Variational Calculus: Finding a Geodesic with EL Equation

[ehrenfest]

Homework Statement

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

[tex] ds/d\tau = \sqrt{\dot{x} + \dot{y}} [/tex]

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

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:

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

How do I get a line from that?

Homework Equations

The Attempt at a Solution

 

[Kreizhn]

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

 

[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.

 

[dynamicsolo]

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.

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

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

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

 

[ehrenfest]

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

 

[dynamicsolo]

Lemme see if I remember how this goes...

Your equation

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

implies that either

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

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

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

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

[tex] \ddot{x} = -\ddot{y} [/tex] , 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.)

 

[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?