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upon the quest of understanding spherical harmonics, I came to the a saying "functions defined on the surface of a sphere".

https://en.wikipedia.org/wiki/Spherical_harmonics

Now what does it mean by "functions defined on the surface of a sphere"? "A function of sphere, f(x,y,z)" is one thing and "functions defined on the surface of a sphere" is surely another thing, isn't it? I am not gettting it quite right.

user3329892
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3 Answers3

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For simplicity let us limit ourselves with 3D and scalar functions. In this context "function defined on a sphere" means that you define the way to correspond a number to each point of the sphere. In mathematical terms it is easier to write using spherical coordinates $\theta$ and $\phi$ (radius coordinate is irrelevant here, as all points of the sphere would have the same radius value). So if you have function $f(\theta, \phi)$ defined on $\theta \in [-\pi/2, \pi/2]$, $\phi\in[0, 2\pi[$ this would be the function defined on the sphere. For example, $f(\theta, \phi)=sin(\phi)cos^2(\theta)$ would be such a function.

Note, that if you want you function to be continuous on the sphere, without 'seam' at $\phi=0$ you should impose condition $f(\theta, 0)=f(\theta, 2\pi)$ (provided your function is defined for $\phi=2\pi$).

Also, as $\theta=-\pi/2$ and $\theta=\pi/2$ correspond to 'north and south poles' of the sphere there should be another condition to avoid singularity there: $f(\pm\pi/2, \phi)=f(\pm\pi/2)$. It means, that function should not depend on $\phi$ at the sphere's 'poles'.

You may define you function in 3D Cartesian space, but then the domain constraints would be more complicated:

$f(x, y, z | x^2 + y^2 + z^2 = 1)$

meaning that function should be defined only for such $x, y$ and $z$ which correspond to unit sphere's surface. It does not mean that it cannot be defined for other points in 3D space, of course, but we are interested only on those on the sphere.

one_two_three
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  • Not any function $f ; [-\frac{\pi}{2}, \frac{\pi}{2}]\times[0, 2\pi] \to \mathbb{R}$ will define for you a map $S^2 \to \mathbb{R}$. There are compatibility conditions you need to impose, e.g. $f(\theta, 0) = f(\theta, 2\pi)$. – Michael Albanese Jan 22 '16 at 13:11
  • Also, what does your last line mean? It is not standard notation. – Michael Albanese Jan 22 '16 at 13:11
  • You are right. Strictly speaking $\phi\in[0, 2\pi[$. There is no intrinsic requirement for function to be continuous on the sphere. The important condition is to avoid ambiguity at the sphere's poles, so for example $f(\pi/2, \phi)$ does not depend on $\phi$. As to the notation for the last expression -- I do not remember what the correct notation would be, this is the one I used when writing down math lectures 20 years ago. – one_two_three Jan 25 '16 at 10:47
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The surface of the sphere is a two-dimensional manifold parametrized by two angular variables ($\theta$ and $\phi$), the best analogue is to think of a function (let's say the scalar temperature or the vectorial wind velocity) defined on Earth's surface, and hence depending on two variables - latitude and longitude.

Shahar Even-Dar Mandel
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  • I am not getting the analogue quite well. Is it something like this?:

    Lets say, there is a perfect polished metal ball who's surface could be defined by the following function:

    f(x,y,z)

    Now lets say I draw a curve on the surface of the ball. Lets say the function of the curve is:

    g(x.y.z)

    can it be said for that curve, that the curve is a "functions defined on the surface of a sphere"?

     – user3329892 Sep 17 '15 at 08:32
  • A general function $g(x,y,z)$ can be used to describe the temperature in every point in space, including point in outer space ($x^2+y^2+z^2>R^2$, where $R$ is the Earth's radius) or inside Earth. However, if you are only interested in temperatures measured on the surface of our planet (NYC or Cairo or a god-forsaken piece of the Atlantic Ocean) then you are limiting yourself to a set of points satisfying $x^2+y^2+z^2=R^2$, there are several ways to use only two parameters to specify a point on this surface, one option is the latitude/longitude set of parameters. – Shahar Even-Dar Mandel Sep 17 '15 at 09:29
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When you mention

"draw a curve on the surface of a sphere",

I do not think this is what the article is suggesting.

When you draw a curve on a surface of a sphere, you are defining a vector valued function $r(t) = x(t)i + y(t)j + z(t)k$ where $i, j , k$ are simply unit vectors in the $x$,$ y$ and $z$ direction. This is because the curve you are drawing is just a set of three dimensional coordinates, and given it is nice enough, we can parameterize it, as I did above. This means I made it a function of some parameter, in my example the parameter was $t$ .

What the article is trying to say is, that the actual DOMAIN of the function is the surface of a sphere. That is, the function is literally a function of the set $S =\lbrace {(x,y,z) | x^2 + y^2 + z^2 = r^2 } \rbrace$

porridgemathematics
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  • In plain lingua, can I say this? :

    Let us assume, some “Functions defined on the surface of a sphere” are f(x), g(x), h(x).

    So f(x) could be x^2+y^2+z^2 = 256, where as g(x) could be x^2+y^2+z^2 = 16, where as h(x) could be x^2+y^2+z^2 = 36.

    Basically any function of a sphere.

     – user3329892 Sep 19 '15 at 07:43
  • The three things you are describing are not in fact functions, but level sets or "surfaces" of the function f(x,y,z) = $x^2 + y^2 +z^2$ . A level set of a function f(x,y,z) is the set of all (x,y,z) such that f(x,y,z)=c where c is some constant. So can you see why what you just desribed werent functions, but level sets of a function? Indeed, these level sets are also geometrically, spheres. Now a function defined on the surface of a sphere is simply a function with a domain being the level set $x^2+y^2+z^2 = c^2$ , indeed this is a sphere of radius $|c|$. – porridgemathematics Sep 19 '15 at 08:19