Variational Calculus: Finding a Geodesic with EL Equation

ehrenfest
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Homework Statement



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?

Homework Equations


The Attempt at a Solution

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

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

Otherwise, the expression you had isn't even correct dimensionally...
 
Yes, that was foolish. But i had it correctly in my calculations so my question remains unanswered
 
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Lemme see if I remember how this goes...

Your equation

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