The point of the Euler-Lagrange equation?

[NihalRi]

When trying to come up with the geodesic equation for a sphere I came across this equation. My question, is this equation just a short cut so we don't have to integrate and differentiate with two variables.
Here is the equation
https://lh5.googleusercontent.com/proxy/Anoym1_CMRu6UeyS26mZmQ6lUIyHNg6uGfJMEY7usCCTrg_NtRCSl2teErq6b-uRGuiqYOWYQD0pMi2GLwYg2HP9fvx3mqPbwLlPonE6blR6Rn2EUaEv5M2u2IR2SjbGCJDinw=w136-h47-nc

 

[haruspex]

When trying to come up with the geodesic equation for a sphere I came across this equation. My question, is this equation just a short cut so we don't have to integrate and differentiate with two variables.
Here is the equation
[PLAIN]https://lh5.googleusercontent.com/proxy/Anoym1_CMRu6UeyS26mZmQ6lUIyHNg6uGfJMEY7usCCTrg_NtRCSl2teErq6b-uRGuiqYOWYQD0pMi2GLwYg2HP9fvx3mqPbwLlPonE6blR6Rn2EUaEv5M2u2IR2SjbGCJDinw=w136-h47-nc[/QUOTE]
I'm not sure what your integration and differentiation alternative looks like. The E-L equation optimises an entire path. Differentiating wrt one variable only optimises wrt that variable.

 

[Ssnow]

I don't know what is precisely your question but if can be useful I want remember you the dependence by the following variables $L=L(t,q(t),\dot{q}(t))$.

 

[lavinia]

When trying to come up with the geodesic equation for a sphere I came across this equation. My question, is this equation just a short cut so we don't have to integrate and differentiate with two variables.
Here is the equation
[PLAIN]https://lh5.googleusercontent.com/proxy/Anoym1_CMRu6UeyS26mZmQ6lUIyHNg6uGfJMEY7usCCTrg_NtRCSl2teErq6b-uRGuiqYOWYQD0pMi2GLwYg2HP9fvx3mqPbwLlPonE6blR6Rn2EUaEv5M2u2IR2SjbGCJDinw=w136-h47-nc[/QUOTE]

Yes. Geodesics are extremals of variations of the arc length function. So if you take a variation of smooth curves with fixed end-points and compute their arc lengths under the metric, solving for critical points tells you that you have a geodesic. This works for Riemannian metrics and although I don't know about Space-Time geometry, I think that timeline geodesics are extremals of proper time where the variation is through time like paths.

In (t,q) coordinates, $<∂/∂t,∂/∂t> = E$ $<∂/∂t,∂/∂q> = F$ and $<∂/∂q,∂/∂q> = G.$

The Lagrangian is then the function, $L(t,q,q') = (E + Fq' + G(t,q)(q')^2)^{1/2}$

- If you parameterize the sphere as a surface in 3 space, then the condition for a geodesic is that its normal vector is perpendicular to the sphere. This is the same as saying the geodesic curvature is zero. Or if you have an equation for geodesic curvature in terms of the metric then solve for geodesic curvature equal zero.

 

[lavinia]

A good exercise - with some tedious algebra - is to derive the differential equations for a geodesic expressed in terms of the metric from the Euler-Lagrange equations. It should not be a surprise that this can be done since the Lagrangian is just the length of the tangent vector to the path and so is expressed in terms of the $g_{ij}$'s. If one looks closely, one sees the Christoffel symbols appearing and then the equation has the usual form.

BTW: It is a theorem that one can also solve the variational problem without the horrid square root. Then the Lagrangian is called the "energy" in analogy with kinetic energy.

The Euler-Lagrange equations show that every extremal is a geodesic but not that every geodesic is an extremal.

BTW: It is not obvious to me that using the Euler Lagrange equations always makes things easier.

 

[lavinia]

If you choose polar coordinates on the sphere then the metric takes on a simple form.

It is $<∂/∂u,∂/∂u> = 1$ $<∂/∂u,∂/∂v> =0$ $<∂/∂v,∂/∂v> = sin^2(u)$

Whenever the metric is of this general form, $du^2 = 1$ $dudv = 0$ and $dv^2 = G(u)$ where G is some positive function that depends only on $u$, one immediately gets that the curves ,$v =$ constant, are geodesics. On the sphere these are great circles through the center of the coordinate system.

One can see this from the Euler-Lagrange equations without solving them in general. One can also see it from the covariant derivative. which is particularly easy to compute. One then uses the symmetry of the sphere to move these coordinates isometrically to any point and conclude that geodesics passing through any point are the radial lines $v =$ a constant i.e. great circles.

- For the covariant derivative computation, choose the orthonormal frame, $∂/∂u$ and $G^{-1/2}∂/∂v$.
The the dual frame is $e_1 = du$ $e_2 = G^{1/2}dv$ and ,therefore,the connection 1-form,$ω_{12}$, is $ω_{12} =1/2 G^{-1/2}G_udv$. The rest of the connection matrix is $ω_{11} = ω_{22} = 0$ and $ω_{21} = -ω_{12}$ by skew symmetry.

So the covariant derivative, $∇_{e_1}e_1 = ω_{12}(e_1)e_2 = 0$.