What's the analogue to spherical coordinates in $n$-dimensions? For example, for $n=2$ the analogue are polar coordinates $r,\theta$, which are related to the Cartesian coordinates $x_1,x_2$ by
$$x_1=r \cos \theta$$ $$x_2=r \sin \theta$$
For $n=3$, the analogue would be the ordinary spherical coordinates $r,\theta ,\varphi$, related to the Cartesian coordinates $x_1,x_2,x_3$ by
$$x_1=r \sin \theta \cos \varphi$$ $$x_2=r \sin \theta \sin \varphi$$ $$x_3=r \cos \theta$$
So these are my questions: Is there an analogue, or several, to spherical coordinates in $n$-dimensions for $n>3$? If there are such analogues, what are they and how are they related to the Cartesian coordinates? Thanks.
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$\begingroup$These are hyperspherical coordinates. You can see an example of them being put to use in this answer.
$\endgroup$ 4 $\begingroup$I was trying to answer exercise 9 of $ I.5. $ from Einstein gravity in a nutshell by A. Zee that I saw this question so what I am going to say is from this question. It is said that the d-dimensional unit sphere $S^d$ is embedded into $E^{d+1}$ by usual Pythagorean relation$$(X^1)^2+(X^2)^2+.....+(X^{d+1})^2=1$$. Thus $S^1$ is the circle and $S^2$ the sphere. A. Zee says we can generalize what we know about polar and spherical coordinates to higher dimensions by defining
$X^1=\cos\theta_1\quad X^2=\sin\theta_1 \cos\theta_2,\ldots $
$X^d=\sin\theta_1 \ldots \sin\theta_{d-1} \cos\theta_d,$
$X^{d+1}=\sin\theta_1 \ldots \sin\theta_{d-1} \sin\theta_d$
where $0\leq\theta_{i}\lt \pi \,$ for $ 1\leq i\lt d $ but $ 0\leq \theta_d \lt 2\pi $.
So for $S^1$ we just have ($\theta_1$):
$X^1=\cos\theta_1,\quad X^2=\sin\theta_1$
$S^1$ is embedded into $E^2$ and for the metric on $S^1$ we have:$$ds_1^2=\sum_1^2(dX^i)^2=d\theta_1^2$$
for $S^2$ we have ($\theta_1, \theta_2$) so for Cartesian coordinates we have:
$X^1=\cos\theta_1,\quad X^2=\sin\theta_1\cos\theta_2,$$\quad X^3=\sin\theta_1\sin\theta_2$
and for its metric:$$ds_2^2=\sum_1^3(dX^i)^2=d\theta_1^2+\sin^2\theta_1 d\theta_2^2$$
for $S^3$ which is embedded into $E^4$ we have($ \theta_1,\theta_2,\theta_3 $):
$X^1=\cos\theta_1,\quad X^2=\sin\theta_1\cos\theta_2,$$\quad X^3=\sin\theta_1\sin\theta_2\cos\theta_3$$\quad X^4=\sin\theta_1\sin\theta_2\sin\theta_3 $
$$ds_3^2=\sum_{i=1}^4(dX^1)^i=d\theta_1^2+\sin^2\theta_1 d\theta_2^2+sin^2\theta_1\sin^2\theta_2\,d\theta_3^2$$
Finally, it is not difficult to show the metric on $S^d$ will be:
$$ds_d^2=\sum_{i=1}^{d+1}(dX^1)^i=d\theta_1^2+\sin^2\theta_1 d\theta_2^2+sin^2\theta_1\sin^2\theta_2\,d\theta_3^2+\cdots+sin^2\theta_1\cdots\sin^2\theta_{d-1}\,d\theta_d^2$$
$\endgroup$ $\begingroup$Just look at n-sphere. A lecture note from Stony Brook is also available.
You can find it in Fock's paper (Fock, V. (1935).)or in some recent papers- like (Howard, S. "Fundamental Solution of Laplaces Equation in Hyperspherical Geometry.") or (Jing-Jing, F., Ling, H., & Shi-Jie, Y. (2011). Solutions of laplace equation in n-dimensional spaces. Communications in Theoretical Physics, 56(4), 623.) for advanced studies.
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