This thing is making me going crazy, mathematicians and physicists use different notations for spherical polar coordinates.

Now during a vector calculus problem I had the following issue: Had to find $d\underline{S}$ for the surface of a sphere of radius $a$ centred at the origin. In all the books I always find that for a parametrised surface $\underline{r}(s,t)$ we have $d\underline{S} = \left(\frac{\partial \underline{r}}{\partial s}\times\frac{\partial \underline{r}}{\partial t}\right)dsdt$ in this order.

For the sphere I have $\underline{r}(\theta,\phi) = a\cos(\theta)\sin(\phi)\underline{i}+a\sin(\theta)\sin(\phi)\underline{j}+a\cos{\phi}\underline{k}$ for $0\leq \theta\leq 2\pi$ and $0\leq \phi\leq \pi$ And hence I get $\frac{\partial \underline{r}}{\partial \theta}\times\frac{\partial \underline{r}}{\partial \phi} = -\underline{r}a\sin{\phi} d\theta d\phi$ which points inwards so I take the opposite of it.

In my notes, they always preserve the order I preserved here (i.e. the first partial on the left (i.e. $\frac{\partial}{\partial \theta}$) is the first component in the brackets of $\underline{r}(\theta,\phi)$). Preserving the order I should always get the correct normal vector. However for some weird reason when in my notes, in the books and online people have to calculate $d\underline{S}$ for a sphere (like here) they always invert the coordinates and write the spherical coordinates as $(r,\theta,\phi)$ for $0\leq \theta \leq \pi$ and $0\leq \phi \leq 2\pi$ and $\underline{r}(\theta,\phi)$ with $\frac{\partial \underline{r}}{\partial \theta}\times\frac{\partial \underline{r}}{\partial \phi} = \underline{r}a\sin{\theta} d\theta d\phi$

why does this happen? It's just a notation convention, however the order of the partial should give the correct normal, although in my example it clearly gives the opposite, while using the other notation, it gives the correct one.

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    $\begingroup$ Actually, I've always written the coordinates in the order you wrote them: first the longitude and then the colatitude. And I'm pretty sure that's how they're usually written. $\endgroup$ – Harnak Dec 27 '16 at 12:44
  • $\begingroup$ @Harnak both in my Vector Calculus course and in my Applied Mathematics course they use the opposite. It is so confusing that there's not a universal way of writing it. However do you know why the sign comes out wrong? Of course that's cause of the antisimmetry of the cross product, but technically mantaining the order should give the correct normal, right? $\endgroup$ – Euler_Salter Dec 27 '16 at 12:49
  • $\begingroup$ The convention here is that of right-handedness or left-handedness of your coordinate system. $\endgroup$ – ÍgjøgnumMeg Dec 27 '16 at 12:58
  • $\begingroup$ @Jahambo99 and which of them is right-handed? $\endgroup$ – Euler_Salter Dec 27 '16 at 12:59
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    $\begingroup$ If you have a parametrization $\phi(u,v)$, the derivatives $\phi_u$ and $\phi_v$ are the tangents to the coordinate curves on the surfaces. So you just have to follow them to see how they're directed and take the product to see where the normal goes. As an example, consider the cylinder $\phi(\theta, z) = (\cos \theta, \sin \theta, z)$: its coordinate curves are the circles and the lines parallel to its axis. The tangents to the circles go anti-clockwise and the tangent to the vertical line go upwards. Taking the product in this order, you get a normal pointing outwards. $\endgroup$ – Harnak Dec 27 '16 at 13:26

You seem to have stumbled onto an example of the right-hand rule. Consider $\frac{d\underline{r}}{d\theta} \times \frac{d\underline{r}}{d\phi}$ vs. $\frac{d\underline{r}}{d\phi} \times \frac{d\underline{r}}{d\theta}$. They will both produce the same magnitude, as well as the same vector (up to sign). That is, $\frac{d\underline{r}}{d\phi} \times \frac{d\underline{r}}{d\theta} = -\left(\frac{d\underline{r}}{d\theta} \times \frac{d\underline{r}}{d\phi}\right)$. This is because the cross product is anticommutative. The correct normal is dictated by the problem or application for the problem. So, if you were asked to find the outward-pointing normal that would be the correct one. It doesn't really matter if you are careful about the order of cross product because identifying if you crossed in the order the problem calls for is relatively easy to do.


There is no "correct normal". On any parameterised surface $(u,v) \mapsto S(u,v)$, the vector $\frac{\partial S}{\partial u}\times\frac{\partial S}{\partial v}$ and its negative are both normal to the surface, and I don't think either one of them is intrinsically more correct than the other.

On simple surfaces like spheres and cylinders, you can distinguish one normal from the other by geometric conditions. So, for a sphere, for example, you might specify that you want the normal that "points towards the center", and for a cylinder you might specify the normal that points "away from the axis". There are still two unit normal vectors at each point of the surface, and neither one is more correct than the other, but at least you have a way of distinguishing one from the other. Note that these normals are independent of how the surface is parameterised.

On complex surfaces, there is often no notion of "center" or "axis", so the geometric approach doesn't work. So, sometimes, the only way to distinguish the two normals is by the fact that the "positive" one is in the direction of $\frac{\partial S}{\partial u}\times\frac{\partial S}{\partial v}$, and the other is in the opposite direction, the direction of $\frac{\partial S}{\partial v}\times\frac{\partial S}{\partial u}$. But by thinking this way, you've introduced a dependency on the surface parameterisation -- if you change the parameterisation, the "positive" normal might flip. That seems to be what's bothering you.

The surface normals are not always interchangeable. Suppose you're trying to offset the surfaces of a solid object, to make it larger, to model some sort of painting or coating process, perhaps. Obviously the direction of offsetting has to be "away from the interior" or "into air". If here's a circular hole in the object, you need to offset along normal vectors that point towards the hole axis. If there's a hemispherical bump, you need to offset away from the center of the hemisphere, and if there's a hemispherical dent, you need to offset towards its center. The "correct" normal is determined by the process you're trying to model (and not by how the surfaces are parameterised). If there's no connection to any physical process or computational algorithm that dictates otherwise, the two normal directions are both equally correct.

  • $\begingroup$ so for any parametrised surface once we calculate the normal (up to a $\pm 1$ factor) we can use either independently? $\endgroup$ – Euler_Salter Dec 27 '16 at 12:58
  • $\begingroup$ @Euler_Salter If you look at a flat plane, "the" normal vector to the plane is usually the vector "pointing upwards", so to speak, that is perpendicular to all vectors in the plane. You can just as easily take the vector "pointing downwards" which has the same property, but you get a factor of $-1$ in front of it. $\endgroup$ – ÍgjøgnumMeg Dec 27 '16 at 13:02
  • $\begingroup$ @Jahambo99 yeah I know that, but read my comment above (comments on the question, not the answer), if we had a more complicated surface parametrised in spherica polar, how would someone pick the correct one? Here is easy to see which one is outwards as it is in the radial direction, but in a more complicated shape it would be harder $\endgroup$ – Euler_Salter Dec 27 '16 at 13:18
  • $\begingroup$ > we can use either independently. Please see additions to my answer. $\endgroup$ – bubba Dec 27 '16 at 14:07

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