# Twin Paradox and experience of speed

I'm sure someone has thought and discussed this before - I just can't seem to find any answers. Maybe the answer is so simple no one bothers to write it anywhere...

The classical twin paradox: Person A stays at Earth. Person B hops on a spaceship, travels around 0.8c to a star 10 ly away. Then B turns around and returns to Earth.

A sees 23 years passing. B sees 11.5 years passing.

So far so good. Except that I can't get my head around how B sees travelling 20 ly in 11.5 years, seeing spaceship going over lightspeed (~1.7 c).

What am I missing here?

• From where did you get the 1.7 $c$ ? Jan 11, 2021 at 13:04
• Length contraction. This is covered in the wiki article. Jan 11, 2021 at 13:05
• @kaylimekay Thanks! I knew I had missed something. Jan 11, 2021 at 13:19
• B doesn't see that. B sees that he has gone to a point, that was 20 ly away before he started, in 11.5 years. But how far does that distance actually look to him while he's moving, because of length contraction? Jan 11, 2021 at 14:47
• Draw a spacetime diagram from the POV of A (forget about the Earth, it is just in the way), there is a lot more going on (that you can check directly) than you indicate in your question. That is what you are missing! Also make sure you understand what "see" really means in this context. Jan 11, 2021 at 16:55

See if this answers your question in a bit more detail. The star is 10 ly away in A's frame. Due to length contraction, at $$v= 0.8c$$ the distance to the star in B's frame is 10 ly $$\times\sqrt{1-v^2}$$ = $$0.6\times10$$ ly = 6 ly. B is traveling at $$0.8c$$ relative to A. And so is also traveling at $$0.8c$$ with respect to the star. B completes the trip in $$\frac{2 \times 6~\mathrm{ly}}{0.8~\mathrm{ly/yr}} = 15$$ years on his clock. Meanwhile, in A's frame, the star is 10 ly away, so A sees the trip take $$\frac{2 \times 10~\mathrm{ly}}{0.8~\mathrm{ly/yr} }=25$$ years. At the end of the the trip the ratio of B's clock to A's clock is $$15/25 = 0.6$$. Their age ratio is the same. This is the correct answer, and didn't need accelerations, changes in frames, etc.