# 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 dont panic ;)

#### PeterDonis

Mentor
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

Mentor
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

Mentor
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

Mentor
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

Mentor
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

Mentor
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

Mentor
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

Mentor
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 travelling 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.

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#### 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 travelling 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

Mentor
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?

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