
#1
Sep2911, 04:24 PM

P: 39

Is T_{ab}[itex]\partial[/itex]_{c}f = T_{ac}[itex]\partial[/itex]_{b}f, where T is a tensor?
Seems to me like you should be able to do this: T_{ab}[itex]\partial[/itex]_{c}f =T_{ab}[itex]\delta[/itex]^{b}_{c}[itex]\partial[/itex]_{b}f =T_{ac}[itex]\partial[/itex]_{b}f Maybe I'm using the Kronecker delta incorrectly. Could someone check this for me? 



#2
Sep2911, 04:52 PM

C. Spirit
Sci Advisor
Thanks
P: 4,941

You can't introduce the kronecker delta in that way or any summation for that matter. You have the index b on the kronecker delta being summed over two other indices which does not conform to the summation convention and makes no sense in that regard. You would have to write something like [itex]T_{ab}\partial _{c}f = T_{ab}\delta ^{d}_{c}\partial _{d}f[/itex]; you can tell that the free indices are the same on both sides.



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