Recent content by cianfa72

Indeed, given the Langevin congruence (i.e. the orbits of the timelike vector field ##\xi^a##) your projection operator on the quotient space is not ##h^a{}_b## , i.e. ##\eta^{ac} h_{cb}## (note in this case ##g^{ab} = \eta^{ab}## since the underlying spacetime is Minkowski). Technically...

The point I was trying to make is technical. The set of smooth tensor fields of a given type on the smooth manifold M isn't a vector space, rather a module over the ring of ##C^{\infty}(M)## functions. So technically ##h_{ab}\xi^a = 0## holds fiberwise. Then one extends this to the tensor...

Yes. The point is that we have a tensor field (the metric tensor field ##g_{ab}##) and a timelike congruence ##\xi^a##. The latter, evaluated at each point, allows us to split both tangent and cotangent space at that point in a direct sum. This decomposition extends to all sort of their tensor...

Yes, agreed. Indeed the projection applies to the set of tensor fields defined on a tensor product bundle fiberwise (in this case twice covariant). This set doesn't carry a vector space structure so the projection is defined fiberwise (i.e. on each tensor product at point p).

I think this depends on the definitions used. For example in Lecture 18: Canonical Formulation of GR I, the metric tensor field ##h_{ab}## transversal to the orbits of the timelike vector fields ##u^a## is defined as ##h_{ab} = g_{ab} + u_au_b##. What is three-dimensional is the pullback of the...

Ah ok, transverse to both of them. Yes, that was my point (where ##i,j## run for 1,2,3). BTW, I believed that the projection of ##g## on the spacelike hypersurfaces was actually 4-dimensional though.

Yes. They are the composite conditions for Lorenz + TT gauge. Namely, in Lorenz gauge $$\partial_{\mu}h^{\mu}{}_{\nu} - \frac {1} {2} \partial_{\mu} h^{\nu}{}_{\nu} = 0$$ the EFEs for linearized gravity in vacuum become ##\Box h_{\mu\nu} = 0##. Take now as solutions the plane waves propagating...

Ok, so that means that even though ##h_{\mu\nu} \left( \frac {\partial} {\partial_t} \right)^{\nu}= 0## , the metric ##g## restricted to spacelike hypersurfaces of constant coordinate time ##t## in TT gauge coordinates is not ##h_{\mu\nu}## (the latter is given in those coordinates as in post #14).

I believe ##h_{\mu\nu}## isn't the transverse part of the metric tensor field ##g## w.r.t. the timelike congruence given by integral curves of ##\frac {\partial} {\partial_t}## vector field in TT gauge coordinates. Indeed for any vectors ##X_p## and ##Y_p## both orthogonal to ##\left( \frac...

Coming back to TT gauge coordinates, the components of the metric tensor field ##g## in that chart are $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$ in which ##h_{\mu \nu}## is solution of the wave equation $$\Box h_{\mu \nu} = 0$$ and ##\eta_{\mu \nu}## is diag(-1,1,1,1).

Let's say we transform from SC exterior chart (that's defined for ##r > 2m, - \infty < t < \infty##) coordinates to Kruskal-Szekeres (KS) coordinates ##(T,X)## (suppressing the inessential others). Now, as far as I can understand your point, instead of restricting the target to the image of the...

Ok, so in the case of SC spacetime geometry, does exist a maximal analytic extension that can be obtained starting from the SC chart and metric in exterior region outside the horizon ?

ok yes. Yes. Just to be clear: the metric isn't stationary (in the spacetime region where the GW passes). The proper length of each arm is defined as the length along a spacelike curve orthogonal to the arm's worldsheet (assuming it has only one spatial dimension). So, what you can get from the...

Ah ok. So for example it is part of Schwarzschild chart's specification that it covers the open region of spacetime outside the horizon. You mean: the attempt to compute the maximum analytic extension of the SC chart that is defined on the open region of SC spacetime outside the horizon (given...

I mean....if I give you an expression for the metric tensor field in some chart and nothing else, you can't say in which region of spacetime such expression applies. Now if the metric components in that (implied) chart have no singularities at all then, in principle, does make sense to extend...