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I am using coordinate notation such that (x,t) is (distance,time)

Alice's frame (S)

Event A: Alice sends a superluminal signal at (0,0)

Event B: Bob receives the superluminal signal at (10,5)

The signal travels 10 lightseconds in 5 seconds (2c) in Alice's frame.

Bob's frame (S')

Event A' = (0,0)

Event B' = (x',t')

Using the Lorentz transformation

[tex]x^{\prime} = \gamma (x - vt)[/tex]

[tex]t^{\prime} = \gamma \left( t - \frac{vx}{c^2} \right)[/tex]

where [tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

Assume Bob is moving at 0.8c relative to Alice and units such that c=1.

[tex]\gamma = \frac{1}{\sqrt{1 - \frac{0.8^2}{1^2}}} = 1.666[/tex]

[tex]x^{\prime} = 1.666 ( 10 - 0.8*5) = 10 [/tex]

[tex]t^{\prime} = 1.666 \left( 5 - \frac{0.8*10}{1^2} \right)= -5[/tex]

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Hi I apprecciated the clear cut presentation of the maths but couldnt help note a certain paradoxical self negation in your final derivation.

It appears to me that if the signal actually went back in time it couldn't possibly arrive at x=10 and conversely if it did arrive at x=10, it by definition was traveling forward in time no matter what the maths say.

There is the negative time interval:minus 5 seconds.