# Understanding gravitational waves (GR)

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• JD_PM
In summary, the conversation discusses the concept of gravitational waves and the behavior of two black holes orbiting each other until they merge. The frequency and amplitude of the waves increase as the black holes approach, and there is a drop in frequency when they merge and create a lower velocity black hole. The conversation also raises questions about the frequency response of detectors and the need to consider modeled data rather than raw data. The concept of proper acceleration and its role in GR is also mentioned.
JD_PM
TL;DR 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.

JD_PM said:
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.

JD_PM said:
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".

JD_PM said:
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
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
JD_PM said:
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:

PeterDonis said:
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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JD_PM said:
Why?

Because that's the definition of free fall.

JD_PM said:
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
PeterDonis said:
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.

JD_PM said:
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.

PeterDonis said:
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".

Let me make the analogy with the magnetic quadrupole moment.

To get the magnetic quadrupole, the system needs to meet the following conditions (if I am not mistaken):

$$\sum_n q_n = 0$$

$$\sum_m \vec p_m = 0$$

Where ##q_n## and ##\vec p_m## are the charges and dipole moments present in the system.

An example is a square with the following configuration:

I know that a changing magnetic quadrupole moment produces electromagnetic radiation (the third time derivative of the quadrupole moment is nonzero).

Is it the whole above analysis applicable for gravitational masses as well? I'd say yes, because to start with we also have:

$$\sum_n q_n = 0$$

$$\sum_m \vec p_m = 0$$

As there are no charges.

PeterDonis said:
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.
Note, there is a relation between proper velocity (a term I hate) and 4 velocity. The former is the spatial part of 4 velocity in an (local for GR) inertial frame.

PAllen said:
Note, there is a relation between proper velocity (a term I hate) and 4 velocity. The former is the spatial part of 4 velocity in an (local for GR) inertial frame.

I agree. The spatial part of the 4-velocity vector is:

$$\eta^µ = \frac{dx^µ}{d\tau}$$

Where ##x^µ## is the displacement 4-vector.

JD_PM said:
To get the magnetic quadrupole, the system needs to meet the following conditions (if I am not mistaken):
This is not correct. Even a system with net charge can have a quadrupole (or indeed, dipole) moment. However, in those scenarios, the net charge (i.e., monopole moment) typically dominates the resulting field at large distances (unless something like charge conservation makes the field from the monopole vanish - like in the case of EM radiation).

In the case of gravitational waves, the "charge" is the mass and it is certainly non-zero. However, energy-momentum conservation leads to the monopole and dipole moments not contributing to the radiation field.

vanhees71 and JD_PM
JD_PM said:
The spatial part of the 4-velocity vector is:
$$\eta^\mu = \frac{dx^\mu}{d\tau}$$
Where ##x^\mu## is the displacement 4-vector.

This isn't the "spatial part" of the 4-velocity vector, it is the 4-velocity vector. ##x^\mu## is all four coordinates, including the time coordinate ##t##.

PeterDonis said:
This isn't the "spatial part" of the 4-velocity vector, it is the 4-velocity vector. ##x^\mu## is all four coordinates, including the time coordinate ##t##.

I see what you mean. Then technically ##\eta## is the spatial part and ##\eta^0## the time part.

Source: Introduction to Electrodynamics by D. Griffiths.

By the way, note that in post #7 I meant the derivative of the proper velocity 4-vector.

PeterDonis said:
Proper acceleration is the derivative of 4-velocity with respect to proper time. "Proper velocity" refers to something different.

So I do not see the difference between proper velocity 4-vector and 4-velocity vector.

I guess you assumed proper velocity not to be a four vector.

I would not use Griffith's as a main reference for SR. In normal SR jargon, nobody uses "proper velocity 4-vector" - just "4-velocity". "Proper velocity" is something different, see the Wikipedia link Peter provided - it is the spatial part of the 4-velocity.

JD_PM
JD_PM said:
I do not see the difference between proper velocity 4-vector and 4-velocity vector.

As you will see if you look at the link I gave for proper velocity, that term usually means the spatial part of the 4-velocity vector. Griffiths uses "proper velocity" to mean the entire 4-velocity vector, but that usage is not very common.

JD_PM
PeterDonis said:
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".

I am aimed at understanding why the changing of the quadrupole moment means that the orbiting black holes emit gravitational radiation.

But first let's delve into the calculation of the quadrupole moment ##Q_{ij}## (it's shown in the paper, page 10).

I don't see why the paper defines the quadrupole moment as:

$$Q_{ij} = \int d^3 x \rho (x) (x_i x_j - \frac{r^2\delta_{ij}}{3})$$

Particularly the usage of the factor:

$$x_i x_j - \frac{r^2\delta_{ij}}{3}$$

I guess it is setting the system's center of mass in the origin of the Cartesian coordinates but that is not enough for me.

JD_PM said:
I am aimed at understanding why the changing of the quadrupole moment means that the orbiting black holes emit gravitational radiation.

But first let's delve into the calculation of the quadrupole moment ##Q_{ij}## (it's shown in the paper, page 10).

I don't see why the paper defines the quadrupole moment as:

$$Q_{ij} = \int d^3 x \rho (x) (x_i x_j - \frac{r^2\delta_{ij}}{3})$$

Particularly the usage of the factor:

$$x_i x_j - \frac{r^2\delta_{ij}}{3}$$

I guess it is setting the system's center of mass in the origin of the Cartesian coordinates but that is not enough for me.
Definitions are not normally subject to debate. A more relevant question is the conclusions about GW power output based on derivatives of that definition that follow from GR. That is a substantive, complex, question. Einstein was the first to derive the relationship using typical (for him) clever, heuristic, arguments. It has been validated more rigorously by many physicists since, and verified by observations such as the Hulse-Taylor binary period change.

JD_PM said:
I don't see why the paper defines the quadrupole moment as

Because that's how a quadrupole moment is defined. The definitions of all multipole moments are well established (basically you transform spherical harmonics into Cartesian coordinates).

JD_PM said:
I guess it is setting the system's center of mass in the origin of the Cartesian coordinates

Yes, that's the standard convention for all multipole moments.

JD_PM said:
I am aimed at understanding why the changing of the quadrupole moment means that the orbiting black holes emit gravitational radiation.

But first let's delve into the calculation of the quadrupole moment ##Q_{ij}## (it's shown in the paper, page 10).

I don't see why the paper defines the quadrupole moment as:

$$Q_{ij} = \int d^3 x \rho (x) (x_i x_j - \frac{r^2\delta_{ij}}{3})$$

Particularly the usage of the factor:

$$x_i x_j - \frac{r^2\delta_{ij}}{3}$$

I guess it is setting the system's center of mass in the origin of the Cartesian coordinates but that is not enough for me.

I'm not sure what paper you were using, not having followed the thread. I believe the best derivatio I ever found was the derivation in Landau & Lifschitz. And that was still pretty hard to follow.

Defintions are not required to be proved, but it is helpful when they are motivated. I believe that the basic motivation for the definition you wrote down is that the GW equations are much more tractable in the transverse traceless gauge. To do this, it's convenient to take the traceless part of the quadrupole moment, to be compatible with the transverse traceless gauge. The ##x_i x_j ## term are the basic quadrupole moment, when you subtract ##\frac{r^2 \delta_{ij}}{3}## you get the traceless part.

Once upon a time I wrote down the wave equations in a general gauge, rather than the transverse traceless gauge, just to satisfy myself that the solution I had for the metric (rather than the emitted power) for a simple binary system was correct far away from the binary system of masses.

The solution wasn't an exact solution of the Einstein field equations, just a solution for the low order terms of 1/r that neglected higher-order terms. I don't remember whether the lowest order terms were 1/r or 1/r^2 anymore.

Unfortunately, I can't find the relevant post. I do remember it was very very messy, and the only reason I was able to get anywhere with it was via the heavy use of computer algebra. You can find the general solution in most texts (I think I used MTW"s gravitation). Once you see the general equation, you'll quickly appreciate why textbook authors quickly move onto the traceless version, using ##\bar{h}_{ij}## rather than ##h_{ij}##.

Sorry this is a bit rambling and not too detailed, but I hope it might be of some help.

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Once upon a time I wrote down the wave equations in a general gauge,

I see that I didn't explicitly write down the wave equations, but the point is that the metric I wrote down was an asymptotic solution to order of 1/r of the EFE for a pair of orbiting circular bodies.

Given the ##h_{uv}##, it should be possible to compute the (asumptotic) GW powers at any given point, but that wasn't the purpose of what I did.

It's also not peer-reviewed. Obviously I didn't deliberately make any errors, but it's not the same as a computation that's been published and checked.

JD_PM
Orodruin said:
In the case of gravitational waves, the "charge" is the mass and it is certainly non-zero.

Mmm I see, thanks.

Orodruin said:
However, energy-momentum conservation leads to the monopole and dipole moments not contributing to the radiation field.

Could someone please recommend a book/paper which explains why this is the case (with the Math of course)?

JD_PM said:
Could someone please recommend a book/paper which explains why this is the case (with the Math of course)?
You do not need a book. It is just that the monopole moment (total mass), and dipole moment (essentially displacement of the center of mass) do not have the appropriate time derivatives to give rise to a radiation field.

Orodruin said:
It is just that the monopole moment (total mass), and dipole moment (essentially displacement of the center of mass) do not have the appropriate time derivatives to give rise to a radiation field.

But why?

In order to have monopole radiation, the time derivative of the monopole moment must be non-zero. Since the monopole moment is conserved, this is not the case.

The same argument goes for the dipole moment but with the second time derivative. By momentum conservation, this second derivative is zero.

The quadrupole moment on the other hand can undergo periodic oscillation and therefore give rise to a radiation field (proportional to the third time derivative of the quadrupole moment squared).

vanhees71 and JD_PM
Orodruin said:
The monopole moment (total mass), and dipole moment (essentially displacement of the center of mass).

OK so the monopole moment represents the total mass and the dipole moment represents the displacement of the center of mass.

What does the quadrupole moment represent though?

JD_PM said:
Mmm I see, thanks.
Could someone please recommend a book/paper which explains why this is the case (with the Math of course)?

MTW (amazon link) talks about it in section $15.2, but I'm not sure that's necessarily the best place to start. Certainly, I'd expect that if you open the book to this section, without reading the prerequistes, you'd most likely wind up confused. And it's midway through a very long, very dense book, so to get the context for the brief section that talks about it, a small number of pages, would be a large amount of reading. The title of$15.2 might provide a clue as to how intelligible it will be without reading the previous 15 chapters.

"Bianchi identity d##\mathcal{R}##=0 as a manifestation of "Boundary of boundry = 0"

Basically, the main issue I see is that there's a a lot of background information that you'll need to understand the explanations. I rather suspect you don't have that background from your questions, but of course it's hard to be really sure of what background you might have.

The reason I'm not sure that MTW is the best book is because it's notation is a bit antiquated. But it's very good about explaining things, and I usually rather like the informal presentation. I'm not really a fan of the comma and semicolon component notation, though. MTW is not too expensive - I've seen some really low-priced ebook versions, but the formatting of the diagrams and long equations strike me as probably being likely to be illegible without the printed verison.

There are more modern books out there, I'm not sure which to recommend. Schutz, for instance is very popular, but most people I talk to on PF who try to learn GR from his book wind up very confused. I don't own it, when I read pieces and snippets of it, though, I usually wind up agreeing with the posters who are asking about said snippets that it's confusing :(.

There are always Carroll's online lecture notes, <<web link>>, which are free, though I don't know if or where Carroll would talk about the particular issue you are concerned at at the moment.

I think I already mention Landau and Lifschitz, "Classical Theory of fields". They had a pretty good derivation of the quadrupole formula as I recall, but I'm not too familiar with them overall. They cover a lot more than just gravity, which is both good and bad. The quadrupole formula is very similar to the "retarded potentials" method for electromagnetism, <<wiki link>>.

I'm pretty sure there is a thread about GR books in general, and I see the task ahead of you as one of picking up a whole lot of background to get to the small piece that you're interested in.

JD_PM
JD_PM said:
I am aimed at understanding why the changing of the quadrupole moment means that the orbiting black holes emit gravitational radiation.

But first let's delve into the calculation of the quadrupole moment ##Q_{ij}## (it's shown in the paper, page 10).

I don't see why the paper defines the quadrupole moment as:

$$Q_{ij} = \int d^3 x \rho (x) (x_i x_j - \frac{r^2\delta_{ij}}{3})$$

Particularly the usage of the factor:

$$x_i x_j - \frac{r^2\delta_{ij}}{3}$$

I guess it is setting the system's center of mass in the origin of the Cartesian coordinates but that is not enough for me.
In Cartesian coordinates the multipole moments are defined as the irreducible moments of the mass distribution. Irreducible moments are defined by the irreducible polynomials of the ##x_i##. Irreducible means that the moments are symmetric and traceless.

For the 2nd irreducible moment you start from ##x_i x_j##, which is already symmetric of course, but you have to get rid of the trace part. The only invariant tensor you can form from the ##x_i## alone is ##x_k x_k \delta_{ij}##. Thus the right moment is given by the factor you quote, ##x_i x_j-x_k x_k \delta_{ij}/3##, and indeed this gives you the quadrupole moment.

This construction gets soon pretty tedious. You can try it for the octopole, starting from ##x_i x_j x_k##.

The general construction is much simpler in terms of spherical coordinates leading to the spherical harmonics ##\text{Y}_{lm}(\vartheta,\varphi)##, which can all be determined fully algebraically using the orbital angular-momentum operator from wave mechanics, ##\hat{\vec{L}}=-\mathrm{i} \vec{x} \times \vec{\nabla}##.

dextercioby, weirdoguy and JD_PM
PeterDonis said:
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
While I have no doubt you are completely correct, that still makes no sense to me. Two object that are attracted to each other are moving closer and if they are closer they are more attracted. In Newtonian terms, they are accelerating towards each other and the acceleration is not constant. As we observe them, are not those two black holes moving faster as time progresses until they collide? Thanks.

bob012345 said:
While I have no doubt you are completely correct, that still makes no sense to me. Two object that are attracted to each other are moving closer and if they are closer they are more attracted. In Newtonian terms, they are accelerating towards each other and the acceleration is not constant. As we observe them, are not those two black holes moving faster as time progresses until they collide? Thanks.
Yes, the attraction changes and in Newtonian physics that means they are applying a gravitational force to each other? So what? This thread is about GR, not Newtonian physics. Things just naturally travel along geodesics (ie are in "free fall")

JD_PM
bob012345 said:
In Newtonian terms,
Unfortunately, GR uses rather different terms.
bob012345 said:
As we observe them, are not those two black holes moving faster as time progresses until they collide?
You can certainly find a way to state that. The problem is that it's not coordinate independent, so it always leads to a thousand "ifs" and "excepts" and "depends on your point of views". For example, the distance between two things whose "surfaces" can't rigorously be described as places in space and don't really have centers in any meaningful sense is a surprisingly difficult thing to pin down. So we use things like quadropole moments which are actually invariants.

Ibix said:
Unfortunately, GR uses rather different terms.
Or fortunately, it depends on your point of view.

JD_PM, phinds and Ibix
Orodruin said:
Or fortunately, it depends on your point of view.
S'all relative, man.

Ibix said:
S'all relative, man.
You mean it is not person-invariant?

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