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This question already has an answer here:

Say we have the earth and a person on it and an astronaut travelling away from the earth in a spaceship in a straight line and then coming back in a straight line at the same speed.

The person on the earth measures the time between when the astronaut leaves and when he comes back as 1 year. This is a proper time since the observer has not moved in his frame of reference. Thus according to the relation $t = \gamma t_o$ the astronaut must have experienced a longer time.

The person on the spaceship measures the time from takeoff to landing as 1 year. Since he's standing at the same point in his frame of reference this is a proper time. Thus according to the relation $t = \gamma t_o$ the person on earth must have experienced a longer time.

The above is contradictory. Either the astronaut experiences a longer time or the person on earth does. The issue is that either the astronaut is thought to be moving relative to the earth or the earth is thought to be moving relative to the astronaut. What's the explanation? Can anyone help me understand?

Okay. I’ve read through the below answers, as well as watched a few videos and the Wikipedia page on it, however, i don’t see any of them explaining it. They claim that the acceleration solves the issue, however, don’t we arrive at the same problem? In the earth point of view the astronaut is rotating, but you might as well see the earth as rotating in the astronauts view. In other words, same problem.

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marked as duplicate by safesphere, WillO, stafusa, Chris, Kyle Kanos Jan 22 '18 at 11:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ en.wikipedia.org/wiki/Twin_paradox $\endgroup$ – safesphere Jan 21 '18 at 5:36
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    $\begingroup$ Possible duplicate of How is the classical twin paradox resolved? $\endgroup$ – safesphere Jan 21 '18 at 5:42
  • $\begingroup$ There are three frames to consider, not two: The frame of the earthbound observer, the frame of the outbound astronaut, and the frame of the inbound astronaut. Call these A, B and C. In Frame A, the astronaut's clock runs slow. In Frames B and C, the earthbound clock runs slow. Frames B and C disagree about the lengths of the outbound and inbound journeys, and about how fast the traveling clock ticks on each part of the journey. Each of the three frames tells a perfectly consistent story. $\endgroup$ – WillO Jan 21 '18 at 6:23
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GR (general relativity) is not necessary.
The twin paradox can be described by SR (special relativity) just noting that SR applies to IRF's (inertial reference frames). Therefore the spaceship reference frame, being subject to acceleration to invert the route, can not be used to describe the complete trajectory of the travelling twin forth and back to earth.
Note that the earth reference frame can be used also to measure the proper time of the travelling twin during the acceleration phase considering the inertial reference frames instantaneously at rest with the spaceship frame.
Further note: a way to convince doubtful practitioners that the two reference frames are not equivalent is that during the acceleration phase the travelling twin experiences a proper acceleration, while the twin on the earth appraises nothing.

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The idea is very simple, one of the frame is inertial but the other one is not. The astronaut changes his direction of motion , so he is experiencing acceleration and cannot use laws of special relativity to make deductions.

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    $\begingroup$ That's not the point. You can solve and understand twin paradox without using general relativity. $\endgroup$ – Andrei Geanta Jan 21 '18 at 7:42
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This is a very famous paradox called the twin paradox (even though it is not a paradox). It is not a paradox since the person on spaceship has to accelerate in opposite direction to come back to the earth. Since he accelerates special theory of relativity does not work the same way like in cases of non accelerating systems. The direction of time rotates for the astronaut.

Here is a video on youtube that explains this quite well:

https://www.youtube.com/watch?v=0iJZ_QGMLD0&t=56s

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    $\begingroup$ Since he accelerates special theory of relativity is no longer valid. This is wrong. It's a common misconception that special relativity cannot handle accelerating objects or accelerating reference frames. $\endgroup$ – Andrei Geanta Jan 21 '18 at 7:44
  • $\begingroup$ Can you please explain. I am also a learner and came across this problem a few weeks ago. So I shared the solution that kind of satisfied me. Edited that sentence a bit though. $\endgroup$ – SR810 Jan 21 '18 at 7:47
  • $\begingroup$ @SRB10 Either SR can handle this or is wrong.... The same as we would need newtonian mechanics to do kinematics. ... $\endgroup$ – Alchimista Jan 21 '18 at 8:55

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