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Suppose that I send my twin away from the earth at 87% the speed of light. From the time dilation equation, since the twin is moving, the time interval between any 2 events, as measured by me, would be greater for the twin than me.

1) Does this mean that if I observe the twin, he will appear in slow-motion? Any action he does should take more time (in my frame) than if I did the same action, right?

2) Who will die first? Suppose that the departure occurs at age=0, and we both are set to die at age=100. Since gamma=2 in this case, the time interval between the same event should be twice as much for the twin, so he should die at 200 in my frame. But the same can be said in the frame of the twin. Who will die first in an absolute sense? Is it even sensible to take about an absolute sense here?

Note: The twin never changes his direction or velocity. This is to avoid any of the problems that arise due to acceleration, which are beyond me at this point.

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    $\begingroup$ the time interval between any 2 events, as measured by me, would be greater for the twin than me. This is certainly false. If it were true you could use it to decide which of you is moving. $\endgroup$ – WillO Nov 30 '20 at 8:07
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    $\begingroup$ "Who will fie first in an absolute sense?" - As a hint, the word "absolute" is the opposite of "relative". $\endgroup$ – Yakk Nov 30 '20 at 14:51
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    $\begingroup$ @will0 Could you elaborate? I dont need this to "decide" which of us is moving. In my frame of reference, I am always at rest. Im basically trying yo say that if a sneeze takes 10s for both of us in the earth frame, when the twin is moving, I will observe the sneeze taking longer than 10s. The twin will observe my sneeze taking longer than 10( by the first postulate of SR). $\endgroup$ – OVERWOOTCH Nov 30 '20 at 14:57
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    $\begingroup$ The twin would appear in very fast motion indeed! 0.87c! $\endgroup$ – Gilbert Dec 1 '20 at 13:12
  • $\begingroup$ Only slightly related, but possibly helpful: physics.stackexchange.com/q/111078 $\endgroup$ – Jonas Dec 1 '20 at 21:23
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Does this mean that if I observe the twin, he will appear in slow-motion? Any action he does should take more time (in my frame) than if I did the same action, right?

Yes. A clock in the ship of the twin will appear to tick at a slower rate. If you want to call it “slow motion” you could.

so he should die at 200 in my frame. But the same can be said in the frame of the twin. Who will fie first in an absolute sense? Is it even sensible to take about an absolute sense here

Yes in your frame. But you will apparently die in 100 years so you will not live long enough to see the other guy die. In the twins frame, 100 years is how long it will take before he dies, but an observer on earth will see this take 200 years. There is no sense in talking about absolute time since such a thing does not exist which is the point of relativity.

Similarly, there is no meaningful concept of “first” since simultaneity is also relative. That is, in a sequence of events, who will come “first, second etc” is relative.

Click here for a descrition of simultaneity which states “Simultaneity is the relation between two events assumed to be happening at the same time in a given frame of reference. According to Einstein's theory of relativity, simultaneity is not an absolute relation between events; what is simultaneous in one frame of reference will not necessarily be simultaneous in another (see Relativity of simultaneity”.

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    $\begingroup$ The only thing I would add to this answer is that it's possible for two events to be sufficiently separated in time and close enough in space that there is no reference frame in which they occur simultaneously. Such events are said to have a timelike separation. So while simultaneity is relative, "not-simultaneous" is not always relative. The same goes for the order of events. As a sanity check, there had better not be a reference frame where my birth and death are simultaneous events, or worse, where my death happens before my birth. $\endgroup$ – Charles Hudgins Dec 2 '20 at 1:37
  • $\begingroup$ is it mathematically impossible for effect cause in any frame of reference? Ik it sounds non-sensical, but then again, so does the rest of SR $\endgroup$ – OVERWOOTCH Dec 2 '20 at 17:26
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  1. yes. You will appear in slow motion to your twin, and your twin will appear to be in slow motion to you

  2. There is no notion of "first". Once you synchronize your clocks at the start of the experiment, there is no further common notion of time. You will die before you see your twin die, and your twin will die before they see you die. The whole reason it is "relativity" is that there is no absolute time. Your death and your twin's death are spacelike-seperated events, and their relative order is reference-frame dependent.

More concretely, your spacetime position at your death is (100 years, 0).

Your twin's position, at their death, in your frame is (200 years, $\sqrt{3} * 100c$ years).

But, as measured by your twin, where you are receding away with a gamma factor of two, it's exactly reversed. Their postion at their death, in their frame is (100 years, 0), and your position, at your death, is (200 years, $-\sqrt{3} * 100c$ years), which is exactly symmetric.

Finally, the bit about spacelike seperation. Let's go back to your frame. The vector that points between the two deaths tells you about how they are spaced apart in spacetime.

It comes out to (100 years, $\sqrt{3} * 100c$ years). It should be pretty clear that the spatial part of this vector is larger than the temporal part. This makes this vector a spacelike vector, which means that the two deaths are actually synchronizable events -- this means that there is a reference frame where the two deaths are simultaneous (unsurprisingly, this is the reference frame moving away from Earth at half the speed of the ship [EDIT: I was lazy writing this, not wanting to add more detail, but see JEB's comment below]). Which event happens first depends on whether you are going faster or slower than this synchronization time.

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  • $\begingroup$ idk If I understand this correctly, but if you are going faster or slower than "that" synchronization time, wouldn't there be a new synchronization time ( corresponding to a frame moving at half of the new velocity? $\endgroup$ – OVERWOOTCH Nov 30 '20 at 5:21
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    $\begingroup$ @OVERWOOTCH what are you synchronizing? A third person's death? We're talking about synchronizing the first two deaths, and showing that there is no absolute answer for "which one of the two people dies first?" And the reason why is "there is a frame where they die at the same time that is neither of their frames" $\endgroup$ – Jerry Schirmer Nov 30 '20 at 5:23
  • $\begingroup$ oh ok. I get it now. thankyou $\endgroup$ – OVERWOOTCH Nov 30 '20 at 5:27
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    $\begingroup$ the frame in which the twins' lives are simultaneous is at half the rapidity, or $v_{sim} = \tanh{(\frac 1 2 \tanh^{-1}v)} = 1/\sqrt 3$ $\endgroup$ – JEB Dec 2 '20 at 4:05
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  1. Does this mean that if I observe the twin, he will appear in slow-motion?

Yes, but how much depends on what you mean by "observe the twin". If you parametrize movement by your own time, the twin will move slowly according to dilation formula. This is however not what you would see directly. The direct observation must also account for the movement and depends on the direction of the movement.

Imagine observer traveling towards the Earth. The Earth is sending signal every second of its own time. In observers frame, this is every 2 seconds. If observer is 1 light minute away from the Earth (in his own frame), he will reach the Earth in 1/0.87=1.15minutes and he counts how many signals does he receive during this time.

The naive formula to compute how many signals observer received in this time is to divide by 2 seconds - there is signal sent every 2 seconds in observers frame and he observes for duration of 69 seconds giving 34 signals received. But this computation is wrong.

The first signal that observer receives is one minute old and on his path to Earth he also catches all the signals that are already on his way from the Earth, but did not reached him yet. So he really detects all the signals sent in an interval of 60+69=129 seconds, giving 64 signals. This means, while the observer observes only for the duration of 69 seconds, he sees events at earth that were happening for the duration of 129 seconds in his own frame.

On the other hand if the twin moves away from the Earth, the signals spend some time catching the observer. If he moves away from the Earth to distance 1 light minute, only signals sent in the first 9 seconds will reach the observer.

So even if all the events on the Earth are slow down by factor of 2, the actual observation will see it slowed down by factor somewhere between 69/64.5=1.07 and 69/4.5=15.33, depending on the configuration of the movement.

  1. Who will die first?

There is no absolute notion of chronology when you are comparing spacelike separated events. So this cannot be decided.

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You have set yourself up for confusion here by trying to simplify the problem:

note: the twin never changes his direction or velocity. This is to avoid any of the problems that arise due to acceleration, which are beyond me at this point

By excluding this you actually have two reference frames--what you see (the twin growing old slower than you do) and what your twin sees (you growing old slower than he does.)

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