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In this answer @PhillS linked to the paper Progress in Lunar Laser Ranging Tests of Relativistic Gravity which has given me a great start. While its main focus is on the Equivalence Principle (EP) they also mention that no local expansion was seen. The last sentence in the abstract is:

" The search for a time variation in the gravitational constant results in G/G ˙ = (4 ±9) ×10 −13 yr − 1 ; consequently there is no evidence for local ( ∼1 AU) scale expansion of the solar system."

Briefly, the data here is 34 years of laser ranging from Earth of a retroreflector array on the moon. While the distance between a given measurement site and the reflector on the moon can vary by as much as 50,000km (due mostly to the Moon's elliptical orbit and the size of the earth) these can be and have been painstakingly modeled. The amazing result is that in 34 years the residual scatter is only about 2cm!!

What I'm looking for is a midrange answer - not (exclusively) high level cosmology, but more than just balloon and raisin cake analogies. Something that will help towards understanding the relationship between the two.

Question: How is a result of no time varying $G$ related to a measurement of no local expansion?

Bonus mini-question: If I understand correctly and the null measurement of expansion is from the earth-moon ranging data, why does the sentence say "local (~1 AU) scale" when the earth-moon distance is only 0.0027 AU?

Lunar Libration image from here

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Lunar Laser Ranging images from here

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    $\begingroup$ The ratio of 400 or so between the Earth-Moon distance and the Earth-Sun distance may seem a lot, but the ratio of the Earth-Sun distance to even the nearest star is about 250000. So for this purpose all distances in the Solar System can be treated as of order 1 AU. $\endgroup$ – John Rennie Apr 24 '16 at 11:07
  • $\begingroup$ Thanks @JohnRennie, I see what you mean. Wish we had a handy term for a few orders of magnitude like "5rder 1 AU" or "ord3r 1 AU" $\endgroup$ – uhoh Apr 24 '16 at 14:36
  • $\begingroup$ @JohnRennie See my answer below for an argument that your comment is not correct; "1 AU" in this case means, mostly, "1 AU". $\endgroup$ – rob Aug 16 '16 at 3:21
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If cosmological expansion applies on the scale of the earth moon system, then in some short period of time $\delta t$ the distance between the earth and moon increases from $r$ to $r+\delta r$. So the force of gravity between the bodies changes to:

$F+\delta F=\frac{GMm}{(r+\delta r)^2}\approx\frac{GMm}{r^2}\left(1+\frac{\delta r}{r}\right)^{-2}\approx\frac{GMm}{r^2}\left(1-\frac{2\delta r}{r}\right)=F\left(1-2\frac{\delta r}{r}\right)$

The same change in gravitational force could also come about via change in $G$:

$F+\delta F=\frac{(G+\delta G)Mm}{r^2}=F\left(1+\frac{\delta G}{G}\right)$

Equating the $\delta F$ terms shows you that the change in the force due to increasing distance is the same as the change in the force due to decreasing $G$ if

$\frac{\delta G}{G}=-2\frac{\delta r}{r}$

Now if our $\delta r$ is due to cosmological expansion being applicable to this distance scale, then for an expansion velocity of $v$, $\delta r=v\delta t$ and by using Hubbles law $v=H_0 r$ where $H_0$ is Hubble's constant, which is around $70kms^{-1}Mpc^{-1}$, which we want to converft to SI units ($ms^{-1}{m^-1}=s^{-1}$) which gives us $H_{0}=2.26\times10^{-18}s^{-1}=7.1\times10^{-11}yr^{-1}$

So plugging that in to our equation gives $\frac{\delta G}{G}=-2\frac{H_{0}r\delta t}{r}$. Simplifying, and turning $\frac{\delta G}{\delta t}$ into $\dot{G}$ we end up with

$\frac{\dot{G}}{G}=-2H_{0}$

This is a simplistic way to do it, but the basic idea is on the right track I believe. A cosmologically-caused increase in the earth moon distance would give a reduction in the mutual force between them the same as a decrease in $G$ would if they stayed at the same distance. Rearranging the maths as above gives you a a direct proportionality between $\frac{\dot{G}}{G}$ and $H_0$, although the constant of proportionality for general relativity is probably not exactly what I've come up with here (and other theories of gravity may give different values).

The paper you mention puts an experimental limit on that constant of proportionality of (roughly) $0.006 \pm 0.012$ (using the value $H_0$ above), as opposed to the 'predicted' value of $-2$

The reason for expressing the result in terms of $\frac{\dot{G}}{G}$ idea that that is directly derived from what they measure independent of any theory. I.e. it is as far as you can go towards putting numerical limits on the expansion without committing to a particular theory or value of $H_0$

(I believe that John Rennie's comment is correct that the are using '$AU$ scale' to indicate that they are talking about effects on the order of the solar system, rather than galactic or cosmological distances.)

Excellent animation of the moon changing through the month BTW.

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  • $\begingroup$ Thank you again @PhillS! This is very helpful. Is it more like one thing is conveniently expressed in terms of another thing, or that they are indistinguishable in this experiment, or more like they are pretty much indistinguishable in a wide variety of cases (imagine instead of orbiting, they are held apart by a giant calibrated thruster or spring). I ask because I'm not sure what happens if more space appears and I want to conserve angular momentum at the same time. $\endgroup$ – uhoh Apr 24 '16 at 15:23
  • $\begingroup$ I've asked a related question. $\endgroup$ – uhoh Jul 16 '16 at 11:21
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Your mini-question is addressed in this review: the gravitational force on the moon due to the earth is only about 40% the gravitational force on the moon due to the sun, so gravity at length scales comparable to the sun-earth distance plays the lion's share of the role in determining the evolution of the moon's position. The model appears to also include perturbations to the earth-moon system from Jupiter, Venus, and perhaps other bodies; I haven't followed the references.

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  • $\begingroup$ Thanks! There is this comment suggesting that in GR context, ~1AU can actually cover at least a few orders of magnitude either way. But your answer is quite helpful. That's an excellent experiential review and discussion and exactly what I need to read. $\endgroup$ – uhoh Aug 16 '16 at 3:17

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