- #1
SiennaTheGr8
- 493
- 193
- TL;DR Summary
- In index notation, the curl can be expressed in a way where the quotient law would seem to "fail." There must be a subtlety that I'm missing.
If I'm not mistaken, the curl can be expressed like this in index notation:
##(\nabla \times \vec v)^i = \epsilon^{i j k} \nabla_j v_k = \epsilon^{i j k} (\partial_j v_k - \Gamma^m_{j k} v_m) = \epsilon^{i j k} \partial_j v_k ##
(where the last equality is because ##\epsilon^{i j k}\Gamma^m_{j k}## is both symmetric and anti-symmetric in ##j## and ##k##). But now with ##(\nabla \times \vec v)^i = \epsilon^{i j k} \partial_j v_k##, doesn't the quotient law say that ##\partial_j v_k## is a tensor, even though it obviously isn't? Clearly I'm wrong. Is there a subtlety to the quotient law that I'm missing?
##(\nabla \times \vec v)^i = \epsilon^{i j k} \nabla_j v_k = \epsilon^{i j k} (\partial_j v_k - \Gamma^m_{j k} v_m) = \epsilon^{i j k} \partial_j v_k ##
(where the last equality is because ##\epsilon^{i j k}\Gamma^m_{j k}## is both symmetric and anti-symmetric in ##j## and ##k##). But now with ##(\nabla \times \vec v)^i = \epsilon^{i j k} \partial_j v_k##, doesn't the quotient law say that ##\partial_j v_k## is a tensor, even though it obviously isn't? Clearly I'm wrong. Is there a subtlety to the quotient law that I'm missing?