Undergrad How Many Gravitons Were Detected in Recent Gravity Wave Observations?

[Paul Colby]

Probably absorption, mostly. Because classical effects do not allow Compton and Raman scattering. (Rayleigh scattering is allowed... but does it create detection in the scatterer?)

What is it you mean by absorption? Can you draw and sum the relevant Feynman diagrams for the radio? This would be somewhat of a challenge but possible, well in principle. I certainly couldn't do it nor would I try. It's not a great way to way to think about the problem.

So, you exclude Raman scattering? If I take an antenna and connect it to ground via a diode and then illuminate it with monochromatic radio waves, it will reradiate a small part of the energy at frequencies other than the illuminating wave because diodes are nonlinear. This is the cause of passive intermodulation or PIM a phenomena which plagues satellites if they put transmit antennas too close to receive antennas and aren't too careful about antenna construction details. Anyway, RF photons go in at $\omega$ and come out at $N \omega$ where $N$ is any integer. If one is broadcasting information then all sorts of different frequencies happen.

But the energy detected by Ligo is not all energy of gravitational waves absorbed in it.

So, I'm struggling with how is this any different than our radio example? In a radio the detected power is amplified to the point it can drive a speaker. In some radios this is done by mixing with local oscillators (AKA lasers for LIGO) and amplifying.

There is one major difference and this is a radio is converting electromagnetic power into more electromagnetic power. The GW detector must convert illuminating gravitational wave power ultimately into electromagnetic power. This is true of all GW detectors including resonate bar detectors since one must at some point detect the bar vibration which is done via electromagnetic means.

 

[Paul Colby]

I think this paper is very relevant to this thread:
https://arxiv.org/abs/1712.04437

This is a nice paper. I would question the assumption that the mirrors have a continuum spectra though. The LIGO mirrors are confined in a finite well so they will have a discrete energy spectrum. Not clear this matters but for tiny things like this it should be noted at least.

 

[Paul Colby]

This is a nice paper.

Starting to read this paper more closely. I think I may need to change my assessment. IMO equation (4), the "energy" of the mirrors, seems bogus. The metric strain given is in the TT gauge. In this gauge, the mirrors are not moving relative to the coordinate system so (4) hasn't its usual meaning as an energy. From here the argument goes down hill.

 

[snorkack]

So, you exclude Raman scattering? If I take an antenna and connect it to ground via a diode and then illuminate it with monochromatic radio waves, it will reradiate a small part of the energy at frequencies other than the illuminating wave because diodes are nonlinear. This is the cause of passive intermodulation or PIM a phenomena which plagues satellites if they put transmit antennas too close to receive antennas and aren't too careful about antenna construction details. Anyway, RF photons go in at $\omega$ and come out at $N \omega$ where $N$ is any integer. If one is broadcasting information then all sorts of different frequencies happen.

That´s overtone generation by multiphoton process, right?
Raman scattering is intermodulation where photons go in at ω and come out at ω-ω₀, where ω₀ is a frequency characteristic to the antenna.

So, I'm struggling with how is this any different than our radio example? In a radio the detected power is amplified to the point it can drive a speaker. In some radios this is done by mixing with local oscillators (AKA lasers for LIGO) and amplifying.

There is one major difference and this is a radio is converting electromagnetic power into more electromagnetic power. The GW detector must convert illuminating gravitational wave power ultimately into electromagnetic power. This is true of all GW detectors including resonate bar detectors since one must at some point detect the bar vibration which is done via electromagnetic means.

I was trying to track down what precisely the "detected power" is, and what the fundamental constraints are on the "detected power" that can be amplified to a detected signal.

Fundamentally, suppose an electromagnetic wave were incident on a laser/maser at a negative temperature, and were to cause stimulated emission.
Stimulated emission should, in principle, be just as detectable as absorption, and existence of stimulated emission under conditions where spontaneous emission is unlikely would be evidence of existence of incident radiation. Yet the "detected power" would be negative.

Could you detect an electromagnetic wave at precisely zero "detected power"?

The electromagnetic wave has conservative properties other than energy. These are momentum and angular momentum.

You might detect an electromagnetic wave by a detector that leaves the energy of the wave precisely unchanged but alters its momentum by altering its direction. The momentum must be received by the detector and become detectable, with amplifiable effects.

If an electromagnetic wave possesses circular polarization, it bears angular momentum. Therefore, a detector which leaves both the energy and direction of a wave perfectly unchanged, but changes its circular polarization, must receive angular momentum and be detectable, with amplifiable effect.

This leaves two other properties of a wave that do NOT carry conserved physical quantities.
These are phase and state of linear polarization.

Detection might, in principle, take place by rearranging conserved physical quantities inside the detector, without transferring any conserved quantities from wave to detector.
But I have a suspicion that there is some rule against this. Is this the case?

 

[Paul Colby]

I was trying to track down what precisely the "detected power" is, and what the fundamental constraints are on the "detected power" that can be amplified to a detected signal.

I feel your pain. This is a question well worth pondering. It has a simple answer for all the cases I know of. Any signal, electromagnetic or gravitational, will need to do work on the detector. The rate of this work is the detected power.

Could you detect an electromagnetic wave at precisely zero "detected power"?

No, never ever.

What you would find is a signal to noise ratio of 0 since nose power is never 0. To be detected this ratio needs to be big enough S/N (5 would be nice). This is why it's so very important to maximize the received power since the majority of noise comes from the first amplification stage and has nothing to do with your signal. In RF systems this is why impedance matching the antenna is important. One maximizes the power coupling from the antenna to the radio by matching the impedance. This means the most you get is half.

This is a question that comes up in every link calculation ever made. Basically, there is no free lunch.

Detection might, in principle, take place by rearranging conserved physical quantities inside the detector, without transferring any conserved quantities from wave to detector.
But I have a suspicion that there is some rule against this. Is this the case?

Interference fringes are like this, right. One is moving energy not creating or losing it. However, one must ask why (and how) have the beams changed. LIGO can be viewed as a parametric amplifier. The parameter in this case is the length, L, of the interferometer arms. As the GW passes it changes this length which takes work. BTW the work taken goes to 0 if the lasers are turned off. The work increases as the laser power in increased. The true problem for LIGO (the minor bit that took like 40 years and 1/3 of the physics community) is maximizing the signal to noise. This was not a simple task.

 

[Paul Colby]

That´s overtone generation by multiphoton process, right?
Raman scattering is intermodulation where photons go in at ω and come out at ω-ω0, where ω0 is a frequency characteristic to the antenna.

So Raman scattering is inelastic scattering off of atoms and molecules, so technically, no. My example was just to show the most primitive radio, a diode with an antenna, will involve harmonics which are, in a complete analysis, are multi photon processes but not technically Raman scattering as far as I know.

 

[PeterDonis]

The parameter in this case is the length, L, of the interferometer arms. As the GW passes it changes this length which takes work

I think this is oversimplifying. The arms are not solid objects. They are just mirrors in free fall (at least for the purposes of analysis here; I realize that the LIGO detectors are actually at rest on the surface of the Earth so they're not in free fall, although future space-based detectors will be). Changing the distance between them by changing tidal gravity, in and of itself, does no work. What does work is changing something that connects the mirrors or travels between them. In your earlier post, you said, correctly, that in the case of LIGO it's the light beams, reflecting off the mirrors, that have work done on them by the GW.

 

[Paul Colby]

I think this is oversimplifying.

Fair enough. At one time a read some papers on the design of LIGOs mirror mounts. They are supported on a rather complicated structure designed to isolate them from seismic noise. The laser mode existing between the mirror apply a force which is balanced by the detent of the support. Think of a pendulum bob. With no laser power the bob hangs vertically. With the laser it presses out till an equilibrium is reached, basically where the forces balance. This isn't a big distance. The total relative movement of the mirrors is less than a proton diameter after all. Basically laser + mirrors is an extended object upon which work is done changing it's length.

 

[Paul Colby]

Think of a pendulum bob. With no laser power the bob hangs vertically. With the laser it presses out till an equilibrium is reached, basically where the forces balance.

While I believe this is true in practice, it's kind of beside the real issue. Even if the mirrors were completely stationary in the TT frame, something I think is essentially true due to their mass, changing the length of the interferometer does work on the light. It's like standing up. My legs do work on my body but also a very very tiny amount of work on the earth. Tiny because I haven't moved the Earth very far from its original position.

 

[snorkack]

I feel your pain. This is a question well worth pondering. It has a simple answer for all the cases I know of. Any signal, electromagnetic or gravitational, will need to do work on the detector. The rate of this work is the detected power.
No, never ever.

What you would find is a signal to noise ratio of 0 since nose power is never 0.

This is a question that comes up in every link calculation ever made. Basically, there is no free lunch.
Interference fringes are like this, right. One is moving energy not creating or losing it. However, one must ask why (and how) have the beams changed.

Is it legal for a detector to receive pure linear momentum, unaccompanied by any power whatsoever, and amplify it into output power?

 

[PeterDonis]

pure linear momentum, unaccompanied by any power whatsoever

There is no such thing.

 

[snorkack]

There is no such thing.

True, because the detector has finite mass.
You can therefore choose a frame of reference where the input power is positive, a frame of reference where input power is negative, or frame of reference where input power is exactly zero.

 

[Paul Colby]

Is it legal for a detector to receive pure linear momentum, unaccompanied by any power whatsoever, and amplify it into output power?

No, momentum is the space component of a 4-vector. You aren't permitted to change the space part without the energy part also changing.

In simple terms, there is no free lunch. For something to be detected, something in the detector must change. For something in the detector to change, work must be done on the detector by some external "signal" or phenomena.

Angular momentum will be the same except it's a component of higher rank tensor and so more confusing.

 

[snorkack]

No, momentum is the space component of a 4-vector. You aren't permitted to change the space part without the energy part also changing.

In simple terms, there is no free lunch. For something to be detected, something in the detector must change. For something in the detector to change, work must be done on the detector by some external "signal" or phenomena.

What makes it illegal for a detection to happen in a frame of reference where the momentum of the detector changes direction, but the energy of the detector is precisely unchanged? Momentum is a vector, energy a scalar.

 

[PeterDonis]

True, because the detector has finite mass.

No, it's because there is no such thing as linear momentum with zero power, period. Even massless objects, like light, can't have linear momentum without delivering power.

You can therefore choose a frame of reference where the input power is positive, a frame of reference where input power is negative, or frame of reference where input power is exactly zero.

Please show your work.

 

[PeterDonis]

a frame of reference where the momentum of the detector changes direction, but the energy of the detector is precisely unchanged?

How do you know there is such a frame of reference? Remember we're not talking about a general possibility, but about the specific case of a gravitational wave detector.

 

[snorkack]

No, it's because there is no such thing as linear momentum with zero power, period. Even massless objects, like light, can't have linear momentum without delivering power.

They cannot have linear momentum without carrying power.
Carrying power is not the same as delivering it.
If a ray of light interacts with a mirror, what is the "power delivered"? The total power of the light which changes direction on reflection? Or only the power which the light loses by getting redshifted, and confers on the mirror?

Please show your work.

Easy.
Since incident and reflected light move in different direction, they undergo different Doppler shift on change of reference frame.
You can choose a reference frame where the reflected light is blueshifted and it is the mirror that is doing work against light pressure to blueshift the incident light.
Or a reference frame where the energy of incident and reflected light are exactly equal, and no work whatever is done by or to light.

 

[Paul Colby]

Momentum is a vector, energy a scalar.

Energy in this case is the fourth component of the energy-momentum 4-vector. This fact matters.

 

[PeterDonis]

Carrying power is not the same as delivering it.

Carrying linear momentum is not the same as delivering it either. An object cannot deliver linear momentum to a detector without delivering power as well.

If a ray of light interacts with a mirror, what is the "power delivered"?

The rate of change of energy transfer from the light to the mirror.

 

[Paul Colby]

What makes it illegal for a detection to happen in a frame of reference where the momentum of the detector changes direction, but the energy of the detector is precisely unchanged? Momentum is a vector, energy a scalar.

Well, the obvious answer is nothing observable changes without some work being done somewhere on the detector. Can you give one complete example where this ever happens? I can't because it would violate general well established principles as I understand them. So help me out here.

 

[snorkack]

Carrying linear momentum is not the same as delivering it either. An object cannot deliver linear momentum to a detector without delivering power as well.

The rate of change of energy transfer from the light to the mirror.

Cannot quite parse that phrase.
So, consider the case of antenna where perfect reflection occurs. At a suitable frame of reference, the reflected wave has precisely the same frequency and amplitude as the incident wave, just different direction of propagation.Therefore the outgoing energy is precisely equal to incoming energy, and the wave does no work. However, the change of momentum is nonzero and is transferred on the antenna.

Is the change in the momentum of antenna, at constant energy, in principle possible to observe and amplify to a detection event, or is it provably impossible to observe and amplify to a detection because there is zero detected power to amplify?

 

[Paul Colby]

Is the change in the momentum of antenna, at constant energy, in principle possible to observe and amplify to a detection event, or is it provably impossible to observe and amplify to a detection because there is zero detected power to amplify?

These kind of questions can't be answered because too much is missing. What is being detected and exactly how is this detection taking place. For LIGO detection is made using CCDs looking at interference fringe shifts. These shifts are due to work done on the laser beam. You must supply these kinds of details in order for it to even be a question.

 

[PeterDonis]

consider the case of antenna where perfect reflection occurs. At a suitable frame of reference, the reflected wave has precisely the same frequency and amplitude as the incident wave, just different direction of propagation.Therefore the outgoing energy is precisely equal to incoming energy, and the wave does no work. However, the change of momentum is nonzero and is transferred on the antenna.

Again, please show your work. With math.

 

[poor mystic]

So gravitational waves are waves of electromagnetic energy which come from some direction, weakly interact with matter, and move on.
You'd think that a massive sphere (say a planet) would focus gravitational waves much as a glass sphere focuses visible light, so that at some distance a concentration of energies might be found.
?

 

[Ibix]

So gravitational waves are waves of electromagnetic energy

No. They are gravitational waves, not electromagnetic waves. We use their effect on laser interferometers to detect them.

You'd think that a massive sphere (say a planet) would focus gravitational waves much as a glass sphere focuses visible light, so that at some distance a concentration of energies might be found.

I believe that we do expect gravitational lensing of gravitational waves, similar to gravitational lensing of electromagnetic waves, if that's what you are referring to. However, our experimental work is in its infancy and we have no direct experimental confirmation.

 

[Paul Colby]

You'd think that a massive sphere (say a planet) would focus gravitational waves much as a glass sphere focuses visible light, so that at some distance a concentration of energies might be found.

With light/radio waves the refractive index of the material is the ratio of the propagation speed in vacuum to the propagation speed in the material. The higher the index the more refraction that takes place. For a GW the mechanical stiffness, or shear modulus of the material determines the index of GW in the material. If one works it out (one being me so be warned) one gets,

$n(\omega) = \sqrt{1+\left(\frac{\omega_o}{\omega}\right)^2}$

where the characteristic angular frequency is,

$\omega_o = \frac{4}{c}\sqrt{\pi G \Gamma}$

where $G$ is Newton's gravitational constant and $\Gamma$ is the shear modulus of the material (which is assumed isotropic). Assume a solid steel planet $\Gamma\approx 70 GPa$. so $\omega_o \approx 1.0\times 10^{-8} Hz$. For the frequency range of interest, $n(\omega)\approx 1$ to more zeros than is useful IMO.

 

[TeethWhitener]

You'd think that a massive sphere (say a planet) would focus gravitational waves much as a glass sphere focuses visible light, so that at some distance a concentration of energies might be found.

Possibly of interest to you:
http://iopscience.iop.org/article/10.1086/312015/fulltext/985903.text.html

 

[snorkack]

With light/radio waves the refractive index of the material is the ratio of the propagation speed in vacuum to the propagation speed in the material. The higher the index the more refraction that takes place. For a GW the mechanical stiffness, or shear modulus of the material determines the index of GW in the material. If one works it out (one being me so be warned) one gets,

$n(\omega) = \sqrt{1+\left(\frac{\omega_o}{\omega}\right)^2}$

where the characteristic angular frequency is,

$\omega_o = \frac{4}{c}\sqrt{\pi G \Gamma}$

where $G$ is Newton's gravitational constant and $\Gamma$ is the shear modulus of the material (which is assumed isotropic). Assume a solid steel planet $\Gamma\approx 70 GPa$. so $\omega_o \approx 1.0\times 10^{-8} Hz$. For the frequency range of interest, $n(\omega)\approx 1$ to more zeros than is useful IMO.

Shear modulus, not compression modulus?
Then superfluids regardless of stiffness cannot refract gravitational waves.
The superfluid interior of a neutron star would only participate in gravitational, frequency-independent bending of the waves, while the frequency-dependent refraction can only happen in the solid crust of the neutron star?

 

[Paul Colby]

Then superfluids regardless of stiffness cannot refract gravitational waves.

That would be my understanding. Fluids only interact with GW through surface traction forces. I believe this is also true for isotropic solids even though they support shear waves. For a proper acceleration of a bit of matter to occur, a net interatomic force must be developed on the bit. If one looks at the stress developed by a passing GW on a solid, points interior to the solid have no net force due to the GW since the divergence of the stress is zero. All effects must propagate inward from the solids boundary where the stress developed has a divergence.

Of course, the background curvature of the star will effect the propagation just as it does with light as far as I know.

[Edit]

Okay, some clarification. The origin of the refractive index calculation uses the constitutive relation for garden variety solids like steel where the stress tensor is related linearly to the strain tensor. Since shear staining a neutron star leads to no stress, it will not refract a GW IMO.