# Variational calculus

1. Oct 18, 2007

### ehrenfest

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

I am trying to find a geodesic with Euler-Lagrange equation by varying the function

$$ds/d\tau = \sqrt{\dot{x} + \dot{y}}$$

EDIT: it should be $$ds/d\tau = \sqrt{\dot{x}^2 + \dot{y}^2}$$

where tau is a parametrization and the dot means a tau derivative.

However, when I plug that into the EL equation for either x or y, I get:

$$(\dot{x} + \dot{y})^{3/2}(\ddot{x}+\ddot{y})=0$$

How do I get a line from that?

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 18, 2007
2. Oct 18, 2007

### Kreizhn

I'm don't know much about variational calculus, but as far as geodesics go, why does it have to be a line?

3. Oct 18, 2007

### ehrenfest

Because the closest distance between 2 points in R^n is and the ds I am using is the differential line element in R^n.

In R^2 the differential line element is: ds = sqrt(dx^2+dy^2)

I just divided both sides by the parameter dtau to get the first equation in my last post.

4. Oct 18, 2007

### dynamicsolo

If I'm understanding what you're saying, shouldn't you get

$$ds/d\tau = \sqrt{\dot{x}^{2} + \dot{y}^{2}}$$ ?

Otherwise, the expression you had isn't even correct dimensionally...

5. Oct 18, 2007

### ehrenfest

Yes, that was foolish. But i had it correctly in my calculations so my question remains unanswered

Last edited: Oct 18, 2007
6. Oct 19, 2007

### dynamicsolo

Lemme see if I remember how this goes...

$$(\dot{x} + \dot{y})^{3/2}(\ddot{x}+\ddot{y})=0$$

implies that either

$$(\dot{x} + \dot{y})^{3/2}=0$$ or $$(\ddot{x}+\ddot{y})=0$$.

For the first term to be zero, you'd need

$$\dot{x} = - \dot{y}$$, which would give you dy/dx = -1 , no?

For the second term to be zero, you'd have

$$\ddot{x} = -\ddot{y}$$ , which takes more antidifferentiation, but I believe also leads to a linear solution. (Not very rigorous, to be sure, but I believe that's basically how the argument runs.)

7. Oct 19, 2007

### ehrenfest

That's rigorous enough for me. Its kind of weird we have that -1 solution though. Did you check that my equation is correct?

Last edited: Oct 19, 2007