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It would seem that LIGO measures wibbles in the metric (not manifold) of spacetime:

How is it that distortions in space can be measured as distances?

It would seem that the expansion of the universe is an expansion of the metric of spacetime:

If space is "expanding" in itself - why then is there redshift?

Imagine for a moment that LIGO is not an interferometer. (So, it just plain times speed changes, rather than using phase shift of orthogonal directions.)

If the ends of one of the arms was indeed receding away from each other, at a speed consistent with the expansion of the universe, is the sensitivity of real-world LIGO of the needed sensitivity of a machine which could measure that shift?

On other words: ideally I'd like to know these two meters:

(A) The left arm of LIGO is about 4km. It was stretched/shrunk (a few times) for roughly .01 seconds by the gravitational wave. How many meters was it stretched/shrunk in .01 seconds?

(B) Assuming the same abstract LIGO arm was affectable and affected by the expansion of the Universe. How many meters is it stretch/shrunk every .01 seconds?

Note - of course, an interferometer is an ingenious device to measure extremely small changes in speed - assuming the changes are orthogonal. Of course, an interferometer, per se, can't at all measure the expansion of the universe since that is uniform in both directions. The sense of my question is, can something that measures distance changes > as accurately as < the LIGO does, measure the expansion of the universe? How big or small is the ongoing expansion of the universe compared to the wibble from the black hole system in question?

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Fattie
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Well, CuriousOne gives the most direct answer, that the universe is not expanding on scales which the gravitational attraction between objects dominates, like on the Earth (indeed, the entire Milky Way). But, let's pretend that we take LIGO, stick it out in space (even away from our local group, to be sure it's in an isolated region), and ask it to measure the expansion of the universe. I'll call this XLIGO to make sure we don't confuse it with reality.

Answer A: I guess you're saying 0.01 s because the frequency of that particular observation was around 100 Hz, but in any case, the maximum strain detected by LIGO for this specific event was around $1\times 10^{-21}$ (see here, look at Figure 2). So, over 4 km that's simply $1\times 10^{-18}~m$.

Answer B: I'm going to do this using the Hubble constant, with tells us how fast two objects are moving away from each other depending on their distance. The approximate value of the Hubble Constant is 75 km/s/Mpc. Read that "For every megaparsec difference, the objects receed at a speed of 75 km/s."

So, over 4 km, $$\rm (75~km/s/Mpc)(1~Mpc/3 \times10^{19}~km)(4~km)\approx 100\times10^{-19}~km/s$$

Or an expansion speed along a single arm of XLIGO of $1\times 10^{-14}~m/s$. Over a time period of 0.01 s that means an absolute shift of $$ (1\times 10^{-14}~m/s)(0.01~s)=1\times 10^{-16}~m$$ So the sensitivity of XLIGO, modeled by the only positive observation, is around 100 times larger then the rough expansion of the universe.

But there is a bigger problem here, and that is that the expansion of the universe is isotropic, identical in every direction. So in order to see the effect, we would have to try to measure the aniotropies (actually, we would need to see the higher-order pole, specifically the quadrapole term). I found a paper (here) discussing that $\Delta H/H\approx 3\%$ is not excluded, so that actually brings us down to the same level of approximate scales.

levitopher
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No. LIGO can not measure the expansion of the universe.

LIGO only detects a specific class of distortions of the space-time, which at least are of spin-2 and with a frequency of the order 100Hz. Gravitational waves are spin-2 (quadrupolar) modes, and the one that was observed this time has the dominant frequency from 35Hz to 150Hz, which is in the sensitive range of LIGO. But the expansion of the universe is a spin-0 (isotropic) mode and of the frequency zero, which can not be detected by LIGO. LIGO uses an interferometer to measure the change of the difference between the lengths of its arms, so it will produce no signal if both arms are stretched equally.

The sensitivity of LIGO is frequency dependent. The best sensitivity $10^{-21}$ is achieved only around 100Hz, and for extremely slow processes like the expansion of the universe (corresponding to 0.1nHz), LIGO basically has no sensitivity. So even if one just wish to use the distance sensitivity, LIGO still can not be used to detect the expansion of the universe, due to the loss of sensitivity outside its frequency range.

Everett You
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If the ends of one of the arms was indeed receding away from each other, at a speed consistent with the expansion of the universe, is the sensitivity of real-world LIGO of the needed sensitivity of a machine which could measure that shift?

As levitopher said, if you interpret LIGO-equivalent sensitivity to mean $10^{-21}/(10^{-2}\text{ s}) = 10^{-19}\text{ s}^{-1}$, then that is sensitive enough to detect a Hubble value of $75\text{ km/s/Mpc}\approx 2.4\times 10^{-18}\text{ s}^{-1}$, though by a factor closer to $25$ than $100$ since they accidentally dropped a factor of $4$.

It doesn't matter, though, because the relative velocity of the ends of the space-LIGO device won't be $H_0 r$. It will be whatever relative velocity your construction/launch process gave to the ends. Lumps of matter don't prefer to recede from each other at $H_0 r$, and won't do it unless the initial conditions were set up that way.

If you were able to prepare the two arm ends with a relative velocity of precisely zero, and arrange for nongravitational forces to be precisely zero, the distance between the ends would drift due to gravity. The gravitational effect can be split into three parts:

  • Self-gravitation of the system, which causes a relative acceleration of $Gm/r^2$ inward.

  • Tidal forces from distant bodies, which result in a relative acceleration of $GMr/R^3$ times a small unitless factor.

  • An outward acceleration of $\frac{Λ}{3} r$ from the cosmological constant. $\frac{Λ}{3}\approx 10^{-32}\text{ m/s}^2\text{/km}$, far too small to matter at ordinary distances.

Either the self-gravitation or tidal acceleration may dominate, but in either case you won't end up with a speed of $H_0 r$.

If you consider the effect of these forces on matter that is homogeneously distributed, whose density is the average density of the universe, and which is expanding at the rate $H_0$, then you'll find that it evolves according to the Friedmann equations and expands at the rate $H(t)$ at later times. But this quasi-LIGO experiment satisfies none of those conditions, so its time evolution is different, even though the forces are the same. Even if you set the initial velocity to $H_0 r$, the velocity at later times still won't be $H(t)\,r(t)$ because the other two conditions weren't satisfied.

If there is a spinning top, and you stop its spinning, there is no experiment you can subsequently do on the top to determine what its angular speed used to be. The same principle applies to Hubble expansion. The matter that makes up LIGO hasn't been Hubble-expanding since it started to coalesce into first-generation stars billions of years ago. It doesn't remember that expansion rate, and never knew the current expansion rate.

benrg
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