Suppose there are two planets, Planet A and Planet B, situated $10$ light years apart from each other. Planet B is currently celebrating its $0^{th}$ year. Two astronauts from Planet A each want to travel to Planet B and arrive on the earliest possible date of Planet B's calendar. One astronaut takes Spacecraft X which travels at a speed of $0.75c$, while the other one takes Spacecraft Y which travels at $0.99c$.

From the equation for time dilation, $\gamma = {1 \over \sqrt{1 - {v^2 \over c^2}}}$ we know Spacecraft X will have a time dilation factor of about $1.512$, and Spacecraft Y will have a time dilation factor of about $7.089$. Then to calculate the date of arrival for each spaceship, we have:

$t = {d \over v}*\gamma$

$t = {10 ly \over 0.75ly/yr} * 1.512 = 20.16 yr$ for Spacecraft X

$t = {10 ly \over 0.99ly/yr} * 7.089 = 71.61 yr$ for Spacecraft Y

If my model is correct, Spacecraft X will arrive on Planet B on a much earlier date than Spacecraft Y, despite travelling at a slower speed, because it makes up for it with a smaller time dilation factor. Following this line of reasoning to its conclusion would suggest ${1 \over \sqrt{2}}c = 0.7071c$ to be the optimal speed for interstellar travel.

So is there an error in my model, or is it truly possible to get there "faster" by going slower?

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  • $\begingroup$ Shouldn't you divide by gamma instead of multiplying? $\endgroup$ – Adrian Howard Jan 15 at 1:58
  • $\begingroup$ I like the use of units in the equation; but making the unconventional step of labeling gamma with "coordinate year per proper time year" would have convinced you to divide by gamma, not multiply. $\endgroup$ – JEB Jan 15 at 2:49

I think you made a mistake in your calculation. Just picture it from planet B perspective: in its frame of reference, it sees two spaceships, one at $0.99 c$, the other at $0.75c$, travel towards it. In this frame of reference, the ships take respectively $10/0.99 \approx 10$ yrs and $10/0.75 = 13.3$ yrs to complete the journey. This is of course compatible with our intuition.

In the frame of reference of the spaceships, what happens? There are two ways to see it. The first is to say that due to length contraction, the ship only has to travel a shorter distance $d' = d/\gamma$. This yields a trip duration (in the reference frame of the ship), of $\tau = d'/v = d/(\gamma v)$.

The other way to see it is to consider the time interval between the departure of the ship from planet A and its arrival at planet B. In planet B's frame of reference, this time is $t = d/v$ as mentionned earlier. Time dilation tells you that the same interval corresponds to a proper time $\tau = t/\gamma$ in the reference frame of the ship. As you see, in both cases, the journey time on the ship is given by $d/(\gamma v)$, and from planet B perspective the corresponding time is $d/v$. So in conclusion, the faster ship does arrive faster than the slower ship, regardless of the frame of reference.

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