Variational Calculus: Finding a Geodesic with EL Equation

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Homework Help Overview

The original poster is attempting to find a geodesic using the Euler-Lagrange equation in the context of variational calculus. The problem involves the differential line element in R^2 and its parametrization.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the Euler-Lagrange equation derived from the differential line element. Questions arise regarding the necessity of a linear solution for geodesics and the dimensional correctness of the expressions used.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the equations and their implications. Some guidance has been offered regarding the conditions under which the terms in the equation could be zero, but no consensus has been reached on the correctness of the original equation.

Contextual Notes

There is a noted concern about the dimensional correctness of the expressions used in the problem setup. The original poster has acknowledged a mistake in their initial formulation but maintains that their calculations were correct.

ehrenfest
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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

 
Last edited:
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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

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

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
 
Last edited:
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.)
 
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:

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