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MathematicalPhysicist
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A rod. having slope m relative to the x-axis of S, moves in the x direction at speed u. what is the rod's slope in the usual second frame S'? (S is at rest realtive to S' which moves along the x direction with velocity v).
well obviously the horizontal length of the rod is contracted or lengthened, depends on your frame:
i think that if L is the length of the rod, and [tex]u'=\frac{u-v}{1-\frac{uv}{c^2}}[/tex] then we have: L'_x=Lcos(theta)/gamma(u')
and the slope in S' is: m'=m*gamma(u') cause the vertical portion of the rod doesn't get change.
the problem is that in the answer key we have:
m'=m*gamma(v)*(1-uv/c^2)
but i don't get this, even with some algebraic manipulations, so i guess I am wrong here, can someone help here?
thanks in advance.
well obviously the horizontal length of the rod is contracted or lengthened, depends on your frame:
i think that if L is the length of the rod, and [tex]u'=\frac{u-v}{1-\frac{uv}{c^2}}[/tex] then we have: L'_x=Lcos(theta)/gamma(u')
and the slope in S' is: m'=m*gamma(u') cause the vertical portion of the rod doesn't get change.
the problem is that in the answer key we have:
m'=m*gamma(v)*(1-uv/c^2)
but i don't get this, even with some algebraic manipulations, so i guess I am wrong here, can someone help here?
thanks in advance.