Curve for great circles and the exponential map

Given a point $p\in S^2=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=a^2\}$ and a vector $v\in T_pS^2$, I want to find a curve $\sigma$ of the great circle which goes through $p$ with velocity $v$.

My idea was to start with the curve $\alpha(t)=(\cos t,\sin t,0)$ and then apply rotations to it so that the plane defined by the resulting great circle has normal vector equal to $p\times v$. Hence, if we write $p\times v = N(\sin \theta_N \cos \phi_N), \sin \theta_N \sin \phi_N, \cos \theta_N$, we had the curve $\gamma(t)=R_z(\phi_N)R_y(\theta_N)\alpha(t)$.

I am not sure that this is correct, and the computations are tiresome. IS there other way to do it?

(Context: I want to know the parametric geodesic $\gamma$ that goes through $p$ with velocity $v$ so that I can determine the exponential map $\exp_p(v)=\gamma(1)$ )

I will consider the unit sphere ($a=1$).

Let $w:=\tfrac{v}{\|v\|}$.

As $p \perp w$, the curve defined by

$$\tag{1}s(t)=\cos(t)p+\sin(t)w$$

• goes, at $t=0$, through point $p$ ($s(0)=\cos(0)p+\sin(0)w=p$).

• belongs to the sphere and describes a great circle on it (consider the orthonormal basis $\{p,w,p \times w\}$; this curve is the unit circle in the "XOY" plane of this basis).

• has the following velocity at $t=0$: $s'(t)|_{t=0}=-\underbrace{\sin(t)p|_{t=0}}_{=0}+\underbrace{\cos(t)w|_{t=0}}_{=1w}=w$, instead of $v$.

We will obtain an initial speed vector equal to $v$ by replacing (1) by:

$$\tag{2}\sigma(t)=\cos(t\|v\|)p+\sin(t\|v\|)w$$

(proof: differentiation of (2) gives $\sigma'(t)|_{t=0}=-\underbrace{\sin(t\|v\|)\|v\|p|_{t=0}}_{=0}+\underbrace{\cos(t\|v\|)\|v\|w|_{t=0}}_{=1v \ \text{by def. of w}}$)

• Ok. You're using the unit-radius sphere, right? So that $p^2=1$. Also: your reparametrization was linear, but it need not be linear, did it? (the trace of the curve would be the same no matter what type of parametrization...)
– Soap
May 17 '17 at 23:16
• Also: you said "clearly describes a great circle": I would say that this is "clear" because the points of the type $s(t)$ are a linear combination of $v$ and $p$ (and are therefore in the plane spanned by these vectors) and $\vert s(t) \vert ^2=1$, meaning that these points are both in that plane and in $S^2$, and the intersection of these sets is precisely the great circle of interest. Right?
– Soap
May 17 '17 at 23:22
• I come back (6:00 am o'clock CET). Your 1st comment: a) yes, I took $a=1$ with your notations, but I should have mentionned it. b) If you want to be compliant with a given speed vector, you need to reparametrize. If only the trajectory is important, no need for reparameterization. For your 2nd comment : yes, that's it ; nevertheless, I will add some explanation to my text. May 18 '17 at 4:14
• Explanations added to the text. May 18 '17 at 4:34
• Just one more thing: was it relevant that the reparametrization is linear?
– Soap
May 18 '17 at 8:49

You need two points to define a line. 2 (non-antipodal) points define a great circle.

lets call them $p_1, p_2.$

$u = p_2 - (\frac {p_1\cdot p_2}{a^2} )p_1$ is a vector in the plane containing $p_1, p_2$ that is orthogonal to $p_1.$

$\frac {p_1}{a}, \frac {u}{\|u\|}$ unitizes these vectors.

$p_1 \cos t + a\frac {u}{\|u\|} \sin t$ will give your parametric equations for your great circle.

$p_1 \cos \frac va t + a\frac {u}{\|u\|} \sin \frac va t$ will give your parametric with velocity v.

• What about two antipolar points? May 17 '17 at 18:57
• I was just thinking about this after I posted as I was going of to lunch. As Riemann defined spherical geometry, each point includes its antipodes. if $p_1$ and $p_2$ are antipodes they are not distinct points May 17 '17 at 21:00