# Linear motion in spherical coordinates, changing origin of spherical coordinate system

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.

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?

• 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)? – Ted Shifrin Nov 19 '19 at 18:14
• $( \theta_\infty, \varphi_\infty )$ 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. – Will Nov 19 '19 at 18:24