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A spaceship moving at .866c compared to a pair of stars. The stars are located at points A and B which appear to be 2 light seconds apart from an observer stationary to the stars at point M (the midpoint of the two stars). The spaceship is flying strait from one star to the other.

Therefore the spaceship will see the two stars separated by a distance of:

L = L0 (1 – (v^2 / c^2))^.5

L = 2ls (1 – ((.866c)^2 / c^2))^.5

L = 2ls * .5

L = 1 light second

Now, at the instant the spaceship is at point M Star A and Star B go nova simultaneously. When will: (a) the spaceship see the stars going nova? (b) the man at point M see the stars go nova?

(a) Since the space ship is at point M when this occurs Star A .5 light seconds away from his position, by definition of midpoint. Now because it is a known fact that light travels at the velocity of c in all reference frames it must be concluded that the spaceship will see the nova from Star A .5 seconds later. The exact same logic and results can be achieved for Star B which also yields a result of .5 seconds later.

(b) By similar logic the stationary man at point M will see the novas of Star A and Star B 1 second after the spaceship passes point M.

Time dilation can also be used to solve this problem. It is possible to find how long it will take the spaceship to see the novas by using the time dilation equation from the stationary mans frame. This can make more since because it is more intuitive that the stationary man will see the novas after 1 second.

t0 = t (1 – (v^2 / c^2 )).^5

t0 = 1s (1 – ((.866c)^2 / c^2))^.5

t0 = 1s * .5

t0 = .5s

As expected the solutions match those found when using length contraction.