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I totally understand the reasons behind why the twin on Earth is aging faster than the travelling twin. However, all the explanations within the twin paradox somehow involves the fact that the travelling twin decelerates and accelerates back up to Earth.

However, I have come up with a twist to this thought experiment. What happens if the travelling twin comes to another planet A and stays there forever (never comes back to Earth). Now, how do you explain that the twin on Earth was still aging faster (before the travelling twin got to the new planet) without using the typical explanation that involves the travelling twin going back to Earth?

Intuitively, the one on Earth should age faster regardless whether the travelling twin comes back or not. I just don't quite know to wrap my head around this new puzzle. Lastly, please explain it to a layman. I only have taken introductory physics courses.

Since this is a thought experiment, any assumption can be made to ask a question that invokes deep thinking answers (maybe there is no real answer, but let try to hear out some thoughts and discuss about it!). Let assume a few things:

  1. Let assume the travelling twin lands on the planet that rotates at the same velocity of the spaceship, so the spaceship does not have to decelerate to land there.
  2. Assume Earth and planet A have identical gravity and planet system, so the time difference will stop growing larger and their time will start clicking at constant rate when the travelling twin got to this planet (I'm only interested on how to explain the time difference during the trip).
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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z♦ Sep 23 '16 at 14:24
  • $\begingroup$ first supposition, a planet with Earth mass rotating at relativistic speed would rip apart $\endgroup$ – Adrian Howard Jun 12 at 8:42
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Time dilation is velocity related. Particles moving faster are aging less.

However, velocity is relative, and for this reason, in order to determine the age difference, a synchronization is necessary. Imagine two spaceships receding from Earth in opposite direction. Acccording to the twin paradox, from the point of view of each spaceship, it is the other spaceship which is receding, and the clock of the other spaceship is observed as running slower. And this principle does apply also to your example, because if your spaceship is receding from Earth, you can consider Earth also as receding from the spaceship. At this moment their difference in aging is not defined because there is no synchronization. For a synchronization, the spaceship must return to Earth or some other kind of synchronization by exchange of signals is required.

Edit with regard to assumption 2:

Assume Earth and planet A are at rest with respect to each other.

There is a problem of synchronization. For showing this it is best to sketch the two spacetime diagrams of Earth and of the spaceship.

We take the following example: A spaceship is traveling at 0,8 c from Earth to the exoplanet A which is at 100 LY from Earth.

From the point of view of Earth we get an absolute distance (there is no Lorentz contraction) of 100 LY and a traveling time of 125 years. We can calculate a proper time of the spaceship (its aging) of 75 years (Lorentz factor $\gamma = 1,6667$). enter image description here Now we sketch the spacetime diagram of the spaceship. As we have calculated, its proper time is 75 years. The distance is contracted by $\gamma$ from 100 years to 60 years. From the point of the spaceship, Earth (and the exoplanet A) are traveling in 75 years 60 lightyears. These data do not correspond to a proper time of Earth of 125 years, but only to 45 years! enter image description here Both diagrams are reflecting a different time lapse. The problem is that the points B and C are not synchronized (the exoplanet at the moment when the spaceship is starting from Earth on one hand, and Earth when the spaceship is arriving at the exoplanet on the other.

If you want, there are two different realities, one of the spaceship and one of Earth. Only synchronization permits to avoid this twofold reality.

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  • $\begingroup$ I feel like this does not really answer my question. Assuming the spaceship does not return ever then what happens? $\endgroup$ – Phu Nguyen Sep 20 '16 at 6:15
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    $\begingroup$ The lack of synchronization is the answer. If you want to compare the age, you must compare them at the same moment, and this is the difficulty, to know what is the same moment on Earth and on the spaceship. This is why the spaceship must come back, or another kind of synchronization must be done such like exchange of signals. $\endgroup$ – Moonraker Sep 20 '16 at 7:52
  • $\begingroup$ They can be "at the same moment" in different planets. Their locations have more to do with spatial than with time dimension. $\endgroup$ – Phu Nguyen Sep 20 '16 at 8:26
  • $\begingroup$ See my edit for the explanation of the synchronization problem. $\endgroup$ – Moonraker Sep 20 '16 at 9:56
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    $\begingroup$ @PhuNguyen 'They can be "at the same moment" in different planets.' There are many different "at the same moment"s, rather than one unique one. This feature of relativity is called "the relativity of simultaneity" and until you come to terms with it you will be forever fooling yourself about stuff like this. The iron rules is: there is no unique time ordering between space-like separated events. $\endgroup$ – dmckee♦ Sep 20 '16 at 17:09
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First of all, if the traveling twin came to another planet and stays there he would not age faster or slower any more. Both twins would age at a same rate of 1 year per year. However the traveling twin would remain to be a little younger, age difference would remain constant from now on.

It's possible to make sure there is an age difference even without going back to Earth. (F.e. the one of the twins could send the other a message like "now I am 20 years old", as soon as message received the other attaches his age to the message "I am 20 years 20 days old" and sends it back, and so on. The message would grow in size and eventually would look like:

1: I am 20 years old

2: I am 20 years 20 days old

1: I am 20 years 30 days old

2: I am 20 years 50 days old ...

and from this list we can see that "speed of time" is identical for both twins but one of the them is older than the other.

The "return back to Earth" step is used to make the paradox more obvious: here are both of them, how come this one is younger, why not the other one? But paradox remains even if they do not meet ever again.

Actually paradox starts here: we have two twins traveling in opposite directions. And from the point of view of each them the other twin is aging slower! Isn't it weird? Theory of relativity says that's ok. Sounds like nonsense. And it's a very natural attempt to construct a paradox to make this nonsense obvious. After all it's impossible that each of the twins is aging slower the other, isn't it? Let's bring them together and just look at them. Each of them expects the other to be younger, but that simply can not be, so the theory of relativity is, well, not correct!

But there was a mistake here. The mistake is done by traveling twin. Yes, from his point of view his brother was aging slower than him. Almost. Something very interesting happened during the "turning back". Let's assume the acceleration was very high, so the whole turning back took a very short time.

Nothing special happened to the traveling twin during this acceleration moment. Neither from his point of view nor from the point of view of his brother. He did not become younger or older during the process. Something strange happened to his home-sitting brother during the acceleration!

Suppose it took several milliseconds to fully stop his spaceship. During these milliseconds the earth suddenly became much-much farther from him. (the imaginary ruler stretched from the Earth in his direction is staying still now and thus the is no length-shortening effect). And time on Earth advanced for several years! Now he accelerates to full speed, the Earth becomes much closer again and time on it advances several more years.

From now on time on Earth goes slower than his time, but as a result of these "jumps" during acceleration when he comes home he would see that his brother is older than him.

Once again the most important point: nothing special happened to traveling twin during acceleration. Something special happened to the rest of the world from his point of view.

So, the relativity theory says: no paradox here. Traveling twin will be younger. When you tried to describe the situation from his point of view and show that the home-sitting brother will be younger you made a mistake.

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  • $\begingroup$ I reworded the question somewhat. I made it clearer now that I'm only interested in the time difference before the travelling twin got to the new planet. <br/> Your solution makes sense except it requires some forms of communication between the twin. Then you rely on the fact the travelling twin turns around and travels back to Earth. I already understood that. I want a solution that does not involve such back to Earth trip. $\endgroup$ – Phu Nguyen Sep 20 '16 at 8:28
  • $\begingroup$ Suppose traveling twin is still traveling with constant velocity. In this case from his point of view he (traveling one) is older, from the point of view of home-sitting twin he (home-sitting one) is older! But you wrote about landing on some ROTATING planet. I do not fully understand the scheme, but as soon as some twin starts rotating his frame of reference is no more inertial. He is accelerating (even though the distance between twins remains constant). And it's not easy to describe the world in his accelerating (rotating) frame of reference. $\endgroup$ – lesnik Sep 20 '16 at 8:40
  • $\begingroup$ The point of that is to make the travelling twin does not have to decelerate. As soon as you invoke deceleration then you take away my twist, and this problem becomes a typical twin paradox again (which I already understand). $\endgroup$ – Phu Nguyen Sep 20 '16 at 8:49
  • $\begingroup$ Basically, my two assumptions are made so that the travelling twin does not need to decelerate when he gets to a new planet (I don't care how it happened; the fact is I assumed it did) and the travelling twin & Earth twin are moving away from each other at a constant velocity without any acceleration starting when the travelling twin got to the new planet. I want a solution that does not involve such back to Earth trip and without any form of communication. $\endgroup$ – Phu Nguyen Sep 20 '16 at 8:50
  • $\begingroup$ No acceleration means that one twin is still flying away from the other with a constant speed. Who is aging faster in this case? It depends from frame of reference. In the frame of reference of one (any) of them, the other one is aging slower. This IS the twin paradox. And (just imagine!) the relativity theory says there is no problem :) $\endgroup$ – lesnik Sep 20 '16 at 9:01
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The deceleration/acceleration story is camouflage devised by Einstein in 1918 (nowadays most Einsteinians admit that the turning-around acceleration is immaterial). The conclusion validly following from Einstein's 1905 postulates is: The moving twin is aging slower as judged from the stationary twin's system, and the stationary twin is aging slower as judged from the moving twin system. "Aging faster" is an invalid conclusion (does not follow from the postulates). The original invalid conclusion (the stationary clock runs FASTER than the moving clock) was advanced by Einstein in 1905:

http://www.fourmilab.ch/etexts/einstein/specrel/www/ ON THE ECTRODYNAMICS OF MOVING BODIES, A. Einstein, 1905: "From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by tv^2/2c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B."

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  • $\begingroup$ I feel like this is a bad answer to my question... $\endgroup$ – Phu Nguyen Sep 20 '16 at 6:19
  • $\begingroup$ "I feel like this is a bad answer to my question..." Let me try to improve it. According to special relativity, at the end of the journey, the traveling twin will conclude that his stationary brother is younger than himself. He will come to this conclusion by checking, all along, stationary clocks against his own clocks - stationary clocks will be found to lag behind his own clocks. The stationary twin will come to the opposite conclusion - moving brother is younger. $\endgroup$ – Pentcho Valev Sep 20 '16 at 7:07
  • $\begingroup$ What you are saying is the paradox itself... The solution to the paradox is well-known. You can look it up if you want. I added a twist to this paradox and asked for a NEW solution. $\endgroup$ – Phu Nguyen Sep 20 '16 at 8:07
  • $\begingroup$ "The solution to the paradox is well-known. You can look it up if you want. I added a twist to this paradox and asked for a NEW solution." There is no OLD solution so there can be no NEW one. The twin paradox is actually an absurdity successfully camouflaged by Einstein in 1918. All (validly deduced) consequences of Einstein's 1905 false constant-speed-of-light postulate are absurd. $\endgroup$ – Pentcho Valev Sep 20 '16 at 9:36
  • $\begingroup$ You should do some research before making wild claim of there is no old solution. $\endgroup$ – Phu Nguyen Sep 20 '16 at 17:03

protected by Qmechanic♦ Sep 20 '16 at 7:48

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