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Using Euler Lagrange in General Relativity

  • Thread starter pixel2001n
  • Start date
  • #1
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:[tex]L = \left( \frac{d\theta}{ds} \right)^2 + sin^2\theta \left( \frac{d\phi}{ds} \right)^2[/tex]

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

Using these, the professor magically gets:

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

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

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 [tex]\theta[/tex] case:

[tex] \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[/tex]

[tex] 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[/tex]

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

The [tex]\phi[/tex] case is even worse:

[tex]
\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[/tex]

[tex]
\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[/tex]

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 [tex]+4sin\theta cos\theta[/tex] 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.
 

Answers and Replies

  • #2
Mentz114
Gold Member
5,424
290
[tex]\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 )[/tex]

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

All your troublesome terms disappear for similar reasons.
 
  • #3
kdv
336
1
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:[tex]L = \left( \frac{d\theta}{ds} \right)^2 + sin^2\theta \left( \frac{d\phi}{ds} \right)^2[/tex]

The Euler Lagrange equation:[tex] \frac{d}{ds} \left( \frac{\partial L}{\partial (dx^\mu/ds)} \right ) - \frac{\partial L}{\partial x^\mu} = 0[/tex]
.
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 [tex] \frac{\partial L}{\partial \dot q} = 0 [/tex] . It's the same type of thing here.

Maybe it would be useful to write the Lagrangian as

[tex] L = \dot{\theta}^2 + \sin^2 \theta (\dot{\phi})^2 [/tex]


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

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

For example,

[tex] \frac{\partial L}{\partial \dot{\theta}} = 2 \dot{\theta} [/tex]
simply.

Then it's a snap to get the correct result from the E-L equations
 
  • #4
And now you must think of [itex] \theta [/itex] and [itex] \dot{\theta} [/itex] as independent variable (same for [itex] \phi[/itex] and [itex] \dot{\phi} [/itex]).
That was my problem. Thanks guys.
 

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