1
$\begingroup$

I’ve recently been trying to wrap my head around the potential behaviours of the Hubble parameter over time for various cosmological models but I’ve run into a little snag when it comes to a coasting universe. So, my understanding is that there are 4 basic cosmological models:

  1. Decelerating universe that re-collapses
  2. Decelerating universe that becomes static after infinite time
  3. Coasting universe
  4. Accelerating universe (dark energy/cosmological constant)

Now the way I’ve been delineating between these scenarios is by considering how the Hubble parameter changes over time. In other words, how the slope of a graph of recessional velocity (V) vs. distance (R) varies with time... i.e. a graph like this:

enter image description here

For scenario 1, the slope starts out positive (expanding) but over time gets less steep, momentarily becoming flat when it crosses the horizontal axis (static), and then eventually becoming more and more negative (collapsing). This would ultimately culminate in a big crunch.

For scenario 2, the slope starts out positive and remains positive (expanding) but is getting less and less steep over time, asymptotically approaching a slope of zero when it becomes flat and coincides with the horizontal axis (static). This would asymptotically result in a universe where all other objects are stationary from your perspective (ignoring peculiar motions of course).

For scenario 4, the situation is a lot like scenario 2, with the slope starting out positive and remaining positive (expanding), but instead of it approaching a slope of zero, it asymptotically approaches some non-zero positive slope (expanding forever) that is dictated by the value of the cosmological constant. This would asymptotically result in a universe where all other objects are uniformly and isotropically moving away from you with some constant ACCELERATION (once again ignoring peculiar motions).

My question is, how exactly does scenario 3 fit into this picture? My best guess is that scenario 3 would correspond to the graph levelling out with a slope of zero but at a constant recessional velocity of V = v (with the caveat of a discontinuity of V = 0 at R = 0, i.e. my location). This is in contrast to scenario 2, where the slope levels out at V = 0 for all values of R. If true, this would asymptotically result in a universe where all other objects are uniformly and isotropically moving away from you with some constant VELOCITY (once again ignoring peculiar motions). This is similar to the asymptotic eventuality of scenario 4, where the universe expands forever, but the recessional velocities are now constant everywhere rather than being proportional to distance. Is this the correct interpretation? If not, then what is the correct interpretation in this context?

Assuming it is the correct picture though, then I have a follow on question... does it even make sense for a universe to asymptotically approach a state like this over time? I ask because in scenario 3, I struggle to envision a "continuous" transformation of the graph over time that doesn't point towards questionable physical interpretations.

P.S. I’m aware that a coasting universe is one in which the expansion rate is usually considered constant, meaning the concept of a changing/“levelling out” Hubble parameter in this context might not entirely be physical. For the sake of the question though, I'm assuming transient behaviour is even possible in the case of an eventual coasting universe.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

For scenario 2, the slope starts out positive and remains positive (expanding) but is getting less and less steep over time, asymptotically approaching a slope of zero

Indeed this description of the evolution of the Hubble rate $H$ is accurate for case 2, case 3, and a subset of cases 4. $$H=\frac{\mathrm{d}a/\mathrm{d}t}{a},$$ where $a$ is the expansion factor. If $\mathrm{d}a/\mathrm{d}t$ drops asymptotically toward zero (case 2), then $H$ does as well. But $\mathrm{d}a/\mathrm{d}t$ could also remain constant (case 3) or continue to grow (case 4), and as long it grows more slowly than $a$, $H$ still tends asymptotically toward zero.

A cosmological constant leads to $\mathrm{d}a/\mathrm{d}t\propto a$, so $H$ approaches a constant value. But any less acceleration than that (due to dark energy with $w>-1$) leads to $H\to 0$ in the infinite-time limit.

Also, if $\mathrm{d}a/\mathrm{d}t$ grows more quickly than $a$ (due to "phantom" dark energy with $w<-1$), then $H$ grows without limit.


Intuitively, in a coasting universe, the recession velocity of a particular object remains constant, but that object is becoming more distant. That's why the slope of the plot of recession velocity against distance continues to drop.

$\endgroup$
7
  • $\begingroup$ Thanks for the clarifications regarding the different types of dark energy! This has definitely cleared up some things for me. So what I'm gathering is that H will always tend towards zero, be it an asymptotically decelerating universe, a coasting universe, or even when there's dark energy (as long as w > -1). That being said, I'm still not sure how you'd differentiate between these various cases via the "final" velocity vs. distance graph alone. They all eventually have a slope of 0... so what is the distinguishing factor between them? Is it the final velocity the flat line settles on? $\endgroup$
    – Scott
    Commented Feb 11, 2023 at 16:38
  • $\begingroup$ For the record, my main motivation behind this question is trying to determine what the relative motion of any one single object is in the infinite time limit for each type of these scenarios. $\endgroup$
    – Scott
    Commented Feb 11, 2023 at 16:41
  • $\begingroup$ @Scott The "final" local Hubble diagram indeed wouldn't help distinguish the many cases where $H$ goes to zero. The slope would be zero and the value would also be zero in all cases. If you were observing over the whole cosmic history, you could distinguish the scenarios by how $H$ depends on time. Alternatively, you could infer the history by tracing the Hubble diagram out to cosmologically large distances (as we do now). $\endgroup$
    – Sten
    Commented Feb 11, 2023 at 16:46
  • $\begingroup$ The "relative" velocity of one object, to the extent that you can clearly define it (it shouldn't be too far away), essentially goes as $\mathrm{d}a/\mathrm{d}t=aH$. In a "coasting" universe, this would be constant for each object. In a decelerating universe, it would drop over time; in an accelerating universe, it would grow over time. The complication when translating this into the behavior of the Hubble diagram is that the object's distance also changes. $\endgroup$
    – Sten
    Commented Feb 11, 2023 at 16:50
  • $\begingroup$ Ahhh I see! So you can only distinguish between each scenario by observing how H changes with time. It's the rate at which H decreases, i.e. dH/dt, that distinguishes between each case. I'm guessing this means an asymptotically decelerating universe tends towards H = 0 the fastest, dark energy cases (w < -1) tend toward H = 0 the slowest, while coasting universes of various "speeds" fall somewhere in between. Is this the correct conclusion? $\endgroup$
    – Scott
    Commented Feb 11, 2023 at 16:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .