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Main Question or Discussion Point
The Lorentz transformation are given by (see the attachment)
x'=(xvt)/√(1v^2/c^2 )
t'=(tvx/c^2)/√(1v^2/c^2 )
Let us transform the event (10^100 m,1sec) in the xframe 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'=(xvt)/√(1v^2/c^2 )
t'=(tvx/c^2)/√(1v^2/c^2 )
Let us transform the event (10^100 m,1sec) in the xframe 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.
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