MTW: Exer. 8.11, pg. 215

1. Sep 6, 2010

Living_Dog

It is the last part that I am having trouble understanding. It says that if u and w are tangent vectors then,

$$\nabla_{\bold{u}}\bold{w} - \nabla_{\bold{w}}\bold{u} = [\bold{u},\bold{w}]$$.​

Now,

$$[\bold{u},\bold{w}] = \partial_{\bold{u}}\partial_{\bold{w}} - \partial_{\bold{w}}\partial_{\bold{u}} = (u^\beta\,v^{\alpha}_{,\beta} - v^\beta\,u^{\alpha}_{,\beta})\bold{e}_\alpha$$.

But,

$$\nabla_{\bold{u}}\bold{w} - \nabla_{\bold{w}}\bold{u} = (u^\beta\,w^{\alpha}_{,\beta} - w^\beta\,u^{\alpha}_{,\beta})\bold{e}_\alpha + (u^\beta\,w^{\alpha} - w^\beta\,u^{\alpha})\bold{e}_{\alpha}_{,\beta}$$.

So how do I reason that the second term disappears? Is the derivative of the vector basis zero b/c this is in a local Lorentz frame? In other words, there are no correction terms??

-LD

2. Sep 6, 2010

Work in coordinate basis $$e_\alpha=\partial_\alpha$$. Use the fact that the Levi-Civita connection is symmetric and
$$\nabla_\alpha e_\beta = ....$$

3. Sep 6, 2010

Living_Dog

[1] By "Levi-Civita connection" do you mean:
a) the "connection coefficients" aka
b) the affine connection aka
c) the Christoffel symbols??

[2] In a coordinate basis, the commutator of the partials is zero. The second term is not a commutator of the partials - but of the components of the respective vectors. So it can't be canceled out.

[3] Finally, $$\nabla_{\alpha}\,\bold{e}_{\beta}$$ is the connection coefficient, $$\Gamma^{\mu}_{\alpha\beta}\bold{e}_{\mu}$$. So then the other term would be: $$\nabla_{\beta}\,\bold{e}_{\alpha} = \Gamma^{\mu}_{\beta\alpha}\bold{e}_{\mu}$$ ... which when subtracted - and since the Christoffel symbols are symmetric in their lower indices - would then make the second term cancel out!

So I was too quick to match dummy indices with these two terms! Yes?? (If so thanks dude!!)

-LD

4. Sep 7, 2010

"would then make the second term cancel out"
Just remeber: if $$A^{\alpha\beta}$$ is antisymmetric and $$B_{\alpha\beta}$$ is symmetric, then $$A^{\alpha\beta}B_{\alpha\beta}=0$$.

5. Sep 7, 2010

Living_Dog

Yes! the (uw-wu) is anti-symmetric in a,b and the e_a,b is symmetric in a,b. That's a good one to remember.

Thanks for all your kind help and patience with me. May God richly bless you, in Jesus' name, amen.

-joe