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Is this statement true?

You can not travel back in time. If I travel fast enough the clock will start to go backwards, but that does not mean I am traveling back in time. It would only mean that the time reference is producing a negative count.

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    $\begingroup$ I took the liberty of editing the title to make it more clear what the question is about. Hope you don't mind. $\endgroup$ – Marek Jul 31 '11 at 8:01
  • $\begingroup$ This is a hypothetical question as relativity prohibits real superluminal travel. $\endgroup$ – Siyuan Ren Jul 31 '11 at 8:50
  • $\begingroup$ @Karsus: relativity does not prohibit it at all and tachyonic solutions can be found e.g. in string theory. What prohibits it is the principle of causality since superluminal travel violates causality (that's also why string theory needs to cope with those solutions by mechanisms such as tachyonic condensation). $\endgroup$ – Marek Aug 4 '11 at 9:00
  • $\begingroup$ see physics.stackexchange.com/questions/574395/… $\endgroup$ – Andrew Steane Aug 20 '20 at 10:09
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One cannot travel backwards in time.

The physics data we have up to now agree completely with the special theory of relativity. This means you cannot travel faster than the velocity of light, and that you can only approach it as a limit. You will not notice a change in your clock in any measurable way within your travelling system.

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  • $\begingroup$ But you agree that you can not travel back in time? $\endgroup$ – Sifimichael Aug 2 '11 at 15:26
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    $\begingroup$ @Sifimichael I think the question was edited so I have to edit my answer. Yes, you cannot travel backwards in time. $\endgroup$ – anna v Aug 2 '11 at 18:08
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Yes, this would be only your impression. E.g. when you are moving away from me then your clock is going slower from my point of view and my is going slower from your point of view. This is because as you are moving away from me the distance between us increases and the light needs longer time to reach you (and this statement is obviously symmetric when we swap me and you). Consequently, the image you'll see of me will be distorted in time and I'll appear to be doing everything slower.

If you would go at the speed of light relative to me (note that this is not physically achievable, but we can discuss this purely theoretically) then all the light from me that could reach you is originating at one moment in my history and consequently it would appear to you that my clock has stopped completely.

Finally, if you were moving superluminally, the light that I emitted as we met and any light I will emit in the future will never reach you (it is simply too slow compared to your speed). On the other hand, as you are moving away, the light that I emitted in the past (and that has already travelled some distance) does have a chance of catching up you. And the further away you move, the older light from me you'll see. Your impression will be that my clock is going backwards. But this is not the end of the story. Because if you were approaching me at the superluminal speeds then it would appear to you that my clock is going forward as usual. Until the point when our wordlines meet. Here's a quick sketch.

enter image description here

The dashed line is you (the superluminal observer), the vertical line is the me (the standing observer) and the 45 degrees lines are light rays I emitted at various events. Note that you'll see me first travelling forward in time upto the event labeled 6 and from the event labeled 7 onwards you'll see me going back in to my past.

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    $\begingroup$ I think this does not work as an answer, because merely receiving signals from someone's past does not imply that person is moving backwards in time. If the signals were sound waves, for example, then as you travel faster than sound you may hear the sound waves from another person in reverse order, but that does not mean the other person is travelling backwards in time. $\endgroup$ – Andrew Steane Aug 20 '20 at 9:23
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Spacetime diagram showing how faster than light travel would allow traveling back in time

Since that would very directly break causality, this is a great argument as to why you can't do it.

So I tried to make a diagram to clarify to myself more precisely how that would go like:

enter image description here

SVG source.

Legend:

  • points:
    • A: Spaceship departs from Earth moving towards C slower than light. Note that the spaceship trajectory from A to C coincides with its time axis (because x = 0 at all times) and is not otherwise indicated.
    • B: Earth sends a tachyon (fantasy particle that travels faster than light) to C. This moves faster than light (inclination between that of the two black light rays) but it appears to move forward in time according to Earth (stays above its blue space axis)
    • C: The tachyon sent from Earth reaches the spaceship, and then the spaceship immediately sends a tachyon back to D. Once again, it appears to move forward in time according to the spaceship (it stays above the red space axis of the spaceship).
    • D: The tachyon send back from the spaceship reaches Earth before Earth sent the first tachyon
  • lines:
    • black: light rays from B and C
    • blue: space and time axes of Earth at B
    • red: space and time axes of the spaceship at C
    • magenta: tachyons sent from B to C and from C to D

So we see that if we were able to send particles faster than light, even if they appear to be moving forward in time to us, it would be possible to send messages back in time (to a timelike spacetime point) from another inertial frame of reference.

Earth observes the spaceship tachyon travel backwards in time from C to D all the way. This is possible due to the fundamental assumption that the laws of Physics are the same for all inertial frames. So if Earth can send tachyons "forward in time according to Earth", then the spaceship must be also able to send tachyons "forward in time according to the spaceship".

Conversely, the spaceship observes that C happened before B, i.e. it sees the tachyon from Earch move backward in time. This is already a simpler causality problem to observe, but I wanted to show the full "send message to past" in a single diagram. To see that simpler causality issue on the diagram, you just have to add a constant time line (parallel to x of the spaceship), this is done for example at: http://www.physicsmatt.com/blog/2016/8/25/why-ftl-implies-time-travel

Variants of the story to make things more dramatic:

  • Earth has a tachyon detector which detonates a nuclear bomb and pulverizes the tachyon sending facility when a tachyon is received. So how can the tachyon be sent, if the facility got nuked in the past?
  • instead of sending a tachyon, you send the entire scientist who pressed the send tachyon button around. They then meet themselves.
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  • $\begingroup$ I don't see how the tachyon sent by the ship is going back to D and not to some D after C. The ship would get the tachyon and then receive photons of the tachyon going backward. Also, after the ship send the second tachyon, later, earth would receive a tachyon and then detecting the movement of the tachyon going backward. $\endgroup$ – Adrian Maire Jul 15 at 13:09
  • $\begingroup$ @AdrianMaire hi, I don't understand very well what you mean. Which photons are you talking about? Do you mean assuming that photons were emitted each time together with the tachyons? Or something else? $\endgroup$ – Ciro Santilli 新疆再教育营六四事件法轮功郝海东 Jul 15 at 13:18
  • $\begingroup$ Which photons are you talking about? I mean whatever particleS are emitted (at c speed) by the tachyon along-side it travel, so that the earth and the ship could measure it travel. $\endgroup$ – Adrian Maire Jul 15 at 13:22
  • $\begingroup$ @AdrianMaire hmm, OK. I don't think the photons helps much conceptually, each reference can just have an infinitely long array of sensors + clocks pre synchronized with en.wikipedia.org/wiki/Einstein_synchronisation so that everything is measured locally for each frame. $\endgroup$ – Ciro Santilli 新疆再教育营六四事件法轮功郝海东 Jul 15 at 13:55
  • $\begingroup$ It would also work but the signal would travel at <c speed along the sensor. The question is: why B is before C and not after? With delay B-C approximatively equivalent to delay C-D. $\endgroup$ – Adrian Maire Jul 15 at 14:22

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