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?

  • 1
    $\begingroup$ There are two answers to this: 1) There is no expansion where we are. Expansion can only happen in systems that are not gravitationally bound. 2) Can a microphone measure continental drift? No it can not because continental drift is too slow. For the same reason LIGO could also not (at least not directly) measure the expansion of the universe, even if it weren't in the wrong place... it's just not sensitive enough for that purpose. $\endgroup$
    – CuriousOne
    Feb 12, 2016 at 19:56
  • $\begingroup$ Hmm, surely the idea is the detector arms were (as I say) "distorted" (bent, stretch squeezed, something) in some way so that the two interferometer paths were different??? $\endgroup$
    – Fattie
    Feb 12, 2016 at 19:56
  • $\begingroup$ from the webster.com on "wibbles" :The word you've entered isn't in the dictionary . Do you mean "wiggles" wiggle is a verb, not a noun, " to move to and fro with quick jerky or shaking motions" $\endgroup$
    – anna v
    Feb 13, 2016 at 15:49
  • $\begingroup$ Anna, there's only one dictionary :) wibble |ˈwɪb(ə)l| verb [ no obj. ] Brit. informal 1 wobble; quiver. It's commonplace in English that the noun is the same as the verb (chant, yawn, run). $\endgroup$
    – Fattie
    Feb 13, 2016 at 15:59
  • 2
    $\begingroup$ Related $\endgroup$
    – rob
    Feb 13, 2016 at 16:43

2 Answers 2


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.

  • $\begingroup$ @levitopher, in the first instance thanks for reminding that LIGO (and XLIGO) is an interferometer - so of course it can't detect anything uniform. I added a note to the question, so, I meant in the sense of "pretend LIGO is an instrument, as sensitive as it is, but it just plain measures linearly". Thanks. $\endgroup$
    – Fattie
    Feb 13, 2016 at 15:45
  • $\begingroup$ @levitopher, thanks a million for your calculations here (thanks a $10^{21}$ ?) To me it seems quite incredible that you have calculated that, as you say "if the arms were receding away from each other at a speed consistent with the expansion of the universe, XLIGO could measure the shift." Just so I've understood correctly, you're saying the needed "factor" is about $10^{18}$ and LIGO is good to about $10^{21}$ ... so LIGO is good enough by about 1000 ?? That seems remarkable: we'd only need something 1/1000th as sensitive as LIGO, to detect the daily expansion of the universe??? $\endgroup$
    – Fattie
    Feb 13, 2016 at 15:50
  • $\begingroup$ In short, I have clarified my two questions about lengths, A and B. Thanks a trillionth to anyone who can answer them! $\endgroup$
    – Fattie
    Feb 13, 2016 at 16:09
  • $\begingroup$ Edited my answer, hopefully this satisfies you. I believe that yes in the fake world where we XLIGO exists we could measure the shift due to the expansion of the universe. There is a reason these guys are sure to get a Nobel Prize - it's an incredible apparatus, and they've made a remarkable measurement. $\endgroup$
    – levitopher
    Feb 13, 2016 at 20:42
  • $\begingroup$ WOW your calculations for A and B are incredible. Where's the button to send money?! Totally amazing. $\endgroup$
    – Fattie
    Feb 13, 2016 at 22:37

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.

  • $\begingroup$ Hi Everett, indeed, sorry again if the question title was misleading; as it says a number of times in the question "of course an interferometer, per se, can't at all measure the expansion of the universe since that is uniform in both directions" - I'm asking if the distance sensitivity of the LIGO is the distance sensitivity required. It would seem that incredibly it is .. by a factor of 1000x. BTW it would be awesome if someone could tell me the two values (A) and (B) :O $\endgroup$
    – Fattie
    Feb 13, 2016 at 19:01
  • $\begingroup$ @JoeBlow Still no. The characteristic frequency of the universe expansion is of the order 0.1nHz. LIGO does not have the required distance sensitivity at 0.1nHz to detect the expansion. $\endgroup$ Feb 14, 2016 at 6:40
  • $\begingroup$ @JoeBlow I have updated the answer. Frequency is important! LIGO does not have the $10^{21}$ sensitivity at 0.1nHz frequency (which is the frequency of the expanding universe). $\endgroup$ Feb 14, 2016 at 6:54
  • $\begingroup$ Hello Everett! Thank you for this ... I am trying to absorb and understand what you say, pls wait! thanks ... $\endgroup$
    – Fattie
    Feb 14, 2016 at 14:39
  • $\begingroup$ Saying it's a spin-0 mode is returning to the original problem of isotropic expansion. Since he was just looking for an estimate of the sizes involved, in my answer I basically assumed we were looking at anisotropic expansion with the same size as $H_0$. Here is a paper that estimates anisotropy at 3%, although presumably that's a dipole effect, not an octopole effect. arxiv.org/pdf/1212.3691v3.pdf $\endgroup$
    – levitopher
    Feb 14, 2016 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.