Curvature in polar coordinates (two dimensions) is \begin{equation} \kappa = \frac{\rho^2 + 2 \rho'(\theta)^2 -\rho \rho''(\theta)}{\left(\rho^2 + \rho'(\theta)\right)^{\frac{3}{2}}} \end{equation}

when we assume $\frac{\rho'(\theta)}{\rho} \ll1$, above expression is written as \begin{align} \kappa = \frac{1}{\rho} + 2\frac{\rho'(\theta)^2}{\rho^2} - \frac{\rho''(\theta)}{\rho^2} \end{align}

what is the corresponding expression for the mean curvature in three dimensions in spherical coordinates?

Typos EDIT: \begin{equation} \kappa = \frac{\rho^2 + 2 \rho'(\theta)^2 -\rho \rho''(\theta)}{\left(\rho^2 + \rho'(\theta)^2\right)^{\frac{3}{2}}} \end{equation} \begin{align} \kappa = \frac{1}{\rho} - \frac{\rho''(\theta)}{\rho^2} \end{align}

  • $\begingroup$ @user10354138, I thought mean curvature is arithmetic average of local principal radius. Therefore, it is not average over $\theta$. see for example math.stackexchange.com/questions/153371/… $\endgroup$
    – alekhine
    Commented Sep 10, 2021 at 14:33
  • $\begingroup$ @ alekhine: Two typos in the question. $\endgroup$
    – Narasimham
    Commented Sep 10, 2021 at 14:35
  • $\begingroup$ @Narasimham, what are the typos? $\endgroup$
    – alekhine
    Commented Sep 11, 2021 at 1:07
  • $\begingroup$ Edited minor typos. In the last equation imho first order derivative can be also neglected. $\endgroup$
    – Narasimham
    Commented Sep 11, 2021 at 1:37

2 Answers 2


Since the surface is defined by $r=f(\theta,\phi)$, we compute an inward normal (not normalised) $$ \mathbf{N}=\nabla (f(\theta,\phi)-r)=-\mathbf{e}_r+\frac1r f_\theta\,\mathbf{e}_\theta+\frac1{r\sin\theta}f_\phi\,\mathbf{e}_\phi $$ so the mean curvature $H$ is \begin{align*} 2H&=-\nabla\cdot\left(\frac{\mathbf{N}}{\left\lvert\mathbf{N}\right\rvert}\right)\\ &=\frac{\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N})-2\nabla\cdot\mathbf{N}}{2(\mathbf{N}\cdot\mathbf{N})^{3/2}} \end{align*} (sign chosen so the mean curvature of a sphere is positive).

But if $\lvert N\rvert\approx 1$ (i.e., $\frac{f_\theta^2}{r^2}+\frac{f_\phi^2}{r^2\sin^2\theta}\ll 1$), then $$ H\approx\frac14\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N})-\frac12\nabla\cdot\mathbf{N} $$

We calculate \begin{align*} -\nabla\cdot\mathbf{N} &=\frac1{r^2}\frac{\partial(r^2)}{\partial r}+\frac1{r\sin\theta}\frac{\partial(-\frac1r f_\theta\sin\theta)}{\partial\theta}+\frac1{r\sin\theta}\frac{\partial(-\frac1{r\sin\theta}f_\phi)}{\partial\phi}\\ &=\frac2r-\frac{f_{\theta\theta}\sin\theta+f_\theta\cos\theta}{r^2\sin\theta}-\frac{f_{\phi\phi}}{r^2\sin^2\theta}\\ &\approx\frac2r-\frac{f_{\theta\theta}}{r^2}-\frac{f_{\phi\phi}}{r^2\sin^2\theta}\\ \mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N}) &=\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N}-1)\approx 0 \end{align*} since every term in $\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N}-1)$ has a factor of $f_\theta$ or $f_\phi$ which are assumed to be negligible. So we end up with $$ 2H\approx\frac2{f}-\frac{f_{\theta\theta}}{f^2}-\frac{f_{\phi\phi}}{f^2\sin^2\theta} $$ or abusing notation, $$ H\approx\frac1r+\frac12\left(-\frac{r_{\theta\theta}}{r^2}-\frac{r_{\phi\phi}}{r^2\sin^2\theta}\right) $$ which you should recognise big parenthesis term is basically the Laplacian of $r$ in the spherical variables.

You can also derive this using a more general result: If $f\colon M^2\to\mathbb{R}^3$ is an embedding and we have a normal vector field given by $\varphi N$, $N$ is the Gauss map and $\varphi$ is some smooth function, then the first variation of the principal curvatures are $$ \delta^{(1)}\kappa_i=\operatorname{Hess}(\varphi)(e_i,e_i)+\kappa_i^2 \varphi $$ (exercise in computing using moving frames, for example) and hence $$ \delta^{(1)} H=-\frac12\Delta\varphi+(2H^2-K)\varphi. $$ (Sign convention: the Laplacian $\Delta:=-\operatorname{Tr}_{I}\operatorname{Hess}$ is minus div grad.)

In our case, $M=S^2$ and we start with the round sphere of radius $r_0:=f(\theta_0,\phi_0)$ for our (local) calculation. If $\nabla f$ is small, then we are making small perturbation of the round sphere $r=r_0$ with $\varphi=r_0-f$ (note $N$ points inwards, to get principal curvatures positive for our starting sphere) and so the first-order approximation we get is $$ H\approx H_0+\delta^{(1)} H=\frac1{r_0}-\frac12\Delta^{\mathbb{S}^2(r_0)}\varphi=\frac1{r_0}+\frac1{2r_0^2}\Delta^{\mathbb{S}^2} f. $$ where the Laplacian on standard unit sphere $\mathbb{S}^2$ is $$ \Delta^{\mathbb{S}^2}=-\frac{\partial^2}{\partial\theta^2}-\frac1{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}. $$ Of course, there is nothing special about the suffix $0$ here except it is the starting point of our calculation, so dropping it altogether recovers the previous formula.

Similarly, for smooth $n$-dimensional hypersurfaces in $\mathbb{R}^{n+1}$ that are described by "$C^1$-approximately-constant $\log r$", we get $$ nH\approx\frac{n}r+\frac1{r^2}\Delta^{\mathbb{S}^n}r. $$ generalising both examples $n=1,2$.

  • $\begingroup$ thanks, the first approach sounds more familiar adn easier way to follow. $\endgroup$
    – alekhine
    Commented Sep 11, 2021 at 10:53

Please note that the following are in polar / cylindrical coordinates ( but not spherical).

We can derive curvatures on the basis of differential lengths/ triangles and curvature as follows. $\angle AOB$ is negligible in comparison to $\angle BOX ,$ and so on. $PB$ is differential arc of meridian. Primes are differentiation w.r.t $\theta. $ Using $r$ instead of $\rho$ for polar coordinate radius vector.

Two dimensional

$(x,y)$ or $( r,\theta) $ are here labeled as $(z,r). OA = r $

enter image description here

$$ k_1 = \frac{d \phi}{ds} = \frac{d \psi+ d \theta}{ds}= \frac {d ( \tan^{-1}\dfrac{r}{r'})+ d \theta}{ds}=$$ $$\frac{r^2+2r^{'2}-r r{''}}{r^2+r^{'2}}\cdot\frac{1}{\sqrt{r^2+r^{'2}}} \tag 1 $$

Three dimensional

$Z$ is the axis of symmetry and $r$ is the radius in cylindrical coordinates.

If H is mean curvature of a surface of revolution in particular, then $ 2 H = k_1+k_2; $

Arc $PB $ on meridian is extended up to Z-axis as $PQ.$ The minor curvature

$k_2=\dfrac{1}{PQ}=\dfrac{1}{R_2} $ can be represented as in the above meridional diagram.

$$k_2=\frac{\cos \phi}{r}=\frac{\cos {(\theta+\psi)}}{r}=\frac{\cos \theta \cos \psi- \sin \theta \sin \psi}{r}$$

$$=\frac{\cos \theta \;r'- \sin \theta\;r}{r \sqrt{r^2+r^{'2}}}\tag 2 $$

$$ 2H=\frac{(r^2+2r^{'2}-r r{''})/({{r^2+r^{'2}}} )+{(\cos \theta \;r'/r- \sin \theta\;)}}{\sqrt{r^2+r^{'2}}} ,\tag 3 $$

Linearizations can next be made on this.

$$ H=\frac{(1- r{''}/r- \sin \theta\;)}{2r}.\tag 4 $$


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