einstein's little train that couldn't
swansont said:
Perhaps it was in response to
swansont: I though you would like to try this one on for size.
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All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.
At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.
To demonstrate the following:
Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.
Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.
At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.
The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1
vtx = -vtx - vt1 – vt1
2tx = -2t1
tx = -t1
Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.