# Using Euler Lagrange in General Relativity

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.

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Mentz114
Gold Member
$$\frac{d}{ds} \left( \frac{\partial}{\partial (d\theta/ds)} \left(sin^2\theta \left( \frac{d\phi}{ds} \right)^2\right)\right ) \equiv \frac{d}{ds} \left( \frac{\partial}{\partial y} \left(sin^2\theta \left( z \right)^2\right)\right )$$

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.

kdv
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:$$\frac{d}{ds} \left( \frac{\partial L}{\partial (dx^\mu/ds)} \right ) - \frac{\partial L}{\partial x^\mu} = 0$$
.
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

And now you must think of $\theta$ and $\dot{\theta}$ as independent variable (same for $\phi$ and $\dot{\phi}$).
That was my problem. Thanks guys.