how Albert Einstien calculates t'=t/srqt(1-c^2/t^2)?

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how Albert Einstien calculates t'=t/srqt(1-c^2/t^2)?

2006-06-06 02:37:42 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics

4 answers

I think you have got it wrong it's v^2/c^2 in denominator.
Explaining how he calculated is a bit lengthy.
You can consult any standard book.
Arthur Beiser's book is a good try.

2006-06-06 02:44:17 · answer #1 · answered by santosh k 3 · 0⤊ 0⤋

With patience of course, and lot of determination to spare!

2006-06-06 09:43:28 · answer #2 · answered by Merlin 3 · 0⤊ 0⤋

well..i guess.. thats a question.. you need to ask him..;)

2006-06-06 09:40:31 · answer #3 · answered by Redweedplace ® 3 · 0⤊ 0⤋

he's a genius!!!

2006-06-06 09:39:29 · answer #4 · answered by Ajescent 5 · 0⤊ 0⤋