LIGO detection question

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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!
 

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  • #2
Ibix
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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".
 
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pervect
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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!
If you look at http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.241103 you'll see the following hint:

The chirp mass [26,45], which controls the binary’s evolution during the early inspiral, is determined very precisely. The individual masses, which rely on information from the late inspiral and merger, are measured far less precisely.
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.
 
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Thank you for your great and detailed response, Pervect- I really appreciate it.

jb
 

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