Geodesics on a cone in flat space

In summary, a cone is flat, you can cut a wedge out of a paper to make it into a cone, and the curvature tensor is flat everywhere but the tip of the cone.
  • #1
WannabeNewton
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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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  • #2
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.
 
  • #3
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:
 
  • #4
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.
 

1. What is a geodesic on a cone in flat space?

A geodesic on a cone in flat space is the shortest path between two points on the surface of a cone that is embedded in a flat three-dimensional space. It is the equivalent of a straight line on a flat plane.

2. How is a geodesic on a cone different from a geodesic on a sphere or a plane?

A geodesic on a cone differs from those on a sphere or a plane because the surface of a cone is curved in one direction and flat in the other. This creates a unique curvature that affects the path of a geodesic.

3. What is the mathematical equation for a geodesic on a cone in flat space?

The mathematical equation for a geodesic on a cone in flat space is known as the "cone equation." It is a second-order differential equation that takes into account the curvature of the cone's surface.

4. How are geodesics on a cone in flat space used in real-world applications?

Geodesics on a cone in flat space have various applications in fields such as engineering, physics, and mathematics. They are used to calculate the shortest distance between two points on a cone's surface, which is important in designing structures such as bridges and tunnels.

5. Are there any real-life examples of geodesics on a cone in flat space?

One real-life example of a geodesic on a cone in flat space is the shape of a suspension bridge. The cables of the bridge form a conical shape, and the shortest distance between the two anchor points is a geodesic on the cone's surface. Another example is the path of water droplets on the surface of a spinning cone, which follows a geodesic due to the surface's curvature.

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