Relativity Crib Sheet 2006

Relativity Crib Sheet 2006 - 1 Physics 200a Relativity crib...

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1 Physics 200a Relativity crib sheet Shankar (2006) Here are some very basic things you should know to do this problem set. X = ( ct,x ) = ( x 0 ,x 1 ) Under a LT x 0 1 = x 1 - βx 0 1 - β 2 (1) x 0 0 = x 0 - βx 1 1 - β 2 , (2) x 0 2 = x 2 (3) x 0 3 = x 3 (4) where β = u c . Let us forget x 2 ,x 3 . Note that we use x 0 = ct and not just t since x 0 has the same units as x and the LT looks nice and symmetric. You can check that X · X X 2 = x 2 0 - x 2 1 = x 0 2 0 - x 0 2 1 = s 2 is the same for all observers, an invariant. A particle of mass m and velocity v has energy E and momentum p given by E = mc 2 q 1 - v 2 /c 2 p = mv q 1 - v 2 /c 2 (5) The energy-momentum vector is P = ( P 0 ,P 1 ) = ( E c ,p ) It is a pity we need to bring in E c and not E but this is to make sure both components have the same units (recall x 0 = ct ) and LT has the same form as for components of X : P 0 1 = P 1 - βP 0 1 - β 2 (6) P 0 0 = P 0 - βP 1 1 - β 2 , (7)
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2 It follows P 2 = P 2 0 - P 2 1 = P 0 2 0 - P 0 2 1 is an invariant. What is this invariant value? You can show by explicit
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Relativity Crib Sheet 2006 - 1 Physics 200a Relativity crib...

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