Is Lorentz invariance is true in curved spacetime?

[kroni]

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 ?

 

[Nugatory]

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.

 

[bcrowell]

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

 

[jambaugh]

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 ds^2=dx^2 + dy^2 + dz^2 -dt^2 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.

 

[kroni]

Tanks for your answer ! I under stand it !