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So if the astronaut is experiencing a constant 1g acceleration, how much time would pass for them before an outside observer would measure their speed at .9999C?

I saw this question and noticed that an astronaut experiences acceleration a little differently than an observer due to relativistic effects. How much would that alter overall acceleration from an observer, as well as affect the difference in time the astronaut experiences during the acceleration?

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If the astronaut is experiencing a constant 1 g acceleration as measured in the astronaut's frame (i.e., proper acceleration, or acceleration as felt by the astronaut), it would take a bit less than 5 years (4.80009 years, to be precise) to achieve a delta v of 0.9999 times the speed of light. This is a trivial result of the relativistic rocket equation at constant acceleration:

$$\frac{\Delta v}{c} = \tanh\left(\frac{aT}{c}\right)$$

or

$$T = \frac{c}{a}\mathop{\text{atanh}}\left(\frac{\Delta v}{c}\right)$$

where

  • $T$ is the proper time (time as measured by the astronaut),
  • $c$ is the speed of light,
  • $a$ is the proper acceleration (acceleration as measured by the astronaut), and
  • $\Delta v$ is the accumulated change in velocity as measured by an inertial observer initially at rest with respect to the astronaut.

Link to the calculation

This is of course unachievable using any known or hypothesized methods of acceleration in space.

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  • $\begingroup$ Would an outside observer notice that the astronaut began to accelerate more quickly as he approached the speed of light? $\endgroup$
    – Dakota Wharton
    Commented Sep 1 at 22:14
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    $\begingroup$ @DakotaWharton -- The other way around. An outside observer would see the astronaut accelerate at an ever lower rate. $\endgroup$
    – David Hammen
    Commented Sep 2 at 7:03
  • $\begingroup$ How much time would pass for a stationary observer? I'm guessing about $4.8 * \gamma(v=0.99c) = 34$ years but doing the exact integral would be more accurate. $\endgroup$
    – ScienceSnake
    Commented Sep 2 at 20:55
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    $\begingroup$ @ScienceSnake A stationary observer -- specifically a non-accelerating inertial observer initially at rest with respect to the astronaut -- would see about 68.5 years pass. The calculation is similar to the one used to calculate proper time: $$t = \frac{c}{a}\sinh\left(\mathop{\text{atanh}}\left(\frac{\Delta v}{c}\right)\right)$$ $\endgroup$
    – David Hammen
    Commented Sep 4 at 13:00

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