<p>One exasecond <strong>is</strong> an SI unit. There is no rule that says that one is not allowed to combine the prefix exa- with second.</p>
<p>Thus the interpretation is straightforward: it's the number of <span class="math-container">$e$</span>-fold expansions that <em>would</em> occur over that timeframe if the expansion rate were constant. Likewise, as a frequency in attohertz, it's the frequency at which such <span class="math-container">$e$</span>-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.</p>

CC BY-SA 4.0
2022-05-04T08:31:11.617Z
2022-05-04T08:31:11.617Z
The_Sympathizer
<p>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, <em>if</em> 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 <em>parameter</em>, which is time-dependent.</p>

CC BY-SA 4.0
2022-05-04T04:56:58.807Z
That's understandable. However question is not about it will contract or not. Question is about what Hubble constant in the form $10^{-18} Hz$ means? Does this have some interpretation or no way out?
Agnius Vasiliauskas
CC BY-SA 4.0
2022-05-04T05:08:18.810Z
Besides Hubble law has similar properties as tangential and angular velocities relationship : $v=\omega r$. This doubles my intuition that Hubble constant has something to do with frequency and periodic processes.
Agnius Vasiliauskas
CC BY-SA 4.0
2022-05-04T06:51:59.753Z
You forget, that for something periodic to happen, we have to measure a full period. So just because you get a result in Hz does not mean that you automatically have something periodic. I'm 46 years old. That's 6.893×10^-10 Hz... Now what?
kruemi
CC BY-SA 4.0
2022-05-04T07:47:46.117Z
@kuemi in what way have I "forgotten" that?
ProfRob
CC BY-SA 4.0
2022-05-04T07:49:39.853Z
@AgniusVasiliauskas if it doesn't contract, it cannot be periodic and the frequency is meaningless other than the inverse of an expansion timescale.
ProfRob
CC BY-SA 4.0
2022-05-02T21:39:05.447Z
2022-05-04T07:51:30.063Z
2022-05-04T07:51:30.063Z
ProfRob
<p>A change in units doesn't change the effect. As @planetmaker says in the comments to the question:</p>
<blockquote>
<p>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...</p>
</blockquote>
<p>A good choice of units can help with intuition.  The standard units for the Hubble parameter <span class="math-container">$H_0 \approx 70$</span> 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 <span class="math-container">$70$</span> km/s faster.</p>
<p>There are many other choices we could have made.  Why not <span class="math-container">$2.3\times 10^{-18}$</span> Hz in base S.I. units.  Or why not <span class="math-container">$70$</span> m/s/kpc or <span class="math-container">$70$</span> 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.</p>
<p>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 <span class="math-container">$700$</span> 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.</p>
<p>The Milky Way belongs to the Local Group of Galaxies, which is about <span class="math-container">$3$</span> Mpc across.  The Local Group is part of the <a href="https://en.wikipedia.org/wiki/Virgo_Cluster#Characteristics" rel="nofollow noreferrer">Virgo Supercluster</a> of galaxies, which is about <span class="math-container">$30$</span> 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 <span class="math-container">$10$</span>s of Mpc.</p>
<p><strong>In that sense the right unit for cosmological expansion is something-per-Mpc</strong>.</p>
<p>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 <span class="math-container">$2.3\times 10^{-18}$</span> Hz means, but I can understand <span class="math-container">$70$</span> km/s/Mpc.</p>
<hr />
<p>If you assume the universe has been expanding at a constant rate of <span class="math-container">$H_0$</span> since it started, then <span class="math-container">$H_0^{-1}\approx 14$</span> 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 <a href="https://en.wikipedia.org/wiki/Lambda-CDM_model" rel="nofollow noreferrer">standard <span class="math-container">$\Lambda$</span>CDM cosmological model</a>.  It turns out that <span class="math-container">$H_0^{-1}$</span> isn't all that bad of an estimate.</p>

CC BY-SA 4.0
2022-05-03T18:51:46.723Z
2022-05-03T18:51:46.723Z
Paul T.
<p>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.</p>

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2022-05-05T06:17:35.430Z
_Hence, the Hubble Constant does not provide info on the frequency of oscillation since it is far from constant_ Does not follow. For, example angular frequency can vary if accelerating/decelerating torque is applied to the rotating system. Similarly if there's a force acting on a universe expansion/contraction, then there can be analog of "angular acceleration", which implies varying frequency. So frequency as per se, doesn't have to be constant, and still it may be somehow related to a Hubble constant.
Agnius Vasiliauskas
CC BY-SA 4.0
2022-05-06T17:37:28.387Z
Right.  The question was is 1/Ho the universal oscillation frequency, and I just meant to say it is not.  I have edited my answer.
eshaya
CC BY-SA 4.0
2022-05-04T21:24:50.333Z
2022-05-06T17:33:35.670Z
2022-05-06T17:33:35.670Z
eshaya
<p>SI units are just Units. They do not necessarily say <strong>what</strong> they are describing. They are like numbers in that sense.</p>
<p>What is a square-second or even second cubed? You can't imagine it but we're using it for <a href="https://en.wikipedia.org/wiki/Acceleration" rel="nofollow noreferrer">acceleration</a> and <a href="https://en.wikipedia.org/wiki/Jerk_(physics)" rel="nofollow noreferrer">jerk</a> all the time (yes, there is a unit of measurment for jerk!)</p>
<p>In some cases, units are reduced. So the <a href="https://en.wikipedia.org/wiki/Specific_impulse" rel="nofollow noreferrer">specific impulse</a> 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.</p>
<p>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.</p>
<p>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...)</p>
<p>So just because we're getting Hz or Seconds as a Result does not mean we're always talking about time or frequencies.</p>

CC BY-SA 4.0
2022-05-04T06:49:14.143Z
2022-05-04T06:49:14.143Z
kruemi
<p>The interpretation of the Hubble parameter as a frequency means that it is the rate at which any existing length (in any of the three dimensions of space) is increased by itself, i.e. is doubled every <span class="math-container">$ \tilde 2 \cdot 10^{18} $</span> seconds.</p>

CC BY-SA 4.0
2024-12-06T09:47:13.830Z
2024-12-06T09:47:13.830Z
Gerold Broser
<p>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 <a href="https://en.wikipedia.org/wiki/Exponential_growth" rel="nofollow noreferrer">standard growth formula</a> 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 %.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2023-01-04T06:53:57.123Z
2023-01-04T07:21:23.200Z
2023-01-04T07:21:23.200Z
Chaz
<p>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.</p>
<p>It's just that the top instead of measuring a distance is measuring an expansion so there's no unit necessary.</p>
<p><span class="math-container">$2.27 \times 10^{-18} / \text{s}$</span> 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.</p>

CC BY-SA 4.0
2022-05-08T09:55:50.547Z
Bad comparison. SI units are well-defined system. Velocity is measured in $m/s$, and frequency - $1/s$ (or full periods per second). These units are not interchangeable like you want to imply. If it walks like a duck, quacks like a duck - then it's probably a duck. Similarly it's frequency if has units of Hz.
Agnius Vasiliauskas
CC BY-SA 4.0
2022-05-08T23:08:15.473Z
An SI dimension can have multiple interpretations.  Something with units of N*m is not necessarily energy.  It might be torque.  We don't require it to be one or the other in every situation.
BowlOfRed
CC BY-SA 4.0
2022-05-09T05:10:19.847Z
Some may have, especially when there are multiple units in expression. But there's not much room for interpretation of 1/s.
Agnius Vasiliauskas
CC BY-SA 4.0
2022-05-09T07:04:57.497Z
And btw, to some degree torque IS work done per 1 radian, as per: $$ W_{rot}=\tau \Delta \theta $$. So torque and work done in a rotating system is "almost" the same thing, thus $[\text{N} \times \text{m}]$ IS energy/work in one way or the other.
Agnius Vasiliauskas
CC BY-SA 4.0
2022-05-06T22:26:20.937Z
2022-05-06T22:26:20.937Z
BowlOfRed
<p>The standard interpretation of Hubble constant <span class="math-container">$\approx 70~\text{km/s/Mpc}$</span> means that each mega-parsec of distance adds <span class="math-container">$70~\text{km/s}$</span> 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 <span class="math-container">$2.27 \times 10^{−18} ~\text{Hz}$</span> 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.</p>
<p>We know that the oscillation period is <span class="math-container">$T=1/f$</span>, so substituting the Hubble constant <span class="math-container">$H_0$</span> would give <span class="math-container">$\approx 13.9~\text{billion years}$</span>, 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 <strong>SI</strong> constant would be the universe oscillation frequency?</p>

CC BY-SA 4.0
2022-05-02T21:34:52.447Z
Why do the units matter?
ProfRob
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2022-05-02T21:36:24.317Z
In experimental physics units always matters, besides this may give new insights.
Agnius Vasiliauskas
CC BY-SA 4.0
2022-05-02T22:02:48.453Z
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. The difference is only how well you have an intuitive feel for the number with respect to its units
planetmaker
CC BY-SA 4.0
2022-05-03T13:42:41.450Z
Don't get too hung up about units. SI units are the basis for many scientific equations/formulae. Using a term such as exa-seconds in conversation or in writing is more convenient than saying or writing 10^18 seconds. If there is a city that is 50 km from where you live, are *always* going to say to *everyone* when taking about the distance the city is that it is 50 thousand meters from where you live or are you going to say it is 50 kilometers from where you live? If you need to use that distance in a formula, yes you would use 50 000 meters, but not in every day conversation.
Fred
<cosmology><universe><hubble-constant>
Interpretation of Hubble constant in SI units
Agnius Vasiliauskas
# Interpretation of Hubble constant in SI units

(Asked by: username_0 on 2022-05-02)

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?

**Tags:** cosmology, universe, hubble-constant
> **Comments:**
> **username_1**: Why do the units matter?  
> **username_0**: In experimental physics units always matters, besides this may give new insights.  
> **username_2**: 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. The difference is only how well you have an intuitive feel for the number with respect to its units  
> **username_3**: Don't get too hung up about units. SI units are the basis for many scientific equations/formulae. Using a term such as exa-seconds in conversation or in writing is more convenient than saying or writing 10^18 seconds. If there is a city that is 50 km from where you live, are *always* going to say to *everyone* when taking about the distance the city is that it is 50 thousand meters from where you live or are you going to say it is 50 kilometers from where you live? If you need to use that distance in a formula, yes you would use 50 000 meters, but not in every day conversation.  

---

## 8 answers

### Answer by username_4 on 2022-05-04. Score: 4 | Accepted answer

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 by username_1 on 2022-05-02. Score: 8

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.

> **Comments:**
> **username_0**: That's understandable. However question is not about it will contract or not. Question is about what Hubble constant in the form $10^{-18} Hz$ means? Does this have some interpretation or no way out?  
> **username_0**: Besides Hubble law has similar properties as tangential and angular velocities relationship : $v=\omega r$. This doubles my intuition that Hubble constant has something to do with frequency and periodic processes.  
> **username_5**: You forget, that for something periodic to happen, we have to measure a full period. So just because you get a result in Hz does not mean that you automatically have something periodic. I'm 46 years old. That's 6.893×10^-10 Hz... Now what?  
> **username_1**: @kuemi in what way have I "forgotten" that?  
> **username_1**: @AgniusVasiliauskas if it doesn't contract, it cannot be periodic and the frequency is meaningless other than the inverse of an expansion timescale.  

---

### Answer by username_6 on 2022-05-03. Score: 2

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](https://en.wikipedia.org/wiki/Virgo_Cluster#Characteristics) 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.

---

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](https://en.wikipedia.org/wiki/Lambda-CDM_model). It turns out that $H\_0^{-1}$ isn't all that bad of an estimate.


---

### Answer by username_7 on 2022-05-04. Score: 2

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.

> **Comments:**
> **username_0**: _Hence, the Hubble Constant does not provide info on the frequency of oscillation since it is far from constant_ Does not follow. For, example angular frequency can vary if accelerating/decelerating torque is applied to the rotating system. Similarly if there's a force acting on a universe expansion/contraction, then there can be analog of "angular acceleration", which implies varying frequency. So frequency as per se, doesn't have to be constant, and still it may be somehow related to a Hubble constant.  
> **username_7**: Right.  The question was is 1/Ho the universal oscillation frequency, and I just meant to say it is not.  I have edited my answer.  

---

### Answer by username_5 on 2022-05-04. Score: 1

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](https://en.wikipedia.org/wiki/Acceleration) and [jerk](https://en.wikipedia.org/wiki/Jerk_(physics)) all the time (yes, there is a unit of measurment for jerk!)

In some cases, units are reduced. So the [specific impulse](https://en.wikipedia.org/wiki/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 by username_8 on 2024-12-06. Score: 1

The interpretation of the Hubble parameter as a frequency means that it is the rate at which any existing length (in any of the three dimensions of space) is increased by itself, i.e. is doubled every $ \tilde 2 \cdot 10^{18} $ seconds.


---

### Answer by username_9 on 2023-01-04. Score: 0

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](https://en.wikipedia.org/wiki/Exponential_growth) 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.


---

### Answer by username_10 on 2022-05-06. Score: -1

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.

> **Comments:**
> **username_0**: Bad comparison. SI units are well-defined system. Velocity is measured in $m/s$, and frequency - $1/s$ (or full periods per second). These units are not interchangeable like you want to imply. If it walks like a duck, quacks like a duck - then it's probably a duck. Similarly it's frequency if has units of Hz.  
> **username_10**: An SI dimension can have multiple interpretations.  Something with units of N*m is not necessarily energy.  It might be torque.  We don't require it to be one or the other in every situation.  
> **username_0**: Some may have, especially when there are multiple units in expression. But there's not much room for interpretation of 1/s.  
> **username_0**: And btw, to some degree torque IS work done per 1 radian, as per: $$ W_{rot}=\tau \Delta \theta $$. So torque and work done in a rotating system is "almost" the same thing, thus $[\text{N} \times \text{m}]$ IS energy/work in one way or the other.  

---

