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Suppose A and B are a long distance apart initially. B takes off in a spacecraft in A's direction at a really high speed. Both were aged 0 when B took off. When B about to cross paths with A, A observes him to be 30 years old (while A is 60). At this point, 60 year old A shoots the 30 year old B. B dies.

Now, from B's frame, the bullet has to be fired by A when B is 120. That's because the event of 'bullet firing' must happen after the event of 'A turning 60', and A turns 60 in B's frame only when B is 120. So B has to die at 120 as seen from his own frame of reference.

How can B die at both 30 years old and 120 years old?

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    $\begingroup$ This is a follow-up question to physics.stackexchange.com/q/486482/123208 $\endgroup$ – PM 2Ring Jun 18 at 15:02
  • $\begingroup$ So will your next question involve the cousin C? $\endgroup$ – Solar Mike Jun 18 at 15:08
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    $\begingroup$ This question has two different answers depending on what you mean by "Both were aged 0 when B took off". Do you mean to assume that this is true in A's frame or in B's frame? $\endgroup$ – WillO Jun 18 at 16:08
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    $\begingroup$ The silver bullet for almost all of SR paradoxes: draw a spacetime diagram. $\endgroup$ – Feynmans Out for Grumpy Cat Jun 18 at 16:51
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    $\begingroup$ @FeynmansOutforGrumpyCat : I have a long history (on this site and elsewhere) of telling people that the best approach to almost every problem in SR is to draw the spacetime diagram. But in the present case, I actually think that the step-by-step reasoning in my answer is likely to be more educational for someone who is confused at the level of this OP. $\endgroup$ – WillO Jun 18 at 17:11
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Case I: When B takes off, A and B are both aged 0 in A's frame.

1) When B passes A, B is 30 (given in your setup).

2) But B's clocks run at half-speed according to A, so A says that B has been traveling 60 years.

3) Therefore A is 60.

4) But B says A's clocks run at half speed, so B says A was born 120 years before the shooting --- that is, 90 years before B was born.

5) So B's story is this: "90 years before I was born, A was born. He aged at halftime, so on my birthday, he was 45. At that time I started my journey to earth, which took 30 years. During that time, A aged another 15 years, so he was 60 when we met. Then he shot me. I died at 30."

Your mistake: You said that "from B's frame, the bullet has to be fired by A when B is 120". That's not correct. The correct statement is "from B's frame, the bullet has to be fired 120 years after A was born". Since B is 30 at the time of the shooting, A must have been born 90 years before B.

Your bigger mistake: You assumed that two different problems (namely this one and the one you asked in your last post) have to have exactly the same answer. In the other problem, A and B were in the same place at the same time when both were born. In this problem they weren't.

Case II: When B takes off, A and B are both aged 0 in B's frame.

1) When B reaches earth, he is aged 30 (given in the problem).

2) According to B, A ages at half-speed. Therefore A is 15, not 60 as you supposed. The 15-year-old A shoots the 30 year old B. Game over for B.

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  • $\begingroup$ Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases. $\endgroup$ – Aaron Stevens Jun 18 at 17:26
  • $\begingroup$ Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time? $\endgroup$ – Aaron Stevens Jun 18 at 17:30
  • $\begingroup$ @AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way. $\endgroup$ – WillO Jun 18 at 17:39
  • $\begingroup$ If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though? $\endgroup$ – Aaron Stevens Jun 18 at 17:44
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    $\begingroup$ @AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens! $\endgroup$ – WillO Jun 18 at 18:06
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You suppose that A and B are far apart "initially" and that they were "both aged 0 when B took off". It means that you assume that simultaneity is an absolute notion, independent of the frame of reference, which is why you come to a paradox. It is exactly the symetric case of two people starting at the same point, diverging fast, and observing each other: each one will see the other one aging slower than him, because they compare moments in non parallel time sections. This is best shown by a drawing, this is the one I have suggested in this thread: link to a video at the second displaying the graph.

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In addition to Willo's answer

As it was being said by @WillO and @PM 2Ring and others, your problem here is that you have assumed:

'Both were aged 0 when B took off'

Well that's simply wrong, because like this problem, you have not specified in which frame?! Of course you could have asked

1.When B takes off, A and B are both aged 0 in A's frame.

or

2.When B takes off, A and B are both aged 0 in B's frame.

or

3.before B takes off (where he is at the rest relative to A, but in different position), A and B are both aged 0 in both frame. Indeed, their frame would be same, because they are at rest at that moment. Then instantly, B accelerate toward A and make his velocity constant after a moment in his frame.

But they are different questions. In case 3 You might think that if B zero y/o (after taking off) sends a signal to A, in his frame, he'd assume that A is zero y/o too but that's 100% wrong, in his frame A will be a lot older than B.

Someone might wondering how this asymmetry possible even though their initial state were same, well in short in accelerated frames like B, time can elapse faster or slower depending on position. i.e in B frame, A's clock will elapse a lot faster than his as long as B is accelerated. And if you assume that B will accelerate instantly, then you will see that in B's frame A ages instantly as well! If you are intested in last case i suggest that you read

Explorations in Mathematical Physics from Don Koks

Accelerated frame section.

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