Let $\Phi(r,\theta,\phi)=(r\cos \theta \sin \phi,r\sin \theta\sin\phi,r\cos\phi)$ be the usual spherical coordinate transformation.

One way to define coordinates systems at each $\theta,\phi$ on the unit sphere is to use the fact $\frac{\partial \Phi}{\partial r}$ has a constant magnitude of $1$, which implies $\frac{\partial \Phi}{\partial r}\perp \frac{\partial \Phi}{\partial \theta},\frac{\partial \Phi}{\partial \phi}$. Another quick calculation shows $\frac{\partial \Phi}{\partial \theta}\perp \frac{\partial \Phi}{\partial \phi}$. So the partial derivatives are always orthogonal.

I want to apply Stokes' theorem to calculate a path integral of a field along the unit circle in the $xy$ plane. To calculate the integral of the curl I need the normal to the sphere in terms of $\theta,\phi$. By definition it's given by $\frac{\partial \Phi}{\partial \theta}\times \frac{\partial \Phi}{\partial\phi}$. I think that by orthogonality this should be exactly $\sin \phi\hat r(\theta,\phi)=\sin \phi(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)$ but when I calculate the mnemonic determinant, I get something else, namely the same vector except the $\hat x$ component goes in the opposite direction.

How can this be?


The Stokes theorem states that $$ \oint_C \vec{F}\cdot d\vec{r} = \iint_S \nabla \times \vec{F}\cdot d\vec{S}=\iint_D \nabla \times \vec{F}(u,v)\cdot \color{red}{\pm} (\vec{r}_u\times \vec{r_v})\;dA $$

where $\vec{r}(u,v)$, $(u,v)\in D$ parametrizes the surface $S$ enclosed by $C$.

There is a $\color{red}{\pm}$ sign to indicate that the orientation depends on the orientation of $C$. So depending on how you compute $\vec{r}_u\times \vec{r_v}$, and how you parametrized $C$, you can obtain a $(-1)$ factor in your integral.

So for your case, the normal to the sphere is $\color{red}{\pm}\frac{\partial \Phi}{\partial \theta}\times \frac{\partial \Phi}{\partial\phi}$, you need to fix the sign such that it is in accordance with the parametrization of $C$, that is must satisfy the following rule (conventionally): if $C$ is orientated counter-clockwise when seen from "above", the normal to $S$ must have a positive third component.

  • $\begingroup$ I just answered myself too, and I think I understood you point! $\endgroup$ – Calc Feb 6 '17 at 15:04

I forgot some signs in the mnemonic determinant. Basically the ordered triple $\frac{\partial \Phi}{\partial r},\frac{\partial \Phi}{\partial \phi},\frac{\partial \Phi}{\partial \theta}$ is the one giving a positive basis. My cross product had the wrong order, so comes with a minus sign. In other words, the result is $$-\sin \phi\hat r(\theta,\phi).$$


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