Version of the twin paradox without accelerations

Question

The main argument used to solve the twin paradox is based on the accelerations and decelerations of the observer considered to be in motion. In the twin paradox of the version described here, accelerations and decelerations are bypassed, thus the main argument is disarmed.

We consider three twin brothers, Mov (moving), Sta (stationary) and Rem (remote). Initially, Mov stands 100 meters behind the Sta and begins to accelerate his movement towards the Sta. The moment Mov exceeds the Sta, its speed stabilizes, their clocks synchronize while at the same time the experiment begins. The final speed of the Mov is not great, while the age difference from the Sta due to the acceleration is negligible. Therefore, from the start of the experiment the acceleration has been canceled and the relation of Mov to Sta is symmetric.

Mov continues his journey for millions of years, moving towards observer Rem. Rem is immobile in relation to Sta, so he was able to synchronize his watch with Sta during Mov's journey. Towards the end of the journey, Mov approaches Rem close enough to discern his features. Once that happens, Mov stops.

At this point, it is obvious that the ages of Sta and Rem are the same. In addition, there was no acceleration or deceleration of Mov throughout the experiment, which means that Mov's symmetry relationship with Sta is not broken. Consequently, the ages of Mov and Sta must have been the same when Mov observed Rem's features on the move. Thus, as the three observers of the experiment were twin brothers, Mov must have found that Rem was exactly the same as himself shortly before he stopped moving. So the age difference is zero.

From the above result it also follows that if a fourth observer had stood at the starting point of Mov's journey and they were synchronizing their clocks before Mov started moving, then the initial acceleration would not make any age difference throughout his journey Mov.

So, when the spaceship of the classic version of the twin paradox returns to Earth, a disgruntled old man will come out of it, as he had believed that his journey would last a few years. No one had warned him that he was likely to spend the rest of his life in solitary confinement. I believe that this result can only be explained in the context of quantum physics.

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An unjustified reply has been deleted, but some clarification is needed:

The main argument is that in a state of absolute symmetry no time dilation can be justified. If you have a counter-argument, then I would ask you to at least outline its main points. This would help a lot.

Answers to comments

Rob: Then the twins will have a common age after their return to Earth.

gandalf61: This problem is technical, it does not change the substance of the arguments. There is no law that forbids the existence of two identical people at any distance at the same time, and that is enough for us.

SamuraiMelon: See it like this: You can think of Mov as immobile and his siblings moving in relation to him. The results of the experiment would be exactly the same according to Relativity.

Charles Francis: Clocks can be synchronized by exchanging light signals. Einstein describes how this is done in his main article. You may notice, however, that I do not use the indications of the clocks in order to draw the final conclusions of the experiment.

Answer 1

The story according to Sta and Rem: When Mov passes Sta, all three clocks say 12:00. When Mov gets to Rem, Sta and Rem's clocks have advanced four hours and say 4:00, but Mov's clock has advanced only two hours and says 2:00. Mov's clock spent the whole time running at half speed.

The story according to Mov: When Mov passes Sta, Mov's and Sta's clocks say 12:00, while Rem's clock says 3:00. When Mov gets to Rem, Mov's clock has advanced two hours and says 2:00 while Sta's and Rem's have advanced only one hour, so Sta's says 1:00 and Rem's says 4:00. Sta's and Rem's clocks spent the whole time running at half speed.

All three agree that when Mov and Sta are next to each other, their watches both say 12:00. All three agree that when Mov and Rem are next to each other, Mov's watch says 2:00 and Rem's watch says 4:00.

What asymmetry do you perceive in this?

Answer 2

The final speed of the Mov is not great

In this case, there is no measurable relativistic effect in a lifetime.

But if the speed is great, the distance between Sta and Rem, measured by Mov is smaller than this distance measured by Sta and Rem. The time measured by Mov's clock is: $$t_m = \frac{d_m}{v}$$, what is smaller than that measured by Sta and Rem:$$t_s = \frac{d_s}{v}$$

We can think that Mov should see Rem getting younger all the time, because the situation is symmetric. In that case why he is younger when they meet?

It is because when Mov passes by Sta and synchronize their clocks, it doesn't mean that the clocks of Mov and Rem are also synchronized. For Mov, the clock of Rem at that moment is $(c=1)$: $t_r = t_m + vd_r$ (his line of simultaneity $t'=0$), because if he is moving with velocity $v$ to Sta and Rem, Rem is coming to him with a velocity $-v$.

Mov will see Rem's clock ticks slower all the time, but the initial time difference explains why he (Mov) and not Rem is younger at the meeting.

Answer 3

I'd strongly advise you to forget everything you think you know about relativity and start over.

Examples of things you should try to forget are "time dilation", "relativity of simultaneity", and the like, because these are all artifacts of trying to fit the physics of special relativity into the obsolete world picture of Newton. Newtonian physics is actually more complicated than special relativity, because it has two things (space and time) where SR has one (spacetime). When you shoehorn SR into the space+time picture it's even worse, because it isn't even a good fit for that picture, which would already be a more complicated picture even if it was a good fit. So try to forget it all.

Here's a different paradox. Two drivers set off on a trip from San Francisco to Chicago. One takes the most direct route, while the other takes a detour through Dallas. When they arrive, their tripmeters show different values. Yet, their trips are symmetrical in the sense that each driver gets farther away from the other in the first half of the trip, and then gets closer in the second half. The maximum distances are even the same.

The resolution of this paradox is that there's a specific well-defined way in which odometers work, and it's simply not true that the somewhat vaguely defined symmetry mentioned above extends to a symmetry of odometer readings. I can't even really see why anyone would expect it to, but at any rate it doesn't.

Now you could make this more precise in the following way: you expect the odometers to measure east-west distance. But actually their readings are "dilated" from east-west distance by a factor of $\sqrt{1+v^2/c^2}$, where $v$ is the vehicle's change of north-south distance between two north-south lines a unit east-west distance apart, and $c$ is a unit conversion constant because we're using different units for north-south and east-west distance for some reason.

That's obviously stupid. Here's a better way: odometers work by counting the number of times the wheels turn, and multiplying by the circumference of the wheels. The operation of the odometer is invariant of heading. No matter which way the vehicle is moving, it measures the same thing: the distance traveled. And one of the routes is simply longer. The longer route is the one with a change of heading in it, because a straight line is the shortest distance between two points.

This is almost the same as the standard twin paradox. The only important difference is that in place of the Pythagorean distance formula $\sqrt{x^2+y^2}$, in spacetime the distance formula is $\sqrt{t^2-x^2}$. This has the consequence that a straight line (inertial motion) in spacetime is the longest distance between two points (events) instead of the shortest.

Quartz watches work very much like odometers: they count the number of times a crystal vibrates, and multiply by the known vibrational period of the crystal to get an elapsed time. The aging of a human body is vastly more complicated, and less precise, but it's still the result of interactions of atoms that proceed at a roughly predictable rate. All of these processes are independent of your heading (velocity) in spacetime. Regardless of heading, they measure the same thing: the length of your worldline. That's what elapsed time is: the length of the worldline. And the worldline of one of the twins in the standard twin paradox setup is longer than the other's.

With that long preamble, here's your setup:

                        Rem
     Portland --------------------/ Minneapolis
                              /
                          /
                      / Mov
                  /
San Francisco /-------------------- Kansas City
                        Sta

You should be able to work out the rest yourself. Seriously, translate everything in your question to Euclidean geometry, imagining that the four cities are at the vertices of a rectangle and all the highways are straight. The equivalent problem in special relativity really is equally trivial. You can make it very complicated if you insist on, for instance, defining two different Cartesian coordinate systems, one aligned with Rem and Sta and the other with Mov, and expressing everything in terms of both coordinate systems, and using different units for north-south and east-west distance, and calling the coordinates of Mov's coordinate system north-south and east-west also, and so on and so forth. But you don't have to do that. It's an easy problem if you don't make it hard.

Answer 4

Mov's symmetry relationship with Sta is not broken. Consequently, the ages of Mov and Sta must have been the same when Mov observed Rem's features on the move.

Your argument by symmetry is interesting, but it fails here. Mov and Sta are indeed symmetric, but your “consequently” statement does not follow from the symmetry. According to Sta, Mov is younger by some amount, and according to Mov, Sta is younger by the same amount. This is symmetric so it shows at least that symmetry is insufficient to make the claim. In addition to being symmetric it is also consistent with the Lorentz transform and all of the resulting relativistic effects.

Furthermore, there is no symmetry whatsoever with Rem. So when Mov observes Rem he sees that Rem is older. According to Mov, Rem and Sta are not synchronized and they are different ages, with Rem being older from the beginning.

Answer 5

Let's say M moves 1,000,000 ly at $\gamma=1000$:

S sees a 1,000,000 yr trip in which M ages 1,000 yr. S & R age 1M yr.

R sees a 1,000,000 yr trip in which M ages 1,000 yr. S & R age 1M yr.

M sees a 1,000 yr trip in which M ages 1,000 yr and S & R age 1 year.

S & R's clocks are always synchronized. Inertial-M disagrees. He reads them as ticking 1,000 times too slowly, with R's clock being 999,999 years ahead of S's.

That was 1/2 the twin paradox. If M turns around and heads home, all of a sudden S's clock is now 999,999 years ahead of R's, and S ages 1 more year on the trip home, while M ages another 1,000 years. The final result is S ages 2,000,000 year and M ages 2,000: The Twin Paradox.