- #1

aawahab76

- 44

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The Lorentz transformation are given by (see the attachment)

x'=(x-vt)/√(1-v^2/c^2 )

t'=(t-vx/c^2)/√(1-v^2/c^2 )

Let us transform the event (10^100 m,1sec) in the x-frame to the x'-frame that is moving in the usual geometry with the speed v=10^(-10) c. Could you see that that t'≈-10^81 sec!! That is negative, right? Am I incorrect? Please help.

x'=(x-vt)/√(1-v^2/c^2 )

t'=(t-vx/c^2)/√(1-v^2/c^2 )

Let us transform the event (10^100 m,1sec) in the x-frame to the x'-frame that is moving in the usual geometry with the speed v=10^(-10) c. Could you see that that t'≈-10^81 sec!! That is negative, right? Am I incorrect? Please help.