Number of gravitons detected

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Main Question or Discussion Point

Gravity waves are very weakly absorbed.
The gravity wave detectors have so far detected 3 events of gravity wave emission: 2 from black hole pairs and 1 from neutron star pair.

The absorption is weak but the detectors are very sensitive.

How much energy did the gravity wave detectors actually absorb from the gravity waves that propagated through Earth?
The frequency was low, peak something like 250 Hz, and therefore graviton energy low. How many quanta of gravity waves did the 3 detections consist of?

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Does the detector, as a matter of principle, make any alteration of the state of the gravity wave when itself undergoing an irreversible change of its state in a detection event?

jedishrfu
Mentor
What are you trying to say?

As the gravitational wave passes through the earth the arms of the detector shrink and expand a tiny amount. The detector doesn’t go into some irreversible state.

Ibix
You can have a machine that extracts energy from a gravitational wave - Feynman's sticky beads does it, at least in principle. And LIGO must generate a small amount of energy to be recorded.

I find it difficult to imagine that this doesn't affect the gravitational wave. You could stack up a sequence of sticky bead generators and produce arbitrarily high power outputs if it didn't - although I note that energy isn't always conserved in GR, so that's not as strong an argument as I'd like.

Note that they are called gravitational waves. Gravity waves are a kinfd of surface wave in water.

Paul Colby
Gold Member
How much energy did the gravity wave detectors actually absorb from the gravity waves that propagated through Earth?
Propagation through the Earth will effect the energy absorbed by the detector only at a level to which the GW are absorbed or scattered by the earth. I expect this is a very small effect when compared to the energy absorbed by the detector itself.

The interferometer mirrors in LIGO are suspended so as to allow inertial for movement along the interferometer arm. A high powered laser beam is then established in the between the mirrors. This beam will constantly impart a small pressure to the suspended mirrors. Because of this pressure, relative movement of the mirrors do work on the laser beam which shifts it's center frequency by a small amount relative to the light in the other interferometer leg. The work done by this relative movement is the energy absorbed by the detector.

PeterDonis
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I find it difficult to imagine that this doesn't affect the gravitational wave.
It does. But for LIGO, the effect is very, very small. @Paul Colby gives a good description of the energy transfer process.

You could stack up a sequence of sticky bead generators and produce arbitrarily high power outputs if it didn't - although I note that energy isn't always conserved in GR, so that's not as strong an argument as I'd like.
It's strong enough. Yes, there are subtleties involved with energy conservation in GR, but TAANSTAFL still applies.

Ibix
It does. But for LIGO, the effect is very, very small.
Extending the analogy with Feynman's beads, it occurred to me that a back of the envelope estimate of the energy absorption would be the peak kinetic energy of the mirror assemblies. The displacement is around 10-21 times 4km, and the frequency looks to top out around 50Hz, which gives a velocity on the order of 10-16m/s. I don't know the mass of the mirror assemblies, but for any plausible answer the energy absorbed is going to be within an order of magnitude or two of 10-32J.
It's strong enough. Yes, there are subtleties involved with energy conservation in GR, but TAANSTAFL still applies.
TANSTAAFL, unless there's a subtlety there I'm missing. I take it that a sphere of 100% efficient gravitational wave absorbers will generate no more than the mass loss of the emission event? So the upper bound on the energy available for absorbtion from the first event (mass loss of $3M_{sun}$, around 400Mpc away) is something like $(A/400^2)3M_{sun}$, where A is the effective area of the detector (order of $(4km)^2$ measured in megaparsecs squared?

PeterDonis
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2019 Award
a back of the envelope estimate of the energy absorption would be the peak kinetic energy of the mirror assemblies
Kinetic energy relative to what? Remember that the mirror assemblies are moving inertially: no work is being done on them (see further comments below). The only work being done is on the laser beam.

"No work is being done on the mirrors" - this is probably the hardest think to grok about gravitational waves: they are waves of spacetime curvature, and spacetime curvature, by itself, does no work on anything. Things moving solely under the influence of spacetime curvature (e.g., gravity) are moving inertially, in free fall. That's why many people in the early days of gravitational wave studies (late 1950s ish) thought that GWs couldn't carry energy, because they couldn't see how they could do work.

In the Feynman's beads scenario, for example, the work isn't done directly on the beads by the GW; it's done by the inter-atomic forces between the beads. Think of the whole setup as a bunch of masses connected by springs, which all start out in equilibrium. The GW moves the masses, which stretches or compresses the springs, which then respond by pulling or pushing on the masses. The springs pulling and pushing are what directly do work. And all of this changes the stress-energy tensor of the mass-spring setup, which in turn changes the spacetime curvature--that's what reacts back on the GW and changes it to carry less energy than it was before (or, put more precisely, reduces its ability to induce stretching and compressing of springs in the next mass-spring setup it comes across).

I take it that a sphere of 100% efficient gravitational wave absorbers will generate no more than the mass loss of the emission event?
Yes.

If a wave passes through detector changing neither wave's energy, momentum, angular momentum nor particle count, but does change wave's phase, can the detector change detector's state irreversibly, or is this illegal?

Ibix
Kinetic energy relative to what?
I was thinking of the rod in the sticky beads case - a non-inertial coordinate system. But I hadn't thought it all the way through, since the energy absorbed in that analysis is the difference between the kinetic energy of frictionless beads versus sticky beads. And in the case of LIGO there isn't (ideally) any interaction between the mirrors and the tunnel walls (which behave much like the rod in the sticky beads case), so that can't be the mechanism.

I'm a little confused where LIGO does absorb energy, then. If the work is done on the lasers, then to first order the energy increase in one beam is offset by the decrease in the other, I think. The arm lengths change by a factor of $\sqrt{1\pm h(t)}$, where $h(t)$ is the gravitational wave amplitude. That makes the energy in each arm change by one upon that factor - but $E/\sqrt{1\pm h(t)}\simeq E( 1\mp h(t)/2)$, so the increase in energy in one arm is the same as the decrease in the other. But surely it has to be absorbing something to make a detection? Are we really relying on the $h^2$ term?
In the Feynman's beads scenario, for example, the work isn't done directly on the beads by the GW; it's done by the inter-atomic forces between the beads. Think of the whole setup as a bunch of masses connected by springs, which all start out in equilibrium. The GW moves the masses, which stretches or compresses the springs, which then respond by pulling or pushing on the masses. The springs pulling and pushing are what directly do work.
That I can understand. The rod resists the length change, so its components are non-inertial, and the beads get dragged out of their inertial paths by friction leading to heating and usable energy.
And all of this changes the stress-energy tensor of the mass-spring setup, which in turn changes the spacetime curvature--that's what reacts back on the GW and changes it to carry less energy than it was before (or, put more precisely, reduces its ability to induce stretching and compressing of springs in the next mass-spring setup it comes across).
And this is similar to an analysis of EM antennae which regards the radiation after an antenna as the sum of the incident radiation and the (slightly out of phase) radiation from the induced dipole in the antenna. Gravitational waves are presumably more fun because they aren't linearly superposed, but I think the principle is the same.
grok
This is a rather jarringly different style of Martian to your avatar...

PeterDonis
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If the work is done on the lasers, then to first order the energy increase in one beam is offset by the decrease in the other, I think.
Off the top of my head, I think either increase or decrease of the arm length should add energy; roughly speaking, the energy added should be proportional to the amplitude of the interference fringes in the detector. But I have not tried to take a detailed look at the math.

Ibix
Off the top of my head, I think either increase or decrease of the arm length should add energy; roughly speaking, the energy added should be proportional to the amplitude of the interference fringes in the detector. But I have not tried to take a detailed look at the math.
Does the "stretch and squish" apply to the tangent space as well? If so, it occurs to me that the EM wave becomes not quite an EM wave, as its component in the plane of the gravitational wave gets stretched-and-squished. That would obviously dissipate energy.

PeterDonis
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Does the "stretch and squish" apply to the tangent space as well?
No. The tangent space at each point of spacetime is a flat Minkowski space.

PeterDonis
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2019 Award
If so, it occurs to me that the EM wave becomes not quite an EM wave, as its component in the plane of the gravitational wave gets stretched-and-squished. That would obviously dissipate energy.
This has nothing to do with the tangent space. It happens in spacetime. I'm not sure what you mean by "not quite an EM wave", though. The frequency and wavelength can change, but the fact that it's in vacuum with no EM sources doesn't change.

pervect
Staff Emeritus
Does the "stretch and squish" apply to the tangent space as well? If so, it occurs to me that the EM wave becomes not quite an EM wave, as its component in the plane of the gravitational wave gets stretched-and-squished. That would obviously dissipate energy.
I believe the "stretch and squish" is basically an artifact of the coordinate choice. By singling out a single point in space , which is a worldline in space-time, the point being an object "at rest in space", the worldline being the worldline of the object "at rest in space", one can eliminate the "stretch and squish" around a single point. The mathematical representation turns out to be less convenient, but it may be more physically intuitive in the new coordinates. Specifically, these are Fermi-Normal coordinates around that point in space (worldline in spacetime).

It's possible to talk about this without using coordinates. There is a local map called the "exponential map" <<wiki link>> that maps point from the tangent space to points on the manifold. Basically, distances in the tangent space from the central point in space to other nearby points are the same in the tangent space as they are in the manifold.

Of course one may need to define exactly what one means by "distances in the manifold". In this case by "distances in the manifold", I mean the length of a curve which is both a space-like geodesic in the associated space-time, and also is perpendicular to the reference worldline of the "stationary point" that is singled out.

Distances between points other than the central point are not necessarily quite the same in the tangent space and on the manifold, though. The best one can do is make the distances from the central point to nearby points match.

Creating such a map of the globe is an interesting thought experiment, one I've often wished I had a graphic of. But I don't have such a graphic. Technically speaking, this visual aid wouldn't be a Fermi-normal coordinate map, it'd just be Riemann normal coordinate map. The fermi-normal coordinates include time in the picture, the simplified graphic would not.. The idea with fermi-normal coordinates and the exponential map is similar though.

One imagines singling out a particular point on the globe (say the north pole), drawing great circles through this point (representing geodesics). Then one uses polar coordinates to define a map from the curved globe to a flat sheet of paper such that the polar coordinate "r" represents the distance along the great circle, and the polar coordinate "theta" represents the longitude of the great circle (which is constant).

This map would be the exponential map, the coordinates r and theta used in the construction would be Riemann normal coordinates. Fermi-normal coordinates are similar, but they also have to deal with time. This isn't all that complicated mathematically, but it may be difficult to visualize without some thought - and it's certainly harder to draw diagrams of.

Imagining such a map, one might see that it would be a good map near the North pole, but grow increasingly distorted way from the North pole. (Having a grahpic one would see this, but I don't have such a graphic, so one needs to imagine it).

Fermi-normal coordinates of a GW certainly exist, but they're a pain to work with. They do provide some physical insight though. The Fermi-normal coordinates have the property that near the reference point all the Christoffel symbols are zero, and their

Wiki's stub on "Fermi normal coordinates is" <<wiki link>>. It's worth noting that the metric near the reference worldline has a Minkowskii metric, and that the Christoffel symbols all vanish, which is what makes it fairly intuitive.

PeterDonis
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2019 Award
I believe the "stretch and squish" is basically an artifact of the coordinate choice.
Not really. It's true that you can choose coordinates such that the coordinate distance between the mirrors in LIGO is constant as a GW passes through; but all that does is put all of the stretch and squish into the metric coefficients.

By singling out a single point in space , which is a worldline in space-time, the point being an object "at rest in space", the worldline being the worldline of the object "at rest in space", one can eliminate the "stretch and squish" around a single point. The mathematical representation turns out to be less convenient, but it may be more physically intuitive in the new coordinates. Specifically, these are Fermi-Normal coordinates around that point in space (worldline in spacetime).
You can do this around one of the LIGO mirrors, but the region covered by these coordinates (in which you can ignore tidal gravity, which is what the stretch and squish really is) will not be large enough to cover LIGO as a whole. So you cannot analyze LIGO as a whole using this method.

Distances between points other than the central point are not necessarily quite the same in the tangent space and on the manifold, though.
The tangent space, strictly speaking, has no distances. It is the space of all tangent vectors at a point. The vectors have lengths (norms), but their lengths do not correspond to distances.

Fermi-normal coordinates of a GW certainly exist
I disagree. The whole point of FN coordinates is to cover a patch on which tidal gravity can be ignored. GWs are waves of tidal gravity; that's what they're "made of". So within a single FN coordinate patch, GWs are not even detectable.

pervect
Staff Emeritus
Not really. It's true that you can choose coordinates such that the coordinate distance between the mirrors in LIGO is constant as a GW passes through; but all that does is put all of the stretch and squish into the metric coefficients.
I can see that my remarks were not too clear, the ambiguity in my remarks is basically about what "stretch and squish" means.

In fact, I find it useful to think of the distance between the mirrors as varying with time when a gravitatioanl wave (GW) passes. This happens in coordinates where the metric coefficients are essentially very close to being Minkowskian - coordinats such as Fermi normal coordinates that make the metric Mikowskii at a point in space (in the neighborhood of a worldline of an observer in space-time). By also setting the first derivatives of the metric to be zero, this coordinate choice makes the metric close to Minkowskii over a rather broad region. Setting the first derivative of the metric coefficeitns to zero is equivalent to making the Christoffel symbols vanish.

The Riemann curvature tensor of the GW is non-zero, and the Riemann curvature tensor can be interpreted in the weak field case as a tidal force.

The opposing point of view has the metric varying , having the GW being a metric pertubation and the mirrors having constant coodinates. It's a valid point of view, but not the one I was espousing. One can also describe this point of view as having "stretch and squish", but in this point of view the stretch and squish happens to the metric coefficients.

Paul Colby
Gold Member
If a wave passes through detector changing neither wave's energy, momentum, angular momentum nor particle count, but does change wave's phase, can the detector change detector's state irreversibly, or is this illegal?
Some numbers are in order.

The power flux in a passing GW is

$P = \frac{c^3}{16\pi G}\omega^2 h^2$​

where $h$ is the amplitude of the wave in the transverse traceless gauge and $\omega$ is the angular frequency, say 200 Hz. $h \approx 10^{-21}$. I get 0.013 watt per square meter. Now the LIGO aperture is like 16 square kilometers so about 208,000 watts of gravitational wave energy are passing through the detector aperture. Clearly, only a tiny fraction of this power is converted into detectable signal.

Your central question I think is how much of this power is converted. Well, the power in the laser beam is $I$ watts at say 700 nm. The photons per second, $N$ are, (h is plank's constant when it is)

$N = \frac{I}{h \nu}$​

The momentum change of a photon bouncing off a mirror is,

$\delta p = 2 h \nu c^{-1}$​

$F = \delta p N = 2 I c^{-1}$​

$L$ being 4 km. In rough terms, the work done in changing the length of the arm is

$W = F L h = 2 I c^{-1} L h$​

What's I? I'm sure one can look it up, but lets assume 1kW. I get $W = 2.7\times 10^{-23}$ watts converted by the detector. The other stuff you're asking about such as radiation back reaction, is a small small small fraction of this number.

Paul Colby
Gold Member
Since the thread title is "How Many Gravitons Detected" I should do the final step and divide $W$ by $h f$ where $f = 200$ Hz to get the graviton rate. I get an estimated 200 million gravitons per second absorbed by the detector. The duration of the signal, again could be looked up, but I recall it's less than a second?

PeterDonis
Mentor
2019 Award
I find it useful to think of the distance between the mirrors as varying with time when a gravitatioanl wave (GW) passes.
You are using "distance" to mean "coordinate distance". But this is coordinate-dependent. As you agree, you could choose coordinates in which the coordinate distance between the mirrors is constant and all of the stretch and squish is in the changes to the metric coefficients.

However, in both of these coordinate systems (and in any coordinate system), the GW passage will cause the LIGO detector to register interference between the beams in the two arms. The "distance" in the sense along each arm is changing in an invariant way.

This happens in coordinates where the metric coefficients are essentially very close to being Minkowskian - coordinats such as Fermi normal coordinates that make the metric Mikowskii at a point in space (in the neighborhood of a worldline of an observer in space-time). By also setting the first derivatives of the metric to be zero, this coordinate choice makes the metric close to Minkowskii over a rather broad region.
Not broad enough to cover all of LIGO.

Setting the first derivative of the metric coefficeitns to zero is equivalent to making the Christoffel symbols vanish.
You can only do this along a single worldline, because you cannot set all of the second derivatives of the metric coefficients equal to zero; those are curvatures and you can't make curvature go away by changing coordinates. And the region over which the Christoffel symbols are negligible (not zero, but small enough not to matter) will not be large enough to cover all of LIGO. (If it were, your equations would predict zero signal in the detector.)

What was the effect of LIGO detection on particle count of gravitational waves it detected?

Photons can be and commonly are detected when they are absorbed. In which case the photon stops existing and its whole energy, momentum and angular momentum are transferred on the detector.

However, photons can also undergo Compton scattering. In which case the photon continues (in a different direction) and some of its energy and momentum are conferred on translational motion of the detecting particle. Also detectable.
And photons can also undergo Raman scattering. In which case the photon also continues in a different direction, but some of its energy is conferred on internal motion of the detecting system - or indeed received from the internal motion of the detecting system.

What is the mode by which Ligo detects gravitons? Absorption, Compton scattering, Raman scattering, some others?

PeterDonis
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What was the effect of LIGO detection on particle count of gravitational waves it detected?
Nobody knows. The gravitational waves detected by LIGO were far too weak for us to detect any quantum aspects of them. So we have no way of counting the particles.

Also, you are assuming that the GWs detected by LIGO were in states that even have a definite particle count. We don't know that that's true (and I suspect it's not, just as most EM waves are not in states that have a definite photon number).

Photons can be and commonly are detected when they are absorbed.
Yes, and in fact the term "photon" is best used to refer to these discrete detection events, not to any property of the EM field before the detection. For example, detectors detect discrete events with laser light, but the state of the EM field in a laser is a coherent state, which does not have a definite photon number. (In fact it's an eigenstate of the annihilation operator, so "removing a photon" from it does not even change the state at all!) So the term "photon" can be highly misleading unless you discard virtually all of your normal intuitions about what it means. (Note that coherent states do not just model lasers; the kinds of EM field states we think of as classical EM waves, like radio waves, are also coherent states.)

What is the mode by which Ligo detects gravitons? Absorption, Compton scattering, Raman scattering, some others?
None of the above. LIGO does not detect gravitions. It detects gravitational waves--fluctuations in tidal gravity. LIGO is not the gravitational analogue of a "photon detector" that gives discrete events. It is the gravitational analogue of an antenna: the passage of a wave creates oscillations in the antenna that are driven by the oscillations in the wave. In short, LIGO is a classical detector, not a quantum one.

Note, btw, that if you put all the things I've said above together, the most likely answer to your original question, what is the effect of LIGO on the particle count of the GWs it detects, is "none" (because of my parenthetical notes about coherent states above).

Nobody knows. The gravitational waves detected by LIGO were far too weak for us to detect any quantum aspects of them. So we have no way of counting the particles.
The angular momentum of the gravitational wave is quantized in units of h - no matter how small the ν, and thereby energy hν. (Actually, a supposed property of graviton is spin 2).
The quantum state of the Ligo detector is also quantized in the units of h.
Also, you are assuming that the GWs detected by LIGO were in states that even have a definite particle count. We don't know that that's true (and I suspect it's not, just as most EM waves are not in states that have a definite photon number).

Yes, and in fact the term "photon" is best used to refer to these discrete detection events, not to any property of the EM field before the detection. For example, detectors detect discrete events with laser light, but the state of the EM field in a laser is a coherent state, which does not have a definite photon number. (In fact it's an eigenstate of the annihilation operator, so "removing a photon" from it does not even change the state at all!) So the term "photon" can be highly misleading unless you discard virtually all of your normal intuitions about what it means. (Note that coherent states do not just model lasers; the kinds of EM field states we think of as classical EM waves, like radio waves, are also coherent states.)

None of the above. LIGO does not detect gravitions. It detects gravitational waves--fluctuations in tidal gravity. LIGO is not the gravitational analogue of a "photon detector" that gives discrete events. It is the gravitational analogue of an antenna: the passage of a wave creates oscillations in the antenna that are driven by the oscillations in the wave. In short, LIGO is a classical detector, not a quantum one.

Note, btw, that if you put all the things I've said above together, the most likely answer to your original question, what is the effect of LIGO on the particle count of the GWs it detects, is "none" (because of my parenthetical notes about coherent states above).
What are then the effects of the antenna on the coherent state of the laser radiation/radio waves/gravitational radiation?
If the perturbed wave state, post-detection, has the same frequency but lower amplitude, it also has a smaller particle count. That´s the mode of absorption.
Compton and Raman scattering have the same particle count post-detection, but energy is transferred to antenna because the perturbed wave ends up redshifted to a lower frequency.

So which of them does the Ligo antenna do?

PeterDonis
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The angular momentum of the gravitational wave is quantized in units of h
No. The graviton is spin 2, but that does not mean a gravitational wave in free space must have quantized angular momentum, any more than the photon being spin 1 means an electromagnetic wave in free space must have quantized angular momentum.

The quantum state of the Ligo detector is also quantized in the units of h.
You appear to have a misconception that everything must be quantized in units of h. That's not correct. You need to improve your understanding of quantum mechanics.

What are then the effects of the antenna on the coherent state of the laser radiation/radio waves/gravitational radiation?
A very, very, very small reduction in amplitude.

If the perturbed wave state, post-detection, has the same frequency but lower amplitude, it also has a smaller particle count.
Wrong. A coherent state does not even have a definite particle count to begin with.

The state post-detection does have a very, very, very slightly smaller expectation value of energy and photon number. But that's not the same as "smaller particle count". (For one thing, the expectation value of photon number in a coherent state is not, in general, an integer.)

Again, you need to improve your understanding of quantum mechanics.