# Hubble Recession vs. Moon Recession

1. Jun 5, 2006

### kmarinas86

Last edited: Jun 5, 2006
2. Jun 6, 2006

### pervect

Staff Emeritus
Nope. Ned Wright addresses this issue in his cosmology FAQ.

http://www.astro.ucla.edu/~wright/cosmology_faq.html#SS

with a reference to the literature

http://xxx.lanl.gov/abs/astro-ph/9803097
(the Cooperstock reference below)

[add]I'm thinking about analyzing this problem in a different way, but I don't want to hijack this post, and I'm not confident of my approach yet - so I may post some more on this in another thread.

Note that the increase in the moon's orbit is well understood - it's due to the spin-orbit coupling between the moon and the Earth. This is due to the tides on the Earth caused by the moon. This effect has nothing to do with GR, it is correctly predited by Newtonian gravity.

see for example
http://curious.astro.cornell.edu/question.php?number=124 [Broken]

Last edited by a moderator: May 2, 2017
3. Jun 8, 2006

### kmarinas86

But according to my calcuation above, either the Hubble constant does not apply over short distances < 1AU, or the moon is affected by the Hubble constant, however ever small of a fraction its contribution. Sure, gravity overwhelms it, that quite obvious and not new. But when I plug in the numbers, assuming a Hubble Constant of 70 km/s/Mpc (see first post), I get something in the range of the centimeters. Could the Hubble Expansion be a co-cause of the Moon Recession? Of course, we know that this would result in the usual conservation of angular momentum.

And why does the literature you cite have a answer that is orders magnitude different than when I approximated by mutiplying initial Hubble Velocity (H*D) with time? If that source is right, then the (H*D) cannot be applied at the scale of the solarsystem... which is weird since the literature you cite assumes that expansion occurs at all scales. Is it that the real function of expansion depends on the energy density, and thus makes (H*D) too simple, or just plain wrong, for cosmologists?

Last edited by a moderator: May 2, 2017
4. Jun 9, 2006

### pervect

Staff Emeritus
The motion of bodies near the solar system can be computed from the actual metric in the solar system. This metric is not the metric used for cosmology. The equations of cosmology such as Hubble's law do not apply to the case of planetary motion. Cosmological equations use an "averaged" metric that is not actually present in the solar system.

If you don't like thinking in terms of metric, think in terms of forces. (The solar system is not particularly relativistic, so thinking in forces does not introudce any serious conceptual errrors.

The forces on the planets in the solar system are just those due to nearby bodies, and are reasonably well-known and well-studied. There are not any "mysterious" forces due to distant bodies (though there are some tiny forces usually neglected, like gravitational effects from alpha-centauri). There are not any "mysterious forces" due to the universe as a whole, or its expansion. Well, there are probably not any such mysterious forces.

If you look at

for more discussion (I didn't want to hijack this thread with my questions in that discussion), you'll see that there are some technical issues with proving this point rigorously. From what I've read, Birkhoff's theorem seems to probably be able to be generalized to include a cosmological constant, but I could be incorrect in this.

I should also add that "dark matter" concentrations could, in principle, cause "mysterious effects", because we can't see "dark matter" (by defintion). Any such dark-matter effects must be tiny, there are not a lot of confirmed anomilies in the motions of planets in the solar system. (The pioneer anamoly is a potentially important counter-example, however people are still looking at the issue).

Note that the papers I cited previously in this thread take a somewhat different approach, but come to very similar conclusion. The biggest effect that anyone suggests using any of the various approaches is "parts in a septillion"

5. Jun 21, 2006

### Jimmy Snyder

I've got good news and bad news

I'll start with the bad news but before I do, I will state an assumption. If you disagree with the assumption, then there is no need for you to read the rest of my post.

Assumption: Hubble flow occurs at all scales.

Under this assumption, after all other effects are accounted for, the Moon should be receding from the Earth at a speed of 2.75 cm/yr as calculated by the OP. However, I don't think you can measure this recession for the following reason. Consider an experment carried out in a remote region of space where forces are nil. Place two neutrons at rest with respect to each other and at a distance from each other equal to the radius of the Moon's orbit. Come back a year later and see where they've gotten to. In my opinion, they will not have moved. My reason is that in order to make it seem that they are at rest wrt each other at the outset, you must actually give them a velocity toward each other of 2.75 cm/yr in order to cancel out the Hubble flow. When you let them go, they will continue to have a velocity that will exactly cancel out the Hubble flow and so the experiment will fail.

Now for the good news. I got this idea from the Cooperstock paper although I did not actually read much of it. The idea is to measure not velocity, but acceleration. Cooperstock essentially starts with a somewhat stationary pair of objects and waits billions of years until Hubble flow takes them far apart from each other enough to measure the displacement effects of the acceleration. I propose to start with objects that are initially moving very fast wrt to each other so that they reach that distance much sooner than billions of years. In fact 80 minutes should suffice. Here is what to do. Place a particle accelerator in a remote region of space, and a particle detector 10 AU away from it (80 light minutes, roughly the distance from the Sun to Saturn). Keep them at rest wrt each other so that their relative velocity exactly offsets Hubble flow. Send an electron at nearly the speed of light from the accelerator to the detector. Measure the speed of the particle at the detector. If the assumption is correct and if the Hubble constant is 70 (km/s)/Mpc then there should be a acceleration equal to 1 part in 10^14

$$70 (km/s)/Mpc \times 10 AU \times 5 \times 10^-12 MPc/AU = 3.5 * 10^-9 (km/s)$$

$$c = 3 \times 10^5 (km/s)$$

This experiment is not practical at the present time. However, it does not require such fantastic resourses as the galaxy-sized accelerators needed for string theory. It is reasonable to think that such an experiment could possibly be carried out within the 21st century.

My thanks to Dr. Yen-Ting Lin of Princeton University and the Spanish Pontifical Catholic University of Chile for help with the equations. I should point out that some of the parameters are variable. For instance, by using a particle of speed 10^-4 * c, you only need find a difference in velocity of 1 part in 10^10. But then the experiment would take 8 * 10^5 minutes or 1.5 years to complete.

Last edited: Jun 22, 2006
6. Jun 21, 2006

### pervect

Staff Emeritus
My \$.02, again:

I think you are insisting on taking the hard apporach to the problem, after I"ve attempted to point out that there is an easier way.

But as long as you get the same answers in the end, it doesn't necessarily matter.

You are basically close to being on the right track here, except that there is something that you have not considered.

a(t) is not a truly linear function of time. If a(t) were linear, you'd be correct.

The next level of approximation is to see what happens when a(t) is modelled with a linear term and a square law term. If assuming that a(t) is linear gives no effect, you need to carry out the analysis to the next order to find out what the actual effect will be.

Exercise: calculate what happens between the neutrons if a(t) is not linear. You should find that they accelerate towards each other if the second derivative of the scale factor d^2a/dt^2 < 0, away from each other if the second derivative is positive

Exercise: a(t) satisfies the Friedman equation. A convenient form of this is, using currently accepted values for the various constants

http://www.astro.ucla.edu/~wright/Distances_details.gif
http://en.wikipedia.org/wiki/Friedmann_equations

$$da(t)/dt = H0 \sqrt{\Omega_m/a + \Omega_{vac}a^2}$$

also written as

$$[da(t)/dt] / a(t) = H0 \sqrt{\Omega_m/a^3 + \Omega_{vac}}$$

You can see that these are equivalent

here a(t) is the changing scale factor of the universe

H0 is the current value of Hubble's constant.

H(t) is, by defintion, (da/dt) / a

H0 is the value of H(t) now.

This gives you numbers that you can actually plug in to find the second derivatve of a(t). The simplest case to analyze in the one that Cooperstock also analyzes. This is the case with no cosmological constant. You can then assume

$\Omega_{vac}=0$ and $\Omega_m=1$

Compare with
http://xxx.lanl.gov/PS_cache/astro-ph/pdf/9803/9803097.pdf
equation 3.1

You will note that in this model, with no cosmological constant, the neutrons accelerate towards each other as previously noted.

Exercise: compare this acceleration with the gravitational acceleration due to the matter density rho in the sphere between the two neutrons using Newton's law. (Matter in the hollow sphere outside the two neutrons won't contribute). You'll need to find the matter density rho via the equation

rho = 3H^2 / 8 Pi G

(This is the assumed "critical" value of matter density equivalent to $\Omega_m=1$, the one that makes space flat, i.e. k=0.)

As a result of this calculation, you should then see that this provides exactly the same answer as the previoius methods did.

If you wish to go beyond this "no-cosmological constant" model to the consensus Lambda-CDM model would be

$\Omega_{m}$, the contribution due to matter is .27
$\Omega_{vac}$, the contribution of the vacuum is .73

I've decided in retrospect to snip discussion of this - better to get one point through than to lose it in a sea of information.

Last edited by a moderator: Apr 22, 2017
7. Jun 22, 2006

### Jimmy Snyder

Good point. Actually, it seems to me that I have only used the constant term, not even the linear one. And it seems that I have shown, at least to my own satisfaction, that the zero order effect cannot be measured even in principle. I wonder if the first and higher order effects are measurable in practice. That is, by staring at the Moon for a really long time to see if anything interesting happens. My uneducated guess is 'no'.

Don't forget, my not so hidden agenda is to measure Hubble flow at small scales and in a practical way. Your discussion raises, not lowers my hopes.

I also wonder if over the time scale of my proposed experiment (80 minutes) that first and higher order effects are also ignorable. My experiment would look for an acceleration (due to the constant term in Hubble flow) on the order of 1 part in 10^14. If I take higher order effects of Hubble flow into account, would this alter the acceleration by an order of magnitude? If so, it could only enlarge it and make it easier to measure.

8. Jun 24, 2006

### pervect

Staff Emeritus
The free-space neutron-neutron acceleration value has been calculated - it's unfortunately too tiny to measure.

See 3.1 of the Cooperstock paper, which puts it at 3*10^-47 m/s^2 for two negligible mass test particles 1 au apart.

Unfortunately, I think one needs a fair knowledge of relativity to perform the calculations I suggested - one at least needs to know how to calculate the conserved quantities for geodesic motion in the FRW metric.

Because the metric is space translation invariant, there will be conserved momenta. The formula itself is not horribly complex

The FRW metric
ds^2 = -dt^2 + a(t)^2 (dx^2 + dy^2 + dz^2)

leads to the following conserved quantity:

a^2 dx/dtau =

a^2 vx / sqrt(1- a^2(vx^2+vy^2+vz^2))

where
vx = dx/dt, vy=dy/dt, vz=dz/dt

This conserved quantity being the conserved x component of the momentum. However, it takes considerable knowledge to justify it. (The form of the y and z components should be reasonably obvious from the above).