[Math] Gradient of a real valued function defined on a sphere

vector analysis

Points on a sphere of radius $R$ is expressed in spherical coordinates as $\left(\varphi,\theta\right)$.

For a real valued, continuous and differentiable function $f:\mathbb R\times\left[0,\pi\right] \to \mathbb R$ evaluated on said sphere centered at $\left(0,0,0\right)$ with radius $R$, what is its gradient written in the form of angles (radians) in the azimuthal and zenith directions?

Additional notes and assumptions:

  • $f\left(\varphi+2\pi,\theta\right)=f\left(\varphi,\theta\right)$;
  • $f\left(\varphi,0\right) = f\left(0,0\right)$, $f\left(\varphi,\pi\right) = f\left(0,\pi\right)$;
  • $\varphi\in\mathbb R$ is the azimuthal angle and
    $\theta\in\left[0,\pi\right]$ is the zenith angle.

By design, the direction of the gradient has to be parallel to the tangent plane at any position, and points in the direction of greatest increase of $f$. Meanwhile, the magnitude of the gradient is the "slope" of $f$ in said direction.

I'm aware that the gradient operator of $\mathbb R^3$ in the spherical coordinate system can be written as a linear combination of the basis vectors, namely
$$
\nabla =
\mathbf{e}_{r} \frac{\partial}{\partial r}
+ \mathbf{e}_{\varphi} \frac{1}{r\sin\theta} \frac{\partial}{\partial \varphi}
+ \mathbf{e}_{\theta} \frac{1}{r} \frac{\partial}{\partial \theta}.
$$

Since $f$ is only mapped from points on the sphere, I removed the term with $\mathbf{e}_{r}$ to arrive at
$$
\nabla f =
\mathbf{e}_{\varphi} \frac{1}{R\sin\theta}\frac{\partial f}{\partial\varphi}
+ \mathbf{e}_{\theta} \frac{1}{R} \frac{\partial f}{\partial\theta},
$$

from where I naively translate the "arc lengths" into their respective "angles" by
\begin{align}
\frac{1}{R \sin \theta} \frac{\partial f}{\partial\varphi}
\to
\frac{1}{R^2\sin^2\theta} \frac{\partial f}{\partial\varphi},
\quad &
\text{because the circle of latitude has radius of }R\sin\theta,
\text{and} \\
\frac{1}{R} \frac{\partial f}{\partial\theta}
\to
\frac{1}{R^2} \frac{\partial f}{\partial\theta},
\quad &
\text{because the circle of longitude has radius of }R.
\end{align}

Therefore, I believe that the gradient described in azimuthal and zenith angles is the following pair:
$$
\frac{1}{R^2} \left(
\frac{1}{\sin^2\theta} \frac{\partial f}{\partial\varphi},
\frac{\partial f}{\partial\theta}
\right).
$$

My questions would be:

  1. Is there a rigorous formulation for this kind of gradient operator that I'm looking for?

  2. What about gradient at the poles? I'm asking because $\sin\theta$ is obviously $0$ when $\theta=0,\pi$.

Best Answer

Rigorous formulation

This involves the definition of the surface gradient operator, which is defined as $$ \nabla_{\Gamma} = \nabla - {\mathbf e}_{r} \left({\mathbf e}_{r} \cdot \nabla\right). $$

The projection of the gradient along the unit normal ${\mathbf e}_{r}$ is evaluated by $$ {\mathbf e}_{r} \left({\mathbf e}_{r} \cdot \nabla f\right) = {\mathbf e}_{r} \left(\frac{\partial f}{\partial r} {\mathbf e}_{r}^2\right) = {\mathbf e}_{r} \frac{\partial f}{\partial r} $$ which when subtracted from $\nabla f$ gives (at $r=R$) $$ {\nabla}_{\Gamma} f = \mathbf{e}_{\varphi} \frac{1}{R\sin\theta}\frac{\partial f}{\partial\varphi} + \mathbf{e}_{\theta} \frac{1}{R} \frac{\partial f}{\partial\theta}. $$

Angular gradient field

Because of $\mathbf{e}_{\theta}\cdot\mathbf{e}_{\varphi} = 0$ and that the longitudinal rotation and the latitudinal rotations are independent of each other, the component of ${\nabla_{\Gamma}}f$ in the direction of $\mathbf{e}_{\theta}$ and $\mathbf{e}_{\varphi}$ can be individually resolved into angular components.

Polar regions

The original extent for asking the case of $\sin\theta=0$ is for numerical purposes. However, it should be noted that the gradient operator cannot be defined on $\sin\theta=0$ due to the choice of coordinate system and boundary conditions.

Related Question