A variation of the twin paradox

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I've read the twin paradox and if I am correct the resolution is that one twin accelerates and decelerates so he comes back younger. But I have a different scenario that I would like to ask:

What if you have two twins equally distant from a point in space and completely at rest relative to each other. At the same moment they both accelerate to 0.99c and then travel at a constant speed to the midpoint. From each of their points of view the other is moving quickly towards him and aging slower. They both decelerate as they near each other and come to a complete stop at the midpoint. They finally pass each other going in the opposite direction and glance at each other through their spaceship windows. How can each see the other as being younger? What if they decelerated and came to a stop at the midpoint? Who is older/younger? Both experienced the same acceleration. Both, then travelling at a constant speed deemed that the other was moving relative to them so each thought the other's clock is slower.
 
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  • #2
Nugatory
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I've read the twin paradox and if I am correct the resolution is that one twin accelerates and decelerates so he comes back younger
That's not right, although many casual explanations will steer you towards that misunderstanding. We have many threads here that explain what's really going on, and you should also give this FAQ a serious read: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

The quick summary is that the acceleration is a red herring. The two twins take different paths through spacetime, the traveler's path is shorter, and less time passes on a shorter path through spacetime - it's quite analogous to two cars driving different routes between the same two points in space and finding that their odometers say that one of them covered more miles than the other. The acceleration only comes into the picture because it's the easiest way of setting them on different paths between the separation and reunion events; saying that the acceleration caused the difference in ages is like saying that turning the steering wheel is why the cars' odometers disagree.

In your variant of the problem the situation is completely symmetric so both paths are the same length and the twins both age by the same amount, as your intuition is suggesting. This is true even though both twins, analyzing the problem as if they are at rest and the other is moving, find that the other twin's clock is slower than theirs. (Note that the the traditional form of the twin paradox also works this way - at every step of the journey the traveller finds that the earth twin's clock is slow, and vice versa, yet earth twin ages more). This apparent paradox is allowed by relativity of simultaneity, and if you are not already familiar with that concept you should google for it - it is the key to just about every apparent paradox in relativity.

In your version, the two twins start their journey at the same time if you use a frame in which they are initially at rest. But if you use a frame in which one or the other is at rest after the acceleration, the two do not start their journeys at the same time; the moving twin starts earlier and the slower clock compensates so that both are equally aged when they meet. The standard version of the paradox has them start at the same point to avoid this problem.
 
  • #3
Ibix
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At the same moment
The same moment in which frame? Frames do not agree on simultaneity for events that are space-like separated, so your question is missing information and cannot be answered (Nugatory is assuming that you mean the same moment in their initial rest frame, which is probably what you did mean). Failing to specify which frame is being used to measure positions and times is where pretty much all homebrew variants of standard experiments come unstuck.
 
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  • #4
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I've read the twin paradox and if I am correct the resolution is that one twin accelerates and decelerates so he comes back younger. But I have a different scenario that I would like to ask:

What if you have two twins equally distant from a point in space and completely at rest relative to each other. At the same moment they both accelerate to 0.99c and then travel at a constant speed to the midpoint. From each of their points of view the other is moving quickly towards him and aging slower. They both decelerate as they near each other and come to a complete stop at the midpoint. They finally pass each other going in the opposite direction and glance at each other through their spaceship windows. How can each see the other as being younger? What if they decelerated and came to a stop at the midpoint? Who is older/younger? Both experienced the same acceleration. Both, then travelling at a constant speed deemed that the other was moving relative to them so each thought the other's clock is slower.
In this symmetric situation, when they both stop and compare their clocks, they will see that their clocks show the same time. Even without stopping, at the moment when they pass near each other they will see that their clocks show the same time.
 
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  • #5
martinbn
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To see that the acceleration isn't the reason for the different age consider the following version. The twins are at rest, then start in the same direction with the same acceleration, then coast with the same velocity. Then twin A decelerates turns back and goes to the original place. Twin B does the same but at a later moment, so he travels a bit further before turning back. All the accelerations and declarations are the same, they just happen at different times. Still the twins will have different ages when they meet.
 
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  • #6
Ibix
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And the opposite extreme of @martinbn's example is a standard twin paradox, except the stay-at-home twin zips backwards and forwards between Earth and Moon at the same speed the traveller uses to go to the other star and back. The ages are the same when they meet up, despite radically different acceleration profiles.
 
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  • #7
Mister T
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I've read the twin paradox and if I am correct the resolution is that one twin accelerates and decelerates so he comes back younger.
When you sketch a spacetime diagram of the situation you see that the twins have different paths through spacetime. Each path is called a worldline. The traveling twin has a kink in his worldline associated with his turn-around.

What if you have two twins equally distant from a point in space and completely at rest relative to each other. At the same moment they both accelerate to 0.99c and then travel at a constant speed to the midpoint. From each of their points of view the other is moving quickly towards him and aging slower. They both decelerate as they near each other and come to a complete stop at the midpoint. They finally pass each other going in the opposite direction and glance at each other through their spaceship windows. How can each see the other as being younger?
They can't! You need a reunion, and you don't have one. The twins need to be at the same place at the same time, but this has to happen twice. They compare their ages at the first union (usually they have the same age) and then again later at the reunion. You don't have a reunion.

What if they decelerated and came to a stop at the midpoint? Who is older/younger? Both experienced the same acceleration. Both, then travelling at a constant speed deemed that the other was moving relative to them so each thought the other's clock is slower.
Whether they observe the other's clock as running slow or not tells you nothing about their comparative ages. Grandpa and grand baby have different ages, but they could each observe the other's clock as running slow.

What you're missing here is an understanding of how one concludes that the other's clock is running slow. You need three clocks, and two of them have to be synchronized and at rest with respect to each other. Failure to understand, or just gloss over the details, is what leads to confusion over the twin paradox.
 
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I've read the link posted above, but I'm confused about simultaneous events when I thought I accounted for that. Traveller A and B start at one point and synchronize their clocks. They agree to travel opposite directions (1 light year) and agree on a time that they will accelerate from rest (a total of 2 light years apart) until they reach a speed close to that of light towards each other. I would think that since they both undergo exactly the same conditions that they will both start towards each other at exactly the same time, accelerate for the same time towards each other and finally coast at the same velocity towards each other. Is there anything wrong with this part of the thought experiment?
 
  • #9
Ibix
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That's just a twin paradox - but with unequal velocities in the outward and inward directions and two travelling twins and no stay-at-home.

The fact remains that "simultaneous" is a frame-dependant concept. For anyone not at rest in the initial frame of reference the two travellers have different speeds - so they have different clock rates and travel different distances in different times before stopping and turning around (they do turn around the same distance from their origin point, but the origin point is moving in such a frame). So they do not start their returns simultaneously in general.
 
  • #10
Mister T
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For anyone not at rest in the initial frame of reference the two travellers have different speeds - so they have different clock rates and travel different distances in different times before stopping and turning around
I was under the impression that the OP had no turn-around in his scenario. The twins start far apart, head towards each other, and meet.
 
  • #11
Ibix
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I was under the impression that the OP had no turn-around in his scenario. The twins start far apart, head towards each other, and meet.
In the original scenario, agreed. But post #8 revises it to include an outbound leg doing slow clock transport.
 
  • #12
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I only changed that in the hope that A & B could synchronize when they would head inbound at the same time. I am getting the impression from the posts following that it is not that simple to do even this?

The problem that I am still asking is if two traveller can simultaneously start 1 light year apart from each other and travel at the same speed, 0.99c, how are they both the same age when they each should be aging slower relative to the other?
 
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  • #13
PeterDonis
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if two traveller can simultaneously start 1 light year apart from each other and travel at the same speed, 0.99c, how are they both the same age when they each should be aging slower relative to the other?
Because, in each one's frame in which the other is "aging slower", the other also started traveling much earlier. This is an example of relativity of simultaneity.

For example, call the traveller on the left A and the other one B. In the frame in which A and B are originally at rest--the frame in which they start 1 light year apart and in which their clocks are originally synchronized--they start traveling simultaneously. But in the frame in which A is at rest while he is traveling--i.e., a frame moving to the right at 0.99c relative to the original frame--B starts traveling much earlier than A does, because of relativity of simultaneity: the events "A starts traveling" and "B starts traveling" are only simultaneous in the original rest frame, not in the frame in which A is at rest while he is traveling. In this frame, B's clock does run slow relative to A's, but B's earlier start time makes up for that, so A's and B's clocks read the same when they meet.

In the frame in which B is at rest while he is traveling, things are simply switched around from the above: A's clock runs slow, but A starts traveling much earlier, so A's and B's clocks read the same when they meet.

Note that the above also implies that, in either of the travelling frames (the frames in which A or B are at rest while traveling), A's and B's clocks are not synchronized before they start traveling.
 
  • #14
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I only changed that in the hope that A & B could synchronize when they would head inbound at the same time.
The only way to unambiguously synchronize is to be co-located.

I am getting the impression from the posts following that it is not that simple to do even this?
If they are co-located, it's trivial. Note that they need not be at rest relative to each other to accomplish this.

The problem that I am still asking is if two traveller can simultaneously start 1 light year apart
If they are separated along the line of relative motion, there is no way to unambiguously establish "simultaneously". The events could be observed to occur in either order, depending on the observer's direction of motion along the line of relative motion.
 
  • #15
Ibix
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The problem that I am still asking is if two traveller can simultaneously start 1 light year apart from each other and travel at the same speed, 0.99c, how are they both the same age when they each should be aging slower relative to the other?
Look up Einstein's train thought experiment. He shows that a simple consequence of the invariance of the speed of light is that two frames do not agree on whether or not two lightning strikes were simultaneous.
 
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PeroK
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I only changed that in the hope that A & B could synchronize when they would head inbound at the same time. I am getting the impression from the posts following that it is not that simple to do even this?

The problem that I am still asking is if two traveller can simultaneously start 1 light year apart from each other and travel at the same speed, 0.99c, how are they both the same age when they each should be aging slower relative to the other?
This is, in my opinion, a tough question to resolve. It's not possible to understand it while you are talking in loose terms. A prerequisite is to formulate the problem precisely using clocks, calculated distances and the various rules of SR, encapsulated in the Lorentz Transformation.

Note that the physical symmetry of the situation: two clocks undertaking symmetric journeys and returning to the starting point ( in the original frame) demands that the clocks read the same time when they are reunited.

There is an additional pitfall regarding simultaneous acceleration: two synchronised clocks, set a distance apart, and simultaneously accelerated in the original frame stay synchronised in that frame. But, they do not stay synchronised in their own frame. In fact, when the acceleration is over, if they are travelling at the same speed, in sync and the same distance apart in the original frame; then they are out of sync and further apart in their rest frame.

I say this because, in analysing this problem, it is easy to make obvious assumptions about acceleration between frames that are false.

I would say, therefore, that this is a problem that requires a certain mastery of SR to resolve.

That said, the key is the relativity of simultaneity and the way simultaneity changes when you change your inertial reference frame by accelerating.
 
  • #17
robphy
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This spacetime diagram on rotated graph paper might help.
To develop physical intuition, we make the arithmetic easier by choosing convenient velocities and displacements so that we can read off calculations by counting.

upload_2018-10-9_5-30-49.png


Here Bob (blue) and Carol (green), starting equidistant in Alice's frame, approach the meeting event O at speed (3/5).
According to Bob, Carol is approaching at (15/17)... since that is [itex]\frac{ (\frac 35)+ (\frac 35)}{1+(\frac 35)(\frac 35)}=\tanh(\rm arctanh(\frac 35)+arctanh(\frac 35))[/itex].

Just after Bob begins his approach at event Q (17 ticks before O on Bob's worldline),
Bob says the event on Carol's worldline just after [itex]Q_{BobSimult}[/itex] (8 ticks before O on Carol's worldline) is simultaneous the event on Bob's worldline just after [itex]Q[/itex], and it is 15 ticks away. (8,15,17 form a Pythagorean triple.)
(Just before Q, Bob accordingly would have identified [itex]P[/itex], which is 25.5=(2*12.75) ticks away. [By similar triangles, 3:4:5 and 12.75:17.00:21.25.]
Assuming a sudden acceleration by Bob, Bob's "spacetime diagram (based on simultaneity)" would develop discontinuities [and, in the other direction, double-counting].)
So, according to Bob, when these final 17 ticks elapses for Bob, only 8 elapses for Alice. Thus, the time dilation factor is 17/8, which is [itex]\frac{1}{\sqrt{1 -(\frac{15}{17})^2}}=\cosh(\rm arctanh(\frac{15}{17})) [/itex]
And Carol measures the same numbers for this identical experiment, as needed by the Principle of Relativity.

If Bob could watch Carol and Alice (because each would broadcast a signal after every tick), everything seems normal but delayed up to event Q.
At event Q, as Bob begins his approach, Bob would see Alice and Carol in fast motion... by a factor of 2 because between each tick of Bob's clock, two of their ticks (following the gridlines) would be received. Think of folks broadcasting one-hour tv show episodes.
(2 is the Doppler factor for 3/5: [itex]\sqrt{\frac{1+\frac 35}{1-\frac 35}}=\exp(\rm arctanh(\frac{3}{5})) [/itex].)
This continues until 4.25=(17/4) ticks before O, when Bob watches Carol move in faster-motion by a factor of 4 because light from event P ( when Carol begins her approach) arrives at Bob's worldline. (4 is the Doppler factor for (15/17). Alice is still at a factor of 2).
[Since Q is not a grid point on the graph paper (at this resolution), study the event 16-ticks-before-O.])
And Carol measures the same numbers for this identical experiment, as needed by the Principle of Relativity.
 

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Special Relativity does account for twin paradox. I get the

following from a book by Paul Davies.



Assume Alpha Centari is exactly 4 light years away, and one twin is
traveling there at 4/5 speed of light. (Using a 3,4,5 triangle I avoid
irrational numbers in my computations.


Traveling at 4/5 the speed of light, from the point of view of the
stay at home twin, the trip will take 10 years, 5 years there, 5 years
back.

Time for the traveller T' = T( sqrt( 1- (v^2/c^2))) = 3/5 T
Likewise, the distance for the traveler, D' = 3/5 D


The traveler on the spaceship sees himself traveling a distance of
4*3/5 = 2 2/5 light years in a time of 3 years, and likewise the 2 2/5
light years back in a time of 3 years, so the traveler will see the trip as lasting 6
years.

Suppose the twins have super telescopes and can see each other throughout
the trip. As long as they are traveling apart, the twins will see each other as
aging at 1/3 speed. As long as they are traveling towards each other,
the twins will see each other as aging at triple speed.
The difference is, the traveling twin will see the stay at home twin
as aging at 1/3 speed for the 3 years to Alpha Centauri,
and at triple speed for the 3 year trip back to earth for a total of
3* 1/3 + 3*3= 1 + 9= 10 years.

The stay at home twin will see the travel age at 1/3 speed for 9
years, the 5 years it takes the traveler to get to Alpha Centauri,
plus the 4 years it takes the light to get back to earth. Since the
total trip will take 10 years, the stay at home twin will see the
traveler age at triple speed during the one year he observes the
traveler coming back to earth. The earth observer sees the traveler
age at 1/3 speed for 9 years, and at triple speed for 1 year, for a total of
1/3*9 + 3*1 =3+3=6 years.

Both observers see each other aging at the same slow rate while moving
apart, they see each other aging at the same fast rate while moving
together. The difference lies in one observer deliberately changes
the relative motion of his rocket from moving away from earth to
moving towards earth, and the other observer remaining passive, and
not seeing the change until the light from the
traveler reaches earth. If the earth could be accelerated like a
rocket ship, and the earthbound observer decided to change his frame
so the rocket appeared to be moving towards him at 4/5 lightspeed
rather that away at 4/5 lightspeed, while the rocket remained in
motion past Alpha Centauri, then it would have been the earth twin who
appeared to age less.

Of course you could have some intermediate situation where BOTH
observers decide to change their relative motion before they see the
other observer change his motion.

.
 
  • #19
Ibix
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Special Relativity does account for twin paradox.
I don't think anyone was saying otherwise (?).
 
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robphy
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Special Relativity does account for twin paradox. I get the

following from a book by Paul Davies.



Assume Alpha Centari is exactly 4 light years away, and one twin is
It would be more useful to address the specific questions posed
rather than to copy-paste
the practically same comment [together with the uncommon "Centari"]
again and again, here and elsewhere.
At least you should quote the original source [from 2006?], even if it is yourself.

Google search for... "Assume Alpha Centari is exactly 4 light years away, and one twin"
 
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  • #21
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There is a simpler way to view this. Each sees the distance to the center to be shortened by the same amount and travels with the same speed; they will therefore have aged by the same amount when they meet at the center.
 
  • #22
Ibix
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There is a simpler way to view this.
That's very much a matter of personal preference. The key point is that there exists a frame in which the travellers are doing the same thing; thus their experiences must be the same and the ages at meetup must be the same. Your answer is one particular expression of that symmetry.

The problem is how other frames than the simple one interpret this. They must reach the same conclusion - but why? The travellers aren't moving at the same speed, so they don't get the same time dilation and length contraction factor. The answer is the relativity of simultaneity - in other frames the two ships don't start at the same time and the difference cancels out the differences in distance and time.
 
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Regarding the second scenario outlined by the OP, where A and B start out at rest wrt one another and are colocal, and then they leave outbound, and each turn around and come back, isn't the "paradox" resolved by the fact that both will see the other "turn around" at a different time then they turn around themselves? That is, the solution is the relativity of simultaneity?

In the head of the OP, who is looking at both twins from the origin, he sees them turn around simultaneously, but each individual twin disagrees about that. They would see the other turn around at a different time than they turn around, which would compensate for each measuring the other with a slower clock, and then that should fully account for the equivalent aging when they meet up.

Am I right on that?


ETA- furthermore, for the reference frame in the center where they meet (at rest), the two would be completely symmetrical: both clocks would run slowly and at the same rate. Is that correct?
 
  • #24
PeroK
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Regarding the second scenario outlined by the OP, where A and B start out at rest wrt one another and are colocal, and then they leave outbound, and each turn around and come back, isn't the "paradox" resolved by the fact that both will see the other "turn around" at a different time then they turn around themselves? That is, the solution is the relativity of simultaneity?

In the head of the OP, who is looking at both twins from the origin, he sees them turn around simultaneously, but each individual twin disagrees about that. They would see the other turn around at a different time than they turn around, which would compensate for each measuring the other with a slower clock, and then that should fully account for the equivalent aging when they meet up.

Am I right on that?


ETA- furthermore, for the reference frame in the center where they meet (at rest), the two would be completely symmetrical: both clocks would run slowly and at the same rate. Is that correct?
The resolution is to analyse the problem "properly". The easiest analysis is in the initial rest frame, where by symmetry the clocks read the same time when they return.

The issue, of course, is that both clocks change their IRF at the turnaround point. This means that you cannot analyse the whole scenario from the reference frame of one clock, without including a change of IRF.

There are a number of approaches, which may shed light on the matter. For example, assume that in the initial rest frame clock A moves to the right (and back), clock B to the left (and back) and clock C stays in the middle. You could analyse the whole scenario from a frame moving in one direction throughout: a frame moving to the right with A on its outbound leg. In that frame, C would move to the left throughout; B would move almost twice as fast as C to the left for a time, then stop; and, A would be at rest for a time, before moving almost twice as fast as C. You could check that they all meet up at some point and calculate the time on each clock.

This would at least break the apparent specialness of the initial rest frame, by giving a consistent analysis in a different IRF.

Another idea is to study just the inbound leg: assume clocks A and B are equidistant from C and all at rest. Clocks A and B then instantaneously accelerate and move towards C. Again, the analysis is simple in C's frame. But, it may be enlightening to analyse the scenario from the frame of A or B and confirm that the results are consistent in this frame.
 
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PeroK
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Another idea is to study just the inbound leg: assume clocks A and B are equidistant from C and all at rest. Clocks A and B then instantaneously accelerate and move towards C. Again, the analysis is simple in C's frame. But, it may be enlightening to analyse the scenario from the frame of A or B and confirm that the results are consistent in this frame.
@Sorcerer
I've found some notes on this. We have clocks A and B initially a distance ##d## apart, with A on the left. Both accelerate quickly/instantaneously to speed ##v## towards each other.

We can analyse the problem in the original IRF and see that the clocks meet after time ##\frac{d}{2v}##, at which point both clocks read ##\frac{d}{2\gamma v}##.

We can also analyse this in a frame moving with ##A## at a speed ##v## to the right. We take the origin as ##A##'s initial position. The event ##E_B##, where ##B## starts to move has coordinates ##(0, d)## in the original IRF. Applying the Lorentz transformation gives the coordinates in ##A##'s frame as ##(-\frac{\gamma vd}{c^2}, \gamma d)##.

Note that this in itself is a good example of the relativity of simultaneity. After ##A## has accelerated, in its new IRF clock ##B## has already been moving for time ##\frac{\gamma vd}{c^2}##. And, it started a distance ##\gamma d## away in this frame. (Whatever happened to length contraction, you may wonder!)

Note that this is the event where ##B's## clock is set to ##0##.

Now, we need the velocity addition rule to get the speed of ##B## in ##A##'s frame, which is: ##u = \frac{2v}{1 + v^2/c^2}##.

And, with a bit of algebra, we have: ##\gamma_u = \frac{1+v^2/c^2}{1-v^2/c^2} = \gamma^2(1 + v^2/c^2)##

First, let's calculate where ##B## has got to at time ##t=0## in ##A##'s frame:

##d_0 = \gamma d - u \frac{\gamma vd}{c^2} = \frac{d}{\gamma (1 + v^2/c^2)}##

And, that's not the length contraction formula either! Good job we are doing this properly with maths and not just talking about it.

Now, we calculate how long it takes ##B## to reach ##A##. We have:

##t_A = \frac{d_0}{u} = \frac{d}{2 \gamma v}##

And that is the reading on ##A##'s clock as expected from the analysis in the original IRF, which is good.

Finally, we calculate the reading on ##B##'s clock. The total time that clock ##B## has been running in ##A's## frame is:

##t_B = \frac{\gamma vd}{c^2} + t_A##

And, as clock ##B## is time dilated by a factor of ##\gamma_u## in ##A##'s frame, we have the reading on ##B##'s clock:

##\tau_B = \frac{t_B}{\gamma_u} = \frac{d}{2 \gamma v}(\frac{2v^2/c^2}{1 + v^2/c^2} + \frac{1 - v^2/c^2}{1 + v^2/c^2}) = \frac{d}{2 \gamma v}##

And, again, this is consistent with the original analysis.

The point about doing the mathematics, I think, is that as the problem becomes more complicated, it becomes harder to say: the paradox is resolved because ... In the end, it is resolved by analysing the problem properly and all the factors play their part: length contraction, time dilation and relativity of simultaneity (which are all packaged up in the Lorentz transformation).
 
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