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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.

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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) 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

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  • $\begingroup$ 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 Jul 8 '13 at 3:35
  • $\begingroup$ Rulers might expand but light would still be travelling at same speed. Hence the changes would still be noticeable. $\endgroup$ – Boltzee Jun 8 '17 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 Jan 31 '18 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 Feb 15 at 11:48
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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.

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  • $\begingroup$ 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 Feb 15 at 12:33
  • $\begingroup$ "drift apart" and "carried along" are two ways of describing precisely the same thing, except that "carried along" makes it seem as if there is something inexorable about it, which is a bit misleading. My aim here is merely to promote a better physical intuition about this motion. $\endgroup$ – Andrew Steane Feb 15 at 13:37
  • $\begingroup$ 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 Feb 15 at 22:38
  • $\begingroup$ Remain at comoving coordinate, hence increasing physical distance: yes, both agreed. Now compare with two small balls released from rest, one above the other, at some height above the Earth. Both are in free fall, both worldlines are geodesics, and their physical separation increases with time. The cosmological case differs in that there the metric is non-static, but as far as the forces are concerned the 'balls released into free fall' gives a fair comparison. It illustrates why other forces can easily overcome the cosmological free fall motion. $\endgroup$ – Andrew Steane Feb 16 at 14:35
  • $\begingroup$ I still don't get it. In the "release the balls" example, the balls experience a different potential at a given time, so it makes sense (I think) that they drift apart. But in the expanding space, the potential is (ideally) the same everywhere. It seems to me that in order for the two particles (for which we by magic removed all forces) to start drifting apart, they should in fact be carried along some cosmic equivalent of tectonic plates. $\endgroup$ – pela Feb 17 at 20:34
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This is a more naive answer, for a more complete one refer to the previous answers!

The whole concept of the Hubble constant comes from considering a homogeneous and isotropic universe, which is an approximation. This is actually a good approximation, but it only works on large scales! Locally (on the scale of the solar system, for example) homogeneity and isotropy are not true at all, so trying to do a local measurement would be like applying classical physics to subatomic particles: out of the range of validity of the theory you're using!

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  • $\begingroup$ This isn't really incorrect, but it's not the full story. If you haven't, I highly suggest reading the other answer to this question. $\endgroup$ – Chris Feb 15 at 11:45
  • $\begingroup$ @Chris Yes, I did! I just wanted to give a more naive, short answer which would hopefully be more understandable with less background. This was by no means intended to depict the whole story. I added a note on this in my answer. $\endgroup$ – Angelo Brillante Romeo Feb 15 at 12:58

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