Interpretation of Hubble constant in SI units

Question

The standard interpretation of Hubble constant $\approx 70~\text{km/s/Mpc}$ means that each mega-parsec of distance adds $70~\text{km/s}$ to a galaxy recession velocity from us (or to a space expansion rate ). But when expressing Hubble constant in SI units directly one gets about $2.27 \times 10^{−18} ~\text{Hz}$ frequency. What does this SI unit version interpretation of the Hubble constant mean? I'm not satisfied with the Wikipedia explanation about expansion rate within 1 exa-second, because exa-second is not strictly an SI unit.

We know that the oscillation period is $T=1/f$, so substituting the Hubble constant $H_0$ would give $\approx 13.9~\text{billion years}$, something close to the age of the Universe. Based on that, does it mean that if the Universe had an oscillating (expansion-contraction) periodic life-time model, then the Hubble SI constant would be the universe oscillation frequency?

Answer 1

Unless the expansion/contraction is periodic then the frequency is just the reciprocal of an expansion timescale. There isn't any evidence that the universe will contract again. However, if it did, then the period of oscillation would almost certainly be much longer than 13.9 billion years, and would be unrelated to the present day Hubble parameter, which is time-dependent.

Answer 2

One exasecond is an SI unit. There is no rule that says that one is not allowed to combine the prefix exa- with second.

Thus the interpretation is straightforward: it's the number of $e$-fold expansions that would occur over that timeframe if the expansion rate were constant. Likewise, as a frequency in attohertz, it's the frequency at which such $e$-fold expansions take place: about 2.27 quintillionths of one happens every second, just as a wheel rotating at 0.5 Hz means that half of a rotation happens each second.

Answer 3

A change in units doesn't change the effect. As @planetmaker says in the comments to the question:

It's inconsequential, whether you drive with 100km/h or 62 mph or 54 knots or 28m/s through town - it's too fast and produces the same impact energy...

A good choice of units can help with intuition. The standard units for the Hubble parameter $H_0 \approx 70$ km/s/Mpc, were choose for precisely this reason. For every additional mega-parsec of space between us and a galaxy, it appears to recede with a velocity $70$ km/s faster.

There are many other choices we could have made. Why not $2.3\times 10^{-18}$ Hz in base S.I. units. Or why not $70$ m/s/kpc or $70$ mm/s/pc? These use the same intuitive interpretation, but are less suited to the scale of the effect. Cosmological expansion only matters on very large scales on the order of the distances between galaxies.

The stellar disk of the Milky Way is ~60 kpc across. When we observe stars 10 kpc away they don't appear to recede at $700$ m/s. Their motion is dominated by the local gravity of the galaxy, not the larger scale cosmological effects. Cosmological expansion has almost no effect on their motion.

The Milky Way belongs to the Local Group of Galaxies, which is about $3$ Mpc across. The Local Group is part of the Virgo Supercluster of galaxies, which is about $30$ Mpc across. The effect of local gravity in our cluster means that the observed apparent recession of our neighbors is about 10% slower than expected. Based on this it seems like cosmological effects start to really matter at the scale of a few to $10$s of Mpc.

In that sense the right unit for cosmological expansion is something-per-Mpc.

Base S.I. units are human scale: meters, kilograms, seconds. They are great for human scale processes. They aren't so intuitive when we apply them to cosmological processes. I can't grok what $2.3\times 10^{-18}$ Hz means, but I can understand $70$ km/s/Mpc.

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If you assume the universe has been expanding at a constant rate of $H_0$ since it started, then $H_0^{-1}\approx 14$ Gyr would be the age of the universe. The universe has not been expanding at a constant rate. So we carefully account for how the expansion changes with time using the standard $\Lambda$CDM cosmological model. It turns out that $H_0^{-1}$ isn't all that bad of an estimate.

Answer 4

In a closed universe, the Hubble Constant would vary from high values, through zero, to negative high values. In an oscillating universe, this would repeat. Hence, the inverse of the Hubble Constant is not proportional to the frequency of oscillation since it is far from constant. In an open universe with very low density and no Cosmological Constant, the Hubble Constant does not vary and the age of the universe can be found by looking at any two galaxies and measuring distance/velocity = 1/Ho. In a universe with just enough density to be closed, and again no Cosmological Constant, its age is 1.5/Ho. In other words, the Hubble constant is the inverse of the age of the universe in an open universe, to order of magnitude, as long as the Cosmological Constant does not dominate.

Answer 5

SI units are just Units. They do not necessarily say what they are describing. They are like numbers in that sense.

What is a square-second or even second cubed? You can't imagine it but we're using it for acceleration and jerk all the time (yes, there is a unit of measurment for jerk!)

In some cases, units are reduced. So the specific impulse of a rocket engines is given in seconds. But it's not a time that is given but basically how efficient the mass of a propellant can be converted into thrust.

So getting Hz when you simplyfy (reduce) a unit does not mean that we're getting a frequency. 1/time always gives you a frequency. But That does not mean that it's a frequency. A frequency would imply some kind of oscillation where you get back at the starting point.

So if a task takes you 1s it does not necessarily mean that you can do that task at 1Hz frequency because you might need to return to the starting point or set up to start again (it's a common misconception managers have when talking about production rate, they take the time for a task and just divide the hours in a working day by the time the task takes and assume that this is the productivity...)

So just because we're getting Hz or Seconds as a Result does not mean we're always talking about time or frequencies.

Answer 6

If it were km/s, you wouldn't think of it as a frequency. Don't think of this as a frequency but a rate.

It's just that the top instead of measuring a distance is measuring an expansion so there's no unit necessary.

$2.27 \times 10^{-18} / \text{s}$ can be interpreted to mean that any region of space expands by that relative amount in one second. Something at rest with respect to an object 1Mpc away would (if hubble expansion were the only effect) find itself about 70 km further away after one second.

Answer 7

This is a good question. It would be best to think of the 2.27×10−18 Hz as a rate rather than a frequency. In this way, we can apply the standard growth formula typically used in calculating growth. F=I(1+r/n)^nt. Your Hubble parameter(r), and the time between initial and final distances(t) would both be in seconds, and the compound rate (n) would be equal to 1. This makes your Expression of the Hubble parameter in units of 1/s useful, and the intuitive interpretation of it is similar to any other rate of growth expressed as a decimal as opposed to a %.

Another thing to consider is expressing it in terms of 3D volume instead of distance by considering spheres of volume V. This would yield a volumetric expansion rate of around 5.59x10-53 Hz, and the above formula could be similarly used.

Additionally, one could consider using the Planck time instead of seconds for the rate, and then compare predictions when using a compounding rate that is quantized (n=1) vs using the continuously compounding formula F=Ie^rt.

Edit: Notably tho, most uses for the above formulas would have to assume the simple case where the Hubble parameter is/has been constant. You can still use the formula F=I(1+r/n)^nt to find the Velocities of receding galaxies (the typical use for the Hubble parameter) by using 1 for n and t. This simplifies The equation to F=I(1+r), So the speed for a galaxy at a distance I would be (F-I)/s= [I(1+r)-I]/s = Ir/s. The speed you get is expressed with the same distance unit in the numerator as the units chosen for measuring, so you wind up with something like 0.00000... lightyears/sec or Mpc/s instead of Km/s.

Regardless, I personally reject the notion that Km/s/Mpc is more intuitive than something like 7.16% per billion years. That said, if the parameter is assumed not to be constant, then working with fundamental units like plancklengths/lightyears*planktimes may be useful.