Using Euler Lagrange in General Relativity

1. Mar 13, 2008

pixel2001n

I'm trying to follow a professor's notes for finding Christoffel symbols for a two-sphere. He gives the following two equations:

The Lagrangian for a two sphere:$$L = \left( \frac{d\theta}{ds} \right)^2 + sin^2\theta \left( \frac{d\phi}{ds} \right)^2$$

The Euler Lagrange equation:$$\frac{d}{ds} \left( \frac{\partial L}{\partial (dx^\mu/ds)} \right ) - \frac{\partial L}{\partial x^\mu} = 0$$

Using these, the professor magically gets:

for $$x^\mu = \theta$$:$$2\frac{d^2\theta}{ds^2} - 2 sin\theta cos\theta \left(\frac{d\phi}{ds^2}\right)$$

for $$x^\mu = \phi$$:$$2sin^2 \theta \frac{d^2\phi}{ds^2} + 4 sin\theta cos\theta \left(\frac{d\theta}{ds}\frac{d\phi}{ds}\right)$$

And the Christoffel symbols can be found with minimal effort. The problem is that I can't follow the derivation to get the equations. I feel like this is a really simple thing, yet I'm having trouble getting the same answer as he showed.

For example, in the $$\theta$$ case:

$$\frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left( \left( \frac{d\theta}{ds} \right)^2\right)\right ) + \frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right ) - \frac{\partial}{\partial \theta}\left(\left( \frac{d\theta}{ds} \right)^2\right) - \frac{\partial}{\partial \theta}\left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right) = 0$$

$$2 \frac{d^2\theta}{ds^2} + \frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right ) - \frac{\partial}{\partial \theta}\left(\left( \frac{d\theta}{ds} \right)^2\right) - 2 sin\theta cos\theta \left( \frac{d\phi}{ds} \right)^2 = 0$$

But I don't see how to drive the other terms to zero.

The $$\phi$$ case is even worse:

$$\frac{d}{ds} \left( \frac{\partial}{\partial (d\phi/ds)} \left( \left( \frac{d\theta}{ds} \right)^2 \right) \right ) + \frac{d}{ds} \left( \frac{\partial}{\partial (d\phi/ds)} \left( sin^2\theta \left( \frac{d\phi}{ds} \right)^2 \right) \right ) - \frac{\partial}{\partial \phi} \left( \left( \frac{d\theta}{ds} \right)^2 \right) - \frac{\partial}{\partial \phi} \left( sin^2\theta \left( \frac{d\phi}{ds} \right)^2 \right) = 0$$

$$\frac{d}{ds} \left( \frac{\partial}{\partial (d\phi/ds)} \left( \left( \frac{d\theta}{ds} \right)^2 \right) \right ) + \frac{d}{ds} \left( 2 sin^2\theta \left( \frac{d\phi}{ds} \right) \right ) - \frac{\partial}{\partial \phi} \left( \left( \frac{d\theta}{ds} \right)^2 \right) - \frac{\partial}{\partial \phi} \left( sin^2\theta \left( \frac{d\phi}{ds} \right)^2 \right) = 0$$

I'm not sure how to simplify it from there, since each time I try a method I get the wrong answer. I especially don't see where his $$+4sin\theta cos\theta$$ came from.

All my attempts have failed. It seems like such a trivial thing, since the professor left it out. And obviously the results are correct since they give the correct Christoffel terms. I would be eternally grateful if someone with a working knowledge of how to combine ordinary and partial derivatives could give me a step by step.

2. Mar 15, 2008

Mentz114

$$\frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right ) \equiv \frac{d}{ds} \left( \frac{\partial}{\partial y} \left(sin^2\theta \left( z \right)^2\right)\right )$$

This term is zero, because $$\left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right )$$
is not explicitly dependent on $$\frac{d\theta}{ds}$$.

All your troublesome terms disappear for similar reasons.

3. Mar 15, 2008

kdv

Mentz114 already gave you the explanation. Are you familiar with the Lagrangian approach in ordinary mechanics? If so, recall that is the time derivative of a variable does not appear in the lagrangian the corresponding conjugate momentum is conserved because $$\frac{\partial L}{\partial \dot q} = 0$$ . It's the same type of thing here.

Maybe it would be useful to write the Lagrangian as

$$L = \dot{\theta}^2 + \sin^2 \theta (\dot{\phi})^2$$

where a dot indicates a derivative with respect to s, obviously.

And now you must think of $\theta$ and $\dot{\theta}$ as independent variable (same for $\phi$ and $\dot{\phi}$).

For example,

$$\frac{\partial L}{\partial \dot{\theta}} = 2 \dot{\theta}$$
simply.

Then it's a snap to get the correct result from the E-L equations

4. Mar 17, 2008

pixel2001n

That was my problem. Thanks guys.