The gradient of a scalar function

Question

I found this definition of gradient of scalar function $\Phi$:

$\nabla \Phi = (g^{ij}\partial_{j}) \vec{g_{i}}$

And I know:

Metric tensor of spherical coordinates

$g_{11} = 1$

$g_{22} = r^2$

$g_{33} = r^2sin\theta$

But I do not know how to use first definition. How can I find gradient of scalar function in spherical coordinates by using metric tensor

Answer 1

You need the inverse of the tensor. Since it is diagonal, you have $g^{11}=1$, $g^{22}=1/r^2$ and $g^{33}=1/r^2\sin^2\theta$ (I think there is a typo in your last component).

The first component of the gradient of $\Phi$ would be $$ g^{11}\partial\Phi/\partial r+g^{12}\partial\Phi/\partial \theta+g^{13}\partial\Phi/\partial \phi=\partial\Phi/\partial r. $$ since the off-diagonal elements of the metric tensor are zero. The second component would be $$ g^{21}\partial\Phi/\partial r+g^{22}\partial\Phi/\partial \theta+g^{23}\partial\Phi/\partial \phi=\frac{1}{r^2}\partial\Phi/\partial \theta. $$ and the third $$ g^{31}\partial\Phi/\partial r+g^{32}\partial\Phi/\partial \theta+g^{33}\partial\Phi/\partial \phi=\frac{1}{r^2\sin^2\theta}\partial\Phi/\partial \phi. $$ Observe that these are the components in the base $\{g_i\}$, which is orthogonal but not orthonormal. If you use unit vectors you get a slightly different expression.