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Unformatted text preview: Tensor exercise - Solutions Problem 1: Although d I dt 6 = 0 in general, show that d I dt = 0 . Solution: Since I = 3 i =1 I i e i e i , and I i = 0, we have d I dt = X i ( I i e i e i + I i e i e i + I i e i e i ) = X i ( I i e i e i + I i e i e i ) = X i I i ( e i ) e i + I i e i ( e i ) . Now contract the left leg of each term with : d I dt = X i I i [ ( e i )] e i + I i ( e i )( e i ) . Each term in the equation is now just a vector. The first term on the right hand side vanishes [because ( e i ) = 0], but the second term does NOT vanish in general. (In the second term, we are taking the dot product of with e i , not with e i !) So d I dt = X i I i ( e i )( e i ) . Since ( e i ) is just a scalar, and ( e i ) = 0, each term on the right hand side of the above equation is a vector perpendicular to , so the whole expression d...
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This note was uploaded on 01/22/2012 for the course PHYS 3202 taught by Professor Tan during the Spring '11 term at Georgia Institute of Technology.
- Spring '11