I'm trying to find the normal unit vector at each point of the sphere $x^2+y^2+z^2 = a^2$ using cartesian and spherical unit vectors, as shown below. I was able to get it using cartesian unit vectors. My problem is with the spherical case.

Cartesian Unit Vectors

For $F(x,y,z)=0$, we have $\hat n=\dfrac{\nabla F}{\|\nabla F\|}$, so:

$$ F(x,y,z) = x^2 + y^2 + z^2 - a^2 \implies \begin{cases} \nabla F=(2x,2y,2z) \\[2px] \|\nabla F\| =2 \sqrt{x^2+y^2+z^2} \end{cases} \\[40px] \bbox[5px,border:1px solid black] { \hat n =\frac{x\, \hat i+ y \, \hat j + z \, \hat k}{\sqrt{x^2+y^2+z^2}}} $$

Spherical Unit Vectors

$$ \begin{cases} x = r\sin \theta \cos \phi \\y = r\sin \theta \sin \phi \\z = r\cos \phi \end{cases} \implies \begin{cases} F(r,\theta,\phi)=r^2-a^2 \\ \theta \in [0,\pi] \\ \phi \in [0,2\pi) \\ \end{cases} \implies \begin{cases} \nabla F=(2r,0,0) \\[2px] \|\nabla F\| =2r \end{cases} \\[40px] \bbox[5px,border:1px solid black] { \hat n =\hat r + 0 \, \hat \theta + 0\, \hat \phi = \hat r } $$

The problem is that the expression with spherical unit vectors does not take into account the coordinates of the point. In other words, $\hat n=(1,0,0)$ for every $(r,\theta,\phi)$. So, my second approach was calculate it via parametrization of the sphere.

Using Parametric Equation (Cartesian)

Below, I used spherical coordinates to write the parametrization, but I'm still using $\hat i$, $\hat j$ and $\hat k$. $$ \begin{alignat}{1} S(\theta,\phi) &=(a\sin\theta\cos\phi,a\sin\theta\sin\phi,a\cos\theta) = a\sin\theta\cos\phi \, \hat i + a\sin\theta\sin\phi \, \hat j + a\cos\theta \, \hat k \end{alignat} $$ $$ \\[30px] \hat n = \frac{\dfrac{\partial S}{\partial \theta} \times \dfrac{\partial S}{\partial \phi}} {\left\| \dfrac{\partial S}{\partial \theta} \times \dfrac{\partial S}{\partial \phi} \right\|}= (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \\[50px] \bbox[5px,border:1px solid black] { \hat n = \sin\theta\cos\phi \, \hat i + \sin\theta\sin\phi \, \hat j + \cos\theta \, \hat k} $$

Using Parametric Equation (Spherical)

Assuming $m \in [0,\pi]$ and $n \in [0,2\pi)$: $$ X(m,n)=(a,m,n)=a\,\hat r + m \, \hat \theta + n \, \hat \phi $$

$$ \begin{array}{c|c|c} \dfrac{\partial X}{\partial m}(m,n)=(0,1,0) \quad & \quad \dfrac{\partial X}{\partial n}(m,n)=(0,0,1) \quad & \quad \dfrac{\partial X}{\partial m} \times \dfrac{\partial X}{\partial n} = (1,0,0) \end{array} $$

$$ \hat n = \frac{\dfrac{\partial X}{\partial m} \times \dfrac{\partial X}{\partial n}} {\left\| \dfrac{\partial X}{\partial m} \times \dfrac{\partial X}{\partial n} \right\|}= (1,0,0) \\[50px] \bbox[5px,border:1px solid black] { \hat n =\hat r + 0 \, \hat \theta + 0\, \hat \phi = \hat r} $$

The same problem happens here.

My questions are:

  1. What is the normal unit vector of the sphere using $\hat r$, $\hat \theta$ and $\hat \phi$? I was expecting something similar to $\hat n= \dfrac{(1,\theta,\phi)}{\|(1,\theta,\phi)\|}$, i.e, for a given point $(r,\theta,\phi)$ the unit normal has components in $\hat \theta$ and $\hat \phi$ directions that take into account the position of the point on the sphere.

  2. What was the mistake/misconception that led me to $\hat n = \hat r$ instead of an expression that also contains $\hat \theta$ and $\hat \phi$?


1 Answer 1


There's nothing wrong with your reasoning. As it is explained here, the new spherical coordinates are dependent on the position. This becomes obvious when you write down $\hat{r}$ in cartesian coordinates: $$\hat{r} = \sin\theta\cos\phi \hat{x} + \sin\theta\sin\phi \hat{y} + \cos\theta \hat{z}$$ Thus, to each pair $(\theta,\phi)$ you have a different versor $\hat{r}$, which has norm ne and points outwards the sphere. This explains why $\hat{n}=\hat{r}$ in sphereical coordinates.

  • $\begingroup$ I think I got it, thanks! The $(1,0,0)$ was leading me to think that my calculations were wrong, because I was thinking that this was telling me that the normal was fixed at the position $r = 1$ and $\theta = \phi = 0$, despite the changes in $\theta$ and $\phi$. $\endgroup$ Sep 23, 2020 at 14:58

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