to define current time in the inbound IRF. In switching these IRFs, there is a sudden jump in what is simultaneous in the traveler's frame. This jump is caused by a change in the "perspective" from the traveler's frame. It is not physical in the sense that no actual jump occurs on the home clock in it's own reference system, it just keeps on ticking there. Furthermore it is not something the traveler actually sees on the home clock as a local observer using a telescope at the turn around. It is just a result of the definition of simultaneity in the two different frames.

If the traveler watches the home clock through a telescope, on the outbound leg, he sees it tick very slowly and on the inbound leg he sees it ticks very fast. The apparent change in home clock's rate is abrupt because he reverses instantly.

What I haven't done are the detailed calculations, but in principle we can define what the simultaneous reading on the home clock is at all points along the traveler's journey including the reading when the traveler returns.

This may not be the best way to look at the problem, but as a novice, I was thinking specifically about inertial frames and how they affect simultaneity as a way to view the paradox.

In the home clock's inertial rest frame, there is no time discontinuity on the traveler's clock. But we can also calculate from the home frame the exact simultaneous reading on the traveler's clock at each moment on the home clock. The two mappings, traveler clock to home clock and home clock to traveler clock are different. The cause of this is non-symmetrical acceleration. The home clock does not accelerate at all.

 

[PeterDonis]

The purpose was to find, for a given time on the traveler's clock, the corresponding simultaneous reading on the home clock as defined by SR in the traveler's frame .

And the answer is "mu": the question is not well-posed, because it assumes that there is a unique, physically given answer to the question. There isn't. Simultaneity is not a physically meaningful thing; it's just a convention. The way you are defining simultaneity is one way to do it, but it's not the only way, it has issues (like not assigning a unique time to all events), and it's no more meaningful, physically, than other ways of doing it.