DIFFERENTIAL GEOMETRY COURSE
Explanation 182 : Suppose a car is riding over a surface, and gravitation is pulling it in the direction of -N. If the driver doesn't need to turn the wheel, the car is riding over a geodesic.
This is the case if and only if its traject curve x(s) is always complying with the requirement that, in linear approximation for Δs → 0, t(s+Δs) - t(s) has the same or opposite direction as N(s), so if the principal normal n of the curve has the same or opposite direction as the surface normal N. In other words: the component x ..tang of the curvature vector on the tangent must be 0.
Another description is the following.
Given a curve on a surface. The tangent planes in the points of the curve are enveloping a developable surface, whereon this curve is lying. When we develop this last surface, the image of the curve on the surface is a planar curve, the socalled trace curve. In other words: draw the curve with wet ink on the surface and roll the developable surface that the tangent planes are enveloping over the floor in such a way that the curve leaves a print on the floor. The print is the trace curve.
If the trace curve is a straight line, we call the original curve a geodesic of the surface. The shortest traject over the surface between two points on the surface is always a geodesic. We call the curvature of the trace curve geodetic curvature.
Example 183: Draw the trace curve of a great circle on the sphere (the corresponding developable surface that the tangent planes are enveloping is here a right circular cylinder),
and draw also the trace curve of a little circle (the corresponding developable surface that the tangent planes are enveloping is then a right circular cone).
So which curves are the geodesics of the sphere?
The next thing we have to do is deducing the equations we need to calculate the geodesics of a surface.
Definition 184: The geodesic curvature of a curve x(s) on a surface is the length of the tangential component of the curvature vector x .., so
κgeod = ||x .. - (x ..).N|| = |κ| ||n - (n.N)N||,
where κ is the curvature of x(s), n the principal normal vector of this curve, and N the normal vector on the surface.
We also have κ2 = k2 + κgeod2, where k is the normal curvature.
Definition 185: We call a curve on the surface a geodesic of this surface if the geodesic curvature is 0 in all points of the curve.
Proposition 186: Define the Christoffel symbols Γijk and γijk by
xuiuj = Γij1 xu1 + Γij2 xu2 + hij N,
γijk = xuiuj . xuk = Γij1 a1 k + Γij2 a2 k .
Then we find the geodesics x(u1(s),u2(s)) by solving
Γ11k (u1.)2 + 2 Γ12k (u1. u2.) + Γ22k (u2.)2 + uk.. = 0 (k=1,2).
Proof: Check by calculation that
x .. = (xu1u1 u1. + xu1 u2 u2.) u1. + xu1 u1.. + (xu1u2 u1. + xu2 u2 u2.) u2. + xu2 u2...
The tangential component of x .. must be 0. This gives the equations.
Proposition 187: Christoffel symbols are intrinsic variables and the concept geodesic is invariant under bending, because
2 γijk = (δ/δui) aj k + (δ/δuj) ai k - (δ/δuk) ai j.
Proof: From ai k = xui . xuk we get (δ/δuj) ai k = γijk + γkji.
Remark 188: The theory of differential equations learns that through any given point on a surface there goes exactly one geodesic in every direction. And through any two points on the surface there goes a geodesic that is the shortest traject over the surface between these two points.
Problem 189: Prove that the geodesics of the cylinder (a cos(v), a sin(v), u) are: the right circular helices, the circles and the rules.
Problem 190: Determine the geodesics on the paraboloid surface of revolution (u cos(v), u sin(v), (1/2)u2).
Problem 191: Determine the geodesics on the right circular cone.
Problem 192: The principal normal of a geodesic is also normal on the surface, so the osculating plane contains the normal on the surface.
Hence prove that an asymptotic line that is also a geodesic must be a straight line (see 142).
Are the circles of latitude of an arbitrary surface of revolution geodesics? And the meridian curves?