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Gradient in polar coords using tensors

  1. Oct 22, 2012 #1
    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.
     
  2. jcsd
  3. Oct 22, 2012 #2
    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.
     
  4. Oct 22, 2012 #3
    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 [itex]\vec{grad}=[\frac{\partial}{\partial r}]\hat{r}+\frac{1}{r}[\frac{\partial}{\partial \vartheta}]\hat{\vartheta}[/itex]. The vectors [itex]\hat{r}[/itex] and [itex]\hat{\vartheta}[/itex] are unit vectors, while [itex]\vec{e_{r}}[/itex] and [itex]\vec{e_{\vartheta}}[/itex] aren't unit vectors. But I can write [itex]\vec{e_{j}}=\hat{u_{j}}||\vec{e_{j}}||=\hat{u_{j}}\sqrt{g_{jj}}[/itex], with [itex]\hat{u_{j}}[/itex] a unit vector.

    Then the formula for the gradient becomes [itex]\vec{grad}=[\frac{\partial}{\partial x^{i}}]g^{ij}\sqrt{g_{jj}}\vec{u_{j}}[/itex] 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...
     
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