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.
\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.