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Dec 15, 2022 at 3:00 history tweeted twitter.com/StackPhysics/status/1603223419016847362
Dec 14, 2022 at 23:55 comment added WillO Draw the picture!
Dec 14, 2022 at 23:54 answer added Eric Smith timeline score: 2
Dec 14, 2022 at 22:08 comment added HTNW I believe the correct interpretation of this statement is: "if you see some object/signal/whatever traveling at $v>c$ in some frame, then you can make a Lorentz transformation to a frame where that object arrived at its destination before leaving its source." It doesn't make sense to talk about a "Lorentz transform with $v\ge c$" because such things don't exist.
Dec 14, 2022 at 22:07 answer added kricheli timeline score: 4
Dec 14, 2022 at 21:45 comment added James @user1551817 i'm a bit uneasy too by arguments by contradiction like "if this happens then reality as we know it would unravel!!"... it's rather more soothing to have an implicit "you can't do it no matter how cunning you are" type of argument i guess?
Dec 14, 2022 at 21:40 comment added user1551817 @James Yes I thought about that too, but I assumed that wasn't quite right as it doesn't explicitly show that t2 < t1.
Dec 14, 2022 at 21:39 comment added user1551817 @TobiasFünke just some university lecture notes. There is no extra information given other than what I have provided. It is just a "note:" in the text.
Dec 14, 2022 at 21:33 comment added James the Lorentz factor $\gamma$ blows up at v=c, for v>c it becomes imaginary, so t′ has become incalculable?
Dec 14, 2022 at 21:20 history edited Qmechanic CC BY-SA 4.0
edited tags; edited title
Dec 14, 2022 at 21:14 comment added Tobias Fünke Which book, author, page...?
Dec 14, 2022 at 21:05 history asked user1551817 CC BY-SA 4.0