# The point of the Euler-Lagrange equation?

1. Jan 9, 2016

### 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 dont have to integrate and differentiate with two variables.
Here is the equation

2. Jan 10, 2016

### haruspex

Last edited by a moderator: May 7, 2017
3. Jan 11, 2016

### 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 $L=L(t,q(t),\dot{q}(t))$.

4. Jan 11, 2016

### lavinia

Last edited by a moderator: May 7, 2017
5. Jan 12, 2016

### 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.

Last edited: Jan 13, 2016
6. Jan 13, 2016

### 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, $∇_{e_1}e_1 = ω_{12}(e_1)e_2 = 0$.

Last edited: Jan 15, 2016