# Can the Hubble constant be measured locally?

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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–  Qmechanic Jul 20 '13 at 5:44
related: physics.stackexchange.com/q/3062 –  Ben Crowell Jul 20 '13 at 15:18
related: physics.stackexchange.com/q/2110 –  Ben Crowell Aug 6 '13 at 5:45
–  Ben Crowell Sep 9 '13 at 21:17

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 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.

General relativity can't describe quantum-mechanical systems such as atoms. A nice way of discussing atoms, nuclei, photons, and solar systems all on the same footing is to note that in geometrized units, the units of mass and length are the same. Therefore the existence of any fundamental massive particle sets a universal length scale, one that will be known to any intelligent species anywhere in the universe. Since photons are massless, they can't be used to set a universal scale in this way; a photon has a certain mass-energy, but that mass-energy can take on any value. Similarly, a solar system sets a length scale, but not a universal one; the radius of a planet's orbit can take on any value. A universe without massive fundamental particles would be a universe without length measurement. It would obey the laws of conformal geometry, in which angles and light-cones were the only measures. This is the reason that atoms and nuclei, which are made of massive fundamental particles, do not expand. Photons, however, do expand -- or at least, this is one possible way of describing cosmological redshifts.

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

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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! –  burgerking Jul 8 '13 at 3:35