Talk:Gravitational wave/Archive 2

Back to the basics

Any good article in a popular encyclopaedia should develop its ideas in proper sequence, beginning with the basics and progressing to ever deeper levels of complexity. What this particular article needs in its introductory level is a very simple mathematical treatment of gravitational radiation. I would supply one if I could find one. Unfortunately, all the expert sources seem pre-occupied with extreme cases, such as black holes and massive binary stars in spiral collapse, which complicates the maths. Can anyone direct me to a site that deals with gravitational radiation from small masses? Yes I know that this modest kind of radiation is too small to be measurable in scientific experiments, but likewise it's not so extreme as to be incalculable for laymen. I have searched the web in vain for hours for some such simple treatment. Lucretius 09:04, 29 October 2006 (UTC)

The mass doesn't really matter, and having black holes actually simplifies the math, if anything. I think that what you're looking for is the radiation emitted by a simple binary, in the approximation that it isn't spiralling inwards because of its loss of energy in the form of gravitational waves. This is what I put in the "Sources of Gravitational Waves" section, describing the Earth-Moon system. I don't really know much about web sources, but the best book I can recommend for this sort of thing is "Gravitation", by Misner, Thorne, and Wheeler, Chapters 35 and 36.
I agree that this article is poorly structured, and wouldn't mind seeing it put in better order. I'll be glad to help with that, if you have a good idea for an outline. As an interesting note, Steven Spielberg's next movie will involve gravitational waves as an important part of the plot. I have no idea when it's coming out, but I'll bet anything that this page gets tens of thousands of views just because of the movie, so it'd be nice to see it in good shape. --MOBle 04:48, 30 October 2006 (UTC)

Thanks for this reply. I have concerns about equations like this one, copied from the text:

${\displaystyle {\frac {G}{c^{4}}}\,{\frac {1}{r}}\,\mu R^{2}\omega ^{2}\ll 10^{-33}\ .}$

Firstly, it seems to me that this equation could be written in more 'user-friendly' terms for a laymen. When I translate the factors such as 'angular frequency' and reduced mass, and then cancel out, I get this:

${\displaystyle {\frac {v^{2}}{c^{4}}}{\frac {GMm}{M+m}}{\frac {1}{r}}}$

The result is a dimensionless number. Speaking as a layman, I would rather see a calculation of the actual energy that is radiated. The metric stuff can come later in a more advanced level of the article. I think we should remember that people who turn to this article are probably looking for a basic explanation of gravitational radiation. The only people who could really understand the article as it is now are people who have little or nothing to learn from it - they know this stuff already. Anyhow, that's how i see it. Lucretius 08:28, 30 October 2006 (UTC)

Your translation of the equation isn't quite right: ${\displaystyle \omega R}$ is only the velocity if one of the objects is stationary. The Earth is roughly 80 times more massive than the Moon, so this is a reasonable approximation, but not exact, and certainly not general. I have made a few changes, though. I think the metric perturbation is a good way of thinking of things, if we point out that this is basically the fractional change in the size of anything through which the wave passes. Still, what I wrote there may be a little too precise. We could maybe get by with more approximations.
In any event, there's a whole lot of restructuring of the article to be done, and I think it would make more sense to do that first, then work out the details. The current "Characteristics" section needs to be totally rewritten, the "Derivation" section needs to be cut down and tidied up a lot, and the rest needs reorganizing. Probably the biggest problem with this article is that most of the individual sentences are right and many are relevant, but they're all in the wrong places. As a first stab at this, here is my suggested outline, please revise as necessary:
• Leader (I think the current one is pretty good)
• Introduction -- just a rough outline of the more "user-friendly" parts of the following article
• The effect of a passing gravitational wave
• Pictures
• Description of the basic effect
• Polarizations
• Sources of gravitational waves
• Discussion of general properties of sources
• The list of sources
• Radiation from the Earth-Moon system
• numbers for this system (masses, separation and orbital frequency)
• energy loss
• comparison of that energy loss to energy radiated by stars, or used by humans
• Gravitational wave detectors
• The general idea
• Laser interferometers
• LIGO, VIRGO, GEO, TAMA, LISA
• Einstein@Home
• Prospects for detection
• Mathematics
• A sentence or two about the metric
• The metric perturbation
• Einstein's equations
• Linearized Einstein's equations
• Wave solutions
• Generation of these waves by a source
• Simulations
MOBle 12:00, 5 November 2006 (UTC)

Thanks again for your reply. I think your overview looks very good. The article at present looks as if a committee worked on it and it certainly needs coherent management. You seem to know quite a bit about the topic and I for one am happy to sit in the back seat and enjoy the ride. I hope you find time to include an equation for gravitational radiation. Yes, the Einstein maths belongs at the end. Lucretius

Okay. I think I'll make this my new Wikipedia project. (Maybe simultaneously cleaning up PSR B1913+16.) I might be slow about it, though. --MOBle 00:37, 5 November 2006 (UTC)

Which Direction

In the sentence "Roughly speaking, they will oscillate in a cruciform manner, orthogonal to the direction of motion. First, east-west separated particles draw together while north-south separated particles draw apart, after which east-west separated particles draw apart while north-south separated particles draw together, and so forth." - what direction does "orthogonal" mean? The images are nice but does the radiation come from the viewpoint of the reader or from left/right/north/south? I think you better need an image in 3D. -- Nichtich 23:48, 3 November 2006 (UTC)

In the pictures, the wave is passing directly through the screen, either from behind or in front. We should try to make this clearer in the rewrite. I would welcome 3D versions of these pictures, if they didn't complicate things beyond understanding, but I'm not going to spend my time on it. --MOBle 01:08, 5 November 2006 (UTC)

massless particle waves

"Massless" must refer to infitesimal mass rather than no mass since E=M*c^2. Adaptron 11:44, 4 November 2006 (UTC)

Well, we don't really have any firm theory for the graviton, but we might by looking at linearized general relativity. If we assume that it travels at the speed of light (which we linearized gravity says it would), then the formula E=M*c^2 wouldn't apply, just as it doesn't apply to the photon. --MOBle 00:37, 5 November 2006 (UTC)

Under construction

Hi MOBle. I finally found a formula for gravitational radiation (power) and maybe you could include it in your reconstruction of the article. I took it from here [1] The formula is as follows:

${\displaystyle {\frac {32}{5}}{\frac {G^{4}}{c^{5}R^{5}}}(Mm)^{2}(M+m)}$

Using

${\displaystyle \omega ={\frac {2\pi }{T}}\approx {\sqrt {\frac {GM}{R^{3}}}}}$

and

${\displaystyle E_{kin}={\frac {1}{2}}mv^{2}\,and\,v=\omega R}$

one can rewrite this formula as

${\displaystyle {\frac {128\pi }{5}}{\frac {m}{M}}\left({\frac {v}{c}}\right)^{5}{\frac {E_{kin}}{T}}.}$

Now, it is quite obvious, that the energy radiated is quite negligible compared to the kinetic energy of the Earth. --84.59.143.136 10:59, 28 June 2007 (UTC)

I don't know how this was derived but I'm hoping you'll know a derivation that is user-friendly for laymen. Is there for example a derivation based on the Lamor formula for electromagnetic radiation?:

${\displaystyle {\frac {2}{3}}{\frac {Kq^{2}}{c^{2}}}{\frac {a^{2}}{c}}}$

where K is Coulomb's constant, q is charge and a is acceleration. Using this as a model, I translate the gravitational radiation formula as follows:

${\displaystyle {\frac {32}{5}}{\frac {GM}{R^{2}}}{\frac {GM}{cR^{2}}}[{\frac {Gm^{2}}{c^{2}}}{\frac {G(M+m)}{c^{2}R}}]}$

Here m is the mass that radiates and M is the mass that imposes a speed on it. The factors are acceleration, acceleration divided by c and, in the square brackets, is the stuff left over, which appears to be the gravitational equivalent of ${\displaystyle {\frac {Kq^{2}}{c^{2}}}}$ and which features a ratio gravitational radius/R. I would be surprised if this is an acceptable derivation, but it gives you an idea of the sort of thing an amateur like me can relate to. Hopefully you can find something along these lines. Lucretius 04:36, 5 November 2006 (UTC)

If the radiated power really depends on the acceleration, regardless how a particle is accelarated, one may accelerate a charged particle by electromagnetic forces, since much higher acceleration can be achieved. Indeed, a highly accelerated electron might emit a considerable amount of its kinetic energy as gravitational radiation. 84.59.58.38 08:33, 29 June 2007 (UTC)
At least equal-mass black hole binaries will radiatiate their kinetic energy rapidly, if their orbit veleocities reach some percent of the speed of light. But, can a black hole really radiate gravitational radiation ? --84.59.133.121 10:45, 2 July 2007 (UTC)
I've never seen a derivation by analogy with the Larmor formula. That's nice. I would have expected a factor of two to come out of nowhere, though. I'll look into it.
I guess, if you like the energy idea, then looking at the energy given off by PSR B1913+16 would be the most interesting thing to do. Despite having a separate page for that binary, it would be good to put something here about the mathematics, for continuity. I've added a little about this to the outline above. --MOBle 00:37, 5 November 2006 (UTC)

Closer to home, if M is the Sun, m is Earth and R the mean distance between them, the radiated energy is approx 200 Watts. That's a couple of lightbulbs worth of radiation. This would be a fine example of the weakness of gravitational radiation relative to the gigantic masses involved.

Regarding a factor of 2, forgive my ignorance but I'm not sure if you mean it should or should not appear out of nowhere. There are these possibilities - if the Earth radiates 200 Watts, then the Sun might radiate another 200 Watts under the Earth's gravitational pull. Or if m=M, then 2 naturally appears. But perhaps the entire system is supposed to radiate 200 Watts, which I suppose would be radiated from the centre of mass. The location of the source of radiation is surely an important issue and maybe you could touch on that also. Or maybe it's a non-issue. I have no idea. Lucretius 02:13, 5 November 2006 (UTC)

You're right. Discussing the Earth and Moon would be more effective on this page, and I'll just put in some discussion on the PSR B1913+16 page about that system. --MOBle 12:00, 5 November 2006 (UTC)
Why, the Moon's orbit velocity around the Earth is quite small, but the moon Io of Jupiter is much faster.

In the section 'Effects of a passing gravitational wave', there is no mention of how quickly the pulse occurs to the ring of particles. If I understand you correctly, this pulse is 'linearly polarized' and the wave travels at the speed of light. In that case, the frequency of the pulse is determined simply by the length of the wave or the pulse, isn't it? The smaller is the length, the more rapid is the pulse - and the higher would be the energy? This 'less is more' quality is typical of an electromagnetic wave, whereas I had thought gravitational lengths were a case of 'more is more'. That's another 'layman' issue that could be cleared up. Gee, I'm glad I'm not writing this article! Lucretius 05:39, 5 November 2006 (UTC)

Also, what polarizes a gravitational wave and why is the angle of polarity different to that of an electromagnetic wave? Is there a simple explanation for this polarization? I hope so. In return for your knowledge I offer you my ignorance, which I think is fairly typical of the general reader and which the article should try to address. Asking questions is hard work, though not so hard as answering them, I guess. Feel free to ignore me. Lucretius 06:30, 5 November 2006 (UTC)

I'll try to answer these questions in the article, except for the polarization angle question. This is actually a deep consequence of the "spin" of the gravitational field. (Its spin is 2.) Now, this is an interesting and important fact, but I'm not sure how to explain this, because there are lots of issues. By spin here, I don't actually mean quantum spin. However, the quantum spin of the graviton is 2, and is a direct result of this type of spin. Anyways, it's complicated, and I don't know how to explain it in simple terms, so I'm not going to try just yet. Maybe there should be a section on quantum theory and gravitational waves.
These are good questions, and this is exactly the kind of interaction that will hopefully make this a good article. The ability to ask questions and pursue the answers is more important than the ability to answer them. Thanks, and keep 'em coming. --MOBle 12:00, 5 November 2006 (UTC)

Hey Lucretius. Now, the reason most of us go into theoretical physics is because we're no good with actual numbers. I checked to make sure that your formula for radiated power is right, and I think that 5 should be a pi. Also, I plugged in the numbers, but I get ${\displaystyle 1.162\times 10^{-5}}$ Watts radiated. Could you check your math again? I'm not saying that I'm right, but I don't see my mistake.

32*G^4/c^5/pi

(M1 M2)^2(M1+M2)

1/R^5

--MOBle 23:11, 5 November 2006 (UTC)

Hi once again MOBle! You must be flat out trying to get too much done because the error in this case is yours. Firstly,the tell-tale sign of too much work is a simple reading error - you'll notice that I mentioned Earth and SUN, not moon. The numbers needed are:
R = 1.5 x 10^11 m
Sun = 2 x 10^30 kg
Earth = 6 x 10^24 kg
Punch those into the formula and the formula will jab you back with approx 200 watts.
Regarding Pi instead of 5, don't ask me. I don't know. However, the site I got it from (linked above) definitely says 5. My expectation is that 32/5 has something to do with angles, same as the 2/3 in the Larmor formula, but I could be wrong.
Regarding people getting into theoretical physics because they make simple errors in maths (sorry, math), phooey! They get into theoretical physics because they are good at maths. But maybe they get so used to complex problems that their basic maths gets a bit rusty in the process. However, in this case your error was in literacy, not maths. Sorry, my error this time - MATH! Lucretius 06:46, 6 November 2006 (UTC)
By the way, in my Larmor translation of gravitational power, these 2 things have the same units kg.m and obviously refer to each other:
${\displaystyle {\frac {Kq^{2}}{c^{2}}}}$
${\displaystyle {\frac {Gm^{2}}{c^{2}}}}$
The following bit can be understood as a ratio of half the Scwharzschilde radius to the distance R, but better still it can be understood to refer to the metric thingy:
${\displaystyle {\frac {G(M+m)}{c^{2}R}}={\frac {v^{2}}{c^{2}}}}$
where v is a speed derived from the total mass of the two bodies separated by the distance R. Apart from this metric thingy, everything in the Larmor formula has its equivalent in the gravitational formula, as far as I can tell. But the question is whether or not the system radiates this energy, or is the source either M or m separately. The exact arrangement of factors depends on what is doing the radiating. This scruple about the origin of radiated energy is relevant to the article (my way of arriving at it via Larmor is quite idiosyncratic and of course should not feature in the article, unless there is a reliable source that makes the same connection). Lucretius 07:34, 6 November 2006 (UTC)
Okay. Now I agree, but for that factor of 5/pi. Kip Thorne says it's pi, so I say we go with him.
Gravitational waves are -- just like E&M waves -- created by the entire field, rather than just the point masses -- or point charges. Even without a solid footing for deriving the equation in analogy to the Larmor formula, it might be useful to draw the analogy just for the sake of familiarity. I'll leave this up to you. I can certainly put a little about the power formula in the maths [ ;) ] section.
As for getting too much done, don't worry -- I only work on these things in spurts. (I'm actually going to LIGO for a week in a couple days, so I'm guessing I won't have any time to do Wikipedia stuff for a while.) --MOBle 08:09, 6 November 2006 (UTC)

Hey MOBle - you've made some wonderful changes, particularly to the section 'Effects of a passing gravitational wave'. The changes bring that section to life and make full use of the given graphic. Even I understand it! The Sun-Earth section is also very good - I admit to being a bit scared by trigonometry but the formula for power is intelligible even to a determined maths simpleton such as myself. Cheers Lucretius 08:39, 6 November 2006 (UTC)

Thanks. Very nice of you to say. I think this interaction is helping the article a lot.

Something else: in the section about waves from other sources it is said of two inspiralling stars that "...their orbit is about 75 times smaller than the distance between the Earth and Sun — which is actually smaller than the Sun itself". I don't understand this -does it mean that the diameter of the Sun is greater than is the distance between those other 2 stars? This would imply that the Sun's diameter is nearly 10^11 meters (ie nearly the distance between Earth and Sun). That's an amazing statistic if it's right. But I'm sure it must be wrong. As you said in the 'passing gravitational wave' section, the distance R is very large and Sun and Earth are 'very small'. But the universe is full of wonders and maybe it's right. I'm not used to scaling things visually where exponents are involved and maybe that's the problem here. Less confusing perhaps would be this: The distance between these two stars is less than the diameter of our own Sun.

Good catch. I lost two factors of two. The distance between the stars is actually three times the diameter of the Sun.

I don't want to make waves but, at the risk of being tedious, can I suggest a further change to the Sun-Earth section? Is it possible to put the power formula before the trig stuff? I say this because I think many readers would be scared off by the trig before they actually get to the bit they can understand. The simplest parts should always come first wherever possible. Trig might be simple to you but it's a foreign language to 99.9% of the human race (this could be a conservative estimate). Lucretius 10:06, 6 November 2006 (UTC)

Check. I think the changes will also help lead in to why detecting them directly is so difficult. --MOBle 10:47, 6 November 2006 (UTC)

Hi, MOBle - yes this is better. The simple formula for power will encourage the reader to keep reading and the trig is then less discouraging. Also, after the trig, comes another pleasantly simple piece of maths, the formula for amplitude, which I think rewards the average reader for persistence. However, you appear to have left G out of the formula for angular velocity and that needs fixing. I won't fix it because I don't want to start fiddling with your math(s) - fiddling with math(s) is for me like a bag of salted peanuts and it's hard to stop once I get started, which could prove disasterous for the integrity of the article.

As an aside, I'm puzzled why they call angular velocity a velocity since it appears to be the inverse of a time. That's another good reason for me not to fiddle since maybe I have misunderstood something regarding definitions. Lucretius 07:21, 7 November 2006 (UTC)

Oh. Yet another nice catch. I always think in geometrized units, so G=c=1 to me. I had to do a lot of searching to find those factors of ${\displaystyle G^{4}/c^{5}}$ and such. (I have no idea what the mass of the Sun is in kilograms, but I can always remember that it's 1.477 kilometers.) Good work.
As for angular velocity, all you have to do is multiply by the distance from the rotation axis to get the regular velocity, ${\displaystyle v=\omega r}$. Angular velocity and angular momentum obey rules similar to the ones for regular velocity and momentum, too. There's even an angular mass -- the moment of inertia -- and the formula for rotational energy looks a lot like ${\displaystyle {\frac {1}{2}}mv^{2}}$. --MOBle 21:35, 7 November 2006 (UTC)

Thanks for this. Regarding your error, it came about because I've been asking you to put things in layman's terms, which you are not used to doing. Lucretius 07:33, 8 November 2006 (UTC)

Still under construction

Regarding my last suggestion, I've fleshed it out a bit. Here is my idea of what the section should look like (but it's just a draft and it's mostly a re-arrangement of your own words!):

In general terms, gravitational waves are radiated by objects whose motion (whether travelling, pulsing or spinning)involves acceleration, provided however that this same motion does not describe a perfect sphere or a disc. More technically and in accordance with general relativity, the quadrupole moment (or some higher multipole moment) of an isolated system must be changing in time in order for it to emit gravitational radiation.

The simplest example of a quadrupole moment changing in time is the spinning dumbbell, tumbling end-over-end (as opposed to spinning around its long axis). The heavier the mass, and the faster it's tumbling, the greater the gravitational radiation it will give off. If we imagine the two weights of the dumbbell to be massive stars like neutron stars or black holes, orbiting each other quickly, then significant amounts of gravitational radiation would be given off.

Some more detailed examples:

• Two objects orbiting each other with angular frequency ${\displaystyle \Omega }$ in a quasi-Keplerian planar orbit, have a time-varying quadrupole moment, so this system will radiate.
[Observers far from the system and in its equatorial plane will observe linearly polarized radiation (aligned with the rod) with frequency ${\displaystyle \nu =\Omega /\pi }$. Observers far from the system and lying on its axis of symmetry will observe circularly polarized radiation].
• A spinning non-axisymmetric planetoid (say with a large bump or dimple on the equator) will define a system with a time-varying quadrupole moment, so this system will radiate.
[Observers far from the system and lying in the plane of rotation will observe linearly polarized radiation. Observers far from the system and near its axis of symmetry will observe circularly polarized radiation].
• A supernova will radiate except in the unlikely event that it is perfectly symmetric.
• An isolated object in "rectilinear" motion will not radiate. This can be regarded as a consequence of the principle of conservation of linear momentum.
• A spherically pulsating spherical star (non-zero monopole moment or mass, but zero quadrupole moment) will not radiate, in agreement with Birkhoff's theorem.
• A spinning disk (nonzero but stationary monopole and quadrupole moments) will not radiate. This can be regarded as a consequence of the principle of conservation of angular momentum. On the other hand, this system will show gravitomagnetic effects.

Lucretius 10:50, 9 November 2006 (UTC)

Incidentally, does a 'disc' refer to a particular type of galexy, or is it a purely imaginary object that meets the needs of a mathematical argument? Lucretius 10:57, 9 November 2006 (UTC)

Here, the disk just refers to a geometric object. A disk galaxy isn't a perfect disk, but still wouldn't really radiate much (as a galaxy) because the speeds are too low, and everything's basically Newtonian.

Hi MOBle. Yes, the latest change to 'Sources' makes good sense to me. I'm a bit sorry to see your explanation of polarization WRT angles left out and maybe this could be fitted in elsewhere. In fact, it belongs in 'Effects of a gravitational wave', perhaps as a sub-subsection titled 'Types of gravitational waves'. Another graphic would be useful in that context, showing 2 inspiralling stars and the points at which an observer would detect different types of gravitational waves.

I've made some cosmetic changes to paragraph spacing throughout the article. Scientific text is less intimidating if it comes in discrete chunks (the layman's brain is quantized!). In the section titled 'Prospects' the conclusion is somewhat mystifying. What does this mean? "By directly studying the details of gravitational radiation given off by these systems, astronomers could potentially learn much which they would not be able to learn from electromagnetic radiation." Lucretius 01:21, 12 November 2006 (UTC)

Your changes look good to me. You thought the sentences about polarizations in the bullet list were good? I don't think I put those in. I just removed them because I thought they broke up the flow and maybe caused confusion, but I would have no problem with them being replaced there.
Also, I still haven't worked on anything at or below the Einstein@Home section. I think that Prospects section is yet another good, though misguided part of the old article. Looking at it now, it occurs to me that it might be a good start for a new main section on the Astrophysics we might hope to learn about with gravitational waves.

Here's an interesting scruple, though I'm not sure it is entirely relevant to the article. I've read that electric charges in a gravitational field (such as the Earth's) should radiate electromagnetic energy because they are accelerating - we should have free electrical power in huge quantities! The fact that charges do not do radiate in a gravitational field is considered a conundrum among some physicists. It occurs to me that the same conundrum applies to gravitational radiation. Imagine a gigantic junkyard in space, not accelerating either linearly or orbitally. It should not radiate gravitational energy. However, the discarded fridges, pots, pans and car tyres that make up the junkyard are all individually accelerating within the collective gravitational field, and therefore these should radiate. Therefore the junkyard should radiate. Is there a scientific explanation for this conundrum? Maybe it's not a conundrum at all - the junk isn't really moving anywhere and therefore it isn't 'really' accelerating. Maybe the same could be said of electrical charges in a gravitational field. I don't know. The conundrum is out there on the web but it could be misinformation.Lucretius 07:20, 12 November 2006 (UTC)

I like to think of electromagnetic radiation as occurring when the charge accelerates with respect to the electromagnetic field in some sense. Now, imagine you have a free charge in a gravitational field with no outside E&M field. Of course, we know that the gravitational field is just a curvature of spacetime, and the charge will just be moving along a geodesic. But then, so will the E&M field — they'll be accelerated along with each other.
An electron sitting in a trash heap, of course, isn't a free charge, but it and the surrounding E&M field have reached an equilibrium so that the trash heap and the planet it rests on don't collapse. This obviously isn't the whole story, but it's close enough, I think. The combination of classical gravity with classical electromagnetism is pretty well understood (where "classical" means: ignore quantum mechanics). --MOBle 13:59, 12 November 2006 (UTC)

Thanks for this. The stuff I read about charges radiating in a gravitational field must be misinformation. However, I'm still puzzled about the gravitational parallel - shouldn't the pots and fridges radiate in the gravitational field of the junkyard, even if they are stationary relative to each other? Therefore the junkyard should radiate. Anyway, what are the pots and fridges accelerating relative to? Is the gravitational field accelerating past them? That doesn't sound right. But they have weight and therefore they must be accelerating. All very curious.

As for polarization WRT angles, I agree that it needed to be removed from the section 'Sources' but I do think there is a place for it in 'Effects of a grav wave', preferably with a simple diagram to eliminate the need for many words. I don't have the computer nous to come up with such a diagram but maybe you do. Wave types is an aspect of the subject that is new to me and it seems quite fascinating. And clearly it is relevant since some waveforms (eg linearly polarized) would be harder to detect than others. The important thing is to explain it simply and concisely so as not to clog the flow of the article, which is now beginning to run quite smoothly.

I'm looking forward to the Astrophysic section you are contemplating. I'm also wondering what can be said about quantum gravitational radiation. There are also topics such as Hawking radiation - that's not really grav radiation but it is gravitationally induced radiation. It's up to you whether or not to pursue those lines of enquiry. You might feel there is enough to do already. Lucretius 07:27, 13 November 2006 (UTC)

Hey Lucretius. I'll think about any good way to illustrate polarizations. This topic really is very analogous to the polarization of light. Hawking radiation isn't really closely related enough to be in this article. On the other hand, it might be useful to provide a link for the main article in the leader like the link about gravity waves.
I've changed the intro a little. I think this is far more approachable than the old one. This section, however, is the one you'd be entirely suited for. Feel free to rip it apart and make any changes you want. I still haven't touched anything in or after the (new) Astrophysics section. I don't know if that will turn into anything interesting. The other thing that obviously needs to be done is a total rewrite of the Mathematics section -- though that should be straightforward -- and probably a deletion of that energy section at the end. Also, I've seen interesting treatments of the graviton in the context of gravitational waves. Maybe I could squeeze that in near the Math section. I think everything else is looking pretty good, though. --MOBle 09:10, 13 November 2006 (UTC)

Sounds

It might be useful to explain the different types (not polarizations) of sources in terms of the different "sounds" their grav. waves would make. Because grav. waves have an amplitude and frequency, just like sound waves, we can consider what they would sound like if the amplitude were high enough. (The difference in speeds doesn't really matter.) If the wave's amplitude were high enough, the output of LIGO could be plugged into a speaker and we could literally hear it. In fact, in the LIGO control room a speaker that actually is hooked up to the output (though not actually in the expectation that a signal will be heard).

For example, a spinning neutron star -- like what Einstein@Home is looking for -- would have the sound of a pure tone. It might be nice to play the tone expected from the pulsar in the Crab Nebula, for example. Merging binaries, however, would have totally different sounds, where the pitch increases to some maximum, and then there's a brief pop, and it's all over. The difference might be illuminating.

I think this might fit nicely into the article, with a little blurb about comparing grav. waves with sound waves just after the bullet points about amplitude, frequency, etc. I've put in some commented text about this. That could then tie into some follow up in the section about sources, where we could link sound files next to the relevant source.

I can make something like the sounds on this page in OGG format, if this would be helpful enough to be worth the trouble. --MOBle 09:10, 13 November 2006 (UTC)

This sounds like a great idea. Unfortunately there's something wrong with my audio and I don't know what the sounds sound like. But I'm sure they sound great. Laymen will love it. Lucretius 09:46, 13 November 2006 (UTC)
The intro is crystal clear, excites interest and needs no input from me. I found a split infinitive ('to directly measure') which I corrected by a simple expedient (I deleted 'directly'). Split infinitives are being used more frequently these days, even by quite respectable news agencies, but they are inelegant and just plain wrong. I speak as a school teacher. [:}] Lucretius 10:01, 13 November 2006 (UTC)

Perpetual motion? (Bear with me)

I recently added the following to the Perpetual motion:

• Global violation of the first law of thermodynamics through cosmic expansion and redshift: Though the laws of physics make a local violation of the first law impossible, the energy lost when a photon is redshifted due to the expansion of space does not go anywhere apparent. It may be that the reverse process is possible: create a contraction of space, possibly using gravity waves (which are, so far, only theoretical) and fill that space with photons that are then blueshifted. This may well be impossible, depending on the as-yet-undetermined nature of gravity waves. If it were possible, a machine to take advantage of it might need to be be truly enormous, using for example supernovae as the source of gravity waves. Still, unlike the previous example, the scale involved would not be greater than a star cluster and would not tend to destroy (or create) anything at the galaxy scale or beyond.

Obviously, I realize that this needs work. I know that the basic premise is solid - cosmic expansion violates the first law, destroying energy through redshift, and there is no a priori reason to be certain that the reverse could not happen. However, it may well be that the proposed mechanism (using gravitational waves) is unsound, and that some process I do not understand would mean that the apparently "created" energy would actually just be bled off from the wave itself. If this is true and well-understood, please create a section here to explain it. (Either way, the sentence above that starts "This may well be impossible" is just my way of trying to hedge on this issue while still sounding authoritative and definitely needs to be replaced by something less star-trek-y.) Any help or even response here would be appreciated. --Homunq 23:25, 14 November 2006 (UTC) ps. I also included a related entry in List of unsolved problems in physics.

Hi Homunq. You are putting forward personal speculation and it doesn't belong here. Your basic premise is wrong - energy isn't destroyed by redshift: it's simply spread out more thinly. You could argue that it is converted into work. But it isn't destroyed. Lucretius 04:11, 15 November 2006 (UTC)

1. You might want to look at Peacock's "Cosmological Physics". Basically, you need to consider the total energy of a system -- both the photon (stress-energy tensor) and the gravitational field (which you neglect, though I don't honestly know of any good treatment of this for an expanding universe). More basically, Lucretius is right.
2. If you devised a way to take energy from a grav. wave and put it in a photon, you would still have the problem of creating the gravitational waves. This would require energy, so you wouldn't have a violation of the first law.
3. You need to have reliable sources for anything you put on Wikipedia. It's good of you to add things you find interesting, but you need to be able to point to an article published in a peer-reviewed journal, or a book from a credible publisher. Unless you have those for these remarks (which I doubt), you should probably revert your edits. (Of course, if you do have them, you should have cited them in the articles anyway.)
--MOBle 11:23, 15 November 2006 (UTC)
Consider this fact also, Homunq: if redshift breaks the first law then so would a police siren. Do you seriously think the police would break the law? Also I don't understand the distinction you make between 'local' and 'global'. Light comes from definite locations. It does not come from the entire universe. I have heard that entropy does not apply to the universe as a whole. Lucretius 12:50, 15 November 2006 (UTC)
The following discussion is an archived debate of the proposal. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the debate was PAGE MOVED per discussion below. There was some old page history located at the target, which is now merged in. -GTBacchus(talk) 07:23, 25 November 2006 (UTC)

Requested move

Gravitational radiationGravitational wave — The article is primarily about the waves, rather than the radiation they carry with them. The title should really deal with the main content of the article. Currently, most pages link to "gravitational wave" or "gravitational waves", which just redirects to "gravitational radiation". MOBle 19:11, 18 November 2006 (UTC)

Survey

Add  * '''Support'''  or  * '''Oppose'''  on a new line followed by a brief explanation, then sign your opinion using ~~~~.

Discussion

Hi Moble. Do you mean you intend moving sections of the article that deal with waves to a new article 'Gravitational Waves'? If so, yes this would make sense to me. Lucretius 08:04, 20 November 2006 (UTC)

BTW, what would you then do with the section 'Wave amplitude from the Sun-Earth system'? Lucretius 08:12, 20 November 2006 (UTC)

Hey Lucretius. No, I've requested that an admin move the whole thing to "gravitational wave", just so the title is really the main content of the article, and there are fewer redirects. The "radiation" name refers to energy carried by the waves (in some sources, also to the momentum and angular momentum), whereas the article is clearly mostly about the waves themselves. This is just some cleanup. The request is found on this page. --MOBle 16:04, 20 November 2006 (UTC)

Yes I see no problems at all with that move. Don't know if anyone else is following the great work you are doing here and I hope that doesn't discourage you. Anyway, you have my vote for this change. Lucretius 08:23, 21 November 2006 (UTC)

The above discussion is preserved as an archive of the debate. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

Thanks for the support, Lucretius. It was a minor detail, but I think it's good to keep things like that tidy. --MOBle 19:32, 25 November 2006 (UTC)

My opinion

A gravitational wave detector (such as the Weber bar, Laser Interferometry) in free fall cannot detect gravitational waves, just as it cannot detect its own motion in the plane of one of its laser arms. This has already been proven in the Michaelson-Morely and Kennedy-Thorndike experiments. Research such is this is like tossing federal or other grant money down a rathole.

The contraction these detectors purport to measure is Lorentz contraction, which could only be measured by a remote detector if instruments could be referenced to an inertial reference frame that is close to (or accelerating with) the source of the gravitational waves. In most cases, and for most purposes, this is impractical. To measure anything significant in terms of gravitational waves at all, you would need instruments that can measure distances between planetary size objects, such as the Earth and its Moon, while subtracting out tidal effects. A sophisticated gravity wave detector based on ranging to massive bodies, parked at one of the Earth-moon LaGrange points might be able to accomplish this, but no other research I have seen done in this field is worth even so much as a wooden nickel in terms of funding. —Preceding unsigned comment added by 70.106.60.44 (talkcontribs)

This is interesting but I assume either you are wrong or the issue is complex enough to be argued for and against (why would intelligent people throw money down an obvious rathole?).
You say that the detectors are in freefall. By this I understand you to mean that the detectors are accelerating in the gravitational field of the wave source and therefore they and their surrounds are all subjected to the same relativistic effects and these effects cannot therefore be measured locally. But the gravitational pull on the detectors has nothing to do with the radiation that is being measured (distance from the source excepted). Perhaps I have misunderstood your argument. Lucretius 03:01, 25 November 2006 (UTC)
Just a few comments to Dan(?):
1. The "unsigned" template is there instead of a signature. To sign, you can create a user account (click the link at the very top right of your page), and put four tildes (~) where the unsigned thingy is currently. Otherwise, you should just leave the unsigned template as is.
2. Whether or not gravitational waves are detectable -- even in principle -- was a subject of much debate years ago. The consensus in the scientific community was that they are detectable in principle. The issue you seem to be raising is whether or not they are strong enough to detect with things like LIGO. This is an issue that needs to be discussed with hard numbers, not claims and opinions. This is done by professional physicists in peer-reviewed literature. (One of the items on the to-do list is to include these references in the article.) Generally, this is not the place to decide such an issue.
3. There is an upcoming book on Classical Mechanics (everything but Quantum, so it includes Relativity) by Roger Blandford and Kip Thorne. You might be able to find it on Google, or buy it when it comes out. It deals with these issues with college-level math. You might find it interesting.
--MOBle 19:32, 25 November 2006 (UTC)

Recent edits

Hey Lucretius (and anyone else who's watching). I've made a bunch of (unfortunately sloppy) changes to the article. I may have screwed up some formatting (losing em dashes, for instance), and there's a lot more yet to be done. Overall, though, I think my changes were worthwhile, and have helped us along. Let me know what you think. Sorry for any screw-ups. --MOBle 19:32, 25 November 2006 (UTC)

I haven't noticed anything amiss. The point brought up by the other bloke - that g waves cannot be detected locally - and your response - that there has been debate within the scientific community about the existence or detectability of g waves - are news to me. The article should certainly find room for some mention of any such debate, preferably early in the article. I have to admit I don't understand the relevance of Lorentz contractions in the measurement of g waves, nor do I understand in what sense the detectors are assumed to be in freefall. In fact, I assumed this was all specious piffle. Lucretius 12:48, 26 November 2006 (UTC)
They're not in free-fall, and Lorentz contractions aren't particularly relevant. The debate about measurability was in different terms and decades ago, when nobody had really done any analysis. Specious piffle. But, I think it's good to point people to the appropriate sources, rather than getting in a debate that probably won't go anywhere. --MOBle 19:11, 26 November 2006 (UTC)

You're a gentleman and a scholar. Lucretius 08:24, 27 November 2006 (UTC)

Wavelength

There was a change in the definition of wavelength from "distance along the wave between points of maximum stretch or squeeze" to "time interval between points of maximum stretch or squeeze". I've changed it back because the original is precise, correct, and more familiar in analogy with other types of wave. The second version is how I would think of it, but then I worry that beginners would be confused about the difference between wavelength and period (inverse frequency). If there is disagreement, we can discuss it here. --MOBle 19:11, 26 November 2006 (UTC)

This looks good to me. Something else - in the section 'Astrophysics and gravitational waves' it is said that electromagnetic waves can be blocked out (eg by a cloud of interstellar dust) but that grav radiation will still get through. Surely that can't be altogether true. I imagine some gravitational energy would be converted into kinetic energy as it passes through the cloud. Surely therefore it is partially blocked and, in extreme cases, could be completely blocked out, same as EM waves. If this doesn't happen, then the pertinent thing about g waves is that they are not converted into other forms of energy. It's a riddle to me. Lucretius 09:19, 27 November 2006 (UTC)
A further thing: I think this description from 'Effects of a passing g wave' is really unnecessary. I quote: "First, east-west separated particles draw together while north-south separated particles draw apart, after which east-west separated particles draw apart while north-south separated particles draw together, and so forth." The animation makes this description unnecessary. Moreover the description makes me go cross-eyed as I try to sort out the convoluted repetitions.
Feel free to brush up the prose. I'm certainly not the best judge of that stuff. I'll keep an eye on accuracy.
WRT the earlier point about g waves not being converted into kinetic energy, I guess the logic is this: particles aren't really moving with the wave but instead the space they are in is expanding and contracting. That does make g waves unique and it deserves comment in the article. I don't know what else the logic might be.Lucretius 10:18, 28 November 2006 (UTC)
In the simplest picture, we can think of dust. (In this context, this means that we assume there is no interaction between grains of dust.) As the g wave passes through the dust cloud, the dust grains will start moving, which will convert grav wave energy into kinetic energy. However, that same energy will then be re-radiated as gravitational radiation, leaving the dust with (basically) no resulting motion. Of course, the effect of the grav waves on the dust is usually so extraordinarily tiny that the absorption and re-radiation doesn't really happen at any appreciable rate.
Another possibility is that the effect is strong, and whatever stuff the wave is passing through is really thick (viscous). In this case, there could be heating of the "stuff", which actually would steal energy from the wave. However, the "stuff" would need to be extremely viscous (like a planet or star), and would have to be extremely close to the source. I don't actually know any numbers for this, but my guess is that it would have to be basically next to an inspiralling binary.
On the other hand, electromagnetic radiation is pretty easy to stop. Dust can do it easily. There was a time billions of years ago when the Universe was in such a state that E&M waves couldn't really get anywhere. Gravitational waves passed right through the Universe at this time with no problem. --MOBle

About the wavelength thing... you're obviously right. I don't know what I was thinking.--Laur2ro | Talk 10:34, 28 November 2006 (UTC)
No problem. We all have the occasional brain fart. --MOBle 21:45, 29 November 2006 (UTC)

One more scruple. If gravitational energy is radiated, what is the reservoir of energy that is being depleted? I assume the energy that is radiated away must be the energy of the gravitational field. However I have never seen a formula for the energy contained in a gravitational field. I assume this energy must be proportional to gravitational potential energy. I really have no idea about this. This is a question that ought to be addressed somewhere in the article because inevitably it must arise in the reader's mind. In my case, it's been at the back of my mind until now, like a nagging sense of some unfinished business. I have heard of something called 'the energy of position' but this was in the context of the Earth's gravitational field and the energy was the energy needed to lift something against Earth's gravity. In the case of 2 orbiting stars, neither star has been lifted out of the other's gravitational pull, so where does the energy of position come from in this context? I suppose it originally came from The Big Bang. It had to come from somewhere. Lucretius 12:10, 29 November 2006 (UTC)
Basically, the energy is coming out of the kinetic energy. This is why a binary system "inspirals": it needs energy to keep moving; when it loses that energy, it falls down.
What you call 'energy of position' is normally called potential energy. One way of calculating potential energy is to calculate how much work it takes to raise one object off of another. We could imagine taking one star, moving it to the surface of the other, and then raising it back up. By measuring how much work it took to raise it would give us the potential energy.
Another way is to calculate how much work could be done by one object falling towards another from infinitely far away. (There's a slight difference between the two, but it's just a constant number and we can ignore that.) This is more similar to what we might think of with the binary. As one star approaches the other from infinitely far away, its potential energy gets converted into kinetic energy.
I'm putting a little discussion of this in the section on radiation from the Earth-Sun system. --MOBle 21:45, 29 November 2006 (UTC)

Thanks for all this. I'll remove the north-south-east-west stuff. The article should include your brief explanation of a g wave passing through a stellar cloud. I hope you'll supply that. I'm still a bit puzzled about the energy source of radiation. If it comes out of kinetic energy, a radiating object should gradually slow down, which clearly does not happen (eg in the case of an inspiral kinetic energy actually increases). The potential energy explanation also still puzzles me since it clearly represents a form of stored energy. Stored in what or stored where? I can only think it must be stored in the g field. I look forward to your explanation in the article and I'll let you know if it silences my doubts.

At the risk of seeming to quibble, I'll add these further scruples. 1) You say energy of a g wave is converted into kinetic energy and back again into a g wave. This is conceptually neat but I'm puzzled about the practicality of it. If the transition to and from kinetic energy is instantaneous it might as well not occur at all. If it is not instantaneous, then how long does the transition take? 2) You say we can calculate potential energy by imagining lifting one star out of the gravitational pull of the other star. But the work in this context is imaginary and it can't be the real source of the radiated energy. I'm not sure if there are answers to these 'quibbles' and I suppose there is no place for them in the article. Grav waves make a fascinating topic, that's for sure! We should try to reference some scholarly articles that identify some of the problematic aspects of gravitational waves. Lucretius 12:44, 30 November 2006 (UTC)

I've added the explanation of propagation of grav. waves through matter to the To-Do list. In some sense, the potential energy is stored "all around us". (Channeling of Yoda is unintentional, here). That is, it's stored in the geometry of spacetime. So, in that sense, the potential energy is stored in the grav. field. That's the best way I know of to understand it.
All of this discussion of kinetic versus potential energy is a little hand-wavy because energy isn't as well defined in GR as it is in Newtonian mechanics. But, I think there is some validity in saying that the grav. radiation energy comes from the rotational kinetic energy of the binary, which allows the binary to draw closer together, which takes potential energy and converts it back into rotational kinetic energy. In this sense, I think it is valid to say that the grav. wave energy comes from kinetic energy. I don't know of any better way to put that without an advanced treatment of GR.
1. Again, this is an artifact of the fact that things like motion aren't perfectly well defined in GR. I guess the most self-consistent way to think about it is to say that the transition is instantaneous. Of course, the resulting wave won't necessarily be exactly the same wave as we had before it encountered the dust.
2. The work of raising turns out to be exactly opposite to the work of lowering. (No surprise.) Imagine we have a binary, and calculate how much energy would be needed to separate it by twice as much. Now take that binary that's separated by twice as much, and calculate how much energy would be given off by lowering it to the original separation. The two energies are equal and opposite. Now, my claim is that calculating potential energy by way of "raising from the surface" is equivalent to "lowering from infinity", except for a constant (which doesn't matter for technical reasons). This "lowering from infinity" energy is certainly not imaginary.
Grav. waves certainly do form a rich subject, but I think dealing with most of these issues in the article would make it way too big, though. --MOBle 07:01, 1 December 2006 (UTC)

I still have some quibbles but they are not really relevant to the work you are doing here. I agree that any explanation should be kept brief in the article. I hope you can think of a concise way to put the explanations you've put here, though I suppose they could just as well be put in another related article and not really feature in G Waves. The work is yours. I'm enjoying the opportunity to help in my own way while learning from your comments. Cheers. Lucretius 07:55, 1 December 2006 (UTC)

Momentum

Hi again Moble. I've been looking at the section about momentum. I quote:

"Taking the example of a binary black hole system, the system will be giving off energy and angular momentum, thus bringing the binary closer together. The waves can also carry off linear momentum, which will leave the binary with a net velocity relative to its inital velocity just like the usual third law of motion. This could have important consequences for Astrophysics, as a binary black hole in a star cluster might be shot out from it while merging, for example."

1) If the waves carry off linear momentum a radiating binary must lose linear momentum, which suggests the phrasing should be: "a smaller net velocity". Either that or you are using technical terms that aren't in my layman's vocab.

2) I assume the linear momentum has nothing to do with the orbital motion of the two black holes but rather concerns the movement of the whole system from point A to point B. Is this the right assumption? If so, it is not clearly worded and it leaves laymen such as myself scratching their heads somewhat.

3) Does the example have to be a binary black hole? Couldn't it be any binary system? If it is true of any binary system it is better to use this as the more general case.

3) The final sentence appears to refer to 2 black holes that merge into a single black hole while they are being 'shot out' of a star cluster. I'd like to know something of the mechanics of this process and I'm wondering if it can be explained concisely in layman terms. I think also the phrasing would be clearer if written more like this: "a binary system of 2 black holes could merge into a single black hole while being shot out from a star cluster."

According to my interpretation of the section the whole paragraph could be better worded as follows:

"Gravitational waves can also carry off energy, momentum and angular momentum. Thus a binary system inspirals because its gravitational waves carry off energy and angular momentum. If the system is travelling from point A to point B, the waves can also carry off its linear momentum, causing the binary to lose velocity in accordance with Newton's third law of motion. This could have important consequences for Astrophysics - a binary system of 2 black holes could for example merge into a single black hole while being shot out from a star cluster." Lucretius 10:24, 3 December 2006 (UTC)

It looks like you pretty much understand what I meant. Sorry that my passage was unclear. Looking at it now, I realize that it wasn't particularly well written. Just to (hopefully) make it a little clearer: Imagine a cluster of stars, with a binary black hole system in the center. The holes orbit each other, but their center of mass doesn't move with respect to the cluster at first. However, as the binary inspirals, the gravitational waves it gives off will carry linear momentum away in some direction. In keeping with Newton's third law, the binary will gain some linear momentum in the opposite direction. Thus, it may be shot out of the cluster.
It is true that this should apply to any type of binary system. I used black holes because they are "cleaner" and because the effect would be strongest for black hole systems. I don't like the phrase "traveling from point A to point B". Maybe the idea could be explained just in terms of this concrete example using the star cluster for a point of reference. Then, it could explained that the concept itself applies more generally. Personally, I think the example would help me learn by visualizing. I'll leave that to you, of course. If you want to take a crack at the revision, go for it. Otherwise, I could try to fix things up. --MOBle 18:17, 5 December 2006 (UTC)

BTW, I think 'radiated' momentum is better than radiated energy as a means of linking g waves with decreases in orbit. Typically angular momentum decreases with decreases in orbit, unlike kinetic energy. What the article needs therefore is a mathematical formulation for 'radiated' angular momentum. On the other hand, angular momentum per unit time is simply energy, which gets us nowhere. I don't suppose there can be a formula for radiated angular momentum which is not simply a restatement of radiated energy. As always, I'm groping in the dark. Lucretius 13:13, 4 December 2006 (UTC)

If radiated energy is angular momentum per square unit of time, the unit of time would need to be something fundamental. A period of some kind perhaps. Is there a formula for radiated energy along these lines? That would be a neat addition to the article as it would draw several threads together. Lucretius 09:32, 5 December 2006 (UTC)

I know there are formulas for radiated angular momentum. (Expressing them in terms of radiated energy would probably be pretty ugly.) I've never actually looked at these formulas for the case of a simple binary, but maybe they're easy enough to include here. They should be roughly as complicated as the formula for radiated power. I'll take a look. --MOBle 18:17, 5 December 2006 (UTC)

Once again many thanks for your comments. I'll have a go at revising the given paragraph, incorporating your own phrasing from your comments here, which are very clear. If you can find a simple formula for radiated angular momentum, that would be great. Cheers. Lucretius 07:20, 6 December 2006 (UTC)

WRT a formula for radiated angular momentum, an obvious possibility is a rearrangement for the formula for radiated energy. The factors merely have to be re-arranged to provide the dimensions of angular momentum divided by time squared. Thus:

${\displaystyle {\frac {G(M+m)}{R^{3}}}[{\frac {G^{3}(Mm)^{2}}{c^{5}R^{2}}}]}$

where the stuff in square brackets has the dimensions of angular momentum and the preceding unbracketed factor has the dimensions of an inverse squared time. The angular momentum is radiated in a time that is intrinsic to the system and in that case it has the dimension of energy. Dividing by the same intrinsic time then gives the dimension of power. Of course I wouldn't put this in the article as it can be considered a trivial rephrasing of the original equation. But it does show that a formula for radiated angular momentum need not be ugly when related to radiated energy. One ugly result however is this - angular momentum here increases with decreases in R, which is decidedly unconventional to say the least! But maybe this is how radiated angular momentum appears - the inspiral quickens and the radiation increases even as the non-radiated angular momentum decreases. I don't know.

The above formula is interesting when applied to the Planck mass. If M and m are both the Planck mass, and if R is the Planck length, the unbracketed factor is basically the inverse of squared Planck time, and the bracketed angular momentum is hbar. Which makes me think the formula is on the right track - you'll probably find something like it in some reliable text. Lucretius 12:03, 7 December 2006 (UTC)

In fact, the square root of the unbracketed factor is a form of what you call 'angular velocity' - the inverse of a time related to orbit (the square root of GM/R^3 is the time it takes any orbiting body to travel an arc equal to R at the gravitational speed imposed on it by M). So I wouldn't be surprised if the energy radiation formula is in fact derived from a formula for radiated angular momentum. That would make sense. Of course I've left out the 32/Pi factor and I suspect that relates to radiated energy rather than radiated momentum. This is all guess work on my part and you might yet prove me to be wrong. If I'm right and if you can find an authoritative text linking the formulae for power with the formula for angular momentum, the article could make the same link. That would solve the riddle of where the radiated energy comes from. Lucretius 08:40, 8 December 2006 (UTC)

In the simplest case where M=m, the bracketed angular momentum is the angular momentum of the system multiplied by (v/c)5. When v=c the angular momentum of the system equals the amount of 'radiated' angular momentum. This would occur when R=GM/c2 and by then the mass of the orbiting objects must have converted to energy. For a pair of Planck masses, the 'angular momentum' at this point would be hbar and it would be radiated away in Planck time (its radiation is simply Planck energy and this energy in turn would be radiated away in Planck time - that's one hell of an explosion!). For larger pairs of masses at v=c, angular momentum exceeds hbar and it would be radiated in a time that exceeds their Compton time (the power however would be the same as for the Planck masses). It's an interesting story but I can't say if it's a correct interpretation of the equation for radiated energy.

I've searched the web in vain for some hours looking for a simple formula for radiated momentum. Sorry I can't be of any help. I've found some sites but the maths looks like hieroglyphs to me. I hope you have more success.Lucretius 06:10, 10 December 2006 (UTC)

I did find this quote in an arXiv paper: "the loss rate of the energy and angular momentum is given by the quadrupole formula," page 4 here [2]. He then cites 2 texts - 'Gravitation' by Misner, Thorne and Wheeler, and a paper by L Blanchet, Living Reviews No. 4 2006. So that does give circumstantial evidence for an angular momentum restatement of the quadrupole formula for radiated energy. If only we could find a simple statement of it in an authoritative text!

BTW, I have found hieroglyphic formulae for radiated energy with Planck power as a term (c5/G), which is circumstantial support for my interpretation of the quadrupole formula above. But I can't find an authoritative replica of my rephrasing of the quadrupole formula in terms of angular momentum. The simplest way to understand the quadrupole formula is this - multiply gravitational potential energy by 'angular velocity' to obtain the dimension of power; square this then divide by Planck power. I can't see any relation to kinetic energy - are you sure there is authoritative support for kinetic energy? Lucretius 09:46, 11 December 2006 (UTC)

Hi Lucretius. Sorry I've been gone for so long. I have been (and still am) traveling. I know indisputable formulas for angular momentum given off in terms of the metric perturbation, so they certainly exist; the problem is that I don't know -- off the top of my head -- what they simplify to in the case of a simple binary. I can find it in the "quintessential" reference on this stuff, but Thorne says that there are errors in that paper. In particular, that factor of pi/5 we ran into earlier is a problem here again. In the paper, Peters says that the rate of loss of angular momentum is
${\displaystyle {\frac {dL}{dt}}=-{\frac {32}{5}}\,{\frac {G^{7/2}}{c^{5}}}\,{\frac {(M_{1}M_{2})^{2}(M_{1}+M_{2})^{1/2}}{R^{7/2}}}\ .}$

It's very unusual to be able to basically wiggle things to get the units right, and get the right answer. Nonetheless, that appears to be the case here. It appears to be a quirk of the simple circular binary; it doesn't work if the binary is elliptic at all, for example. In your analysis above, you separated the terms into a factor for the angular momentum, and a factor for one over time squared. Take the square root of that time-squared term, and you get the same answer as this one. (We want angular momentum given off per unit time.)

My point is: it's going to be a simple formula -- basically the one given above -- but I can't swear by the constant term just yet. I probably won't have much time to work that out until after the New Year. --MOBle 03:53, 19 December 2006 (UTC)

Thanks MOBle. I've read that grav radiation gradually removes eccentricities from orbits so that they become more circular over time. If this is true then circular orbits are not just an ideal. Aristotle would define them as the end cause of imperfect orbits. I'm not a mystic but I think Aristotle has a point - we live in a universe where complex relations reduce to simpler ones, otherwise we'd know nothing about anything. Anyhow I think a circular orbit is perfect for the article, not just for logican reasons, but also for the sake of non-experts such as myself. The formula you quote here could then be linked to the formula for grav radiation. That's for you to decide of course. You're trained in this stuff. I'm just tossing up ideas like a pig digging up carrots.

Hope you enjoy your travels. Don't plan them too carefully. I was in Europe last year. Best part was ascending Mount Hymettos from the eastern side. Don't know why I did it but it seemed like a good idea at the time.Lucretius 10:26, 19 December 2006 (UTC)

Resources

Hi Moble! There is a comprehensive list of resources (journal papers, books, websites etc) here [3]. I've now added this to the reference section at the end. I've also added an internal link to Hawking radiation, which has a kind of tangential relevance to grav radiation. Lucretius 11:37, 17 December 2006 (UTC)

I also found this reference: [4] titled 'Mathematical and Physical Perspectives on Gravitational Radiation'. It explains the background and significance of terms such as 'radiation reaction' 'wave zone' etc. Lucretius 22:43, 17 December 2006 (UTC)

Nice references. I didn't know Resource Letters existed. That's handy. --MOBle 03:53, 19 December 2006 (UTC)

Hi Moble. Here is a suggested change to the section 'Power radiated by the Earth-Sun system'. At present, the section sources radiated energy in kinetic energy. I think it could be more coherently sourced in angular momentum. Thus:

'The energy radiated by the Earth-Sun system can be understood to come from angular momentum. Angular momentum is radiated away by gravitational waves according to a formula that is similar to the one above for radiated energy:

${\displaystyle {\frac {dL}{dT}}=-{\frac {32}{\pi }}{\sqrt {\frac {G(M+m)}{R^{3}}}}{\frac {G^{3}(Mm)^{2}}{c^{5}R^{2}}}}$

where the square root factor is the angular velocity of a simple Keplerian system. Angular velocity has the dimension 1/time and here it can be interpreted as the frequency with which angular momentum is radiated. Since angular momentum per unit of time can also be understood to have the dimension energy, the square of the angular velocity returns us to the previous formula for radiated energy or energy per unit of time.

The angular momentum of the Earth-Sun system decreases with decreases in R. The radiated angular momentum however increases. In theory, this loss of angular momentum through radiation should cause the Earth to spiral into the Sun. However, the present angular momentum of the system is approximately 1039 kg.m2/s, whereas the angular momentum radiated over about a year is approximately 1013 kg.m2/s - at this rate, the Sun would use up its supply of fuel long before the Earth could spiral into it. Lucretius 09:28, 4 January 2007 (UTC)

In the article the symbol G appears in a number of equations. If I am right it is the "Gravitational constant". I think it would be useful not only to mention this but also provide a link to the constant in Wikipedia. M.H.Nederlof Netherlands

Thanks for this. I have provided the link as requested. Lucretius 09:54, 4 February 2007 (UTC)

New Physics

The following paragraph seems in the text seems baroque in view of Cahill's interpretations of Michelson-Morley and D. C. Miller experimental data; there are now eight independent experiments showing consistent detection of gravitational radiation [5]. Time for a rewrite of the article? -- 74.98.142.235 21:17, 10 March 2007 (UTC)

Although gravitational radiation has not yet been directly detected, it has been indirectly shown to exist. This was the basis for the 1993 Nobel Prize in Physics, awarded for measurements of the Hulse-Taylor binary system.
Thanks for this. The article has been well crafted by MOBle to reach a wide audience, educating amateurs without offending experts. I hope this approach will continue. Experts value caution so please take care! Maybe the best option is to reference the tests you mention and to add something like "Recent tests appear to offer further confirmation of the existence of gravitational waves." Lucretius 23:35, 10 March 2007 (UTC)

Vandal's edits

Someone added 'This is incorrect' in a couple of places in the article. If that someone knows better than others, he/she should justify the changes here, rather than use the article itself to assert superior knowledge. I am assuming from the nature of the 'edit' that that person doesn't even know much about good manners let alone anything about physics. If you haven't learned good manners, you haven't learned anything. Lucretius 08:30, 21 March 2007 (UTC)

I also undid the changes made by 207.210.23.114. That contributor replaced MOBle's binary black hole scenario with a more banal scenario. I don't see the reason why this was done. MOBle's scenario was about black holes being shot out of a star cluster, whereas the banal scenario was simply a restatement of inspiralling binaries. Lucretius 08:20, 22 March 2007 (UTC)

I undid another anonymous revision that looks to have been perpetrated by the previous 'contributor', concerning the inspiral. I wish that person would justify the changes here before making them. Lucretius 09:43, 29 March 2007 (UTC)