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The tachyonic antitelephone paradox is, roughly, this:

Alice is in a space station and watches Bob go past at relativistic speed in a spaceship. Some time later she sends him a message at superluminal velocity. Due to time dilation, she expects it to arrive at Bob when Bob has experienced less subjective time than she has when she sent it. Bob then replies - and expects his reply to arrive at Alice when she experienced less subjective time than when he sent it. Since $t_{Asend}>t_{Brecieve}=t_{Bsend}>t_{Arecieve}$ we have Alice receiving Bob's message before she sent hers.

Suppose Charlie is in a space station (at rest relative to Alice) at the point that Bob receives Alice's message. Charlie also receives Alice's message and also sends a response. It should be obvious that Charlie's response arrives at Alice at the same time that Bob's does, but also that Charlie's response arrives after Alice sends her message.

Because the paradox relies on considering Bob in Alice's reference frame and then Alice in Bob's reference frame (but not on the velocity of the message - Charlie and Bob would still get different results if using lightspeed communication) it seems like the core of the issue is changing reference frames.

Now, many people smarter than me have examined this problem and determined that it is a genuine paradox, so what am I missing?

Qmechanic
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  • No, the core of the paradox is a physical fact: if Alice sends a message when her clock shows 8 o'clock, what does Bob's clock show when he receives it? If it's anything less than 8 o'clock then a paradox will ensue (because the situation is symmetric, so if Bob repeats the message back to Alice her clock will show something smaller still when she receives it). The only way to make sure that Bob's clock shows something later than Alice's, regardless of the speed at which they're moving apart, is for the message to be sent at light speed or below. – Eric Smith Oct 08 '22 at 16:41
  • This answer has helpful diagrams: https://physics.stackexchange.com/a/432251/123208 – PM 2Ring Oct 08 '22 at 16:45
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  • @EricSmith Bob's clock is in Bob's reference frame. Given that time dilation is symmetric (provided that inertial reference frames don't change), if at 8:00 Alice sends a message that will take 1 hour to reach Bob at a velocity of 2c, Bob's clock will read 7:30 in Alice's reference frame when he gets it. But when Bob gets it it will be 9:00 in his reference frame. My contention is that the Tachyonic Antitelephone switches to Bob's reference frame at 7:30 (because that's when Alice thinks he gets it) rather than at 9:00 (when he actually gets it). – Spitemaster Oct 08 '22 at 21:42
  • @PM2Ring No. I've seen those things before (actually read that when writing this question), and I'm not asking "are there issues with FTL travel?" but rather "Is this actually about time travel or is it a consequence of switching reference frames?" – Spitemaster Oct 08 '22 at 21:43
  • @Spitemaster It's always a good idea to mention your prior research, and to link related questions that you've already looked at, and explain why they don't adequately answer your question. – PM 2Ring Oct 08 '22 at 22:08
  • Yes, it's about reference frames. SR (special relativity) says that if two events A & B have spacelike separation, then they can't be uniquely ordered in time. In some frames, A & B are simultaneous, in some frames A occurs before B, in some frames B occurs before A. https://i.stack.imgur.com/AtqPQ.gif – PM 2Ring Oct 08 '22 at 22:13
  • The time shown on Bob's clock as he receives the message is a physical fact which must be the same in all reference frames (assuming the clock is co-located with Bob). So if Alice calculates it to be 7:30, then that is indeed what Bob will observe also, as will everyone else watching Bob through telescopes. That's where the paradox comes from -- Alice sends a message at 8:00, Bob receives it at 7:30. Note that this can only happen with FTL messages. – Eric Smith Oct 08 '22 at 22:45

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