0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – Ted Shifrin Nov 19 '19 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 Nov 19 '19 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.