# Compare 2d metric to metric on sphere-puzzling result

I calculated the infinitisimal distance squared $$(\mathrm{d}s)^2$$ in 3d polar coordinates and 3d cartesian coordinates in an example to show that they were identical. I chose $$P(r,\theta,\phi)$$ as well as $$\mathrm{d}r$$, $$\mathrm{d}\theta$$ and $$\mathrm{d}\phi$$ and got $$(\mathrm{d}s)^2$$ via $$(\mathrm{d}s)^2=(\mathrm{d}r)^2+ r^2(\mathrm{d}\theta)^2 + r^2\sin^2\theta (\mathrm{d}\phi)^2$$

I got the corresponding cartesian differentials via \begin{align} \mathrm{d}x&=\frac{\partial x}{\partial r}\mathrm{d}r+\frac{\partial x}{\partial \theta}\mathrm{d}\theta +\frac{\partial x}{\partial \varphi}\mathrm{d}\phi \\&= \sin\vartheta \cos\varphi\ \mathrm{d} r+r\cos\vartheta \cos\varphi\ \mathrm{d}\theta-r\sin\vartheta \sin\varphi\ \mathrm{d} \phi \end{align} $$\mathrm{d}y$$ = similar; $$\mathrm{d}z$$ = similar

And via $$(\mathrm{d}s)^2=(\mathrm{d}x)^2+(\mathrm{d}y)^2+(\mathrm{d}z)^2$$ I got the identical result as above.

Then I wanted to go to the 2D surface of a unit sphere by setting $$r=1$$ and $$\mathrm{d}r=0$$ in all my formulas (the metric as well as the differentials). And, surprising enough, $$(\mathrm{d}s)^2$$ (cartesian) is still $$(\mathrm{d}s)^2$$ (spherical surface), which puzzles me, since the surface of the sphere is not flat while cartesian space should be flat.

What does the cartesian path calculate?

• Hi, welcome to Physics SE! I've edited the mathjax here as an example. Look at this Math SE meta post for a quick tutorial. You seem to be familiar with most of it; the only change is that we surrond math with \$ symbols, and we use some slightly fancy stuff for the alignment (it's there in the linked tutorial/reference). – user191954 Oct 1 '18 at 16:07

The $$\text{d}s^2$$ calculated the (squared) length of an "infinitesimal path" composed from coordinate differentials $$\text{d}x^i$$ at some poit $$P$$. Your two metrics are exactly the same object, just expressed in different coordinates.
Now if you want to restriuct yourself to the surface of the unit sphere, that will gve a relation between $$\text{d} x$$, $$\text{d} y$$, and $$\text{d} z$$ depending on the point you're at, which is simplified by usisng spherical coordinates, where it's just $$\text{d} r=0$$. However, the metrics are still the same object in different coordinates.
What you seem to be confused about is about the length of a path along the sphere vs. straight across 3d space, which should clearly be different. To get the length of a finite path, you have to specify the path and integrate the $$\sqrt{\text{d} s^2\,}$$ along it. Again, you can use Cartesian or spherical coordinates to specify the path, and the length will not change.
On the other hand, you seem to envisage different paths: One is along the sphere with $$r=\text{const}$$ (most conveniently expressed in spherical coordinates), while to other is a straight line (convenient in Cartesian coordinates). For simplicity, assume you're starting at the equator and moving "up". Then the two paths will be, using $$\lambda$$ as parameter, $$(r=\text{const},\lambda,\phi=\text{const})$$ and $$(x=\text{const},y=\text{const},\lambda)$$. Then the tangent vectors at the starting point are actually the same, so you get the same $$\text{d} s^2$$.