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Metric compatibile connection

  1. Oct 13, 2004 #1

    hellfire

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    To have a connection in GR one imposes several conditions. Among others, one condition is that the connection shall be torsion free, which leads to symmetric Christoffel symbols. My understanding is that this condition is imposed to maintain the theory as simple as possible. The absence of this condition would imply some physical phenomenon (spacetime with torsion, additionally to curvature), which is not observed (?).

    Further, one imposes metric compatibility, which leads to a vanishing covariant derivative of the metric. I assume that the reasons which make this property desirable are the same as above, but I cannot imagine any physical phenomenon related to the absence of this condition. So, my question is: what is the physical interpretation of a connection which is not compatible with the metric? I read in Sean Carrolls GR notes (p. 91), that to introduce fermions in GR one has to make use of a connection which is not metric compatible (called spin connection), why?
     
    Last edited: Oct 13, 2004
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  3. Oct 13, 2004 #2

    Fredrik

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    I don't know these things very well, but I think that if the connection isn't metric compatible, you can parallell transport two vectors and not preserve the "angle" (definied by the metric) between them. I can't think of any physical consequences of that right now.

    I also think that a spin connection is something more complicated. I don't think it's a connection on a manifold. Isn't it a connection on some kind of fiber bundle? The words "spin bundle" comes to mind. I don't even know what that is. :smile:
     
  4. Oct 13, 2004 #3

    Stingray

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    That's right. You can have a connection without a metric, but if you do have a metric, then the one compatible with it is preferred for this reason (presuming no torsion). Preserving inner products is sort of the whole "point" of parallel transport. Certainly derivative operators would act very counter-intuitively if this were not true. Maybe that's not as deep an answer as you were looking for?
     
  5. Oct 14, 2004 #4

    hellfire

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    Well, thanks for your answers. I think I did understood more or less which are the technical reasons which lead to the conditions mentioned above. For a metric compatible connection the invariance of inner products is indeed a reason (equivalent to stating that the covariant derivative of the metric shall vanish everywhere) which makes the whole mathematical formalism easier.

    But I would like to understand whether there are physical phenomena underlying these properties (or the absence of these properties). The spin connection I mentioned above is introduced by Carroll (http://arxiv.org/gr-qc/9712019) when he explains non-coordinate bases (p. 91). In the relation between the connection coefficients of the metric compatible connection and the spin connection a term appears with the partial derivative of the non-coordinate basis vectors. But this is behind my knowledge: I just wanted to get a feeling before trying to understand this part of the notes.
     
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