# Would order 1AU metric expansion of space begin if the solar system were not inside a galaxy?

In this question I describe the >30 years of laser ranging between lasers on Earth and the retroreflector arrays on the Moon. Amazingly, after comparing this data to simulation of all of the orbital mechanics and tidal effects, the residual is only a few centimeters.

If $\text{H}_D$ is about 70 $\text{km}\ \text{s}^{-1} \text{Mpc}^{-1}$ (or about 2.3E-18 $\text{sec}^{-1}$), then with a semi-major axis of 3.84E+08 meters, over 37 years the effect would be of order 1 meter. Since these measurements are consistent with zero to the level of a few centimeters, this is taken as experimental evidence that metric expansion is not taking place locally compared to the rate seen between galaxies.

If I understand correctly this is "suppression" is due to the large amount of mass in our galaxy, even though it is thousands of light years away. So I am wondering - if a similar experiment were done in a similar solar system associated with an isolated star alone in an intergalactic region, what does current theory predict - would metric expansion be detected, or would the mass of the one star be enough to suppress it?

Then what if it were just a planet and a moon without the mass of the star, or even two smaller masses?

I'm looking for an answer at a level similar to this answer and this answer, where time was taken to note the the specific relevant concepts and work from the paper linked in the first answer.

• I'm not sure why you keep asking variants of the same question, which is does space expansion expand matter. There's no "suppression" going on, but the solar system (or any bound system for that matter) simply does not obey the FLRW assumptions, so there is no reason to expect to see space expansion locally. Jul 16, 2016 at 11:32
• @ACuriousMind OK so if two galaxies were far from each other but gravitationally bound, there would be no expansion, but if they were at the same distance from each other but not bound, then there would be expansion? What if they were just on the edge between bound and unbound? Can you give this as an answer? Thanks!
– uhoh
Jul 16, 2016 at 11:43
• @ACuriousMind in a real distribution of (say 100) galaxies with significant relative motion, the multi-body dynamics is such that there may not even be a clear definition of which pairs are bound and which pairs are unbound. I don't think "bound" is really boolean here.
– uhoh
Jul 16, 2016 at 11:52
• I don't think so either. What matters is how good the approximation by the FLRW assumption from which the space expansion is derived holds. Certainly, a star with some planets around it it neither "homogeneous" or "isotropic" in any sense. The top answer about space expansion and matter already says that: Space expansion only works on the scales where galaxies are microscopic and behave like fluid particles. As you zoom in, the model becomes worse. I don't know what you are asking that is not already answered there. Jul 16, 2016 at 12:07
• @ACuriousMind OK thanks. Actually it's easier for me to understand your explanations that the other answer. So if the Earth and Moon were all alone in intergalactic space, is it really for sure, without question, that there would be no expansion observed, and if so, would that be because a) their mass is inhomogeneous, or b) because they are gravitationally bound? Or in this particular case, is the answer actually "we don't really know."
– uhoh
Jul 16, 2016 at 12:30

These notes put some numbers on @ACuriousMind 's answer: one needs to be looking at length scales of 100 Mpc and greater for the FLRW metric to be a realistic description of reality. That's a staggering distance, and equivalent to timescales amounting to the whole Mesozoic era, comprising the rise and fall of the Dinosaurs! So one cannot expect the scalefactor expansion of spacetime to apply to our Earth-Moon system simply because the system doesn't fulfill the assumptions that justify the FLRW metric.

Perhaps a more addressable variant to your question would be to ask what difference a positive cosmological constant makes to a metric that does describe the Earth-Moon system. This is a question that can be answered, and it is reasonably straightforward to go through the derivation of the Scwharzschild metric but with a positive cosmological constant. One finds that the metric changes as follows:

$$g_{t\,t} =c^2\left( 1-\frac{r_s}{r} -\frac{r^2\,\Lambda}{3}\right)$$ $$g_{r\,r} = \frac{c^2}{g_{t\,t}}$$

and the cosmological constant, if small enough, does not change the basic character of geodesics; it will however shift the radiusses of stable orbits. These notes sketch how to work through the computation; the radially symmetric system with nonzero $\Lambda$ being Problem 23. in Chapter 23 of Moore, Thomas A., "A General Relativity Workbook".

So there is no ongoing spacetime expansion in this system: orbits are just a little bigger than they would be with $\Lambda=0$ and some weakly bound orbits would become unbound with positive $\Lambda$. Therefore, we would not expect the Moon to be drifting away any faster that it is owing to nonrelativistic effects.

• Thank you very much for taking the time to look at my question, and then to write out a thoughtful answer, and also for choosing some helpful notes and linking to them. This will keep me occupied for quite a while. I'll use simplified words here - for the earth-moon cloned system in intergalactic space, drift (due to this effect) won't happen mostly because a) the mass distribution is non-uniform, b) they are gravitationally bound, or c) the scale is too small, or some or all of the above. I see all three of these used in different places in forums, and b) bothers me the most.
– uhoh
Jul 16, 2016 at 15:39
• This is wrong. It is not true that orbits never show any secular trend due to cosmological expansion. The secular trend depends on $(d/dt)(\ddot{a}/a)$, which just happens to vanish for the vacuum-dominated cosmology you picked. See physics.stackexchange.com/a/70056/4552 .
– user4552
Aug 17, 2018 at 21:39
• This answer is correct. The previous comment by user4552 (Ben Crowell) is incorrect, for the reason stated in this answer: the FLRW geometry is realistic only at much larger scales. See physics.stackexchange.com/a/601323 for a rebuttal to the argument that he linked. Aug 31, 2022 at 1:42