Consider $\textbf{R}^3$ as a manifold with the flat Euclidean metric, and coordinates {$x,y,z$}. Introduce spherical polar coordinates {$r,\theta,\phi$} related to {$x,y,z$} by

$x = rsin(\theta)cos(\phi)$



so that the metric takes the form

$ds^2=dr^2 + r^2d\theta^2+r^2sin^2(\theta)d\phi^2$.

a) A particle moves along a parametrized curve given by

$x=cos(\lambda), y=sin(\lambda), z=\lambda$.

Express the path of the curve in the {$r,\theta,\phi$} system.

So, that is the question. I found this integral for the path, but I don't know how to do it and I don't know if it is right:


  • $\begingroup$ The equation $x=cos\lambda, y=sin(\lambda), z=\lambda$ is the path of a helix with a radius of $1$. Hence in spherical coordinates, $r=1$, $\sin(\theta)=1 \implies \theta=\pi/2 \implies \cos(\theta)=0 \implies \lambda=0\pm2k\pi$. This post has nothing to do with general relativity. $\endgroup$ Mar 26, 2020 at 10:42
  • $\begingroup$ FYI - when typesetting trig functions use \sin and \cos instead. It makes the post more readable. $$ \begin{matrix} x=r \sin(\theta)\cos(\phi) & \text{vs.} & x=r sin(\theta)cos(\phi) \end{matrix}$$ $\endgroup$ Mar 26, 2020 at 13:59
  • $\begingroup$ And the appropriate coordinate system for a helix would cylindrical - not spherical. An easy way to find the arch length of a helix would be to use cylindrical coordinates and the Frenet frame. $\endgroup$ Mar 26, 2020 at 20:22
  • $\begingroup$ @ja72: if you look at code I wrote I did use \sin and \cos. The code I copied form the OP didn't. $\endgroup$ Mar 26, 2020 at 20:32

1 Answer 1


Consider the invese transformation from Spherical polar coordinate to Cartesian Coordinate: $$r^2=x^2+y^2+z^2$$ $$\phi=tan^{-1}\left(\frac{y}{x}\right)$$ $$\theta=cos^{-1}\left(\frac{z}{r}\right)$$ Using the parametric form

We have $$r^2=1+\lambda^2$$ $$\phi=\lambda$$ $$\theta=cos^{-1}\left(\frac{\lambda}{\sqrt{1+\lambda^2}}\right)$$ You can eliminate $\lambda $ to find the trajectory. $$cos\theta=\frac{\phi}{r}.$$

  • $\begingroup$ this answer is wrong (if $\theta=0$ then $x=y=0$ throughout the trajectory). Also - giving a detailed solution to homework assignments is discouraged $\endgroup$
    – user245141
    Mar 26, 2020 at 9:18
  • $\begingroup$ Sorry I will be aware next time. $\endgroup$ Mar 26, 2020 at 9:29

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