JMz said:
Well, I think this thread is well past the point of worrying about a mere few orders of magnitude. ;-)
Anyway, I envisioned some larger "local" region than an Earth-based lab.
Lol fair enough. I'm not quite sure what sort of experiment could be done though, even of the extreme sort. The effect would be so tiny compared to all sorts of other environmental "noise" that it is hard to image what could be done. But ok, let's actually calculate the acceleration and see what we are dealing with.
Jorrie said:
To follow on with some simple math:
When talking about both masses comoving (with the Hubble flow) at large scales, there is no gravity (potential gradient) involved. At the present time, with the cosmological constant, there will be a local coordinate acceleration between each mass and the local nano chain of**
$$dD^2/dt^2 = D H_0^2(\Omega_\Lambda-\Omega_m/2)/2$$
in opposite directions. ##D## is the comoving distance between the two masses, ##\Omega_\Lambda## the cosmo-constant and ##\Omega_m## the present matter density parameter. Some form of energy extraction at each mass will enter as another negative term inside the brackets. As long as the factor inside the brackets remains positive, there could in principle be continuous energy extraction, not so? If so, it can even increase as ##\Omega_m## decreases over long periods. Or is it all wishful thinking?
Sounds ok, but this form is not so great to use I think since we have to fiddle with H_0 and critical densities and such. I think we can just use the second Friedman equation more directly:
$$\frac{\ddot{a}}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$
Let's then just ignore all matter and focus on the dark-energy dominated case. This is maximum acceleration we are ever going to get with a constant ##\Lambda##. So then we can ignore the density and pressure terms, and we are left with just
$$\frac{\ddot{a}}{a} = \frac{\Lambda c^2}{3} $$
##a(t)## is the scale factor which defines the ratio between proper distance between two comoving objects now ##r(t)##, relative to separation at some reference time in the past ##r_0## so
$$r(t) = a(t) r_0$$
Now, I forget the exact meaning of "t" in these coordinates, so I am not totally confident that the acceleration I am about to calculate is the correct one. If anyone sees that it needs fixing then please feel free :). I guess it is the proper time of comoving observers, though our masses are no longer comoving (although the centre of the bound system is comoving, so maybe if we measure the proper acceleration relative to the comoving point of our system then this is correct)
Anyway, suppose that we have a strong cable that just holds the two masses at fixed proper distance. I think the acceleration of the scale factor should then just give us the acceleration we want (up to my possible misuse of the time coordinate), which is the acceleration of the comoving coordinates relative to our proper separation. So, then, taking two deriviatives with respect to ##t## we have
$$\ddot{r}(t) = \ddot{a}(t) r_0$$
We can then substitute these expressions for ##a## and ##\ddot{a}## back into the second Friendman equation:
$$\frac{\ddot{r}(t) / r_0}{r(t) / r_0} = \frac{\Lambda c^2}{3} $$
$$ \ddot{r}(t) = r_0 \frac{\Lambda c^2}{3} $$
Now, ##\Lambda \sim 10^{-52} \mathrm{m}^{-2}## according to Planck data. ##c^2/3## is about ##3 \times 10^{16} \mathrm{m}^2 \mathrm{s}^{-2}##. So for 1m separation we are looking at ##\ddot{r} \sim 10^{-36} \mathrm{m s}^{-2}##, or ##10^{-37} g##. Thus to get to ##1\mathrm{g}## of acceleration we need ##r_0 = 10^{37} \mathrm{m}##. The universe is only about ##10^{26} \mathrm{m}## in diameter so this is pretty big. This is why I suggested going to large masses instead; the Earth is about ##10^{24}## kg, so for ##1##m separation the Earth would pull on the rope with about ##10^{24} \times 10^{-36} = 10^{-12}## N. 1 Lightyear is ##10^{15} \mathrm{m}##, so setting our separation at 1 lightyear gets us about ##10^{15} \times 10^{-12} = 10^3## N of tension, or 1 kN, which is certainly measureable. However, the mutual gravitational attraction of the two Earths is bigger than that I think, even at 1 lightyear separation (##G M^2 / r^2 = 10^{-11} \times (10^{24})^2 / (10^{15})^2 = 10^7 \mathrm{N}##). But the ##1/r^2## kills this pretty quickly so we only need to go a few more tens or hundreds of lightyears apart. Oh and actually I guess we only need one heavy mass, not two. So let's suppose we have the Earth and a 1kg mass. Then the gravitational attraction is just ##G M m / r^2 = 10^{-11} \times 10^{24} / (10^{15})^2 = 10^{-17} \mathrm{N}##. So ok we are fine at 1 Ly separation as long as one of the masses is small. Although hmm, that's probably no good, we need both masses to be heavy I think. If one mass is small then the Earth will just follow its comoving path and easily drag the small mass along for the ride. So we only really get tension from accelerating the smaller mass, and lose our factor of ##10^{24}##. So ok, back to two large masses, with a few hundred lightyear separation.
On the other hand, if we want to do some smaller experiment with small masses, then we lose this enormous factor of ##10^{24}## which helped us out a lot. The accelerations are super miniscule so I am not sure there is any hope of ever measuring them on small masses.
