An orbit feels the acceleration and never the rate

A cosmological redshift is a change of scale, and the scale is still changing — fast enough that a galaxy’s redshift drifts while it is being watched. The obvious next question is what that change does to things that are not galaxies drifting apart. If the space between galaxies is growing, why is the space between the Earth and the Sun not growing with it? The usual answer is that gravity holds the Solar System together against the stretching, as a knot holds against a pulled rope. That answer implies a contest, and a contest would leave a residue: an orbit held against the expansion should be very slightly larger than it would otherwise be, and should keep growing at some tiny rate.

Neither happens, and the reason is not a contest. The expansion rate does not appear in the equation of motion of a bound orbit at all. What appears is the acceleration of the expansion, and of that, only one part reaches inside a system held together by its own gravity: the part contributed by the cosmological constant, which acts as a small, fixed outward force proportional to distance. An orbit feels that force once, settles at a very slightly different radius, and stays there.

The rate is a velocity, and a velocity is not a force #

In a universe that is the same everywhere, the distance $r$ between two freely moving bodies, measured at one instant, obeys an equation that looks exactly like Newton’s:

$$\ddot r = \frac{\ddot a}{a}\, r, \qquad \frac{\ddot a}{a} = H_0^2\left[\Omega_\Lambda - \frac{\Omega_m}{2a^3} - \frac{\Omega_r}{a^4}\right].$$

Put a mass $M$ between them and add its pull, and the equation becomes $\ddot r = -GM/r^2 + (\ddot a/a)\,r$. The expansion rate $H = \dot a/a$ is not in it. It cannot be, because an equation of motion relates accelerations to positions, and $H$ is a velocity per unit distance.

The rate enters only through where things start. Two bodies set moving apart at the local Hubble rate keep moving apart, because nothing stops them; that is what a galaxy far from any other is doing, and the recession of distant galaxies is the persistence of velocities they were given long ago. A planet is not moving apart from its star at the Hubble rate. It is moving sideways at its orbital speed, and the fact that distant galaxies are receding has no more effect on it than the speed of a train has on a ball thrown inside a carriage. The Hubble flow is a pattern of velocities, and a pattern of velocities does not push.

What can push is a change in that pattern — an acceleration — and the first figure shows that the acceleration term has two parts of very different character. The matter-and-radiation part is the gravity of the universe’s mean density, which was enormous early on and has thinned as the cube of the scale factor. Inside a galaxy or a planetary system there is no smooth cosmic mean density to be felt as a separate thing: the matter that is actually there is what is there, and it is already counted in the mass doing the holding, many million times over. What is left is the cosmological constant’s part, $\Omega_\Lambda H_0^2$, which does not thin at all.

What is left is a tide #

A force proportional to distance from a centre, and the same at every centre, has a familiar name in celestial mechanics. It is a tidal field — the difference between the pull on one side of a system and the pull on its middle — and a tide is always a difference, stretching a body along one direction rather than moving it as a whole. The cosmological constant’s contribution is a tide of an unusually simple kind: isotropic, outward, and uniform in space and time, with a strength of $\Omega_\Lambda H_0^2 \approx 3\times10^{-36}$ per second squared per unit distance.

A tide deforms an orbit; it does not make it grow without limit. The equation for a circular orbit with angular momentum $L$ becomes $L^2/r^3 = GM/r^2 - \Omega_\Lambda H_0^2\, r$, and to first order the outward term simply moves the radius at which the balance holds.

Twelve picometres, once #

The Earth’s orbit is larger than it would be in a universe without a cosmological constant by twelve picometres — about a tenth of the diameter of a hydrogen atom — and it is larger by that amount now, was larger by that amount a billion years ago, and will be larger by that amount a billion years from now. The shift grows as the cube of the orbit’s radius, which is why it is 22 orders of magnitude below the dashed line at the Earth and reaches it only at 111 parsecs from a star of one solar mass.

The Moon’s orbit round the Earth is shifted by about $2\times10^{-16}$ metres, a fifth of the radius of a proton. The Moon is in fact receding from the Earth by 3.8 centimetres a year, measured by laser ranging to reflectors on its surface, and every bit of that is the tide the Moon raises on the Earth’s oceans transferring the Earth’s spin to the Moon’s orbit. A literal expansion of the Earth–Moon distance at the Hubble rate would add about two and a half centimetres a year to that, and it is not there.

The same test is sharper for the planets. If the Earth’s orbit were carried along with the universe it would grow by about ten metres a year, since the Hubble constant is about seven parts in a hundred billion per year. Radar and spacecraft ranging to the inner planets track their distances to within metres over decades, and no such growth has ever been seen. The measured orbits are exactly what a static shift of picometres predicts: indistinguishable from no shift at all.

The fractional shift has a second reading that makes its size less mysterious. It is exactly the ratio of the two accelerations: the push at the Earth’s distance from the Sun is $5\times10^{-25}$ metres per second squared, and the Sun’s pull there is $6\times10^{-3}$, and their ratio is the same $8\times10^{-23}$. The Pioneer spacecraft, far beyond Neptune, once appeared to be decelerating by $9\times10^{-10}$ metres per second squared more than gravity allowed, and a cosmological origin was among the explanations proposed. The push at their distance is thirteen orders of magnitude smaller than that, and the anomaly was eventually traced to heat radiated unevenly from the spacecraft’s own power source.

The same equation holds for anything held together, with the binding force in place of gravity, and the answer only becomes more extreme. An electron in a hydrogen atom is held by an electric pull so much stronger than the push, at so small a distance, that the atom is larger than it would otherwise be by about two parts in $10^{69}$. A rock, a person or a planet is held by forces between atoms, and none of them is expanding. Nothing that is held together by anything is carried along; the expansion is a description of what happens to things that are not held.

The radius inside which nothing expands #

Because the shift grows as the cube of the radius, it becomes important somewhere, and where it does is easy to find. The outward push per unit mass, $\Omega_\Lambda H_0^2\, r$, equals the inward pull, $GM/r^2$, at

$$r_\Lambda = \left(\frac{GM}{\Omega_\Lambda H_0^2}\right)^{1/3},$$

a radius sometimes called the zero-gravity radius. Outside it the push wins and no orbit round the mass exists; inside it gravity wins, and far inside it the push is a perturbation.

The radius has a simpler meaning than its formula suggests. Setting the pull equal to the push is the same as asking when the mass, spread uniformly through a sphere of radius $r_\Lambda$, has a mean density equal to twice the density of the cosmological constant’s energy. That density is about $6\times10^{-27}$ kilograms per cubic metre, so a mass is safe from the push wherever its mean density exceeds the equivalent of about seven hydrogen atoms per cubic metre — whatever the mass is and however it is arranged.

Every system held together by its own gravity is far denser than that. The Solar System’s planets sit at a millionth of the Sun’s radius. A globular cluster at three tenths of a per cent of its own, and a Milky Way halo of a trillion solar masses, whose extent is inferred from rotation curves that refuse to fall, at 18 per cent. A rich galaxy cluster, weighed three independent ways, sits at 18 per cent. These are not coincidences of the examples. A system that has collapsed and settled under its own gravity ends up a couple of hundred times denser than the mean density of the universe, and the mean density of the universe is itself comparable to the cosmological constant’s, so every settled system is safely inside its line by construction.

The exception is the Local Group, at 57 per cent. Its two large galaxies, the Milky Way and Andromeda, are bound to each other, but the group has not settled: it is still falling together for the first time, and it is the one system on the list where the push is not negligible.

A supercluster is not a system #

The same figure, given a different list, shows what happens to structures that are not bound.

A wide binary, whose two stars take hundreds of thousands of years to orbit each other, is as secure against the push as the planets are. An open cluster like the Pleiades is secure too; it will be dispersed by the Galaxy’s tides long before anything cosmological could matter. The Virgo cluster, the nearest rich cluster, sits at 19 per cent.

Laniakea is different in kind. It is the supercluster that contains the Local Group, defined in 2014 not by what holds it together but by which way its galaxies are flowing: the region inside which the peculiar velocities of galaxies point towards a common centre of attraction. Its mass is spread so thinly that it lies at 155 per cent of its own Λ radius. Nothing holds it together. Its galaxies are falling towards each other less fast than the expansion carries them apart, and in the far future its member groups and clusters will be isolated from one another, each inside its own line, with the distances between them growing exponentially — the process that eventually puts every other galaxy out of reach.

The Local Group itself is about sixteen megaparsecs from the Virgo cluster, and Virgo’s own radius, from its 19 per cent, is about twelve. The Local Group is falling towards Virgo in the sense that its recession is slowed by Virgo’s pull, but it is outside Virgo’s line and will never arrive. The largest structures in the universe that will ever be bound are already defined, and they are groups and clusters, not superclusters.

The last orbit the push allows #

Inside $r_\Lambda$ a circular orbit exists, but existence is not the same as stability. An orbit is stable if a small nudge makes it oscillate about its radius rather than drift away, and the frequency of that oscillation, the epicyclic frequency, is what the balance between angular momentum and gravity produces. The outward push weakens it.

The epicyclic frequency squared for an orbit of angular speed $\Omega$ is $r^{-3}\, d(r^4\Omega^2)/dr$, and with $\Omega^2 = GM/r^3 - \Omega_\Lambda H_0^2$ that is $GM/r^3 - 4\,\Omega_\Lambda H_0^2$. It vanishes where the mean density inside the orbit is four times the push’s equivalent, at $4^{-1/3}$, or 63 per cent, of the Λ radius. The factor of four is a property of the inverse square law and the linear push together, and it is the same for every mass.

So a star of one solar mass could in principle hold a stable planet out to 70 parsecs, and a halo of a trillion solar masses a satellite out to 702 kiloparsecs. In practice neither limit is ever reached, because something else intervenes first. At the Sun’s position in the Galaxy, the tidal field of the Galaxy’s own disc is more than a million times stronger than the cosmological constant’s, and it is the Galaxy’s tide, not the universe’s, that strips comets from the outer Oort cloud. The cosmological limit is the one that applies only to a system with nothing around it.

A decelerating universe would not have stretched an orbit either #

It is tempting to connect all this to the history in the first figure — to suppose that bound systems were compressed while the expansion decelerated and began to be stretched when it started to accelerate, 7.7 billion years after the big bang. That is not what the equation says. The sign change is in the sum, and the part of the sum that changed is the mean-density part, which does not reach inside a bound system. The cosmological constant’s push was present at the same strength throughout; it merely came to dominate the universe’s mean density at a redshift of 0.63. An orbit round the Sun 10 billion years ago felt exactly the push an orbit feels now.

A universe with no cosmological constant at all would have left orbits with no outward push whatever, and the question of whether bound systems expand would have had a simpler answer than this one: not even by a picometre. The acceleration that was discovered in 1998 changed what the universe does on the largest scales. What it changed inside the Solar System is twelve picometres, applied once.

The same answer without Newton #

The argument above uses the Newtonian limit of an expanding universe, which is accurate for any system much smaller than the Hubble radius, and the question has also been settled without that approximation. In 1945 Einstein and Straus showed that a spherical region can be cut out of an expanding universe of dust, its matter condensed into a central mass, and the region’s interior replaced by the ordinary solution for the space round a mass — with the exterior universe continuing to expand exactly as before and the interior unaffected by it. The construction extends to a universe with a cosmological constant, where the interior becomes the solution for a mass in a space with a cosmological constant; its orbits are the ones drawn in the last figure.

The expansion of the universe, in that picture, is not a property of space that everything inside it has to share. It is a statement about how the average separation of freely moving matter grows, and it applies where matter is spread thinly enough to be moving freely — which is to say, between systems and not within them. The Hubble constant is a rate for that average, and asking what it does to an orbit is like asking what the average speed of traffic in a country does to a parked car.

What the argument leaves out #

It assumes the dark energy is a constant. If the dark energy’s density changes with time, its contribution to the local acceleration changes with it, and a density that grows would eventually overwhelm bound systems from the largest to the smallest. Nothing in the data requires that, and the push drawn here is for a density that stays fixed.

It treats each system as isolated. A real group sits inside a web of other masses whose tides are generally far stronger than the cosmological constant’s, which is why the Galaxy’s tide, not the push, sets the edge of the Oort cloud.

And the Λ radius is for a point mass. An extended halo has a different radius at each distance, set by the mass enclosed there, and the line drawn is the radius at which the whole mass, concentrated, would balance the push. For the systems on these figures the difference does not move any of them across the line.

Still open: the one group that is close to its line #

The Local Group is at 57 per cent of its Λ radius, and it is the only system in which the cosmological constant’s push is a correction worth making. Its two large galaxies are approaching each other at about 110 kilometres per second. They cannot always have been approaching: at the big bang they were together, and they must once have moved apart, stopped, and turned round. That history — a single radial orbit begun at the beginning of time — is enough to weigh the group, and the push changes the answer by a measurable amount.