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 Exercise [14.28] 
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Joined: 03 Jun 2010, 15:18
Posts: 136
Post Exercise [14.28]
Attached my solution.
It makes use of a technique (Euler-Lagrange equations for variational problems) not mentioned in the book; but I was not able to find any other way to solve the problem.


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05 Jul 2010, 10:29

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [14.28]
Roberto, you got the math right, but were slightly off the mark in actually proving the argument. In particular, you can't use the result of [14.26], because that's the very thing we're trying to demonstrate!

Instead, look at the equation at the top of your page 2 (of the attachment).
As you later point out, if u is chosen to be an affine parameter, R is a constant along the curve and thus the RHS of this equation vanishes, leaving you with an equation defining t^a\partial_a t^b (=dt/du) in terms of t, g, and derivatives of g.

Now consider that along the affine geodesic \nabla_t t is zero. Expanding the expression for the covariant derivative (connection derivative) of a vector t along t yeilds:

\nabla_t t = \partial_t t + t^a t^b \Gamma^c_{ab}

and since that is zero, you can substitute your g expression to replace the partial derivative term, yielding the definition of \Gamma, just as it is in [14.26], as required!

[Actually, there's a little subtlety here: What you actually get is an expression for t^a t^b \Gamma^c_{ab} that's true for any vector t at any point (since you can always construct an affine geodesic through any point in any direction).

It's a simple matter to show, however, that this is sufficient to completely define \Gamma when it's symmetric in it's two bottom indices: Just note that due to the symmetry, given any two arbitrary vectors s and t, st\Gamma = uu\Gamma-vv\Gamma where s = u + v, t = u - v.]


30 Oct 2010, 19:02

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [14.28]
Because of the known problem with Latex, your formulas cannot be read, but I think to have understood anyway your point.
The exercise is to show that metrics defines a connection, and my wording was not clear; before the equation I called (1'') instead of:
"Using result of exercise [14.26], the equation of minimum length line is ..."
I should have said:
"Defining the factor that multiples t^a t^b as the connection, and this agrees with result of exercise [14.26], the equation of minimum length line is ...".
In any case you showed that metrics implies a unique connection, what I missed out.


03 Nov 2010, 14:52
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [14.28]
Roberto
If you want to see the Latex formulae all you need to do is click on "quote" at the top right of the post and you can see the Latex and copy it onto your computer and look at it from there.
Vasco


03 Nov 2010, 20:26

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [14.28]
Just to clarify, the metric implies a unique connection, given the assumption that there's no torsion, yes.


04 Nov 2010, 00:12
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