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There is something I don't understand about the Hubble parameter $H$, as it seems to clump two concepts together that I can't quite unify in my head. On the one side, we have

$$V = D H$$

which means that for a given distance $D$, there is a certain amount of new space created over time - and $H$ is simply the factor that makes this relationship work. So, for example, say we have two points 1 Mpc apart, this would mean they recede at about 70 km/s away from each other (given our current approximation of $H$).

Now the thing I can't wrap my head around is that

$$T =\frac{1}{ H}$$

is also the age of the universe. Contrary to claims made on, say, Wikipedia this means that $H$ cannot possibly have been a constant throughout the past 13 billion years, because mathematically $1/H$ means that $H$ must be continually shrinking as the universe ages.

So if $H$ did start out as some huge value and is now shrinking over time, doesn't this mean the expansion of the universe is slowing down? Because if $H$ is shrinking, I'll get a lower value of $V$ today than I'll get tomorrow. Shouldn't the notation then be more like

$$V = DH(t)$$

So which one is it? If $1/H$ is simply the solution for $D=0$, how can we use it as the expansion-velocity-per-unit-of-distance at the same time? What's worse, how can literature say $H$ has probably been more or less constant forever and simultaneously assert that $1/H$ is the current age of the universe? What am I missing?

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I think what fundamentally needs to be explained here is this:

The physical interpretation of the Hubble time is that it gives the time for the Universe to run backwards to the Big Bang if the expansion rate (the Hubble "constant") were constant. Thus, it is a measure of the age of the Universe. The Hubble "constant" actually isn't constant, so the Hubble time is really only a rough estimate of the age of the Universe.

(source, emphasis added) You can verify this mathematically: if the Hubble time $1/H$ really did track the age of the universe (ignoring the general relativistic complications of what "the age of the universe" really means), then it must be the case that $H(t) = 1/t$. Given the definition of the Hubble parameter as $\dot{a}(t)/a(t)$, you can write

$$\frac{\dot{a}(t)}{a(t)} = \frac{1}{t}$$

This differential equation for $a$ has the solution $a(t) = ct$, which indicates that the universe would be expanding linearly in this case.

In reality, of course, the universe does not expand linearly, at least not always. But the available evidence suggests that it's been expanding pretty nearly linearly for a long time, $\dot{a}(t) \approx \text{const.}$ for the last 10-12 billion years, which is why the Hubble time is so close to the age of the universe as estimated from other methods (well, method - WMAP data).

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  • $\begingroup$ I apologize in advance for my persistent stupidity, by the way. So 1/H does track the age of the universe, yes? For example at t=6.8 billion years, H would then have been about twice as large as today, a value of about 140km/s/Mpc. That means any (not gravitationally interacting) galaxy at that time would have receded from us at twice the speed compared to what it would today at the same distance - which can't be right, can it? $\endgroup$
    – Udo
    Commented Dec 15, 2011 at 3:28
  • $\begingroup$ Oh my goodness, it just dawned on me that all my life I misunderstood what linear expansion actually means! It's not the distance between any two points that scales linearly, is it, it's the only size of the whole universe itself that goes up by roughly the same distance per unit of time. This means the scaling effect between two arbitrary points in space actually becomes smaller over time. This is why I couldn't work out how H works. Oh dear. Everything is totally different from what I imagined ;-) $\endgroup$
    – Udo
    Commented Dec 15, 2011 at 3:39
  • $\begingroup$ Actually, linear expansion does mean that the distance between any two objects (that's $a(t)$) increases linearly with time. The thing is that $1/H$ is only equal to the age of the universe if the expansion is linear, as I calculated. $\endgroup$
    – David Z
    Commented Dec 15, 2011 at 6:42
  • $\begingroup$ Oh, so I'm back where I started then. Let me rephrase my problem in this context: Isn't linear expansion a pretty big assumption? When Hubble presented his work, why did it appear so self-evident to everyone that expansion was linear this whole time? What if that hadn't worked out? Sure, today we might have the observational data to support it (though not really, as expansion does in fact appear to speed up), but in the early 20th century? I don't get how it wasn't just dumb luck that this lined up with the age of the universe so well. $\endgroup$
    – Udo
    Commented Dec 15, 2011 at 10:34
  • $\begingroup$ Linear expansion was just the simplest assumption, and it seemed roughly consistent with the idea that the universe started from an explosion. Of course the expansion doesn't actually work out to be linear in the ΛCDM model, which predicts deviations from linear expansion in the early universe and in the future, but the point is that for most of the universe's current history the expansion has been roughly linear, and that's why the Hubble time is approximately equal to the age of the universe. (If you'd like to continue this discussion, let's take it to Physics Chat.) $\endgroup$
    – David Z
    Commented Dec 15, 2011 at 14:29
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I think there can be confusion when one interprets the Hubble constant in such a manner. Perhaps a better way to think about it would be to look at the definition of Hubble parameter.

$H=\dot{a}(t)/a(t)$

Where $a(t)$ is the scale factor (please see Friedman equations for details). Basically the scale factor gives us information about the expansion of the Universe.

Now, when someone is talking about an expanding Universe, they mean $\dot{a(t)} > 0$, while an accelerated expansion means $\ddot{a}(t) > 0$. So based on the above definition of the Hubble parameter, it is possible for its value to be decreasing while the expansion is still accelerating.

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  • $\begingroup$ Defining H as the evolution of the scale factor over time doesn't really help, I think my inability to comprehend 1/H runs much deeper and starts with implications of the Hubble parameter itself. For example, take two galaxies separated by the distance D. You multiply this distance by H and get their velocity delta V. My problems with H are now manyfold. For example, even without assumptions about the gravitational effects you're describing, 1/H=age_of_universe means this V is decreasing as the universe ages. That doesn't make sense. $\endgroup$
    – Udo
    Commented Dec 15, 2011 at 2:21
  • $\begingroup$ Even more problematic: assume both galaxies have been drifting apart since the dawn of time. That means you could theoretically calculate the age of the universe by playing their motion backwards until D=0. Only the formula V = D*H doesn't allow for this, they'll simply can't get to D=0 in finite time. $\endgroup$
    – Udo
    Commented Dec 15, 2011 at 2:23
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    $\begingroup$ @Udo: remember that D increases over time, so even if H is decreasing, V can still increase. ($a(t)$ is D and $\dot{a}(t)$ is V.) $\endgroup$
    – David Z
    Commented Dec 15, 2011 at 2:37
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    $\begingroup$ @Udo I see. Something else you should be aware of is comoving and proper distance. The distance in Hubble's law is proper distance, which changes with time. However, because $H$ is decreasing, when we observe something at a /fixed/ distance and it is moving with given velocity, then another object going past the same point at a later time will be moving slower than the one before. As for calculating the age of Universe from $H$, that can be model dependent and an estimate based on uniform expansion. Other parameters such as inflation need to be taken into account. $\endgroup$
    – Omar
    Commented Dec 15, 2011 at 2:54
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    $\begingroup$ In GR, one has to be careful what we mean by "non-accelerated expansion". In the "explosion in flat space" model, we would mean that $v=Hd$ is constant. In this model, we may just calculate the time, $d/v$ (because $s=vt$), when the distances were zero, and we get $1/H$. That's a possible interpretation in GR, too. However, that's different from a "constant $H$". In our Universe, $H$ is ultimately going to be nearly constant. But this corresponds to exponential, not linear, growth of distances, $d\sim \exp(Ht)$ because $\dot d = v = Hd$ where $H$ is fixed. $\endgroup$
    – Luboš Motl
    Commented Dec 15, 2011 at 10:01
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1/H gives the Hubble Sphere, and not te age of the universe. The reason for the confusion is that the Hubble time by accident is now nearly equal to the age of the universe. But the Hubble constant (H) is no constant, and varies over time. For example 6 billion years ago, when the universe was 7.5 billion years old, the Hubble constant was about 100 km/s/Mpc, what means the Hubble Time was 9.78 billion years. When the universe is 24 billion years of age, H will be 60 km/s/Mpc, and the Hubble Time will be 16.3 billion years. Not even close to the age of the universe.

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  • $\begingroup$ Lightyears are a measure of distance, not time. The Hubble time has units of time, not distance. $\endgroup$
    – Kyle Oman
    Commented Jan 22, 2015 at 18:13
  • $\begingroup$ Close, though--the radius of the Hubble sphere is given by $c/H$. In "natural units" where c=1, the age of a universe with linear expansion would have the same value as the radius of the Hubble sphere. $\endgroup$
    – Hypnosifl
    Commented Jan 22, 2015 at 18:27

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