Schrödinger's Cormorant

(A2-13) ① We know Cormorant's velocities relative to Nefertiti (frame N) on both legs of the journey. Outbound, he's in reference frame C1, moving with a velocity of +0.5c in frame N, but for the return he has switched to C2, and is moving with a velocity of -0.5c in frame N.

We now want to find Cormorant's new velocity in frame C1. This was, obviously zero on the outbound leg, where this was his stationary reference frame, but what will it be when he now heads off in the opposite direction? For this we need to use the velocity addition formula (section A2-3):

Use this relationship to find his new velocity, in frame C1, as he returns to Earth.

② Here’s what we now know about the state of the universe, from the perspective of frame C1, at the beginning of the second leg:

Cormorant is at Mars and his position in frame C1, xC1(C) = 0. From ①, we now know that his new velocity in C1 is -0.8c.
Nefertiti is still on Earth so her position in frame C1, xC1(N) = -4.33 (length-contracted: we know this from A2-11) and her velocity in C1 continues to be -0.5c.

This is summarised in this picture:

The diagram highlights that on this second leg, both Nefertiti and Cormorant are moving in the negative direction in frame C1. Nefertiti has a head start but is moving more slowly, so at some point Cormorant will catch her up. This will be the point where Cormorant has returned to Earth.

To find the time (as measured in frame C1) it takes for this little drama to play out, all you need to do is write down expressions for the distance covered by Nefertiti and Cormorant, as a function of time in this second leg (all in the same frame of rerefence, so no relativistic corrections needed). Then figure out the time at which these distances become equal.

Do this.

③ So now we know the time that will elapse in frame C1 during this second leg of the journey but neither of our protagonists is stationary in this frame. Anyone who happened to be stationary in frame C1 would be a moving observer of Nefertiti (stationary in N) and of Cormorant (stationary in C2), so the time experienced these two would actually experience would be dilated:

Recall from section 2-5, that the time dilation relationship is:

where t1 is the time experienced by the moving observer.

Use this to figure out the time that Cormorant and Nefertiti will each actually experience during this return leg. You can then complete the table.

④ Just for reassurance, to make sure this all ties together properly, we should verify the position where Nefertiti and Cormorant are, in Nefertiti’s frame of reference, when they are reunited.

We know, from ②, that both Nefertiti and Cormorant experienced a time of 14.43 minutes, in frame C1, during the return leg of the journey. We also know that at the start of this leg, their positions were xC1(C) = 0 and xC1(N) = -4.33. The velocities in C1 will be -0.5c for Nefertiti and -0.8c for Cormorant.

First, use this information to the position reached by the two, in frame C2, at the end of their return journey.

Now, since C1 is moving relative to frame N, this distance will be contracted in Nefertiti’s frame of reference. Use the usual length contraction formula to find this contracted distance and hence see if, as she should be, Nefertiti is still at her original position in frame N.

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