Is Lorentz invariance is true in curved spacetime?

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Hello,
I am re-reading a book about quantum physics and general relativity. To introduce representation of the lorentz group, they explain the definition of lorentz group as the group of transformation that let x² + y² ... -t² unchanged.
But in cuved space the distance is not the same as in minkowsky space, it's the integral of the metric * dx. Is the lorentz invariance still aviable ? and the related decomposition ?
 

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  • #2
Nugatory
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Hello,
I am re-reading a book about quantum physics and general relativity. To introduce representation of the lorentz group, they explain the definition of lorentz group as the group of transformation that let x² + y² ... -t² unchanged.
But in curved space the distance is not the same as in minkowsky space, it's the integral of the metric * dx. Is the lorentz invariance still aviable ? and the related decomposition ?

Lorentz invariance is still locally true - the metric around any given point is arbitrarily close to the Minkowski metric within a sufficiently small area around that point.
 
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bcrowell
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Another way of putting it is that there's a tangent space at each point. Things like momentum vectors live in the tangent space. This paper is also nice: Nowik and Katz, Differential geometry via infinitesimal displacements, http://arxiv.org/abs/1405.0984
 
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jambaugh
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Curved space-time is locally Minkowski and so there is a local lorentz group among the group of coordinate diffeomorphisms leaving an event point invariant. There is local Lorentz invariance etc. This is why in GR you work with Tensor *fields*. (tensor valued functions of position. (tensors being more general representations of the local orthogonal group, i.e. the Lorentz group))

As you noted, the local Lorentz group leaves the differential quantity [itex] ds^2=dx^2 + dy^2 + dz^2 -dt^2[/itex] unchanged... that is provided your coordinates are orthonormal with respect to the local space-time geometry. To be more general we express this invariant as a quadratic form of the differentials in whatever coordinates you are using and the quadratic form coefficients are the components of your metric tensor.

As you can imagine the math gets a bit hairier to express all this in general.
 
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Tanks for your answer ! I under stand it !
 

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