Say we want a particle to travel a certain path along a sphere, always travelling a certain direction (namely an angle from the equator). For example, starting at the origin and travelling a north-east direction, would be an angle of $\pi/4$. If we let then angle be $\alpha$, then is there a way to parameterize the equation of the path?
What I've tried: In spherical coordinates we can write the equation of a sphere of radius R as $$\textbf{r}(\theta,\phi)=<R\sin{\theta}\cos{\phi},R\sin{\theta}\sin{\phi},Rcos{\theta}>\text{where } 0\leq\theta\leq2\pi\text{ and }0\leq\phi\leq\frac{\pi}{2}$$ However, I don't think we can simply say that one of the angles is $\alpha$, as the vector from which we're measing $\alpha$ is constantly changing. If the angle is $\pi/4$ or similar, Ive tried making the argument that $\theta$ and $\phi$ would change at the same rate, therefore you can let them both equal $t$ and proceed from there, but I'm not sure if that's true.
Any help would be appreciated, cheers!
There are several conflicting conventions for spherical polar coordinates among mathematicians, physicists, geographers and astronomists. In this answer, I will stick to following parametrization of the sphere:
$$[0,2\pi] \times \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \ni (\theta,\phi) \quad\mapsto\quad (x,y,z) = (R\cos\phi\cos\theta,\,R\cos\phi\sin\theta,\,R\sin\phi ) \in \mathbb{R}^3$$
i.e. $\theta, \phi$ corresponds to the longitudes and latitudes in geography and $\phi \in [0,\frac{\pi}{2}]$ corresponds to the northern hemisphere.
The question at hand is find a curve whose tangent is making an constant angle $\alpha$ with the circles of latitudes. In above parametrization, the metric on the sphere is given by $$ds^2 = R^2 (\cos^2\phi \, d\theta^2 + d\phi^2)$$
What this means is:
if we move along the line $\phi = \text{const}$ for a small amount $\delta\theta$, the distance traveled $\approx R \cos\phi \delta\theta$.
if we move along the line $\theta = const$ for a small amount $\delta\phi$, the distance traveled $\approx R \delta\phi$.
In order for the curve to make a angle $\alpha$ with the circles of latitudes, we need to make displacement $(\delta\theta, \delta\phi)$ such that
$$\frac{R \delta\phi}{R \cos\phi \, d\theta}\approx m \stackrel{\text{def}}{=} \tan\alpha$$
This implies the curve satisfies following ODE:
$$\frac{1}{\cos\phi}\frac{d\phi}{d\theta} = m \quad\iff\quad\frac{d\sin\phi}{1 - \sin^2\phi} = m \, d\theta $$
If the curve start at a point $(\theta_0,\phi_0)$, we can solve above ODE and get: $$\frac{1 - \sin\phi}{1 + \sin\phi} = \frac{1 - \sin\phi_0}{1 + \sin\phi_0} e^{-2m(\theta - \theta_0)} $$ For example, if we start the curve at $(\theta_0,\phi_0) = (0,0) \iff (x,y,z) = (R,0,0)$, we get $$ \begin{align} \frac{1-\sin\phi}{1+\sin\phi} = e^{-2m\theta} &\iff \sin\phi = \tanh(m \theta)\\ &\iff \begin{cases} x &= R \frac{\cos\theta}{\cosh(m\theta)}\\ y &= R \frac{\sin\theta}{\cosh(m\theta)}\\ z &= R \tanh(m\theta) \end{cases} \end{align} $$ Such curves are called Rhumb line, there are more details on its wiki entry.
At the end is a picture of $6$ rhumb lines. All of them start at $(\theta_0,\phi_0) = (0,0)$ and their $\alpha$ vary from $10^\circ$ (the red curve) to $60^\circ$ (the cyan curve). As one can see, they sort of "spiral" around the north pole as they approach it.
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