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  Derivative of quantities with internal indices

+ 2 like - 0 dislike
1077 views

In the context of the 3 + 1 decomposition of spacetime needed for a Hamiltionian formulation of general relativity, quantities with so called internal indices are introduced (in the book I am reading on p.43). For such quatities $G^i$ , some kind of a "covariant derivative" is defined:

$D_aG^i = \partial_a G^i + \Gamma _{aj}^iG^i$

Using this derivative, a corresponding "curvature tensor" $\Omega_{ab}^{ji}$ is then calculated by

$D_aD_b - D_bD_a = \Omega_{ab}^{ji}G^i$

My quastions about this are:

1) Why is $\Gamma _{aj}^i$ called spin connection; it has to do with the spin of what ...?

2) How is the so called curvature of connection $\Omega_{ab}^{ji}$ related to the "conventional" curvature tensor ?

asked Mar 10, 2012 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
Which book are you using? (and by any chance does this have to do with the spin networks of loop quantum gravity?)

This post imported from StackExchange Physics at 2014-03-17 03:27 (UCT), posted by SE-user David Z
Ha ha @DavidZaslavsky, I`m reading this children`s book :-P and they introduce the notation and concepts I`m asking about around p.43.

This post imported from StackExchange Physics at 2014-03-17 03:27 (UCT), posted by SE-user Dilaton
... the link in my above comment does not work as I expected it; You have to scroll down to the bottom of the article to see why it is a children`s book :-)

This post imported from StackExchange Physics at 2014-03-17 03:27 (UCT), posted by SE-user Dilaton

1 Answer

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1) The spin connection allows you to define covariant derivatives of spinors in curved spacetime. For example, to do this, you want a set of gamma matrices which are covariantly constant, so you use the combinations

$\Gamma^i=\gamma^a E_a^i$

where $\gamma^a$ are the usual flat space gamma matrices and $E_a^i$ are the tetrad components, i.e.

$E_a=E_a^i\partial_i$

where $E_a$ is the tetrad basis for the tangent space.

2) The differential geometric relations between the vielbein formalism and the "standard" one is described in detail here. Section IV B describes the curvature relationship I think you're looking for.

This post imported from StackExchange Physics at 2014-03-17 03:27 (UCT), posted by SE-user twistor59
answered Mar 11, 2012 by twistor59 (2,500 points) [ no revision ]
Thanks @twistor59 this is very helpful; and the whole paper looks interesting. Although I`ll probably have to brush up my knowledge about the tetrade formalism (from my relativity demystified book) a bit first to fully understand it ... :-)

This post imported from StackExchange Physics at 2014-03-17 03:27 (UCT), posted by SE-user Dilaton

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