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