Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B LIGO detection question

  1. Jun 15, 2016 #1
    Re: the second LIGO detection, from Symmetry: “Because of their lighter masses compared to the first detection, they spent more time—about one second—in the sensitive band of the detectors.”

    As an absolute (albeit deeply fascinated) novice here, I'm unclear as to why lighter masses would allow for the waves' greater time in the sensitive band of the detectors.

    Thanks in advance!
  2. jcsd
  3. Jun 16, 2016 #2


    User Avatar
    Science Advisor

    These were lower mass objects, so everything happens more slowly. Orbital speeds are lower and orbital decay happens slower. Since the frequency of the gravitational waves depends on the time it takes the black holes to go round each other, the result is that the frequency change is slower. So it takes longer for the signal to rise from "too weak to detect" to "too high frequency to detect".
  4. Jun 16, 2016 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    If you look at http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.241103 you'll see the following hint:

    Now, if you look at the first Ligo paper, http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102, you can see how the chirp mass is defined and how it controls the evolution of the inspiral. The formula is:

    $$\mathcal{M}^{\frac{5}{3}}= k \, f^{\,-\frac{11}{3}} \, \dot{f}$$

    Here f is the observed instantaneous frequency of the gravitational wave (i.e the chirp), ##\dot{f}## is it's time derivative, and k is some constant given in the paper. By letting ##\dot{f} = df/dt## you can find an integral for the time t to pass through the "sensitivity band"

    $$dt = \int_{f_{low}}^{f_{hi}} \, \frac{k \, f^{\,-\frac{11}{3}} \, df}{\mathcal{M}^{\frac{5}{3}}}$$

    So the time to pass through the "sensitivity band" from ##f_{low}## to ##f_{hi}## is given by the above integral, which is inversely proportional to ##\mathcal{M}##. Thus a lower chirp mass means a greater time to pass through the band.

    There is a reference for where the formula for the "chirp mass" was derived in the paper, but I don't actually know the details of the derivation. But given the existence of the formula, you can see how you can compute ##\dot{f}## given the value of ##f## and ##\mathcal{M}##, and thus controls the evolution of the inspiral (at least during the early phase). You can also see the importance of this parameter in determining how the masses were computed from the observed data.
    Last edited: Jun 16, 2016
  5. Jun 16, 2016 #4
    Thank you for your great and detailed response, Pervect- I really appreciate it.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted