I have a particle with trajectory $P(t)$ describing a straight line. I am working with spherical coordinates (physicist's convention): $$x = r \sin \theta \cos \varphi, \quad y = r \sin \theta \sin \varphi, \quad z = r \cos \theta $$ Writing down $P(t) = ( P_r(t), P_\theta(t), P_\varphi(t) )$ in this coordinate system, at a given point (say $t=0$), I know the partial derivatives: $$ \left . \frac { \partial P_\theta(t) } { \partial t } \right |_{t = 0}$$

$$ \left . \frac{ \partial P_\varphi(t) } { \partial t } \right |_{t=0} $$

Given this information, I am interested in computing the $\theta$ and $\varphi$ components of the trajectory at infinity (say $\theta_\infty$, $\varphi_\infty$), which amounts to describing the $\theta$ and $\varphi$ coordinates of the straight line trajectory in a new spherical coordinate system centered at $P(0)$.

For the simpler situation of polar coordinates $(r,\varphi)$, I proceeded by computing the polar coordinates of a parametrised straight line, expressed in Cartesian coordinates as:

$$ P(t) = r_0 ( \cos \varphi_0, \sin \varphi_0 ) + t ( \cos \theta_{\infty}, \sin \theta_{\infty} )$$

Computing derivatives, after some algebraic manipulations I obtained the simple expression

$$ r_0 \left . \left ( \frac{ \partial P_\varphi } { \partial P_r } \right ) \right |_{t=0} = \tan \left ( \varphi_\infty - \varphi_0 \right ) $$

which allows the computation of $\varphi_\infty$ from $r_0$, $\varphi_0$ and the position derivatives at $t=0$.

I was hoping to find similar formulas for the case of spherical coordinates, but so far I have not managed as the algebra turned out too complex when using this same approach.

I realise that the above essentially boils down to computing the Jacobian of the transformation which moves the origin of the spherical coordinate system to $P(0)$: the components of the particle velocity are tangent vectors in the original coordinate system, and I want to obtain the tangent vector in the new coordinate system centered at $P(0)$. Yet I still find myself a bit swamped by the algebra, and am hoping for a simple formula that looks similar to the one I provided above for the case of polar coordinates.


1 Answer 1


Making some progress with the algebra, for the problem in spherical coordinates I've managed to obtain the expression:

$$ \tan \left ( \varphi_\infty - \varphi_0 \right ) = \frac{r_0 \sin \theta_0 \left . \frac{\partial P_\varphi}{\partial P_r} \right |_{t=0}}{ \sin \theta_0 + r_0 \cos \theta_0 \left . \frac{\partial P_\theta}{\partial P_r} \right |_{t=0}} $$

If we restrict our particle to equatorial motion, taking $\theta_0 = \pi / 2$, this formula recovers the formula in the OP for polar coordinates:

$$ \tan \left ( \varphi_\infty - \varphi_0 \right ) = r_0 \left . \frac{\partial P_\varphi}{\partial P_r} \right |_{t=0}$$

After a significant amount more algebra, I believe I have also arrived at a formula for $\theta_\infty$:

$$ \cos \theta_\infty = \frac{ \cos \theta_0 - r_0 \sin \theta_0 \left . \frac{\partial P_\theta}{\partial P_r} \right |_{t=0} } { \sqrt { 1 + \left ( r_0 \left . \frac{\partial P_\theta}{\partial P_r} \right |_{t=0} \right ) ^2 + \left ( r_0 \sin \theta_0 \left . \frac{\partial P_\varphi}{\partial P_r} \right |_{t=0} \right ) ^2 } } $$

As of yet I haven't checked the correctness of these formulas. Has anyone come across such formulas, e.g. when computing the Jacobian for a translation in spherical coordinates?

  • $\begingroup$ I guess I'm confused. Shouldn't we determine $(\theta_\infty,\varphi_\infty)$ by taking the spherical coordinates of the direction vector of the line (as a point on the unit sphere)? $\endgroup$ Commented Nov 19, 2019 at 18:14
  • $\begingroup$ $ ( \theta_\infty, \varphi_\infty ) $ are indeed spherical coordinates of the line's direction vector. I'm trying to express them in terms of the partial derivatives at $t = 0$ of the position $P(t)$, with respect to spherical coordinates. In my situation I am numerically solving equations of motion in a spherical coordinate system, so I know the values of those derivatives, and want to compute the direction vector from them. $\endgroup$
    – Will
    Commented Nov 19, 2019 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .