The Hubble constant, which roughly gauges the extent to which space is being stretched, can be determined from astronomical measurements of galactic velocities (via redshifts) and positions (via standard candles) relative to us. Recently a value of 67.80 ± 0.77 (km/s)/Mpc was published. On the scale of 1 A.U. the value is small, but not infinitesimal by any means (I did the calculation a few months ago, and I think it came out to about 10 meters / year / A.U.). So, can you conceive of a measurement of the Hubble constant that does not rely on any extra-galactic observations?

I ask because, whatever the nature of the expansion described by the Hubble constant, it seems to be completely absent from sub-galactic scales. It is as though the energy of gravitational binding (planets), or for that matter electromagnetic binding (atoms) makes matter completely immune from the expansion of space. The basis for this claim is that if space were also pulling atoms apart, I would naively assume we should be able to measure this effect through modern spectroscopy. Given that we are told the majority of the universe is dark energy, responsible for accelerating the expansion, I wonder, how does this expansion manifest itself locally?

Any thoughts would be appreciated.


5 Answers 5


Everything doesn't expand equally because of cosmological expansion. If everything expanded by the same percentage per year, then all our rulers and other distance-measuring devices would expand, and we wouldn't be able to detect any expansion at all. Actually, general relativity predicts that cosmological expansion has very little effect on objects that are small and strongly bound. Expansion is too weak an effect to detect at any scale below that of distant galaxies.

Cooperstock et al. have estimated the effect for systems of interest such as the solar system. For example, the predicted general-relativistic effect on the radius of the earth's orbit since the time of the dinosaurs is calculated to be about as big as the diameter of an atomic nucleus; if the earth's orbit had expanded according to the cosmological scaling function $a(t)$, the effect would have been millions of kilometers.

To see why the solar-system effect is so small, let's consider how it can depend on $a(t)$. There is a cosmology called the Milne universe, which is just flat, empty spacetime described in silly coordinates; $a(t)$ is chosen to grow at a steady rate, but this has no physical significance, since there is no matter that has to expand like this. The Milne universe has $\dot{a}\ne 0$, i.e., a nonvanishing value of the Hubble constant $H_o$. This shows that we should not expect any expansion of the solar system due to $\dot{a}\ne 0$. The lowest-order effect requires $\ddot{a}\ne 0$.

For two test particles released at a distance $\mathbf{r}$ from one another in an FRW spacetime, their relative acceleration is given by $(\ddot{a}/a)\mathbf{r}$. The factor $\ddot{a}/a$ is on the order of the inverse square of the age of the universe, i.e., $H_o^2\sim 10^{-35}$ s$^{-2}$. The smallness of this number implies that the relative acceleration is very small. Within the solar system, for example, such an effect is swamped by the much larger accelerations due to Newtonian gravitational interactions.

It is also not necessarily true that the existence of an anomalous acceleration leads to the expansion of circular orbits over time. An anomalous acceleration $(\ddot{a}/a)\mathbf{r}$ just acts like a slight repulsive force, which is equivalent to reducing the strength of the gravitational attraction by some small amount. The actual trend in the radius of the orbit over time, called the secular trend, is proportional to $(d/dt)(\ddot{a}/a)$, and this vanishes, for example, in a cosmology dominated by dark energy, where $\ddot{a}/a$ is constant. Thus the nonzero (but undetectably small) effect estimated by Cooperstock et al. for the solar system is a measure of the extent to which the universe is not yet dominated by dark energy.

The sign of the effect can be found from the Friedmann equations. Assume that dark energy is describable by a cosmological constant $\Lambda$, and that the pressure is negligible compared to $\Lambda$ and to the mass-energy density $\rho$. Then differentiation of the Friedmann acceleration equation gives $(d/dt)(\ddot{a}/a)\propto\dot{\rho}$, with a negative constant of proportionality. Since $\rho$ is currently decreasing, the secular trend is currently an increase in the size of gravitationally bound systems. For a circular orbit of radius $r$, a straightforward calculation (see my presentation here, sec. 8.2) shows that the secular trend is $\dot{r}/r=\omega^{-2}(d/dt)(\ddot{a}/a)$. This produces the undetectably small effect on the solar system referred to above.

In "Big Rip" cosmologies, $\ddot{a}/a$ blows up to infinity at some finite time, so cosmological expansion tears apart all matter at progressively smaller and smaller scales.

Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," http://arxiv.org/abs/astro-ph/9803097v1

  • $\begingroup$ Well answered. The math is a bit foreign to me, but intuitively it makes a lot of sense. That ArXiv paper is exactly what I was looking for! $\endgroup$
    – burgerking
    Commented Jul 8, 2013 at 3:35
  • $\begingroup$ Rulers might expand but light would still be travelling at same speed. Hence the changes would still be noticeable. $\endgroup$
    – user29978
    Commented Jun 8, 2017 at 16:21
  • $\begingroup$ for quantum mecanical systems the extra forces will not matter, except as gravitational corrections to the binding potential, no? A type of fine structure. $\endgroup$
    – anna v
    Commented Jan 31, 2018 at 7:30
  • $\begingroup$ @Boltzee If all rulers expanded equally over time, it is quite likely our understanding of physics would have a speed of light that varies in cosmological time, rather than a constant speed of light. $\endgroup$
    – Chris
    Commented Feb 15, 2019 at 11:48

We do measure the Hubble constant locally: everything that we know about it comes from observations of light in the vicinity of our telescopes. But if you restrict the experiments to a room with opaque walls, then no, it can't be measured locally, because it merely quantifies the average motion of galaxies on large scales, and there's nothing in the room that will tell you that. Note that earlier answers to this question, including the accepted answer, are wrong, inasmuch as they all suggest that it could be measured locally in principle if not in practice. The paper by Cooperstock et al is also wrong.

The error in Cooperstock et al is easy to explain. They assume that the cosmological FLRW metric is accurate at solar system scales. You can plug the FLRW metric into the Einstein field equations (or the Friedmann equations, which are the Einstein equations specialized to FLRW geometries) to see what this implies about the stress-energy tensor. What you'll find is that they've assumed that the solar system is filled uniformly with matter of a certain density and pressure. The force that they calculate is simply the local gravitational effect of the matter that they assumed was present. But it isn't actually there. It's elsewhere: it collapsed into stars and planets. When they treat the cosmological force as a perturbation on top of the usual solar-system forces, they're double-counting all of the matter, once at its actual location and once at the location where it hypothetically would be if it hadn't clumped. Matter only exerts a gravitational influence from its actual location.

General relativity is different from Newtonian gravity, but it's not as different as many people seem to imagine. It's still a theory of gravity: a force between massive objects that's mediated by a field. It's not a theory of test particles following geodesics on meaningless spacetime backgrounds. The FLRW geometry is not a background; it's the gravitational field of a uniform matter distribution. It could be roughly described as a bunch of Schwarzschild patches stitched together and then smoothed. In real life, there is no smoothing, and no FLRW geometry; there is only the (approximately) Schwarzschild local patches. There is no universal scale factor evolving to the ticks of the absolute, true and cosmological time; there is only local motion of ordinary gravitating objects. That this averages out, on huge scales, to a FLRW-like shape with local bumps is known to us, but irrelevant to nature, which only applies local physical laws independently in each spacetime neighborhood.

Measuring the Hubble constant in a sealed room is no different from measuring the abundance of helium in a sealed room. It will only tell you what's in the room. The abundance in the room won't tend to 25% over time. There's no subtle residual effect of 25% abundance that you can measure locally. The universe is about 25% helium because most of the helium from the first three minutes is still around, not because there's a local physical process that regulates the amount of helium.

What about dark energy? Dark energy, by assumption, doesn't clump at all. You can measure its gravitational effect in the room because it's present in the room. The acceleration you'll measure is not $\ddot a/a$, because $\ddot a/a$ incorporates the averaged effect of all matter, not just the stuff in the room. In the distant future, as $Ω_Λ$ approaches $1$, the acceleration you measure will approach $\ddot a/a = H^2$, but there's no way for you to know that unless you look outside of the room and observe that there's nothing else out there. If dark energy clumps (by little enough to evade current experimental limits) then the amount in the room may be smaller or larger than the average. In that case you'll measure the effect of what's actually in the room, not the effect of the average that you're taking it to be a perturbation of. Nature doesn't do perturbation theory.

Same with dark matter. There may be some of it in the room, depending on what it's made of. If there is, the density will probably be larger than the universal average, but it could be smaller, or about equal. In any case, what you'll measure is what's actually in the room, not what would be there if dark matter didn't clump.

Here are some comments on specific parts of other answers.

For two test particles released at a distance $\mathbf{r}$ from one another in an FRW spacetime, their relative acceleration is given by $(\ddot{a}/a)\mathbf{r}$.

That's correct. Assuming the F(L)RW geometry in GR is equivalent to assuming a $(\ddot{a}/a)\mathbf{r}$ field, or $(\ddot{a}/a)\mathbf{r}^2/2$ potential, in Newtonian gravity. By Poisson's equation that implies uniform matter of density $\ddot{a}/a = -\tfrac43 πGρ$ is present everywhere.

Within the solar system, for example, such an effect is swamped by the much larger accelerations due to Newtonian gravitational interactions.

That's incorrect. The effect is absent in the solar system because the matter that would have caused it is absent. This is obviously true in Newtonian gravitation; it's also true in GR.

The actual trend in the radius of the orbit over time, called the secular trend, is proportional to $(d/dt)(\ddot{a}/a)$

I think that this would be correct if the $(\ddot{a}/a)\mathbf{r}$ force actually existed.

Note, though, that if the force existed, it would be due to, and proportional to, the mass located inside the orbital radius, so you may as well say that the trend is proportional to $dM/dt$. This holds regardless of the nature of the mass; it could be a star losing mass to solar wind and radiation, for example. For a circular orbit $mv^2/r=GMm/r^2$, which gives $dr = d(GM)$ if you hold $v$ constant, so this seems reasonable.

If you added FLRW matter to the solar system, you wouldn't get this trend, because it would clump on much smaller time scales. To follow the Hubble expansion over long time scales it would have to behave totally unphysically: gravitationally influencing other matter but entirely uninfluenced by it, just sedately expanding independent of everything else. This happens when the FLRW matter is the only matter in the universe, since there's nothing to break the symmetry; otherwise it makes no sense.

if you write down the Einstein equation for the case of a simple cosmological-constant dominated universe and a spherically symmetric matter source [...] you [...] get an instability in orbits whose radius is greater than some value $r_∗$, which is proportional to $1/(ΛM)$. This outermost instability represents the expansion of the universe starting to dominate over objects orbiting very far from the star [...].

It represents the dark energy, which is present locally, starting to dominate. As you go to larger radii, the total contained dark energy goes up roughly as $r^3$, and $r_*$ is the radius at which the repulsive force from that equals the attractive force of the central mass. Mass outside that radius can be neglected by the shell theorem/Birkhoff's theorem. This doesn't tell you the Hubble constant or the scale factor; it only tells you the local density of dark energy, which as I mentioned before can be measured inside the opaque room.

  • $\begingroup$ This is an interesting answer, but I still don't quite get why it works out. General relativity is at least linear under a small perturbation, isn't it? So given two weak influences, why can't one treat the result as a superposition of their individual contributing effects? You even seem to use that superposition process in supposing it can be decomposed as a "bunch of Schwarzschild metrics" (presumably integrating it with infinitesimal mass over the whole bulk mass distribution). "Nature" may not "do perturbation theory", but that is irrelevant to whether that perturbation theory $\endgroup$ Commented Dec 17, 2020 at 20:28
  • $\begingroup$ can or cannot usefully be applied in a particular instance. Why can't it be here? $\endgroup$ Commented Dec 17, 2020 at 20:28
  • $\begingroup$ E.g. if we assume two particles of extremely, extremely tiny mass so their gravitational attraction is negligible compared to the FLRW expansion, why would they not expand at all? You say it would have to be "entirely" uninfluenced by other matter - but what is the mathematical reasoning for this exact zero? I won't say it can't be so, but I want to see more justification for it. $\endgroup$ Commented Dec 17, 2020 at 20:30
  • 1
    $\begingroup$ But the universe is mostly dark matter and dark energy, neither of which clumps. Why is "the solar system is visible matter, with a perturbative background of diffuse dark matter and dark energy" not valid? In particular, since dark energy is, as far as we can tell, a uniform $\Lambda$ with no structure? $\endgroup$ Commented Jan 18, 2021 at 23:09
  • 1
    $\begingroup$ @JerrySchirmer All I'm saying is that there's nothing special about cosmology. Approximating the widget distribution as visible widgets plus uniform widgets at the average density of visible+invisible widgets over a larger area is exactly as accurate as it would be in a non-cosmological context. I think the local dark matter density is probably orders of magnitude higher than the local dark energy density, so there probably is a local dark matter force but it's nothing like the ä/a force that Cooperstock et al calculated. $\endgroup$
    – benrg
    Commented Jan 19, 2021 at 3:53

Ben Crowell's answer is right, but I am adding a point in order to emphasize it, because this issue keeps coming up. Here is the point:

The cosmological expansion is FREE FALL motion.

What this means is that the clusters of galaxies on the largest scales are just moving freely. They are in the type of motion called 'free fall'. It means they are going along, with their velocity evolving according to whatever the net average gravity of the cosmos as a whole says. There is no "inexorable space expansion force" or anything like that. They are not being carried along on some cosmic equivalent of tectonic plates. They are just falling. In technical language, their worldlines are geodesic. This should help you to understand why forces within galaxies, and within ordinary bodies, will hold those galaxies and those bodies together in the normal way. It is not essentially different from objects falling to Earth under the local gravity: Earth's gravity offers a tiny stretching/squeezing effect, but this is utterly negligible compared to all the ordinary forces within materials.

If you could somehow switch off the gravitational attraction within the solar system and the galaxy and the local cluster, and all the electromagnetic and other forces, then, and only then, would the parts of the solar system begin to drift apart under cosmic free fall motion, commonly called the expansion of space.

  • $\begingroup$ But if particles are not being "carried along" with expansion, what would make them start to drift apart when switching off all forces? $\endgroup$
    – pela
    Commented Feb 15, 2019 at 12:33
  • 1
    $\begingroup$ But it seems to me that these two ways imply two opposite things: Say two particles are separated by 1 cm, and that, by magic, there are no forces between them. If they are not "carried along on some cosmic equivalent of tectonic plates", they should remain at a distance of 1 cm from each other forever. But in the standard description of the expansion of space, they will remain at their comoving coordinates, which means that they will start increasing their physical distance from each other, i.e. drift apart. Right? $\endgroup$
    – pela
    Commented Feb 15, 2019 at 22:38
  • 1
    $\begingroup$ Let's see what it means to 'release' balls in cosmic case. It would mean seeing to it that their initial velocities matched the cosmic comoving ones at their locations. So they are launched on a pair of initially close geodesics which have a non-zero rate of separation.That's why they drift apart. $\endgroup$ Commented Feb 18, 2019 at 21:42
  • 1
    $\begingroup$ Let's see what it means to 'release' balls in cosmic case. It would mean seeing to it that their initial velocities matched the cosmic comoving ones at their locations.” - No it would not. Releasing means they keep the relative velocities they had. If two balls were stationary attached by a solid rod and you remove it and ignore their gravitational attraction (and the mystical “repulsive dark energy”), then the balls would remain stationary to each other. So a galaxy with gravity and rotation suddenly switched off would not drift apart. Space expansion without acceleration is not a force. $\endgroup$
    – safesphere
    Commented May 16, 2020 at 8:20
  • 2
    $\begingroup$ There is no experimental evidence that space expands instead of galaxies simply flying apart on inertia given the initial momentum (like in the Milne model). These two interpretations are currently equivalent. Even if we could falsify one on the global scale, they are non-falsifiable locally on a galactic scale. This is another proof that a galaxy would not drift apart. $\endgroup$
    – safesphere
    Commented May 16, 2020 at 8:35

The heart of it is that the gravity of the star is just so much stronger than the expansionary shear of the expansion of the universe that you can just ignore the shear completely.

That said, if you write down the Einstein equation for the case of a simple cosmological-constant dominated universe and a spherically symmetric matter source, you do get a metric that is different from the Schwarzschild metric, modified by the cosmological constant term. In addition to the well-known instability at $r=6M$ (with its location modified by the $\Lambda$ term, you also get an instability in orbits whose radius is greater than some value $r_{*}$, which is proportional to $1/(\Lambda M)$. This outermost instability represents the expansion of the universe starting to dominate over objects orbiting very far from the star (remember that $\Lambda$ is typically very small relative to other physical quantities).

  • $\begingroup$ So expansion of the universe DOES, in fact, disrupt orbits? Interesting. I was always wondering that such an effect should take place. However, it also seems like this disruption stays constant through time and is only really dependent on the cosmological constant. What if the cosmological constant were 0? Would a very slight "weakening" of gravity still be felt locally? $\endgroup$
    – Max
    Commented Nov 14, 2019 at 15:32
  • $\begingroup$ @Max: it does depend on the Hubble constant, which is deceptively time-variant. the above is a discussion about a known exact solution where exact answers can be worked out. For a dynamic expanding universe, you'd have similar effects, but they would have to be worked out numerically. You still have the general rule that the gravity of the star is still MUCH stronger than the shear effects are. $\endgroup$ Commented Nov 14, 2019 at 16:05

Can the Hubble constant be measured locally?

To answer the question, certain distinctions must be made. What does one mean by "locally". Strictly locally means a small region. Measurement of the Hubble constant means making a comparison between truly local measurements (e.g rulers, or radar within the near solar system) and measurements on larger scales. So, on the smaller scales, i.e. the strict meaning of "locally", one would only be making a comparison between local measurements and themselves. Of course local measurements are equal to themselves, and so one cannot measure expansion.

Within a gravitationally bound system, such as a galaxy or a gravitationally bound cluster, there is no expansion. There is no measurement of the Hubble constant within the Local Group of galaxies, for example. On a larger scales, we measure the Hubble constant for receding galaxies. On cosmic scales, this might still be regarded as "local". Personally I would not be comfortable with that definition. The fact remains, the Hubble constant is measured for recession of galaxies within a neighbourhood of the Sun, but outside of the local group, and outside of gravitationally bound systems. It is a matter of fundamental principle that this is so, and from this point of view there can be no "local" measurement of the Hubble constant.

Nevertheless, it remains that we measure the Hubble constant from recession of galaxies which we can see. That is to say, galaxies within the horizon, or within the light cone. I would call this a neighbourhood, rather than a locality, but language is not generally so precise that I would expect everyone to agree. The Hubble constant applies within this neighbourhood, and outside of our immediate locality.

On larger scales, we usually apply the Hubble constant to the expansion of the universe as a whole. But this requires a massive additional assumption, namely the cosmological principle, that on large enough distance scales the universe is homogeneous and isotropic. While the cosmological principle is, at least superficially, extremely reasonable and difficult to dispute, it is only intended as approximation and is clearly not applicable to small scales where matter is not distributed uniformly. It gives no actual indication of how large a distance scale is necessary for its correct application. Consequently it is quite possible (and I am sure it is also true) that the cosmological principle is only correct on distance scales greater than the observable universe. The implication is that Hubble's constant may be true for recession of galaxies within the observable universe, but that it does not give a measure of the rate of expansion of the universe as a whole. I have written a paper in which I argue that we should distinguish the Hubble constant for the local recession rate of galaxies from the Le Maitre constant, for the rate of expansion of the universe as a whole (with a factor of roughly 2 for the difference between this rates).

  • $\begingroup$ I think "local" usually refers to a region of spacetime that's causally-separated from the remainder of the universe or multiverse, typically by the gravitational collapse associated with large stars whose complete expenditure of their nuclear fuel has left them without radiation pressure to withstand such collapse: As at least half of all stars are in binary pairs, the collapse of one partner in such pairs has left at least 90 of them visibly continuing to follow the elliptical orbit that characterizes all binary pairs. The invisible partners are considered to have become black holes. $\endgroup$
    – Edouard
    Commented Jan 17, 2021 at 21:51
  • $\begingroup$ It's the concentration of a locality's gravitational field that produces a time dilation so extreme that whatever happens in a BH's interior will outlast all decoherence of particles outside it. (Corny phrases like that interior's "expansion to infinity" usually dramatize the result, although any spatial expansion involved may actually happen outside it.) $\endgroup$
    – Edouard
    Commented Jan 17, 2021 at 22:01
  • 1
    $\begingroup$ @Edouard, "local" is formally defined in mathematics. It means a region small enough effects due to curvature are not detectable. $\endgroup$ Commented Jan 17, 2021 at 23:06
  • $\begingroup$ Coincidentally, I'd just run across an item, at mail.google.com/mail/u/0/?tab=wm#inbox/…, describing something I'd forgotten about, which is that the earth's rotation is slowing perceptibly, but only over periods of millions of years: I think that such a slowing might provide a link between the math you've mentioned and the curvature in cosmological physics, which is generally described as overwhelmingly temporal. (Rotation of our observable region is so subtle that every study of its possibility seems to bring different results.) $\endgroup$
    – Edouard
    Commented Jan 18, 2021 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.