- #1

B3NR4Y

Gold Member

- 170

- 8

## Homework Statement

Find the geodesics on the cone whose equation in cylindrical-polar coordinates is z = λρ [Let the required curve have the form φ=φ(ρ)] check your result for the case λ→0

## Homework Equations

[itex] \frac{\partial F}{\partial y} - \frac{d}{dx} (\frac{\partial F}{\partial y'}) = 0 [/itex] and [itex] ds^{2} = d\rho^{2} + \rho d\theta^{2} + dz^{2} [/itex]

## The Attempt at a Solution

The integral of ds gives the length of a curve, a geodesic is a curve with the shortest length between two points on a curved surface. So the Euler Equation is necessary.

[itex] s = \int_{a}^{b} \sqrt{d\rho^{2} + \rho d\theta{2} + dz^{2} } = \int_{a}^{b} \sqrt{1 + \rho \theta ^{'2} + z^{'2}} d\rho [/itex], I have an equation for z so z' is easy to find, it's just λ. [itex] \int_a^b \sqrt{1 + \rho \theta^{'2} + \lambda^{2} } d\rho [/itex].

From here I just used the Euler-Equation, noting that the lack of a θ means that ∂F/∂θ' = constant by the Euler-Equation, so [itex]\frac{\rho \theta ^{'}}{\sqrt{1+\rho \theta^{'2} + \lambda^{2}}} = C [/itex], some algebra gets me to [itex] \theta^{'} = \sqrt{\frac{C+C\lambda^{2}}{\rho^{2}(\rho^{2} - C)}} [/itex] which integrating with respect to rho yields [itex]\theta =\int \sqrt{\frac{C(1+\lambda ^2)}{\rho ^{2}(\rho ^{2}-C})} d\rho[/itex]

I'm not sure if this is correct or not. I assume if they want me to check the result for the case λ=0, that case will look familiar to me. It doesn't.

I made a boo boo, rho is squared in the line element. I'll work it out, but I feel like I still need help.

So I fixed my mistake, and have decided [itex] \theta =\int \sqrt{\frac{C(1+\lambda ^2)}{\rho ^{2}(\rho ^{2}-C})} d\rho [/itex]. Boy is this a hard integral to do, wolfram spits out some pretty crazy stuff. Where should I start integrating this?

Last edited: