Geodesics on a cone in flat space

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So if you take a sphere with coordinates (r, [tex]\theta[/tex],[tex]\phi[/tex]) and keep [tex]\theta[/tex] constant you get a cone.
The geodesic equations reduce to (by virtue of the euler - lagrange equations):

[tex]\frac{\mathrm{d} ^{2}r}{\mathrm{d} s^{2}} - r\omega ^{2}\frac{\mathrm{d} \phi }{\mathrm{d} s} = 0[/tex] where [tex]\omega[/tex] = sin[tex]\theta _{0}[/tex]

and [tex]\frac{\mathrm{d} \phi }{\mathrm{d} s} = k / r^{2}[/tex]
where k is an arbitrary constant gotten from the fact that the derivative of the lagrangian with respect to a cyclic coordinate is a constant.

These reduce to [tex]\frac{\mathrm{d} }{\mathrm{d} \phi }(\frac{\mathrm{d}r }{\mathrm{d} \phi }\frac{1}{r^{2}}) = \frac{\omega ^{2}}{r}[/tex] and if you solve by substituting p = 1 / r then the equation has a solution of the form:

[tex]1 = Arcos(\omega \phi ) + Brsin(\omega \phi )[/tex] where A and B are arbitrary constants. How can you tell from this equation for the geodesic that it follows straight lines on the cone if the cone were to be flattened out?
 
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Of course they're straight -- they're geodesics! I guess what you're really asking is, why is the cone "flat", which follows from the fact that r, φ are plane polar coordinates.
 
Hi WannabeNewton! :smile:

(have a theta: θ and a phi: φ and an omega: ω :wink:)
WannabeNewton said:
… How can you tell from this equation for the geodesic that it follows straight lines on the cone if the cone were to be flattened out?

Flatten the cone by slitting in along one generator …

then the angle on that plane will be ωφ, and so any straight line will have rcos(ωφ + constant) = constant …

which is the same as your geodesic equation! :wink:
 
Basically a cone is flat. Physically, you can cut a wedge out of a paper and fold it up into a cone. Mathematically, you then have a "map" from every point on the flat piece of paper (minus the wedge) to a point on the cone.

However, one has to exclude the tip of the cone to get a proper manifold structure. When this is done, the curvature tensor is flat everywhere it is defined, which is everywhere but the tip of the cone.