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MathematicalPhysicist
Sep15-11, 02:24 AM
I have next question, we have in the next printcreen the equation:
\frac{dU^b}{d\tau}=U^a \nabla_a U^b
why is this right?
(the printscreen is from the book of Woodhouse in GR, page 34)

Thanks.

MathematicalPhysicist
Sep15-11, 06:41 AM
Ok, I think I can see why this is right, we pick a frame where U^a=(1,0,0,0).

Clumsy me.
:-)