I saw a stack post a couple of years ago about the metric on the two-sphere, S2 in cartesian coordinates, x, y and z. I can’t find the post.
Below I sketch out what I remember – my apologies for these being just fragments.
There are actually two questions:
Can someone provide a derivation of the metric on S2 in terms x, y and z based on my steps 1-5 below? X, y and z would be the orthogonal axes with origin at the center of the sphere, z going upwards.
I’m also seeking clarification on what it actually means to say a metric is considered "Euclidean." Which surfaces have a metric considered Euclidean and which have one that is not? Having the metric on S2 in terms of cartesian coordinates makes for an easy comparison to $ds^2 = dx^2 + dy^2 + dz^2$
The metric on S2 in the stack post here in spherical coordinates, is considered "Euclidean." Is this metric Euclidean because it is induced from an R3 embedding? Is it Euclidean because the surface (my understanding is this applies to all Riemannian manifolds) can be divided into flat area elements dA using calculus, and the standard Euclidean metric used for arc length (because the metrics are essentially equivalent in the infinitesimal)?
This is what I recall from that post:
- S2 of radius R was defined as: $z = (R^2 – x^2-y^2)^{1/2}$
- Partial derivatives of the surface with respect to x and y were calculated. I think they were then squared, the way E and G are in the First Fundamental form.
- The squared partials were then multiplied by differentials $dx^2$ and $dy^2$ to produce the final metric.
- I recall an expression $(v + dv) ● (v + dv)$ but I can't remember how that was incorporated.
- I do not recall a norm squared expression using tangent vector v being in the post.
The link here shows the term $Fdudv$ in the section on Arc Length. This is an example, if $F <> 0$, of a metric that would not be Euclidean. I don't know if the fact that the coefficients of $dx^2$ and $dy^2$ are partial derivatives with an exponent of -1/2 (see steps 1 and 2 above) means the final metric would not be considered Euclidean.
To answer your second question, it seems the OP in the post you linked used the word "euclidean space" to differentiate it from spacetime, which indeed has a different metric. The Euclidean metric is usually referred to as the one induced by the Euclidean inner product $$\langle x,y \rangle=\sum_{i=1}^n x_iy_i$$ seeing as it is the most popular metric used on Euclidean spaces ($\mathbb{R}^N$ or spaces isomorphic to it).
Nevertheless you raise a good point. It is precisely correct that the usual metric bestowed onto the sphere is inherited by the surrounding $\mathbb{R}^3$ metric space, with the euclidean metric. In that sense one could call it the (submanifold) Euclidean metric.
As for your first question, you seem to be a bit shaky on differential geometry. The book Differential Geometry of Curves and Surfaces, by Manfredo do Carmo, is a great place to start or to clarify/ review some concepts. Nevertheless, the construction of the metric you're after usually goes somewhat like this:
A general regular surface in $\mathbb{R}^3$ is defined, loosely speaking, as a subset of it such that there are local diffeomorphisms between that surface and open sets of the plane. This essentially means that you can always specify a function which "builds up" (locally) the surface by bending smoothly, without creasing or tearing, patches of the plane. So if you want to calculate a metric you must choose such a specific function. There are in principle many many functions you can choose; the one that gives the metric in the form you want is $$f(x,y)=(x,y, \sqrt{R^2-x^2-y^2})$$ (we disregard where this function is actually defined seeing as you are merely seeking the algebraic steps). $R$ is of course the radius of the sphere and $x,y$ can be understood as projections of the sphere onto the xy plane. This is of course the graphic of (the upper half of) the sphere. The procedure is to take the partial derivatives, , like so $$\frac{\partial f}{\partial x}=\left(1,0,-\frac{x}{z}\right) \hspace{5mm} \frac{\partial f}{\partial y}=\left(0,1,-\frac{y}{z}\right)$$ where $z$ is as you defined in your question, and taking the inner products, the metric is $$G(x,y)=\begin{bmatrix} 1 + \left(\frac{x}{z}\right)^2 & \frac{xy}{z^2} \\ \frac{xy}{z^2} &1 + \left(\frac{y}{z}\right)^2 \end{bmatrix}$$ You can also obtain these coefficients by expressing the coordinate functions' differentials in terms of the "plane variables", $$dX=dx,\hspace{3mm} dY=dy, \hspace{3mm} dZ=-\frac{x}{z}dx-\frac{y}{z}dy$$ Square and add them and you'll arrive at the same coefficients.
As for your questions about flatness and what is Euclidean; Euclidean means what I started by writing in this post; it's the distance function given by that inner product. Now, flatness is a whole another story. Generally, the notion of flatness has nothing to do with the metric; it's an extra structure associated with the choice of what's called a connection on the surface.
However, what one generally does, and that is the case in standard general relativity, is to choose a connection, the Levi-Civita connecion, which is completely determined by the metric. If we choose that connection, then the Euclidean metric is flat. But the way in which it is flat has nothing to do with the algebra of the coefficients of the metric; as you could observe, the induced euclidean metric on the sphere has two vastly different representations, each as much curved (with respect to the LC connection) as the other. By contrast, the euclidean metric on the whole space (the one in the first link you posted) in spherical coordinates is entirely flat.
I suggest you study differential geometry more in depth. It's hard to get a grip of what's going on if you're only going off physics stack posts. Nevertheless, if you want, ask another, this time, self contained, question about what it means for a metric to be flat. But I reiterate: all these notions are precisely formulated in any good introduction to differential geometry like the one I mentioned.