Was thinking about interstellar travel and the ability to provide artificial gravity by using a smooth acceleration and deceleration across the journey, changing from acceleration to deceleration at the halfway mark.

If we ignore relativistic effects, with smooth acceleration of 9.81 ms-2, you’d be going 3.1e8 ms-1 after the first year (3.2e7 s), if I’m not making a mathematical blunder. That’s more than the speed of light at 3.0e8.

My main question, and the one that I initially came here to ask, is: if their ship continues applying the force that, under classical mechanics, was enough to accelerate them at 9.81 ms-2, would the people inside still experience Earth-like artificial gravity, even though their velocity as measured by an observer is now increasing at less than that rate?

A second question that I thought of while trying to figure this out myself as I wrote it up, is… My understanding is that a trip taken at the speed of light would actually feel instantaneous to the traveller, while taking distance/speed of light to a stationary observer. In the above scenario, would the final time taken, as measured by the traveller, be the same as if they were to ignore the speed that they are travelling at according to an outside observer, and instead actually assume they are undergoing continuous acceleration?

  • @cynar
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    89 months ago

    It’s also worth noting that it also experiences zero distance. If you’re willing to tie your brain in knots, a photon doesn’t exist. Instead, space-time flexes so that 1 point touches another, momentarily. Energy is transferred, and space-time recoils back. That flex would be mathematically identical to a photon traversing the intervening space-time.

    There’s a reason we use photons however. Such twisted space is effectively impossible for our brains to usefully comprehend.

    • ZagorathOP
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      19 months ago

      I don’t know if that analogy works, because from the perspective of an observer, a photon doesn’t travel instantaneously. It travels at the speed of light.

      • @cynar
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        19 months ago

        That’s why I said space-time, not just space. Generally worked with in the form of [X,Y,Z,iT] to make them all behave space like. Basically 2 4D positions become the same position. The fact that the 2 positions are displaced in time is almost incidental. The rules for the transformation however still have to collapse down to the same underlying measurements, so it’s a lot more complex than 2 arbitrarily points.