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This seems like a straightfoward question but I cannot find the answer anywhere.

I need to implement the Hessian matrix of a real scalar function f (an Hamiltonian, to be specific) in spherical coordinates (1,$\theta$,$\phi$) on the unit sphere and evaluate it at a turning point where the first derivatives vanish ($\frac{\partial f}{\partial \theta} =\frac{\partial f}{\partial \phi}=0 $).

My first try was $$ H= \begin{pmatrix} \frac{\partial^2 f}{\partial \theta^2} & \frac{\partial^2 f}{\partial \theta \partial \phi} \\ \frac{\partial^2 f}{\partial \theta \partial \phi} & \frac{\partial^2 f}{\partial \phi^2} \end{pmatrix}, $$ which is what I found in many (physics) papers where $\theta$ is conveniently set to $\frac{\pi}{2}$.

However, what I observed numerically is that because the elementary displacement on the sphere $\delta \vec{m}$ caused by a variation d$\phi$ depends on $\theta$, this is valid at the equator and then gradually breaks down closer to the poles because the $\frac{\partial^2 f}{\partial \phi^2}$ and $\frac{\partial^2 f}{\partial \theta \partial \phi}$ terms tend to $0$ when they shouldn't.

This problem is corrected in the spherical gradient by the $\frac{1}{\sin \theta}$ factor on the $\hat{e}_{\phi}$ component, ie: $$ \nabla f = \frac{\partial f}{\partial \theta}\hat{e}_{\theta}+\frac{1}{sin \theta} \frac{\partial f}{\partial \phi}\hat{e}_{\phi}.$$

From various sources (incl. wikipedia), the Hessian is sometimes defined as the Jacobian of the gradient, which in spherical coordinates would be $H=J(\nabla f)= \big[ \frac{\partial \nabla f}{\partial \theta} ,\frac{\partial \nabla f}{\partial \phi} \big]$. If I apply this at a turning point where the first derivatives vanish, I end up with $$ H= \begin{pmatrix} \frac{\partial^2 f}{\partial \theta^2} & \frac{\partial^2 f}{\partial \theta \partial \phi} \\ \frac{1}{\sin \theta} \frac{\partial^2 f}{\partial \theta \partial \phi} & \frac{1}{\sin \theta} \frac{\partial^2 f}{\partial \phi^2} \end{pmatrix}. $$

This is not symmetric anymore, which bothers me because in every book I looked, it says the Hessian is symmetric as long as the mix derivatives commute.

Does this mean this definition does not apply in spherical coordinates? Is there a more general definition I can use? Additionally, I found this post which defines $H=\nabla \nabla^T$. This yields

$$ H= \begin{pmatrix} \frac{\partial^2 f}{\partial \theta^2} & \frac{1}{\sin \theta} \frac{\partial^2 f}{\partial \theta \partial \phi} \\ \frac{1}{\sin \theta} \frac{\partial^2 f}{\partial \theta \partial \phi} & \frac{1}{\sin^2 \theta} \frac{\partial^2 f}{\partial \phi^2} \end{pmatrix} $$

which is a lot more intuitive, but it clashes with the previous Jacobian of the gradient definition

Any help/clarifications or suggestions greatly appreciated.

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merrinee
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1 Answers1

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It sounds like you want the Hessian of a Riemannian manifold. (If you just want the Hessian in spherical coordinates in $\mathbb{R}^3$, then you should just use the chain rule on the regular Hessian).

In this case, the Hessian is written: $$ \mathcal{H}[f] = (\partial_{ij}f-\Gamma^k_{ij}\partial_kf)[dx^i\otimes dx^j] $$ In local coordinates, the unit sphere is given by: $$ \Psi(\theta,\phi)=(\cos(\theta)\sin(\phi),\sin(\theta)\sin(\phi),\cos(\phi)) $$ So the metric tensor is written: $$ g=\begin{bmatrix}\sin^2(\phi) & 0 \\ 0 & 1\end{bmatrix} $$ with Christoffel symbols given by (e.g. see here): \begin{align} \Gamma_{11}^1 &= \Gamma_{22}^1 = \Gamma_{12}^2 = \Gamma_{22}^2=0\\[2mm] \Gamma_{12}^1 &= \frac{\cos(\phi)}{\sin(\phi)}\\[1mm] \Gamma_{11}^2 &= -\sin(\phi)\cos(\phi) \end{align} which gives me the result: $$ \mathcal{H}[f] = \begin{bmatrix} \partial_{\theta\theta}f + \cos(\phi)\sin(\phi)\partial_\phi f & \;\partial_{\theta\phi}f-\dfrac{\cos(\phi)}{\sin(\phi)}\partial_\theta f\\ \partial_{\theta\phi}f-\dfrac{\cos(\phi)}{\sin(\phi)}\partial_\theta f & \;\partial_{\phi\phi}f \end{bmatrix} $$ Apologies if there are any computation errors. See also: [1], [2].

One check is that the Laplace-Beltrami operator is the trace of the Hessian, which is written as $$ \text{tr}(\mathcal{H}[f]) = g^{ij}\mathcal{H}[f]_{ij} = \frac{1}{\sin^2(\phi)}\left[ \partial_{\theta\theta}f+\sin(\phi)\cos(\phi)\partial_\phi f \right] + \partial_{\phi\phi}f $$ From here, it is given by: $$ \Delta_g f = (\sin(\phi))^{-1}{\partial}_{\phi}\left( \sin(\phi)\partial_\phi f \right) + (\sin(\phi))^{-2}\partial_{\theta\theta}f $$ which corroborates it.

user3658307
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    Looks OK. I also agree on mentioning Riemannian manifolds and covariant derivatives. Might I add that on an arbitrary manifold, without a Riemannian metric, the Hessian only makes sense at critical points. In that case, if $f\colon M\to \mathbb R$ and $df(p)=0$, then the Hessian of $f$ at $p$ is defined as the bilinear operator $Hf(X_p, Y_p)=X_p(Y(f))$. This definition is used in Morse theory. – Giuseppe Negro Jul 31 '17 at 15:54
  • @GiuseppeNegro Thanks for the confirmation and the additional information. – user3658307 Jul 31 '17 at 15:58
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    Note that this answer switches from physics coordinate conventions (used in the question) to math, i.e., interchanges $\theta \leftrightarrow \phi$. – Jess Riedel May 04 '18 at 18:05
  • Also note that mathematicians and physicists both take $\phi \in [0, \pi]$, whereas cartographers take $\phi$ to run from $90^{\circ}$ north to $90^{\circ}$ south, so a cartographer would take the $z$-coordinate to be $\sin \phi$ and make other changes accordingly. – David E Speyer Aug 24 '22 at 13:37