A Understanding gravitational waves (GR)

JD_PM

Summary
I want to understand how does GR predict the existence of gravitational waves.
I am reading the following paper on the basic physics of a binary black hole merger: https://arxiv.org/ftp/arxiv/papers/1608/1608.01940.pdf

Imagine two black holes orbiting each other until a point they merge. As you can see in Figure 1, the wave period is decreasing and thus the frequency of oscillation increases. Then, around time 0.43 s there is a drop of the frequency. I interpret this drop as the acceleration of the system being zero (the two orbiting black holes merge and the resulting black hole has a lower velocity).

This is what I have in mind: two black holes orbit each other and they are being accelerated. Thus they will emit gravitational radiation; spacetime is getting curved more and more as time passes and that means that waves of greater amplitude and frequency are being generated (these are the gravitational waves if I am not mistaken). It makes sense for me that the frequency increases as they approach; the waves become more energetic (shorter wavelength greater frequency. The same happens with light; for instance, gamma rays are more energetic than X-Rays and thus higher frequency).

My question is:

Why aren't Livingston and Hanford detectors showing a more progressive increase in frequency? Shouldn't we expect a final clearly higher peak (when the two black holes are just about to merge and thus where the spacetime gets more bent)?

I am starting with GR and gravitational waves so please point out any mistakes you see in my interpretation above.

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PeterDonis

Mentor
two black holes orbit each other and they are being accelerated
No, they aren't. They are in free-fall orbits about their common center of mass. "Acceleration" in GR means proper acceleration; objects in free fall have zero proper acceleration.

Thus they will emit gravitational radiation
They will emit gravitational radiation because the quadrupole moment of the system is changing with time. More precisely, the third time derivative of the quadrupole moment is nonzero. Note that this has nothing to do with "acceleration".

spacetime is getting curved more and more as time passes and that means that waves of greater amplitude and frequency are being generated (these are the gravitational waves if I am not mistaken)
The greater amplitude and frequency of the waves is because the black holes are spiraling inward towards each other, and they are spiraling inward towards each other because they are emitting gravitational waves and therefore losing energy. "Losing energy" for objects orbiting about their common center of mass means the orbits become tighter--shorter orbital radius and faster orbital speed.

I'm not sure spacetime curvature is actually increasing in this scenario, but at any rate spacetime curvature per se is not the key parameter to look at to explain the increasing amplitude and frequency of the gravitational waves.

I think you need to re-think your questions in the light of the above.

• JD_PM

Ibix

An observation about instrument response - all detectors have limits to what they can detect. Looking at the end of the high-amplitude section of the signal, the graph is showing one positive point followed by one negative one, joined by straight lines. That looks to me like an instrument reaching the limits of its frequency response, so I'd be doubtful that the trace to the right of that reflects the actual form of the gravitational wave. In particular, a frequency drop may well be due to aliasing, where the detector simply cannot respond to the changing signal fast enough so starts to slow down. Also, obviously the power output will drop off at some point, but this graph only shows the received power. If the signal is outside the detector's response then this is not the same as the emitted power.

If you are trying to understand gravitational waves, I'd suggest you look for papers that model them rather than looking at data. Our sensors are so primitive at the moment that the received signal is quite likely to be very different from the emitted signal. After that, you can look into detector design and understand how the response follows from the signal.

• JD_PM

pervect

Staff Emeritus
Summary: I want to understand how does GR predict the existence of gravitational waves.

...

As you can see in Figure 1, the wave period is decreasing and thus the frequency of oscillation increases. Then, around time 0.43 s there is a drop of the frequency. I interpret this drop as the acceleration of the system being zero (the two orbiting black holes merge and the resulting black hole has a lower velocity).

This is what I have in mind: two black holes orbit each other and they are being accelerated. Thus they will emit gravitational radiation; spacetime is getting curved more and more as time passes and that means that waves of greater amplitude and frequency are being generated (these are the gravitational waves if I am not mistaken). It makes sense for me that the frequency increases as they approach; the waves become more energetic (shorter wavelength greater frequency. The same happens with light; for instance, gamma rays are more energetic than X-Rays and thus higher frequency).

My question is:

Why aren't Livingston and Hanford detectors showing a more progressive increase in frequency? Shouldn't we expect a final clearly higher peak (when the two black holes are just about to merge and thus where the spacetime gets more bent)?

I am starting with GR and gravitational waves so please point out any mistakes you see in my interpretation above.
As pointed out in the paper you cite, and also by some other posters, rather than think about the "acceleration" causing gravitational waves, it's better to think about the change in the quadrupole moment as being the source. You'll note some apparent (but not actual) discrepancies here - total power emission is proportional to the third time derivatve of the quadrupole moment, the peak strain in Ligo (as mentioned in the paper you cite) is proportional to the second time derivative of the quadrupole moment. There's no real discrepancy here, just people discussing different things.

There's an interesting approximation in one of my textbooks (MTW's Gravitation) that makes some approximations, and relates the appropriate derivative of the quadrupole moment to power flow. The total emitted power in gravitaitonal waves is roughly proportional to the square of this power flow, divided by a large constant, the "unit power" in geometric units. While interesting and relatively easy to understand (less abstract than the quarrupole formula that you'll find everywhere), it appears to only be an approximation. But it is similar to your thinking, in that "acceleration" enters the picture. It enters the picture, though, in terms of the "power flow".

Another thing you are omitting in your very broad picture is the effects of gravitational redshift. As the two black holes get closer together, there is an increasing gravitational redshift. In terms of Schwarzschild coordinates, that would tend to slow down the inspiral, lengthining the period, and weaken the gravitational waves that were emitted as observed by a distant observer.

It's a very hard problem to get the exact waveform - numerical simultaions had to be used, and it was a significant effort to get them to the point where they would converge, and the answers were trusted. I don't know much of the details, but I can say it's not a good place to start learning General Relativity or gravitational waves.

In very general terms in the papers I've seen, the signal is broken down into three phases. Insiprial, merger, and ringdown.

The inspiral phase is close to the picture you describe. When the two holes actually start to merge, other effects (such as the gravitational redshift I mentioned) come into play. Finally, after the merger is complete, the event horizon is commonly described as "vibrating", and that also emits gravitational waves. I'm not sure how to rigorously justify this, though I've heard the "membrane paradigm" referred to in some of my reading.

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JD_PM

Let me go step by step:

objects in free fall have zero proper acceleration.
Why? This is what I think:

The first thing that came to my mind was to make the analogy with the twin paradox; one twin gets on the outbound sidewalk ($\bar{S}$ frame) and the other stays on Earth ($S$ frame). Wrt $\bar S$ frame and going from $\bar S$ to $S$ one gets the following LTs (I set $\bar x = x = 0$ just before the departure of one of the twins):

$$t/\gamma = \bar t$$
$$x = \bar x = 0$$

Then, as $\bar x = 0$ (i.e. an observer on the rocket is at rest on it), the proper acceleration is indeed zero (it is just the second derivative of $\bar x$ wrt proper time).

Do you agree with the analogy?

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PeterDonis

Mentor
Because that's the definition of free fall.

Do you agree with the analogy?
No. Proper acceleration is the reading on an accelerometer attached to the object. It is not the second derivative of position with respect to time; that's coordinate acceleration. Coordinate acceleration is irrelevant to the emission of gravitational waves; it's not even an invariant since it depends on your choice of coordinates (as its name implies).

JD_PM

No. Proper acceleration is the reading on an accelerometer attached to the object. It is not the second derivative of position with respect to time; that's coordinate acceleration.
Oh. I have been thinking that proper acceleration is:

$$a = \frac{dµ}{d \tau}$$

Where µ is the proper velocity and $\tau$ the proper time.

PeterDonis

Mentor
I have been thinking that proper acceleration is:
$$a = \frac{dµ}{d \tau}$$
Where µ is the proper velocity and $\tau$ the proper time.
Proper acceleration is the derivative of 4-velocity with respect to proper time. "Proper velocity" refers to something different.

However, that still isn't the same as what you did in post #5. You were calling $\bar{t}$ "proper time", but it isn't; it's coordinate time in the frame in which the outgoing observer is at rest. It happens to be numerically equal to proper time for that particular observer, but that doesn't mean you can treat them as identical. Proper time is an affine parameter along the worldline, not a coordinate.

"Understanding gravitational waves (GR)"

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