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In differential geometry, the usual curl operation that we are familiar with from elementary calculus is generalized to [itex] \,^*dA [/itex] (where A is a one-form). In three-dimensions, this gives back a one-form.
Now, the components of this one-form are [itex] \sqrt{g} \epsilon_{ijk} \partial^j A^k [/itex].
however, the corresponding contravariant components are [itex] \frac{1}{\sqrt{g}} \epsilon^{ijk} \partial_j A_k [/itex].
Now, to obtain the formula that we learned in elementary calculus, it is the second form that must be used. Why is that the case?
On the other hand, if one looks at the generalization of the gradient, it's the formula for the covariant components of [itex] d \phi = \partial_i \phi [/itex] that one must use to get the usual formula we have learned for the gradient.
So what is the rationale behind choosing one form over another? Maybe one must pick the form that differentiates with respect to the [itex]x^i [/itex] so that one must have the index on the partial derivative downstairs?
I am sure there is something fundamental going on here that I am obviosuly completely missing.
Thanks
Patrick
Now, the components of this one-form are [itex] \sqrt{g} \epsilon_{ijk} \partial^j A^k [/itex].
however, the corresponding contravariant components are [itex] \frac{1}{\sqrt{g}} \epsilon^{ijk} \partial_j A_k [/itex].
Now, to obtain the formula that we learned in elementary calculus, it is the second form that must be used. Why is that the case?
On the other hand, if one looks at the generalization of the gradient, it's the formula for the covariant components of [itex] d \phi = \partial_i \phi [/itex] that one must use to get the usual formula we have learned for the gradient.
So what is the rationale behind choosing one form over another? Maybe one must pick the form that differentiates with respect to the [itex]x^i [/itex] so that one must have the index on the partial derivative downstairs?
I am sure there is something fundamental going on here that I am obviosuly completely missing.
Thanks
Patrick
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