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?

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    Commented Oct 1, 2018 at 16:07

1 Answer 1


TLDR: (a) You're expressing the same metric in differnet coordinate system, so you get the same result. (b) The sphere locally looks like its tangent plane.

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 $\text{d} s^2$.


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