TrickyDicky
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What exactly is the physical meaning of the fact that the covariant derivative of the metric tensor vanishes?
The discussion revolves around the physical meaning of metric compatibility in the context of covariant derivatives and its implications in general relativity. Participants explore theoretical aspects, mathematical definitions, and potential physical consequences related to gravitational fields and the nature of geodesics.
Participants express differing views on the necessity and implications of metric compatibility, with some arguing for its importance while others suggest alternative connections may be valid. The discussion remains unresolved regarding the broader implications of metric compatibility and its physical interpretations.
Some participants note that the discussion involves complex mathematical concepts that may not be fully accessible to all, indicating a potential gap in understanding the implications of metric compatibility.
LAHLH said:We choose this property of metric compatibility as it's known when we specify our connection. One could I believe choose different connections to Christoffell for which you wouldn't have metric compatibility, but then these are non physical. The fact that you geodesics are not just curves that parallel transport their tangent vectors wrt connection, but that they are also paths of extremal distance in spacetime is owing to the Christoffell connection being the connection we chose. Metric compat means lots of nice properties too, such as preservation of angle between two parallel transported vectors, or equivalently that geodesics can't change their character from null/timelike/spacelike.
I'm sure someone else will be able to give deeper and maybe more physical reasons still though...
The vanishing of the covariant derivative of the metric—the condition of
metric compatibility—is sometimes introduced perfunctorily in texts on general
relativity, but Schr¨odinger was right to call it “momentous” (Schr¨odinger 1985,
106). It means that the local Lorentz frames associated with a space-time point
p (those for which, at p, the metric tensor takes the form diag(1,−1,−1,−1)
and the first derivatives of all its components vanish) are also local inertial
frames (relative to which the components of the connection vanish at p).10 If
the laws of physics of the non-gravitational interactions are assumed to take
their standard special relativistic form at p relative to such local Lorentz charts
(the local validity of special relativity), then metric compatibility implies that
gravity is not a force in the traditional sense—an agency causing deviation from
natural motion—, in so far as the worldlines of freely falling bodies are geodesics
of the connection.
The full physical implications of the non-metric compatible connection in
Weyl’s theory remain obscure in the absence of a full-blown theory of matter.
Weyl’s hints at a solution to the Einstein objection seem to involve a violation of
minimal coupling, i.e. a violation of the prohibition of curvature coupling in the
non-gravitational equations, and hence of the local validity of special relativity.
But it seems that the familiar insight into the special nature of the gravitational
interaction provided by the strong equivalence principle—the encapsulation of
the considerations given in the previous paragraph—is lost in the Weyl theory.
Right, that's what's been said.bcrowell said:A covariant derivative is a derivative from which we've removed any contribution that is purely a result of the choice of coordinates.
I'm not sure what you mean here, the second derivatives are the ones determined by the metric.bcrowell said:The first derivatives of the metric are the way they are purely as a result of the choice of coordinates.
TrickyDicky said:bcrowell said:The first derivatives of the metric are the way they are purely as a result of the choice of coordinates.
I'm not sure what you mean here, the second derivatives are the ones determined by the metric.
bcrowell said:Since a covariant derivative like \nabla_a g_{bc} is supposed to exclude variation in g that occurs only because of the choice of coordinates, it has to be zero.
bcrowell said:I've never come across a clear explanation of what metric compatibility means (preferably both mathematically and physically), and why we care about it.