Gradient in polar coords using tensors

[ohmymymymygod]

Using tensors, I'm supposed to find the usual formula for the gradient in the covariant basis and in polar coordinates. The formula is \vec{grad}=[\frac{\partial}{\partial r}]\vec{e_{r}}+\frac{1}{r}[\frac{\partial}{\partial \vartheta}]\vec{e_{\vartheta}} where \vec{e_{r}} and \vec{e_{\vartheta}} are the covariant basis vectors.

In the contravariant basis with \vec{e^{r}} and \vec{e^{\vartheta}} , we know that \vec{grad}=[\frac{\partial}{\partial x^{i}}] \vec{e^{i}}. But from index gymnastics, \vec{e^{i}}=g^{ij}\vec{e_{j}}. So \vec{grad}=[\frac{\partial}{\partial x^{i}}]g^{ij}\vec{e_{j}}.

In polar coordinates, the inverse metric tensor is g^{11} = 1, g^{12}=g^{21}=0, g^{22} = \frac{1}{r^{2}}.

So this gives \vec{grad}=[\frac{\partial}{\partial r}]\vec{e_{r}}+\frac{1}{r^{2}}[\frac{ \partial}{\partial \vartheta}]\vec{e_{\vartheta}}. And lo and behold, this is wrong.

 

[oli4]

Hi,
the gymnastics you are talking about allow you to raise or lower indices of a tensor, but in this case what you want to do is express the gradient in different coordinates which is a different matter.

 

[ohmymymymygod]

I'm not using the raising/lowering indices operations to switch from rectangular to polar coordinates, I'm using them to switch, in polar coordinates, from a contravariant basis to a covariant basis.

Actually, I think I know what I got wrong. The correct formula is \vec{grad}=[\frac{\partial}{\partial r}]\hat{r}+\frac{1}{r}[\frac{\partial}{\partial \vartheta}]\hat{\vartheta}. The vectors \hat{r} and \hat{\vartheta} are unit vectors, while \vec{e_{r}} and \vec{e_{\vartheta}} aren't unit vectors. But I can write \vec{e_{j}}=\hat{u_{j}}||\vec{e_{j}}||=\hat{u_{j}}\sqrt{g_{jj}}, with \hat{u_{j}} a unit vector.

Then the formula for the gradient becomes \vec{grad}=[\frac{\partial}{\partial x^{i}}]g^{ij}\sqrt{g_{jj}}\vec{u_{j}} which I think is more right. (The r factor in √g₂₂ compensates the 1/r² in g²² for the correct 1/r.) But the indices don't seem to balance out in the way they are supposed to in the Einstein notation...