Sciencemadness Discussion Board - Gravitational waves: breakthrough discovery after two centuries of expectation

Gravitational waves: breakthrough discovery after two centuries of expectation

hissingnoise - 11-2-2016 at 12:49

This is Earthshaking news . . . figuratively and literally?

https://www.theguardian.com/science/2016/feb/11/gravitationa...

crystal grower - 11-2-2016 at 13:31

Thanks for info, that's really amazing.
But I think the most cool thing about it is that A. Einstein has known it a century ago.
(Just curiosity: Has Einstein ever got wrong?)

Eddygp - 11-2-2016 at 13:36

Yes, he HAS!!!

Marvin - 11-2-2016 at 13:52

He included a constant to account for the universe being stationary. That constant is now being used to account for the universe accelerating. He also refused to accept quantum theory. So not a perfect record.

I was really really hoping gravity waves would turn out to be false.

The maths behind GR is mind melting.

annaandherdad - 11-2-2016 at 14:31

These were relatively small black holes that combined, about 29 and 36 solar masses for the pair, plus or minus about 4 on each figure. It means both black holes probably formed from the collapse of massive stars, perhaps in a binary star system, some time in the past. The event was a billion light years away. It means that black hole coalescence is now observable via gravity waves within a sphere centered on the earth of about that radius. The wave forms in Louisiana and Washington state differed in arrival time by 7 milliseconds, due to the 2000 km separation between the detectors. This means that the waves came from the southern hemisphere. In the future if more than two detectors are active when a signal is received, it will be possible to give more accurate data on the direction from which the signal came. The wave form is chirped, that is, the frequency and amplitude increase at first, as the two black holes spiral in toward one another, and their orbital frequency increases. The frequency of the gravitational wave is twice the orbital frequency. After the merger there is a ring-down as the two black holes radiate away their quadrupole and higher moments and settle into a new and larger, stable rotating black hole. The system radiated a total gravitational energy of 3 solar masses, with a peak power of 200 solar masses (times c^2) per second. Various quantitative arguments exclude the possibility that the two objects were a pair of stars (neutron or otherwise), or a black hole and a star; only two black holes fits the data.

Rather than correct the above, let me point out that the two detectors are separated by more than 2000km. The 2000km figure is the light travel distance at 7ms. This is the distance between the detectors, times the cosine of the angle between the line joining the detectors, and the direction of the incoming gravitational waves. This determines the direction of the incoming waves to lie on a cone of known opening angle, centered on the line between the detectors. It is known that the cone points to the south because the signal was detected first in Louisiana. If three or more detectors were to receive the signal, you would have overlapping cones that would select out a unique direction.

[Edited on 11-2-2016 by annaandherdad]

careysub - 11-2-2016 at 15:02

Quote: Originally posted by Marvin

He included a constant to account for the universe being stationary. That constant is now being used to account for the universe accelerating. He also refused to accept quantum theory. So not a perfect record.

I was really really hoping gravity waves would turn out to be false.

The maths behind GR is mind melting.

The Cosmological Constant is actually a component of the formal solution of GR, not an arbitrary addition - but its value is not set by the theory. Einstein debated whether it was 'real' and changed his mind more than once. So his original formulation, with all of its details (including the CC) appears to be correct in every aspect.

Yep, he could never adapt to quantum theory - even though he helped create it.

I am happy that GR is proving to be correct (just like I am happy that quantum theory as originally formulated also appears correct).

But we knew that gravitational waves were real already, the energy loss of black hole binaries already measured showed they were radiating gravitational waves, exactly as GR predicts. The question was whether we could detect them.

XeonTheMGPony - 11-2-2016 at 15:39

https://www.youtube.com/watch?v=4UY5A3NJjls < Rated R for language

We're geting there, now we need to hurry up with warp drive!

careysub - 11-2-2016 at 23:10

One thing about gravitational waves I rarely see brought up is that gravitational waves can be extremely "bright", that is, have a lot of energy in them; yet they are extremely difficult to detect. It is because of the ratio of field strength between gravity and electromagnetism: 10^-43!

Roughly speaking, a gigawatt of incident gravitational energy incident on a detector (or you) would generate a 10^-34 watt signal! This is why it was doubted that they could ever be detected. Thank gods for coalescing black holes! Those suckers are the strongest source of radiation emissions in the Universe.

How intense? About a billion times brighter (in the gravity spectrum) than the brightest quasar in the Universe (the blazar 3C 454.3) is in the electromagnetic spectrum!

Its probably a good thing we cannot easily interact with their emissions, but if they weren't so incredibly intense we would never pick up any gravitational waves.

3C 454.3 reached its all time recorded peak brightness of 13.4 magnitude (7 magnitudes dimmer than what the human eye can see) in June 2014. If we could see a major black hole coalescence at the same distance (7.7 billion light years, over half the way back to the beginning of the Universe), it would flash at about -10 magnitude (approaching the brightness of the full moon, which is -13 magnitude).

Check out this paper:
http://arxiv.org/pdf/1602.02872v1.pdf

[Edited on 12-2-2016 by careysub]

phlogiston - 12-2-2016 at 01:24

Your comparison helps a lot to gain some feeling for the effects relative to electromagnetic signals, thanks. But while full moon equivalent is indeed astounding given the distance, it is not particularly bright in an absolute sense. A tiny LED flashlight at armslength easily outshines the entire moon. This should be true for locally generated gravitational waves too.

Given the extreme distances between cosmic events and earth and the resulting 'dilution' of the signal with distance, I am still a little puzzled that local signals of low absolute intensity are not far more easily detected.

I am wandering if perhaps the frequency comes into play here. Given their size, I suspect LIGO detecters are optimal for detecting waves with kilometer wavelengths, wheres local events from small objects accelerating will generate much higher frequency waves. But LIGO has to be large because the relative change in the length of the arms of the detector is still incredibly tiny. If the arms were smaller, the change in size would probably be undetectable.

[Edited on 12-2-2016 by phlogiston]

Fulmen - 12-2-2016 at 02:44

Quote: Originally posted by careysub

So his original formulation, with all of its details (including the CC) appears to be correct in every aspect

As I understand it the original CC was introduced to fit the current accepted static universe as a rather ad-hoc addition. That would mean that he was right, but for the wrong reason (which is pretty close to being wrong).

Quote:

But we knew that gravitational waves were real already, the energy loss of black hole binaries already measured showed they were radiating gravitational waves, exactly as GR predicts

There is a huge difference between measuring energy being lost and actually measuring the energy in the predicted form. Prior to this we didn't know GW were real, although we were pretty confident.

careysub - 12-2-2016 at 06:39

Quote: Originally posted by Fulmen

Quote: Originally posted by careysub

So his original formulation, with all of its details (including the CC) appears to be correct in every aspect

Quote:

But we knew that gravitational waves were real already, the energy loss of black hole binaries already measured showed they were radiating gravitational waves, exactly as GR predicts

There is a huge difference between measuring energy being lost and actually measuring the energy in the predicted form. Prior to this we didn't know GW were real, although we were pretty confident.

The Cosmological Constant is a natural part of the most general form of the GR equations.

A very good analogy can be drawn to the constant that appears when performing an integration. The most general solution of an integral equation contains a constant, which is arbitrary. Usually it gets ignored in practice, but it really is part of the solution (and is a reflection of the fact that when you do differentiation constants disappear).

If you work an integration problem on a test or in homework, and this constant is not included you probably lose points.

Yes there is a big difference between knowing that GWs exist and actually detecting them (the first already generated a Nobel and the second will generate another). But we did know they existed before direct detection.

An analogy with the neutrino can be drawn. We knew neutrinos existed before we detected them since we could detect the energy loss when they carried energy away. But with GWs the case was much stronger since the observed loss exactly matched the predictions of the existing theory (unlike neutrinos).

[Edited on 12-2-2016 by careysub]

careysub - 12-2-2016 at 07:07

Quote: Originally posted by phlogiston

Your comparison helps a lot to gain some feeling for the effects relative to electromagnetic signals, thanks. But while full moon equivalent is indeed astounding given the distance, it is not particularly bright in an absolute sense. A tiny LED flashlight at armslength easily outshines the entire moon. This should be true for locally generated gravitational waves too.

Given the extreme distances between cosmic events and earth and the resulting 'dilution' of the signal with distance, I am still a little puzzled that local signals of low absolute intensity are not far more easily detected.

I am wandering if perhaps the frequency comes into play here. Given their size, I suspect LIGO detecters are optimal for detecting waves with kilometer wavelengths, wheres local events from small objects accelerating will generate much higher frequency waves. But LIGO has to be large because the relative change in the length of the arms of the detector is still incredibly tiny. If the arms were smaller, the change in size would probably be undetectable.

[Edited on 12-2-2016 by phlogiston]

The frequency issue is part of it. Another part is how often these events occur.

If by "local" you mean the Local Group of galaxies for example (Milky Way, Andromeda, Triangulum, plus dwarf galaxies) then the volume of the whole observable Universe is 100 billion times bigger. We are far more likely to see very distant events.

Note that the example I gave for GW brightness was assuming it was an extremely distant one. The one we detected is much closer.

I just calculated the actual brightness of this particular event, which was at 1.2 billion light years. The gravitational wave radiation field was 0.1 watts/per square meter! Sunlight at Earth's orbit (i.e. without any absorption) is 1400 watts/per square meter.

0.1 watt per square meter is about the illumination intensity of an iPhone LED light at one meter.

If the event were 10 million light years away the energy would have equaled solar illumination. If it has occurred in the Andromeda galaxy it would have been 10 times brighter than the Sun. If it had occurred in the center of our galaxy it would have been 100,000 times brighter than the Sun.

Also I should point out the blazar I used as an example is so extremely bright (but still astronomically dimmer than the GW burst) because it has an emission jet pointed directly at us. It is not omnidirectional, unlike the GW event, making the brightness of the black hole merger even more incredible.

[Edited on 12-2-2016 by careysub]

Fulmen - 12-2-2016 at 07:21

Careysub: The point was that he introduced a value specifically to achieve a static universe, rather than predicting a dynamic one. As I understand it Einstein abandoned the idea in 1929 after Hubble's discovery, calling it the biggest blunder of his life. If he's willing to call it a blunder I have no problem taking his word for it ;-)

careysub - 12-2-2016 at 07:36

Quote: Originally posted by Fulmen

Careysub: The point was that he introduced a value specifically to achieve a static universe, rather than predicting a dynamic one. As I understand it Einstein abandoned the idea in 1929 after Hubble's discovery, calling it the biggest blunder of his life. If he's willing to call it a blunder I have no problem taking his word for it ;-)

My point is that there is a subtlety about what the "error" actually was.

The most general form of GR includes a CC naturally, as I said it is not ad hoc (although any particular value is).

Einstein set a value for this to make a static universe, which he believed in (and was mistaken).

With the discovery of the Hubble Law he dropped the CC, believing that including it was in error, since the Universe was not static as he believed.

But he was in error in dropping the CC, since it really is part of the true solution of GR, just not in the way he imagined (no one expected cosmic acceleration, AFAIK). The Universe is stranger than even Einstein imagined.

You might say he was in error twice - once for believing in a static Universe, and once in dropping the CC.

His original form of the equations was the correct one however.

[Edited on 12-2-2016 by careysub]

hissingnoise - 12-2-2016 at 08:12

How the ancient merging of massive bodies produced this "tsunami" in the space-time continuum is a question dogging me since the detection was announced.

Was it caused by the act of two massive bodies colliding or by the loss of mass in the huge release of energy, or was it a combination of both those things?

Fuck!

http://thinkforyourself.ie/2010/02/21/wha%E2%80%99-is-the-st...

careysub - 12-2-2016 at 08:41

Quote: Originally posted by hissingnoise

How the ancient merging of massive bodies produced this "tsunami" in the space-time continuum is a question dogging me since the detection was announced.

Was it caused by the act of two massive bodies colliding or by the loss of mass in the huge release of energy, or was it a combination of both those things?

It is the natural result of two extremely massive, maximally dense bodies in the process of merging which converts ~10% of their mass into gravitational wave energy. The merger involved them spinning around each other at a large fraction of the speed of light.

Similar to the way oscillating charged particle naturally emit energy as electromagnetic waves.

solo - 12-2-2016 at 09:29

Reference Information

Observation of Gravitational Waves from a Binary Black Hole Merger
B. P. Abbott et al.
PHYSICAL REVIEW LETTERS
12 FEBRUARY 2016
DOI: 10.1103/PhysRevLett.116.061102

Abstract
On September 14, 2015 at 09:50:45 UTC the two detectors of the Laser Interferometer Gravitational-Wave Observatory simultaneously observed a transient gravitational-wave signal. The signal sweeps upwards in frequency from 35 to 250 Hz with a peak gravitational-wave strain of 1.0 × 10−21. It matches the waveform predicted by general relativity for the inspiral and merger of a pair of black holes and the ringdown of the resulting single black hole. The signal was observed with a matched-filter signal-to-noise ratio of 24 and a false alarm rate estimated to be less than 1 event per 203 000 years, equivalent to a significance greater than5.1σ.

Thesourceliesataluminositydistanceof410þ160 Mpccorrespondingtoaredshiftz1⁄40.09þ0.03. −180 −0.04
In the source frame, the initial black hole masses are 36þ5M and 29þ4M , and the final black hole mass is −4⊙ −4⊙ 62þ4M , with 3.0þ0.5M c2 radiated in gravitational waves. All uncertainties define 90% credible intervals. −4 ⊙ −0.5 ⊙
These observations demonstrate the existence of binary stellar-mass black hole systems. This is the first direct detection of gravitational waves and the first observation of a binary black hole merger.

Attachment: Observation of Gravitational Waves from a Binary Black Hole Merger -B. P. Abbott.pdf (914kB)
This file has been downloaded 513 times

hissingnoise - 12-2-2016 at 10:34

“It’s such a huge relief to get to share this with the world,” Lisa Barsotti, a principle research scientist at LIGO said.

“I felt like time between September and February was just being stretched out.”

[Edited on 12-2-2016 by hissingnoise]

Fulmen - 12-2-2016 at 17:01

Quote: Originally posted by careysub

My point is that there is a subtlety about what the "error" actually was.

I agree that it is subtle. But there is still good reason to call it a blunder. The original GR predicted a non-static universe, while the current assumption was the opposite. But this wasn't supported by any evidence, just the lack of it. And while a CC would fix that it must have been a precarious and delicate balancing act.
Considering the incredible "act of faith" it took Einstein to follow logic past common sense makes it that much worse.
That later findings has reintroduced the CC doesn't really vindicate his earliest use of it, being right for the wrong reasons isn't far from being wrong.

At least that's how I understand it. Not that I hold it against him, it just underlines just how brief his time was. Few has ever made a similar contribution to science, but the time period this was limited to is just as remarkable. Within a decade he had completely revolutionized science, then nothing. I don't know if there was more for him to do, it takes time before enough observations accumulate to produce new discoveries of this magnitude. But his failure to accept quantum mechanics pretty much eliminated the possibility of him producing more of real significance.
Perhaps the future will vindicate him on that point as well, perhaps we will in time "solve" QM into something more predictable. But does that make him right? Considering the available evidence, did he really have any reason to doubt QM?

wg48 - 12-2-2016 at 19:46

Quote: Originally posted by careysub

One thing about gravitational waves I rarely see brought up is that gravitational waves can be extremely "bright", that is, have a lot of energy in them; yet they are extremely difficult to detect. It is because of the ratio of field strength between gravity and electromagnetism: 10^-43!

Roughly speaking, a gigawatt of incident gravitational energy incident on a detector (or you) would generate a 10^-34 watt signal! This is why it was doubted that they could ever be detected. Thank gods for coalescing black holes! Those suckers are the strongest source of radiation emissions in the Universe.
snip
[Edited on 12-2-2016 by careysub]

That,s not a fair calculation. The 10^43 ratio is the ratio between the gravity force between subatomic particles and the electromagnetic force between them. For example electrons or protons. So unless the detector is a single hydrogen atom.. Unfortunately a Weber type detector (cryo bar) of 10^43 protons or neutrons is a very very large bar . One Weber detector used 1000kg mass which is only about 10^27 neutrons and protons so its still not easy.

But don't forget that the gravitation constant was determined by measuring the gravity force between lead balls way back in the 1700s.

Perhaps a big part of the detection problem is that the waves effect all masses the same. So you can not use a simple torsion balance with a lead ball.

careysub - 12-2-2016 at 20:13

Quote: Originally posted by wg48

Quote: Originally posted by careysub

I said "roughly"... to give an indication of how difficult the detection is.

In fact gravitational wave detectors do not use the coupling between gravity and electromagnetism at all, but instead detect the distortion of space directly. This was true of Weber detector also. The space distortion caused by the major detected event was on the order of 10^-21.

If the coupling were high (in the vicinity of 1) you could hope to capture all, or most of the energy in the wave, which electromagnetic detectors commonly approximate.

wg48 - 13-2-2016 at 03:18

Quote: Originally posted by careysub

In fact gravitational wave detectors do not use the coupling between gravity and electromagnetism at all, but instead detect the distortion of space directly. This was true of Weber detector also. The space distortion caused by the major detected event was on the order of 10^-21.

snip .

Can you explain what the 10^-21 space distortion means.
ie what can be measured or compared to detect the distortion?

My limited understanding of the Webber detector is it compares the force on side of the mass (loosely speaking) if they are not the same the resultant force distorts the mass causing it to ring which is then detected.

j_sum1 - 13-2-2016 at 04:07

A couple of curious questions. Although these might be betyer suited to Randal Munro of whatif fame...

How close to the converging black holes could you get without the massive energy expulsion doing indescribably destructive things to your body?

And how close woukd you need to get in order to observe it with your normal senses?

Marvin - 13-2-2016 at 04:48

I think it is a fair comparison, if electromagnetism was weaker we'd be using kilograms of ions to detect it. We'd also have to be stuck together by a totally different physics, of course.

Another issue accessible to chemists here is that gravity wave coupling is quadrupolar.

careysub - 13-2-2016 at 09:13

BTW, the usual ratio quoted for the strength of fundamental forces is
Strong Force = 1
Electromagnetism = 1/137
Weak = 10^-6
Gravity = 6*10^-39

Giving a direct force strength ratio of 8*10^-37.

My reason for using 10^-43 is this page at Caltech discussing an oscillating charge mass:
http://ned.ipac.caltech.edu/level5/ESSAYS/Boughn/boughn.html
a system, that emits 10^43 times more energy in electromagnetic waves compared to gravity waves. I was postulating that you used such a system as a gravity energy collector, seeking to collect the excitation of gravity into electromagnetism.

I was thinking just this morning about how large a space distortion effect it might take to be dangerous. Odd thing to imagine, very Star Trekkie. One imagines that since this detection involved a distortion of 10^-21 that an event 10^10 times closer (inverse square law) would produce one of 10% which ought to be impressive. That would 0.2 light years - or pretty close for such a (clearly infrequent) event.

The Weber Bar detector relied on space dimensional distortion just like LIGO. The passing of a wave would cause the cylinder to change in length by 10^-16 which would cause it vibrate at 1660 Hz.

[Edited on 13-2-2016 by careysub]

annaandherdad - 13-2-2016 at 09:46

Hi careysub, the power flux falls off as the inverse square, but the amplitude of the wave falls off as the inverse power of the distance. The 10^{-21} quoted is the amplitude of the wave. So you'd have to be 10^{20} times closer to get an amplitude of 10%. So that would be 10^{-11} light year or 3 x 10^{-3} light second or 1000 km, if my rough calculation is right. This sounds about right, the amplitude is basically 1 at the coalescence of the black holes.

hissingnoise - 13-2-2016 at 10:35

Hawking says in this article that he would like to use gravitational waves to test his area theorem: that “the area of the final black hole is greater than the sum of the areas of the internal black holes.” He adds: “This is satisfied by the observations.”

The comments are a read, too . . .

phlogiston - 13-2-2016 at 16:38

Quote: Originally posted by careysub

Quote: Originally posted by phlogiston

Given the extreme distances between cosmic events and earth and the resulting 'dilution' of the signal with distance, I am still a little puzzled that local signals of low absolute intensity are not far more easily detected.
[Edited on 12-2-2016 by phlogiston]

Good point, but no, I meant way more local. Consider the adjacent room. Sorry for being vague.
I understand that in the much smaller volume of 'local space' there will be fewer detectable events.
What puzzles me is why it is not possible to detect the gravitational effect of accelerating masses moving very near LIGO, ie meters away instead of billions of lightyears.
For electromagnetic signals, it is trivial to produce an artificial signal many times stronger than any 'natural source', including the sun, using simple means (big lamp, laser), if measured near enough to the source.
This should hold for gravitational waves too. Would smashing two large blocks of osmium together meters away from one of the detector arms not yield a stronger signal than a black hole merger billions of light years away?

[Edited on 14-2-2016 by phlogiston]

Oscilllator - 13-2-2016 at 18:59

phlogiston if you read careysub's answer he mentioned that ~10% of the mass of the black hole is converted to gravitational wave energy. When you smash two blocks of osmium together, 0% of the energy is converted to gravitational waves. That's probably why.

j_sum1 - 13-2-2016 at 19:53

two blocks of osmium sounds like an expensive way to go. W or even Pb is a more likely option for gravity experiments.

But yes, Oscillator is right. You need something that will make some sort of gravitational wave. Playing with normal sized masses might not do it.

annaandherdad - 13-2-2016 at 20:37

Here are some contributions to the discussion of the comparison between electromagnetic radiation and gravitational radiation. I'll start with some atomic physics.

Electromagnetic radiation arises when charges are accelerated. An electron orbiting in an atom is accelerated, so the atom radiates. As it does so, the electron moves closer to the atom (a lower energy state). As the size of the electron orbit shrinks, the orbital frequency increases, as does the velocity of the electron. However, the total energy of the atom, kinetic plus potential, decreases because of the emitted radiation.

As far as this goes, this is very similar to what happens with the orbiting black holes. The main difference is that the energy states of the atom are quantized, which prevents the electron from spiraling ever closer to the nucleus and emitting ever higher frequency radiation. This is a "catastrophe" predicted by classical physics which is prevented in reality by quantum mechanics. Since the atom has a ground state, the inspiral of the electron stops there. In the case of the orbiting black holes, what prevents this catastrophe is that when the two black holes are close enough together that their event horizons merge, then they combine to form a single, larger black hole, and radiation ceases.

Another important difference between the electromagnetic and gravitational cases is that electromagnetic radiation is dipole, while gravitational is quadrupole. More exactly, the electromagnetic radiation from a localized source can be decomposed into a sum, dipole+quadrupole+octupole+ ... but if the source (the "antenna") is "small" (in a manner described precisely below) then the quadrupole contribution is much smaller than the dipole, etc. In the case of gravity, the radiation sum is quadrupole+octupole+ ..., that is, without the dipole term. Again, if the "antenna" is "small", then the leading term (the quadrupole) dominates. Here I am glossing over the difference between electric and magnetic poles, and the analogous distinction in gravitational waves.

In atoms, sometimes the dipole term is forbidden by selection rules, so one only sees the quadrupole term. In this case, the intensity of the spectral line is much weaker than other transitions of a comparable frequency, which are allowed as a dipole transition. For example, in hydrogen, the 3p->1s is allowed as an electric dipole transition, but 3d->1s only as a quadrupole. The quadrupole line is roughly 10^4 times weaker than the dipole line; both have almost exactly the same frequency. The factor of 10^{-4} is physically (v/c)^2, where v is the electron velocity, multiplied by various factors of order unity.

It is shown in atomic physics that the transition rate (in sec^{-1}) for an electric dipole transition between states A and B in a single-electron atom is
$$w=\frac{4}{3} \frac{\omega^3 e^2}{\hbar c^3}
|\langle A \vert {\bf x} \vert B \rangle|^2. $$

These formulas are in Gaussian units; if you prefer SI, replace e^2 by e^2/4 pi epsilon_0. Here omega is the frequency of the photon emitted. If you multiply this by hbar*omega, the energy of the photon, you get the power emitted by the atom:

$$P=\frac{4}{3} \frac{e^2 \omega^4}{c^3}
|\langle A \vert {\bf x} \vert B \rangle|^2. $$

This formula is derived via quantum mechanics, but it is closely related to the classical Larmor formula for the power radiated by a charge e undergoing harmonic oscillation of frequency omega and amplitude l:

$$P=\frac{2}{3} \frac{e^2 \omega^4 \ell^2}{c^3}$$

Although this formula applies to a charged particle undergoing harmonic oscillation, apart from the prefactor 2/3 the same formula applies for a charged particle in a circular orbit, where l is the radius and omega is the frequency of the orbit.

The atom is a little antenna, for radiating electromagnetic waves. Now there are two kinds of antennas, small ones and big ones. An antenna is small if its length is small compared to the wavelength of the radiation, otherwise it is large. An atom is a small antenna, since its size is about a thousand times smaller than the wavelength of the light emitted (in simple electronic transitions). And the Larmor formula also only applies to a small antenna, where
$$\ell \ll \lambda$$

Here lambda is the wave length,

$$\lambda = 2\pi \frac{c}{\omega}$$

To bring out the dimensionless (and small) ratio
$$\frac{\ell}{\lambda}$$
we can rewrite the Larmor formula as
$$P= \frac{e^2\omega^2}{c} \Bigl(\frac{\ell}{\lambda}\Bigr)^2$$
where I'm dropping all numerical factors like 2/3 and 2*pi.

An important point is that the dimensionless ratio (l/\lambda) is the same as (v/c) for the charged particle undergoing harmonic or circular motion, apart from factors of 2, 2*pi, etc. So if you have a single charged particle radiating by undergoing some motion, circular or harmonic for example, the antenna will be "small" if the velocity of the particle is much less than the speed of light (the particle is nonrelativistic). Then these simple formulas for the power radiated apply.

Antennas that are not "small" are much harder to analyze, because at a fixed time the phase of the wave is not constant over the extent of the antenna.

Now this is electric dipole radiation. The power radiated by electric quadrupole radiation is similar, except the dimensionless ratio (l/lambda) is raised to the 4th power (and the numerical factors in front are different). That is, there is another factor of (v/c)^2 in the formula for the power radiated. So for electric quadrupole radiation, the power is

$$P= \frac{e^2\omega^2}{c} \Bigl(\frac{\ell}{\lambda}\Bigr)^4$$

The only thing that has changed is the exponent of (l/lambda) which is essentially (v/c).

Gravitational radiation does not exist in dipole form, the leading order multipole is the quadrupole. And if you take the formula above for the electromagnetic power emitted in quadrupole radiation and just replace e^2 by GM^2, for two equal masses M in circular orbit around each other, where G is Newton's constant, then you get the power radiated by gravitational waves:

$$P=\frac{GM^2\omega^2}{c} \Bigl(\frac{\ell}{\lambda}\Bigr)^4$$

Here l is the radius of the orbit, and if you reexpress lambda in terms of omega and c, this formula becomes

$$P = \frac{GM^2\omega^6 \ell^4}{c^5}$$

Here omega is the orbital frequency, and the antenna is "small" if the radius is small compared to the wavelength. But since gravitational waves travel at the speed of light, this means that the antenna is small as long as the orbital velocity is small compared to the speed of light.

For the orbiting black holes in the recent event, their orbital velocities are only a fraction of the speed of light up until the final stages of the merger. When the merger finally starts to take place and the event horizons are starting to merge, then the orbital velocities are becoming comparable to the speed of light and the gravitational quadrupole version of the Larmor formula, quoted above, ceases to be valid. But it works ok up until the final moments.

More later.

[Edited on 14-2-2016 by annaandherdad]

[Edited on 14-2-2016 by annaandherdad]

careysub - 13-2-2016 at 20:46

Quote: Originally posted by phlogiston

Refer to my posts above, and check out the Caltech link I included. An ordinary oscillating mass produces gravitational wave energy in the order of 10^-43 watts. But the LIGO detection detected a pulse with the intensity of 0.1 watts/M^2.

That is why, Detection takes a signal with a lot of energy, Ordinary oscillating masses produce incredibly small amounts of energy.

j_sum1 - 13-2-2016 at 21:30

I can think of another issue with an experimental setup involving conventional masses.

Imagine you had a large dumbell mass and were spinning it at a few thousand rpm to try to produce a gravitational wave. Your detector would necessarily have to be close. It would be near impossible to filter out the comparatively enormous effect of mechanical shaking that would be detected along with any gravitational signal generated.

[Edited on 14-2-2016 by j_sum1]

Marvin - 14-2-2016 at 04:48

The practical problem there is that you could potentially measure gravity, but unless you can spin the masses at a large fraction of the speed of light you'd never measure the delay caused by the distance.

careysub - 14-2-2016 at 10:22

Thanks annaandherdad, keep it coming!

annaandherdad - 14-2-2016 at 11:28

You're welcome. I edited the post above to make it more comprehensible. I'll add more later.

phlogiston - 14-2-2016 at 11:58

The Fermi gamma-ray space telescope recorded a gamma ray burst that arrived just 0.4 seconds after GW150914.

http://gammaray.nsstc.nasa.gov/gbm/publications/preprints/gb...

The region of sky where it happened also matches with that of the source of GW140914. It is very likely that the photons and the gravity wave were produced in the same event.

careysub - 14-2-2016 at 12:14

Quote: Originally posted by phlogiston

The Fermi gamma-ray space telescope recorded a gamma ray burst that arrived just 0.4 seconds after GW150914.

http://gammaray.nsstc.nasa.gov/gbm/publications/preprints/gb...

The region of sky where it happened also matches with that of the source of GW140914. It is very likely that the photons and the gravity wave were produced in the same event.

Holy Cow! We really are in a new era of high energy astronomy!

annaandherdad - 14-2-2016 at 13:19

Quote: Originally posted by phlogiston

http://gammaray.nsstc.nasa.gov/gbm/publications/preprints/gb...

This is fantastic and incredible news. The article is a bit disappointing, because it doesn't speculate much on how there could be an electromagnetic signal associated with the black hole merger. But that's the point---there was no reason to expect one, and that group just made the observation. It's hard to believe that the gamma ray signal is not associated with the black hole merger.

Black holes often emit x-rays, but this comes from the accretion disk, matter that is spiraling inward toward the black hole and heating up in the process so hot that it emits x-rays. As the article said, no one expected there to be accretion disks around these small black holes. (Small in the sense that they are not supermassive.)

Now there's this very interesting question that every theorist in the world has already jumped on, what could give rise to an electromagnetic signal when black holes merge?

annaandherdad - 15-2-2016 at 10:20

You can take the formula for the gravitational power radiated and play some games. You can plug in numbers, for example, for two pyramids of Giza orbiting around each other at a distance of 500m and a velocity of 1 km/sec.

Here is a slightly different approach. Suppose you have two lead balls of some mass M connected by a cable. The cable has some tensile strength, say, it's enough so that when the force on the cable exceeds 100 Mg, where g=9.8 m/sec^2 is the acceleration of gravity, the cable will break. In other words, the cable can support 100 times the weight of one of the balls before breaking. We'll ignore the mass of the cable. Then
$$\ell \omega^2 = 100g$$

Now with this constraint, do we get more gravitational power radiated if the cable is long or short? (The velocity of the masses is adjusted to make the tension on the cable just short of the breaking tension.) To answer this, we use the relation above to eliminate omega from the power formula, which gives,

$$P=\frac{GM^2}{c^5}(100g)^3 \ell$$

So longer cables, at constant tensile strength, generate more power. Suppose we are limited by the velocity of the lead balls, at v_max. Then

$$\omega\ell = v_{\rm max}$$

or, with the constraint on the tension,

$$\ell = \frac{v_{\rm max}^2}{100g}$$

so

$$P=\frac{GM^2}{c^5} (100g)^2 v_{\rm max}^2$$

When I plug in M=1000kg and v_max =100m/sec, I get a power of about 3 10^{-15} watts. Not encouraging for generating gravitational radiation via terrestrial sources.

By the way, when the gravitational wave passes through the earth, it causes small motions in the body of the earth. These in turn radiate their own gravitational waves, which are the scattered wave. Although the earth is very massive, the numerical factors are very small, and the scattered radiation is very small. In fact, gravitational waves pass through ordinary matter almost completely unscathed. This is part of what may make them useful in astronomy, they can be detected from places where light cannot get out.

One way to see why these numbers are so small is to express the power in terms of (v/c) of the mass, which is the same as (l/lambda). The quantity P/omega is the energy emitter per cycle; it is

$$\frac{P}{\omega} = \frac{GM^2}{\ell} \Bigl(\frac{v}{c}\Bigr)^5$$

It is the factor (v/c)^5 that makes this number so small for terrestrial radiators. The prefactor, GM^2/l, is the Newtonian gravitational potential of the two bodies at a distance of l.

By arranging a collection of radiators in a phased array, you can make the power radiated grow as the square of the number of radiators, but you still won't get much power in ordinary terms.

[Edited on 15-2-2016 by annaandherdad]

neptunium - 15-2-2016 at 16:39

No detection of GR would have been a major blow to general relativity and physics . this is indeed welcome news. The emission of light in the form of GRB is a curious phenomenon .
Maybe have its source in the quantum world. Large release of energy are often accompanied or generate light.I cannot wait to see what LIGO and VIRGO have in store waiting for the next generation of GW detectors! I've just listen to an interview with the LIGO team who said they could also "see" the ringing of the final black hole dissipating ... Isn't that amazing?

[Edited on 16-2-2016 by neptunium]