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Given that the Cartesian coordinates can be parametrized for spherical coordinates by

$$\left \{ \begin{array}{l}x = r\sin(\theta)\cos(\phi)\\y = r\sin(\theta)\sin(\phi)\\ z = r\cos(\theta) \end{array}\right.$$

We have that

$$\left \{ \begin{array}{l}\hat{r} = \sin(\theta)\cos(\phi)\hat{x} + \sin(\theta)\sin(\phi)\hat{y} + \cos(\theta)\hat{z}\\\hat{\theta} = \cos(\theta)\cos(\phi)\hat{x}+\cos(\theta)\sin(\phi)\hat{y}-\sin(\theta)\hat{z}\\ \hat{\phi} = -\sin(\phi)\hat{x}+\cos(\phi)\hat{y} \end{array}\right.$$

Then if we note that

$$\nabla_r := \frac{\partial}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial}{\partial \theta}\hat{\theta} + \frac{1}{r\sin(\theta)}\frac{\partial}{\partial \phi}\hat{\phi}$$

Then we have that

$$\nabla_r x = \hat{x}$$

My question is: Why the above equation is true? How can I derive the second system showed ? Why is true that, for exemple, $\partial \hat{r}/\partial \theta = \hat{\theta}$?

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1 Answer 1

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I'll use $\rho$ instead of $r$, to make things clearer:

In spherical coordinates, we have $\mathbf{r} = \left< \rho\sin\theta\cos\phi, \rho\sin\theta\cos\phi, \rho\cos\theta \right>$.

Let's say we change $\rho$: this induces a change in $\mathbf{r}$ in the "$\rho$-direction." The best way to describe this rate of change with derivatives, so we have

$$\mathbf{\hat{\rho}} = \frac{\frac{\partial\mathbf{r}}{\partial\rho}}{\left\Vert \frac{\partial\mathbf{r}}{\partial\rho} \right\Vert} = \left< \sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta \right>$$

We can do the same thing with the other parameters, and you should end up with the second set of equations.

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