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For constant dark energy, Hubble value will eventually become asymptotic. If dark energy were dynamic and gently decreasing, what will the value of Hubble eventually become - will it asymptote or keep decreasing?
Trivial as in?the relationship between the Hubble parameter (##H##) and the dark energy density (##\rho##) will be trivial:
The equation immediately below that is the relationship.Trivial as in?
So the Hubble value will keep on decreasing without asymptotating?The equation immediately below that is the relationship.
The Hubble parameter will asymptotically approach that equation, provided it decreases more slowly than ##1/a^2##. If the density approaches zero, then so will the Hubble parameter.So the Hubble value will keep on decreasing without asymptotating?
Does this depend on the theory of gravity? (Apparently not GR, since we are contemplating a non-constant DE.) Or does this hold in any "reasonable" one, e.g., requiring only a homogenous & isotropic universe based on SR and matching Newton in the weak-gravity limit?It really depends upon how quickly it decreases. If it always decreases more slowly than ##1/a^2##, then it will remain the dominant energy density, and the relationship between the Hubble parameter (##H##) and the dark energy density (##\rho##) will be trivial:
$$H^2 = {8\pi G \over 3} \rho$$
If, however, the dark energy density at some point decreases more rapidly, then eventually any spatial curvature will instead become the dominant effect at late times:
$$H^2 = {k c^2 \over a^2}$$
The eventual fate would depend upon the sign of ##k##.
A non-constant dark energy is perfectly fine in GR and spatially isotropic and homogeneous FLRW universes. In a different context, seeDoes this depend on the theory of gravity? (Apparently not GR, since we are contemplating a non-constant DE.)
In the simplest models, inflation is driven by a single scalar inflaton field ##\phi \left(t \right)## that (assuming spatial homogeneity and isotropy of Friedmann-Lemaitre-Robertson-Walker universes) depends only on cosmological time. Initially, the "kinetic energy" ##\dot{\phi}^2/2## is much smaller than the potential energy term ##V\left( \phi \right)## (the form of ##V\left( \phi \right)## depends on the particular model). According (2.3.27) and (2.3.28) of Daniel Baumann's (Cambridge) excellent cosmology lecture notes
http://www.damtp.cam.ac.uk/user/db275/Cosmology/Lectures.pdf
this means that density and pressure are almost related by ##\rho = -p##, i.e., that during inflation, the universe is dominated by a large cosmological (almost) constant. Eventually, the inflaton field "rolls down" to near the minimum of the potential energy. For many models, the kinetic energy term ##V\left( \phi \right)## then dominates, and the universe acts briefly like a matter-dominated FLRW universe; see the discussion around (2.3.49) in Baumann. In order to get Standard Model stuff out, the Lagrangian contains interaction terms between the inflaton field and the Standard Model fields.These cause the universe to transition to a radiation-dominated Standard Model universe.
The Friedmann equations can also be derived in identical form from Newtonian gravity, with the caveat that Newtonian gravity has no concept of anything other than matter being a gravitational source. So the Newtonian form only includes normal matter and the ##k## term. In the Newtonian case, ##k## is not interpreted as spatial curvature, but instead as a value which is determined from some initial relationship between expansion and density.Does this depend on the theory of gravity? (Apparently not GR, since we are contemplating a non-constant DE.) Or does this hold in any "reasonable" one, e.g., requiring only a homogenous & isotropic universe based on SR and matching Newton in the weak-gravity limit?
And, presumably, has Minkowski space as its tangent space. (Even if not required, I think it's safe to say we do live in such a universe, whether or not GR is the right theory of gravity. :-)So, yes, any theory of gravity which matches Newtonian gravity in the weak-field limit and also replicates current gravitational lensing observations must necessarily produce the same results.
I guess my question (in light of your embedded quote) is then, If a non-pressure/non-density contribution appears in the GR field equations, then all derivatives are zero for a constant, but not otherwise -- so how do relatively moving observers describe the difference that they see?A non-constant dark energy is perfectly fine in GR and spatially isotropic and homogeneous FLRW universes. In a different context, see
Effects beyond expansion/contraction of scale would come down to what the moving observer interprets as an equal-time slicing of the universe. If they interpret a different equal-time slicing than a co-moving observer, then they would interpret a smooth density gradient based upon how that equal-time slicing differs from that derived from co-moving observers.That is, some observers might see a purely temporal effect (expansion/contraction of scale a(t)), but relatively moving observers will see something different: What? (I realize that this may be a basic inflation question, but I don't have my copy of MTW on hand to thumb through.)
I think it's a slightly different case than the CMB.Thanks, @kimbyd. Presumably, my (inertial) 2nd observer does not see an isotropic universe, unlike the first one. I imagine that would seem to them like non-equal times. (I'm reasoning by analogy to the CMB: If I were moving relative to its rest frame, I would see some directions that appear hotter, which I could interpret as younger.)
But what would that density gradient look like? Would #2 observe an increasing ρ in the "younger" direction, or am I misreading your explanation?
I traveled all day, and I write this from a hotel room. With me I have one cosmology book, one novel, one daughter, and one wife. None of these is of much use (with respect to this!), so I will nick another post.Thanks, @kimbyd. Presumably, my (inertial) 2nd observer does not see an isotropic universe, unlike the first one. I imagine that would seem to them like non-equal times. (I'm reasoning by analogy to the CMB: If I were moving relative to its rest frame, I would see some directions that appear hotter, which I could interpret as younger.)
But what would that density gradient look like? Would #2 observe an increasing ρ in the "younger" direction, or am I misreading your explanation?
This related to what @kimbyd has written.In the Friedmann-Lemaitre-Robertson-Walker standard cosmological models that have been discussed in this thread, special spatial hypersurfaces are picked out in a coordinate-invariant manner, as the span of the six spacelike (clearly not all linearly independent) Killing vectors that express the spatial homogeneity and isotropy of the standard cosmological models. These hypersurfaces are orthogonal to the timelike conformal Killing vector ##a\left(t\right) \partial / \partial t##.
Is this still true in a dark energy dominated universe (i.e., one in which the expansion is accelerating)? I remember seeing this (rather counterintuitive) result derived, but I can't remember whether it applied to that case.any non-accelerating observer will, over time, slow down with respect to the Hubble flow
Doesn't matter. It's a feature of the uniform expansion. Any inertial observer will asymptotically approach the worldline of some co-moving observer, as long as the expansion continues (the reverse would be true in a collapsing universe). By keeping the rate of expansion from falling, dark energy makes this asymptotic convergence faster.Is this still true in a dark energy dominated universe (i.e., one in which the expansion is accelerating)? I remember seeing this (rather counterintuitive) result derived, but I can't remember whether it applied to that case.