Calculus of Var, Euclidean geodesic

cpsinkule
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Homework Statement


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


Homework Equations


standard euler-lagrange equation

The Attempt at a Solution


I was able to reduce the euler-lagrange equation to \frac{d^{2}r}{dθ^{2}}-rλ=0 where λ=\sqrt{(\frac{dr}{dθ})^{2}+r^{2}} 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.
 
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Were you asked to solve it? Most geodesic equations cannot be solved analytically and require numerical analysis.
 
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.
 
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.
 
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
 
I haven't had to do this stuff in a long time, but I computed my geodesic equations to be

\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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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