# Homework Help: Calculus of Var, Euclidean geodesic

1. Jul 12, 2012

### cpsinkule

1. The problem statement, all variables and given/known data
Calculate the geodesic for euclidean polar coordinates given ds$^{2}$=dr$^{2}$+r$^{2}$dθ$^{2}$

2. Relevant equations
standard euler-lagrange equation

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 13, 2012

### Kreizhn

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

3. Jul 13, 2012

### 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.

4. Jul 13, 2012

### 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.

5. Jul 13, 2012

### 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

6. Jul 13, 2012

### Kreizhn

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

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

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