A Universe of Sorts

§ Geodesic equation, Extrinsic

The geodesic on a sphere must be a great circle. If it's not, so say we pick a circle at some fixed azimuth, then all the velocities point towards the center at this azimuth, not at the center of the sphere! But towards the center of the sphere is the real normal plane. So we get a deviation from the normal.

§ How do we know if a path is straight?

Velocity remains constant on a straight line.

So it has zero acceleration.

If we think of a curved spiral climbing a hill (or a spiral staircase), the acceleration vector will point upward (to allow us to climb the hill) and will the curved inward into the spiral (to allow us to turn as we spiral).

On the other hand, if we think of walking straight along an undulating plane, the acceleration with be positive/negative depending on whether the terrian goes upward or downward, but we won't have any left/right motion in the plane .

If the acceleration is always along the normal vectors, then we have a geodesic.

§ Geodesic curve

Curve with zero tangential acceleration when we walk along the curve with constant speed.

Start with the

(u, v)

plane, and map it to

R(u, v) \equiv (R_x, R_y, R_z)

. Denote the curve as

c: I \to \mathbb R^3

such that

c

always lies on

R

. Said differently, we have

c: I \to UV

, which we then map to

\mathbb R^3

via

R

So for example,

R(u, v) = (\cos(u), \sin(u)\cos(v), \sin(u)\sin(v))

and

c(\lambda) = (\lambda, \lambda)

. Which is to say,

c(\lambda) = (\cos(\lambda), \sin(\lambda)\cos(\lambda), \sin(\lambda)\sin(\lambda))

Recall that

e_u \equiv \partial_u R, e_v \equiv \partial_v R \in \mathbb R^3

are the basis of the tangent plane at

R_{u, v}

Similarly,

\partial_\lambda c

gives us the tangent vector along

c

on the surface.

Write out:

\begin{aligned} &\frac{dc}{d \lambda} = \frac{du}{d\lambda}\frac{dR}{du} + \frac{dv}{d\lambda}\frac{dR}{dv} &\frac{d}{d\lambda}(\frac{dc}{d \lambda})\\ &=\frac{d}{d\lambda}(\frac{du}{d\lambda}\frac{dR}{du} + \frac{dv}{d\lambda}\frac{dR}{dv}) \\ &=\frac{d}{d\lambda}(\frac{du}{d\lambda}\frac{dR}{du}) + \frac{d}{d\lambda}(\frac{dv}{d\lambda}\frac{dR}{dv}) \\ &= \frac{d^2 u}{d\lambda^2}\frac{dR}{du} + (\frac{du}{d\lambda} \frac{d}{d\lambda} \frac{dR}{du}) \frac{d^2 v}{d\lambda^2}\frac{dR}{dv} + (\frac{dv}{d\lambda} \frac{d}{d\lambda} \frac{dR}{dv}) \end{aligned}

How to calculate

\frac{d}{d\lambda} \frac{dR}{ddu}

? Use chain rule, again!

\frac{d}{d\lambda} = \frac{du}{d \lambda}\frac{\partial}{\partial u} + \frac{dv}{d \lambda}\frac{\partial}{\partial v}

§ Geodesic curve with notational abuse

Denote by

R(u, v)

the surface, and by

R(\lambda)

the equation of the curve. So for example,

R(u, v) = (\cos(u), \sin(u)\cos(v), \sin(u)\sin(v))

while

R(\lambda) = R(\lambda, \lambda) = (\cos(\lambda), \sin(\lambda)\cos(\lambda), \sin(\lambda)\sin(\lambda))