- #1
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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?
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?