Asymptotic Hubble constant

  • #1
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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?
 

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  • #2
kimbyd
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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##.
 
  • #3
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the relationship between the Hubble parameter (##H##) and the dark energy density (##\rho##) will be trivial:
Trivial as in?
 
  • #4
kimbyd
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Trivial as in?
The equation immediately below that is the relationship.
 
  • #5
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The equation immediately below that is the relationship.
So the Hubble value will keep on decreasing without asymptotating?
 
  • #6
kimbyd
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So the Hubble value will keep on decreasing without asymptotating?
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.
 
  • #7
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Thank you for helping to get to the end of it.
 
  • #8
JMz
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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##.
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?
 
  • #9
George Jones
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Does this depend on the theory of gravity? (Apparently not GR, since we are contemplating a non-constant DE.)
A non-constant dark energy is perfectly fine in GR and spatially isotropic and homogeneous FLRW universes. In a different context, see

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.
 
  • #10
kimbyd
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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?
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.

The difference between Newtonian gravity and General Relativity for this equation rests upon the fact that General Relativity allows other things besides mass to source gravity. This effect is observed by observing the deflection of light rays by gravitational sources (gravitational lensing): General Relativity predicts twice the deflection of Newtonian gravity, and the observations match the General Relativity prediction.

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.
 
  • #11
JMz
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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.
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. :-)
 
  • #12
JMz
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A non-constant dark energy is perfectly fine in GR and spatially isotropic and homogeneous FLRW universes. In a different context, see
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?

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.)
 
  • #13
kimbyd
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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.)
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.
 
  • #14
JMz
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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?
 
  • #15
kimbyd
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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 think it's a slightly different case than the CMB.

With regard to the CMB, the anisotropy is an observational effect due to the relativistic Doppler shift from the observer's motion relative to the CMB rest frame. I don't think there's necessarily a conclusion that the universe in one direction appears younger or older, or that this relativistic Doppler shift corresponds to any sort of difference in inferred equal-time slices.

Ultimately, I don't know whether the inferred equal-time slices would differ with a non-co-moving observer. I suspect they would, as difference in simultaneity is a fundamental cornerstone of special relativity. But I haven't worked through the details. I'm just saying that if there is an additional impact, it would come from that effect.
 
  • #16
George Jones
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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.

Any FLRW universe (e.g., no dark energy, constant dark energy, time-varying dark energy) is spatially homogeneous and isotropic only for a special set of observers.

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##.
This related to what @kimbyd has written.

I have to attend (for work) an all-day meeting tomorrow and another all-day meeting Saturday. I will be back home late on Sunday.
 
  • #18
kimbyd
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As an aside, any non-accelerating observer will, over time, slow down with respect to the Hubble flow. So you might start off seeing something rather different from a co-moving observer, but over time these effects will disappear as you effectively become a co-moving observer.

This is because no matter which direction you move, you'll end up catching up to things moving away from you, so that your relative velocity compared to the surroundings will decrease over time.
 
  • #19
PeterDonis
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any non-accelerating observer will, over time, slow down with respect to the Hubble flow
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.
 
  • #20
kimbyd
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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.
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.

Presumably you're thinking of the fact that if you have two co-moving observers separated by some distance, dark energy will tend to cause them to move away from one another at an accelerating rate. But that's not the situation that's being examined.

Consider the following scenario: an object is currently at the origin (origin being a specific co-moving location), moving in some direction at speed v. The rate of expansion is constant (i.e., dark energy-dominated). As measured from the origin, let's initially assume that the inertial object just moves with constant speed. Once it has moved a distance d = v/H, its speed will be identical to the recession velocity at that location. As the co-moving observers at a distance d are inertial, it will have to accelerate away from the origin just like other objects at that location do.

Naturally if one were to actually do the math, one would have to find a smooth transition from the receding object to the situation where it is being carried away by the expansion. I think this could be done by assuming that dark energy creates a constant acceleration on the object which is a function of distance, an acceleration proportional to distance. This would be consistent with the Newtonian limit for a cosmological constant.

Note that this argument does ignore local gravity dynamics: if the inertial observer manages to find itself gravitationally-bound to some overdensity (e.g. a galaxy cluster), its motion will no longer be governed by the expansion.
 
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  • #21
kimbyd
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Update:

I decided to have fun and went through what would happen if we assume the Newtonian limit for a cosmological constant, and no other forces acting on the object. In this case, the object would obey the following differential equation:

$$\ddot{d} = {\Lambda \over 3} d$$

This is based on https://arxiv.org/abs/gr-qc/0004037

Now, if we're simply assuming a universe with only a cosmological constant, the Hubble parameter is a constant:

$$H^2 = {\Lambda \over 3}$$

So we can write:

$$\ddot{d} = H_0^2 d$$

As this is a linear differential equation, a sum of solutions is also a solution. We can see pretty easily by inspection that ##d(t) = c_1 e^{H_0 t}## and ##d(t) = c_2 e^{-H_0 t}## are both valid solutions. As a second-order differential equation, only two solutions are possible, so the sum of these two fully-defines possible motions. Initial conditions determine the rest. Thus the general formula for an inertial object in a universe dominated by a cosmological constant is:

$$d(t) = c_1 e^{H_0 t} + c_2 e^{-H_0 t}$$

The above formula must also work for co-moving objects, which follow the equation:

$$d(t) = d_0 a(t)= d_0 e^{H_0 t}$$

This is indeed a possible form of the general solution above.

If we now consider an inertial object moving past the origin, such that ##d(0) = 0## and ##\dot{d}(0) = v_0##, then it's possible to do a bit of math to show that the following solution fits:

$$d(t) = {v_0 \over 2 H_0}\left(e^{H_0 t} - e^{-H_0 t}\right)$$

As time increases, the second term will decrease, eventually approaching the path of a co-moving observer which starts at ##d_0 = v_0/2H_0##. The rate of convergence depends upon the rate of expansion: faster expansion = faster convergence.
 

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