I GR: Attractors & Liouville's Theorem

atyy
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Does GR have attractors?
In classical Hamiltonian mechanics, because of Liouville's theorem about the volume of phase space being preserved by time evolution, there are no attractors.

Naively, I think of the Raychaudhuri equation in GR as showing a shrinking volume. However, I guess Raychaudhri's equation does not deal with the phase space of GR.

Does GR have attractors?
Is there something like the Liouville theorem in GR, especially since GR has a Hamiltonian formulation?
 
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Just a guess, but given that very little is known about the global properties of solutions to the initial value problem, I would say that it is a very hard question.
 
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atyy said:
Is there something like the Liouville theorem in GR, especially since GR has a Hamiltonian formulation?
Every Hamiltonian system obeys the Liouville theorem. GR is not an exception. Note, however, that the phase space in field theories is infinite dimensional. In particular, the 3-dimensional volume of a lump of matter (which may shrink due to gravitational collapse) has nothing to do with the infinite-dimensional phase-space volume.
 
atyy said:
Naively, I think of the Raychaudhuri equation in GR as showing a shrinking volume.
GR is invariant under the time reversal. Raychaudhuri equation can describe also the expansion. In the black-hole context, such an expanding solution is called a white hole. Such solutions are usually discarded because they require unrealistic initial conditions, or equivalently, because for such solutions entropy decreases with time.
 
Demystifier said:
Every Hamiltonian system obeys the Liouville theorem. GR is not an exception. Note, however, that the phase space in field theories is infinite dimensional. In particular, the 3-dimensional volume of a lump of matter (which may shrink due to gravitational collapse) has nothing to do with the infinite-dimensional phase-space volume.
Are you sure the theorem holds in field theories?
 
martinbn said:
Are you sure the theorem holds in field theories?
Formally yes, but I guess it's not fully rigorous due to the infinite number of degrees of freedom.
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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