Calculus of Var, Euclidean geodesic

[cpsinkule]

Homework Statement

Calculate the geodesic for euclidean polar coordinates given ds[itex]^{2}[/itex]=dr[itex]^{2}[/itex]+r[itex]^{2}[/itex]dθ[itex]^{2}[/itex]

Homework Equations

standard euler-lagrange equation

The Attempt at a Solution

I was able to reduce the euler-lagrange equation to [itex]\frac{d^{2}r}{dθ^{2}}[/itex]-rλ=0 where λ=[itex]\sqrt{(\frac{dr}{dθ})^{2}+r^{2}}[/itex] is the Lagrangian itself (namely the linear element)

My main concern is that I have the correct differential equation, I'm curious because I can't possibly imagine how this author expects me to solve that if it is indeed the correct DE for the lagrangian.

 

[Kreizhn]

Were you asked to solve it? Most geodesic equations cannot be solved analytically and require numerical analysis.

 

[cpsinkule]

I would think it would be easy to solve because it is obviously a straight line, that's why I think I have done something wrong to get this non-linear second order DE.

 

[Kreizhn]

I do not necessarily think it would be easy. Given an arbitrary straight line that does not cross through the origin, how would you write this in polar coordinates? It is actually a very difficult thing to do.

 

[cpsinkule]

so does that mean I have the correct DE? I'm not too worried about actually solving it, just that I have the right one and applied the E-L equations correctly to this problem

 

[Kreizhn]

I haven't had to do this stuff in a long time, but I computed my geodesic equations to be

[tex]\frac{d^2 r}{dt^2} - r \left( \frac{d\theta}{dt}\right)^2 = 0, \qquad r^2 \frac{d^2 \theta}{dt^2} + 2r \frac{dr}{dt} \frac{d\theta}{dt} = 0 [/tex]