- #1
ohmymymymygod
- 2
- 0
Using tensors, I'm supposed to find the usual formula for the gradient in the covariant basis and in polar coordinates. The formula is [itex]\vec{grad}=[\frac{\partial}{\partial r}]\vec{e_{r}}+\frac{1}{r}[\frac{\partial}{\partial \vartheta}]\vec{e_{\vartheta}}[/itex] where [itex]\vec{e_{r}}[/itex] and [itex]\vec{e_{\vartheta}}[/itex] are the covariant basis vectors.
In the contravariant basis with [itex]\vec{e^{r}}[/itex] and [itex]\vec{e^{\vartheta}}[/itex] , we know that [itex]\vec{grad}=[\frac{\partial}{\partial x^{i}}] \vec{e^{i}}[/itex]. But from index gymnastics, [itex]\vec{e^{i}}=g^{ij}\vec{e_{j}}[/itex]. So [itex]\vec{grad}=[\frac{\partial}{\partial x^{i}}]g^{ij}\vec{e_{j}}[/itex].
In polar coordinates, the inverse metric tensor is [itex]g^{11} = 1, g^{12}=g^{21}=0, g^{22} = \frac{1}{r^{2}}[/itex].
So this gives [itex]\vec{grad}=[\frac{\partial}{\partial r}]\vec{e_{r}}+\frac{1}{r^{2}}[\frac{ \partial}{\partial \vartheta}]\vec{e_{\vartheta}}[/itex]. And lo and behold, this is wrong.
In the contravariant basis with [itex]\vec{e^{r}}[/itex] and [itex]\vec{e^{\vartheta}}[/itex] , we know that [itex]\vec{grad}=[\frac{\partial}{\partial x^{i}}] \vec{e^{i}}[/itex]. But from index gymnastics, [itex]\vec{e^{i}}=g^{ij}\vec{e_{j}}[/itex]. So [itex]\vec{grad}=[\frac{\partial}{\partial x^{i}}]g^{ij}\vec{e_{j}}[/itex].
In polar coordinates, the inverse metric tensor is [itex]g^{11} = 1, g^{12}=g^{21}=0, g^{22} = \frac{1}{r^{2}}[/itex].
So this gives [itex]\vec{grad}=[\frac{\partial}{\partial r}]\vec{e_{r}}+\frac{1}{r^{2}}[\frac{ \partial}{\partial \vartheta}]\vec{e_{\vartheta}}[/itex]. And lo and behold, this is wrong.