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According to Frankel's book "The Geometry of Physics", the components of a contravariant gradient vector can be obtained from the inverse of the metric tensor as follows (in section 2.1d, Page 73):

$$ (\nabla f)^i = \sum_j g^{ij} \frac{\partial f}{\partial x^j}, $$

while the metric sensor is:

$$ g_{ij} = \bigg\langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j} \bigg\rangle. $$

Take the spherical coordinate as example,

\begin{align} x &= r\sin\theta\cos\phi \\ y &= r\sin\theta\sin\phi \\ z &= r\cos\theta \end{align}

and let $J$ be the Jacobi matrix

$$ J = \begin{pmatrix} \sin\theta\cos\phi & -r\sin\phi\sin\theta & r\cos\phi\cos\theta \\ \sin\theta\sin\phi & r\cos\phi\sin\theta & r\sin\phi\cos\theta \\ \cos\theta & 0 & -r\sin\theta \end{pmatrix}, $$

the metric tensor can be obtained as

$$ \left( g_{ij} \right) = J^TJ = \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2\sin^2\theta & 0 \\ 0 & 0 & r^2 \end{pmatrix}. $$

Finally, the contravariant gradient vector can be obtained as follows according to the first equation:

$$ \nabla f = \begin{pmatrix}\displaystyle{ \frac{\partial f}{\partial r} \\ \frac{1}{r^2\sin^2\theta} \frac{\partial f}{\partial \phi} \\ \frac{1}{r^2} \frac{\partial f}{\partial \theta}} \end{pmatrix}. $$

However, the correct answer is

$$ \nabla f = \begin{pmatrix}\displaystyle{ \frac{\partial f}{\partial r} \\ \frac{1}{r\sin\theta} \frac{\partial f}{\partial \phi} \\ \frac{1}{r} \frac{\partial f}{\partial \theta}} \end{pmatrix}. $$

I don't know why I cannot get the correct answer using metric tensor. Could you please help me figure it out? Thank you!

TJH
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1 Answers1

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In your first equation for $\nabla f$, $$ \nabla f = \Big( \frac{\partial f}{\partial r}, \frac{1}{r^2 \sin^2 \theta} \frac{\partial f}{\partial \phi}, \frac{1}{r^2} \frac{\partial f}{\partial \theta} \Big), $$ what you've written down is an expression for $\nabla f$ with respect to the coordinate vectors $\tfrac{\partial}{\partial r}, \tfrac{\partial}{\partial \phi}, \tfrac{\partial}{\partial \theta}$. That is, your equation means $$ \nabla f = \frac{\partial f}{\partial r} \frac{\partial}{\partial r} + \frac{1}{r^2 \sin^2 \theta} \frac{\partial f}{\partial \phi} \frac{\partial}{\partial \phi} + \frac{1}{r^2} \frac{\partial f}{\partial \theta} \frac{\partial}{\partial \theta}. $$ Note that these coordinate vectors are not unit vectors; if we use $\boldsymbol{\hat{r}}, \boldsymbol{\hat{\phi}}, \boldsymbol{\hat{\theta}}$ to denote the associated unit vectors, then $$ \boldsymbol{\hat{r}} = \frac{\partial}{\partial r}, \qquad \boldsymbol{\hat{\phi}} = \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi}, \qquad \boldsymbol{\hat{\theta}} = \frac{1}{r} \frac{\partial}{\partial \theta}. $$ Your second equation (the standard one written in physics) is written in terms of these unit vectors; it states $$ \nabla f = \frac{\partial f}{\partial r} \boldsymbol{\hat{r}} + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \boldsymbol{\hat{\phi}} + \frac{1}{r} \frac{\partial f}{\partial \theta} \boldsymbol{\hat{\theta}}. $$ It should now be clear that there is no inconsistency between these equations.

mollyerin
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