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According to this chart of the Lorentz factor as a function of speed:

Lorentz factor

If a spacecraft neared (roughly) 0.85c, would it appear to be traveling at 1.7x the speed of light from the perspective of those on board - i.e. covering 0.85c distance, but in half the time amount of time?

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marked as duplicate by DavePhD, user10851, JamalS, John Rennie, Danu Jun 2 '14 at 9:51

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There is a spacecraft moving with velocity $v = 0.85c$ in a frame of reference where, helpfully, someone has put out 'mile' posts exactly 1 light-second apart (in that frame), along the direction of the spacecraft's travel.

In this frame of reference, the spacecraft just passes 85 mile posts in 100 seconds according to clocks at rest in this frame.

However, according to the spacecraft, the mile posts are less than 1 light-second apart (length contraction) and, in fact, measure

$$\sqrt{1 - (0.85)^2} = 0.527... $$

light-seconds apart.

So, even though the spacecraft's clock reads an elapsed time of just about 52.7 seconds (time dilation) when passing the "85" mile post, the distance covered, according to the spacecraft, is just

$$d = 85 \cdot 0.527$$

light-seconds and thus, as must be the case, the spacecraft observer calculates a relative speed of $0.85c$ too.

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  • $\begingroup$ @nate: A. Centauri wrote: "according to the spacecraft, the mile posts are less than 1 light-second apart." Obviously not true. Dilatation and contraction are opposite relations. If Δt′>Δt then Δx′<Δx. So if the stationary observer sees his time as "larger" (more seconds than the traveler) then he must see his distance as smaller (less meters). That's simple maths. See J. Rennie's answer: physics.stackexchange.com/q/109776/43402. He follows similar logic, but is equally wrong. They both didn't assume which frame is primed and which is unprimed, so they can switch at will in the meantime $\endgroup$ – bright magus Jun 2 '14 at 13:29

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