Post #528492

@dougmerritt@mathstodon.xyz We can even do a "twin paradox" scenario: I step off the ledge on the 100th floor (t,z)=(0,1) for those Moving People. √3/2 ≈ 0.866 years later the 50th floor ship catches up with me (√3/2,1) Assuming I somehow survive landing in that net (delta-v is a ruinous 0.866c; velocity angle 𝜑 ≈ 1.317, which represents a passage of time of 𝜑 years on the 100th floor and 𝜑/2=0.658 years on the 50th floor), I can then spend 5−𝜑 years on the 50th floor, then have them fire me upwards at 0.866c, just fast enough that after another 0.866 years in free fall, the 100th floor people can grab onto me, but for them a total of 10 years will have passed since I first stepped off. For me, total time passed will be 5−𝜑+√3 ≈ 5.415 years. 2/2