As I understand it, in special relativity, when two bodies are moving away from one another, there is no absolute sense in which one is moving away from the other (e.g. when you jump in the air it is equally true to say that the Earth is moving away from you as it is to say that you are moving away from the Earth; or when a spaceship is going deep into space it’s just as true to say that the Earth is moving away from the spaceship as it is to say the spaceship is moving away from Earth).

This has interesting implications when it comes to the more funky aspects of special relativity (i.e. time dilation, length contraction). Because this means that if Bob is moving close to light speed relative to Jane, Bob will perceive Jane as experiencing length contraction and time dilation, but Jane will not experience these things. From her point of view, it is Bob that is experiencing length contraction and time dilation. So both will always experience the other as experiencing these things, because from their point of view it is always the other person moving at near light speeds. So special relativity is symmetrical this way.

As I understand it though, this symmetry breaks when it comes to acceleration. This is how you can have a scenario where e.g. Bob ages a lot compared to Jane (because he accelerated or decelerated more).

So my question is: why does this symmetry in special relativity break when it comes to acceleration?

  • Sasha [They/Them]@lemmy.blahaj.zoneEnglish
    1·
    2 days ago

    I don’t think this is true, you’re right that what’s considered simultaneous changes*, but it’s not related to distance and that’s not how redshift works.

    At best Alice could use the redshift to work out how fast she’s moving relative to Bob, if she’s moving towards and away from him at the same speed, she’ll always get the same result. She’d actually think the same moments are simultaneous regardless of direction.

    The time difference is only accumulated during her acceleration so it can only be measured during it.

    *but only when her speed is different

    • AbouBenAdhem@lemmy.worldEnglish
      2·
      3 days ago

      Instead of Alice (aka Jane) turning around, say Alice and Jane are two non-accelerating observers moving toward and away from Bob, who pass each other four light years out. Would you agree that, at the moment they pass each other, Alice and Jane would see different points in Bob’s history as being simultaneous in their respective reference frames? It would be measurable even though no acceleration is occurring.

      And it’s related to distance because the further they are from Bob, the greater the discrepancy in their calculations of Bob’s relative time will be.

      You’re right that this isn’t how Alice/Jane would think of it in practice, because everyone would convert their own calculations into a stationary reference frame that would take their own time dilation into account. But that doesn’t illustrate OP’s question about the reference frame symmetry as well.

      • Sasha [They/Them]@lemmy.blahaj.zoneEnglish
        1·
        2 days ago

        Would you agree that, at the moment they pass each other, Alice and Jane would see different points in Bob’s history as being simultaneous in their respective reference frames?

        I wouldn’t say that’s necessarily true, no. It’s only true if their speeds are different, the direction they’re travelling doesn’t factor into it. In either case, their lines of constant time are the same because space is rotationally symmetric. I’m not thinking of them adjusting their calculations to some “true” sense of simultaneous, if that’s what you meant in your edit, because there isn’t one by definition.

        And it’s related to distance because the further they are from Bob, the greater the discrepancy in their calculations of Bob’s relative time will be.

        That really depends on what you mean by time shift in your original comment. If you mean the shift in what they perceive as simultaneous, then it’s not, but it seemed to me that’s what you meant. If you mean the difference in their age then I honestly can’t remember how it factors in for the accelerating case, I haven’t had to think about SR problems in a while.

        • AbouBenAdhem@lemmy.worldEnglish
          2·
          3 days ago

          Consider the Lorentz boost to convert events from Alice’s reference frame to Jane’s. The time of an event (like Bob’s current point in Alice’s frame) will differ in Jane’s by a factor of vx/c2 (times the Lorentz factor). Since v (their velocity relative to each other) and x (the distance to Bob) are both nonzero, the times of events in Bob’s history will differ between Alice’s frame and Jane’s by a nonzero amount. And the time shift is proportional to x, which is why the distance to Bob matters.

          • Sasha [They/Them]@lemmy.blahaj.zoneEnglish
            2·
            2 days ago

            Cool yeah, dug through the maths and took the time to understand the situation you were describing and I understand now.

            I thought you were describing a something else, and then slightly confused myself by only considering the metric and not the global picture. I was trying to abuse some old heuristic techniques and they don’t quite work for this case, though it was fine for what I was picturing, where Jane and Alice are symmetric about Bob.

            Thanks for taking the time to convince me.

            Actually reading back over this is hilariously dumb, forgive my bad reading comprehension, that is how redshift works.