Gravitation waves from two objects

In summary, we discussed a two body gravitating system and the gravitational potential energies for each object. It was determined that the total energy of gravitational waves emitted would be twice the change in gravitational potential energy per object as they approach each other. However, this does not necessarily produce gravitational waves and the conditions for gravitational wave production involve a quadropole moment and the luminosity of emitted power. It was also noted that GR does not have an exact equivalent to potential energy, but the energy at infinity can be used to study systems in the Schwarzschild geometry.
  • #1
kmarinas86
979
1
Consider a two body gravitating system:

The gravitational potential energy for [itex]m_1[/itex] is:

[itex]Gm_1 m_2/r[/itex]

The gravitational potential energy for [itex]m_2[/itex] is:

[itex]Gm_2 m_1/r[/itex]

So each of their gravitational potential energies are the same.

Therefore, wouldn't the total energy of gravitational waves emitted be twice the change of gravitational potential energy per object as they approach each other?

This would make:

[itex]Energy\ of\ Gravitational\ Waves = \Delta 2Gm_2 m_1/r[/itex]

Am I right?
 
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  • #2
You seem to be calculating the quantity of kinetic energy that the bodies gain (by falling closer together).
 
  • #3
kmarinas86 said:
Consider a two body gravitating system:

The gravitational potential energy for [itex]m_1[/itex] is:

[itex]Gm_1 m_2/r[/itex]

The gravitational potential energy for [itex]m_2[/itex] is:

[itex]Gm_2 m_1/r[/itex]

So each of their gravitational potential energies are the same.

Therefore, wouldn't the total energy of gravitational waves emitted be twice the change of gravitational potential energy per object as they approach each other?

This would make:

[itex]Energy\ of\ Gravitational\ Waves = \Delta 2Gm_2 m_1/r[/itex]

Am I right?

No, you can not count the gravitational potential energies twice. The potential energy of a two-body system is just [itex]-Gm_1 m_2/r[/itex]
 
  • #4
And, in any case, just "approaching" each other will not produce gravitational waves.
 
  • #5
notknowing said:
No, you can not count the gravitational potential energies twice. The potential energy of a two-body system is just [itex]-Gm_1 m_2/r[/itex]

Thank you for clearing that up!
 
  • #6
HallsofIvy said:
And, in any case, just "approaching" each other will not produce gravitational waves.

But wouldn't that imply that a change in gravitational potential energy does not necessarily produce gravitational waves? If so, under what conditions may gravitational waves be produced?

http://en.wikipedia.org/wiki/Gravitational_radiation

wikipedia said:
In Einstein's theory of general relativity, the force of gravity is due to curvature of spacetime. This curvature is caused by the presence of massive objects. Roughly speaking, the more massive the object is, the greater the curvature it causes, and hence the more intense the gravity. As massive objects move around in spacetime, the curvature will change. If the objects move around in the right way, ripples in spacetime can spread outward like ripples on the surface of a pond. These ripples are gravitational waves.

Ok... So what exactly is the change in gravitational potential energy equal to (besides the change in -GMm/r) ??

Is it:

Change in -GMm/r = 2 * the change in -KE for each of the two objects with respect to an observer at the center of mass + the change in -the energy of gravitational waves

Am I missing something important in that equation? Have I added something that's not supposed to be there?
 
  • #7
Gravitational waves will be produced any time a system has a quadropole moment. The following is taken from MTW, chapter 36.

The quadrapole moment can be written in cartesian coordinates as

[tex]Q_{jk} = \int \rho \left (x_j x_k- \frac{r^2 \delta_{jk}}{3} \right) d {V} [/tex]

using the definitions of MTW.

The luminosity (power) of gravitational waves emitted will be for the small signal case and can be approximated by

[tex]\frac{1}{5} \left( \frac{d^3 Q} {dt^3} \right) ^2 \frac{c^5}{G} [/tex] where Q is the magnitude of the quadrapole

c is the speed of light
G is the gravitational constant

c^5/G converts from "geometric units" to standard units (note that I've added this rather than explain geometric units, if you read the text you'll have to follow geometric units).

The above is just a rough approximation (but it's a lot easier than solving the field equations to get an exact answer).

More useful and easier to understand (but less general) are the equations for the gravitational radiation emitted by a pair of orbiting masses.

This is : radiated power = [tex]\frac{32}{5} \frac{\mu^2 M^3} {a^5} \frac{c^5}{G}[/tex]

where [tex]\mu[/tex] is the reduced mass, m1 m2 / (m1+m2)
M is the total mass M = m1 + m2
a is the semi-major axis

There may be some largish (30x) discrepancies between the two formula - the second formula is more accurate as well as being easier to understand (though less general).

I've ignored some correction terms for the eccentricity of the oribt.

Note that all of this has very litttle to do with "gravitational potential energy" which is a concept that's Newtonian in nature anyway.

For human sized source, it might be useful to know that a 490 ton steel bar of radius 1 meter, length 20 meters, spinning at 28 radians / second (placing it close to its yield point) radiates gravitational waves at approximately 10^-23 ergs/second (an example from MTW).

For really precise answers for things like colliding black holes, etc. more sophisticated methods numerical simulations need to be done..
 
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  • #8
If you calculate the total energy of the two objects, kinetic and potential, for two different distances, their difference should equal the energy emitted ba gravitational waves. Of course only if you make sure the system did not lose energy due to other reasons.
The calculation should be quite accurate for r>>r0 (the Schwarzschild radius).
 
  • #9
Ich said:
If you calculate the total energy of the two objects, kinetic and potential, for two different distances, their difference should equal the energy emitted ba gravitational waves. Of course only if you make sure the system did not lose energy due to other reasons.
The calculation should be quite accurate for r>>r0 (the Schwarzschild radius).

GR does not quite have anything equivalent to "potential energy", though it has a few ideas that come close. Genreally, though, energy, in any of its several forms which it is defined in GR, is not divided up into different terms representing "potential" and "kinetic" as it is in Newtonian theory.

If we take the simple example of an object of negligible mass following an orbit in the Schwarzschild geometry, if we ignore gravitational waves, the "energy at infinity" is the appropriate concept of energy to use. (Actually in this simple case, one can use several different concepts, all of which give the same answer.) This energy stays constant (it is a constant of motion for the object) and one can also say that the object follows a geodesic.

The emission of gravitational waves causes an object to deviate slightly from a geodesic, and also causes the energy at infinity of the object to not be strictly constant.

The appropriate notion of energy (GR has several, they are all closely related but slightly different) to use to study the emission of gravitational waves is the Bondi energy (rather than the energy at infinity).

One can say that in an asymptotically flat space-time that the Bondi energy is conserved, and that the energy of the system decreases by the amount of gravitational radiation.

See for instance (you'll only get the first page without going to the library - but it is rather technical in any event)

http://links.jstor.org/sici?sici=0080-4630(19620821)269%3A1336%3C21%3AGWIGRV%3E2.0.CO%3B2-C

This analysis requires an asymptotically flat space-time though, in order that the Bondi energy can be defined. One can think of an asymptotically flat space-time as being a space-time that represents an "isolated system" in a classical vacuum that is not interacting with other systems.
 
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  • #10
Second time derivative of quadrupole moment tensor

Hi, kmarinas,

kmarinas86 said:
Consider a two body gravitating system:

The gravitational potential energy for [itex]m_1[/itex] is:

[itex]Gm_1 m_2/r[/itex]

The gravitational potential energy for [itex]m_2[/itex] is:

[itex]Gm_2 m_1/r[/itex]

So each of their gravitational potential energies are the same.

Therefore, wouldn't the total energy of gravitational waves emitted be twice the change of gravitational potential energy per object as they approach each other?

This would make:

[itex]Energy\ of\ Gravitational\ Waves = \Delta 2Gm_2 m_1/r[/itex]

Am I right?

No: you are trying to use a Newtonian computation in the context of gtr, which doesn't make much sense, particularly in the context of gravitational radiation!

See for example Schutz, A First Course in General Relativity, for an excellent introduction to the quadrupole approximation in the theory of weak-field gravitational radiation (this uses the linearized-gtr approximation, as Schutz explains).

HallsofIvy said:
And, in any case, just "approaching" each other will not produce gravitational waves.

In general, if you have two massive objects, isolated from other mass, which are falling toward one another--- perhaps spiralling around each other, as in the late state merger of a neutron star-neutron star binary--- this changes the quadrupole moment of the system, which WILL result in gravitational radiation.

pervect said:
Gravitational waves will be produced any time a system has a quadropole moment.

pervect forgot to add a crucial modifier: the second time derivative of the quadrupole moment tensor must be nonzero in order for gravitational radiation to be produced. The rate at which energy is carried away is proportional to a kind of "square" of the third time derivative; see (36.1) in MTW.

pervect said:
One can say that in an asymptotically flat space-time that the Bondi energy is conserved, and that the energy of the system decreases by the amount of gravitational radiation.

Very roughly speaking, the ADM mass is evaluated "at spatial infinity" and is a constant for a given (asymptotically flat) spacetime model, reflecting the "total mass-energy" in this model. The Bondi mass is evaluated "along null infinity" and measures how much energy is lost due to "radiation by the system of some mass-energy off to null infinity".

I know that pervect just made a slip in the post quoted above, but I would like to recommend some reading for anyone not familiar with Bondi or ADM mass. The textbook by D'Inverno offers an excellent introduction to the Bondi radiation theory. The textbook by Carroll offers a very readable introduction to the Komar integrals which can be used to evaluate the mass and angular momentum of a model of an isolated object (e.g. the Kerr vacuum solution).

Chris Hillman
 
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  • #11
For a popular level article on the topic, I would suggest

http://www.nasa.gov/vision/universe/starsgalaxies/gwave.html

which describes how Nasa scientists have numerically solved for the theoretical prediction of gravity waves emitted by merging black holes. (I haven't read the details , but they are probably inspiralling black holes rather than a direct "head on" collision).

On another note, I just found out that Bondi papers on gravitational radiation are publically accessible online.

It looks like this free access is good only till the end of december, 2006.

The complete archive will be FREE online at www.journals.royalsoc.ac.uk[/URL] until the end of December 2006. Following this period it will continue to be free as part of any of the Royal Society's new journal subscription packages.
[/quote]

I could have saved myself a few bucks if I had known this - you can also get this article through JSTOR at a library, but at 10 cents a page it was several bucks to print the whole thing out. (Also, I find it easier to keep track of pdf files than paper).

Note that several other interesting GR papers have been published at the royal society, including some of the Dixon mass papers mentioned in another thread.

I probably should have said that one can find the total energy in gravitational energy emitted from a gravitating system via the change in the Bondi mass rather than what I did say. Of course, this still isn't as accurate as Chris Hillman's description, but it's probably more accessible to the lay reader. Note that this procedure will still only work correctly if the Bondi mass is defined, which still requires certain preconditions.

For an electromagnetic analogy (which is easier to understand), I would suggest reading about the Larmor radiation equation (though the referenced section in MTW does go into this). Note that even in Newtonian theory, if you have an electric charge at potential P1, and it drops to potential P2, you cannot say that it radiates away an energy equal to P2-P1. Rather, some of the potential energy difference goes into kinetic energy, and only some fraction of this difference gets turned into electromagnetic radiation. The Larmor radiation equation provides a quantitative way of estimating the amount of radiation emitted in this situation. The Larmor radiation equations involve the dipole moment (or rather one of its time derivatives). The equations I posted from MTW involving the time derivatives of the quadropole moments are the GR equivalent to the Larmor radiation equation. There isn't any contribution to gravitational radiation from the dipole moment terms, because they are zero. The first non-zero terms contributing to gravitational radiation are the quadropole terms.Note that one can also, in principle, solve Maxwell's equations for any specific case (usually a numerical solution) to find the total energy converted into electromagnetic radiation, as an alternative to using the Larmor radiation equation. This is more akin to what the scientists did in the Nasa article (but for gravity rather than E&M, they solved the Einstein field equations instead of Maxwell's equations). It's probably the most accurate technique, when done correctly, but it's also the most work.
 
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  • #12
Oh gee, I just wrote a reply but the system logged me off unexpectedly and my work was lost.

Time permitting, I'd like to try to explain what Bondi did and why it is so important, but in the mean time, interested readers can see the textbook by D'Inverno, which has an excellent introduction. I'd also like to try to explain a bit about Lie group analysis and why
http://www.arxiv.org/abs/gr-qc/0611109 is potentially interesting, in the context of searching for exact solutions modeling an axisymmetric collision of two nonrotating black holes.

Chris Hillman
 
  • #13
Chris Hillman said:
Oh gee, I just wrote a reply but the system logged me off unexpectedly and my work was lost.

Chris Hillman

I hate that when it happens. I do have a suggestion. When the system seems sluggish in accepting a post (or all the time, if you are really paranoid), right click on the window containing your text, hit "select all", right click again, and hit "copy" - assuming, of course, you are running WinDoze(tm) and not some other (probably better) OS.

This puts the entire text of the article into the paste buffer. You can then save it in a file if the article doesn't make it through (or leave it in the paste buffer for a while if you aren't sure).

If you're fast on the mouse, you can often save the text before it disappears when you sense something is going wrong because of the long delays. Also, in some circumstances, the "back" button on the browser will work and you'll get your text back (other times, it won't).
 
  • #14
Hi, pervect, I am not running Windoze--- what an idea!

Good to know I am not the only one who has been caught this way. I have found that opening a new browser window to log back in avoids this problem. On this occasion I just got lazy and I paid the price.

Chris Hillman
 
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  • #15
Corrections to a previous post

pervect said:
Gravitational waves will be produced any time a system has a

a quadropole moment whose third time derivative is nonzero

The following is taken from MTW, chapter 36.

The quadropole moment can be written in cartesian coordinates as

[tex]Q_{jk} = \int \rho \left (x_j x_k- \frac{r^2 \delta_{jk}}{3} \right) d V [/tex]

using the defintions of MTW.

The luminosity (power) of gravitational waves emitted will be for the small signal case and can be approximated by

[tex]\frac{1}{5} \left( \frac{d^3 Q} {dt^3} \right) ^2 \frac{G}{c^5}[/tex] where Q is the magitude of the quadropole

I believe that [tex]Q = \sum_{i=1}^{3} Q_{ii}[/tex]

though I don't have time to confirm this at the moment.

c is the speed of light
G is the gravitational constant

sanity check:

Q has units of kg*m^2
So we have units of (kg*m^2/sec^3)^2*(G/c^5)

Google calculator confirms this is watts

Sanity check #2:

If we have a massive steel beam of radius r=1 meter, length L=20 meters, density = 7.8 gm/cm^3, total mass 490 tons, rotating just below the point where it breaks up at 28 radians/second, we should get

2.2*10^-22 erg/sec radiated power (MTW pg 979).

The above is just a rough approximation (but it's a lot easier than solving the field equations to get an exact answer).

More useful and easier to understand (but less general) are the equations for the gravitational radiation emitted by a pair of orbiting masses.

This is : radiated power =

[tex]\frac{32}{5} \frac{\mu^2 M^3} {a^5} \frac{G^4}{c^5}[/tex]

where [tex]\mu[/tex] is the reduced mass, m1 m2 / (m1+m2)
M is the total mass M = m1 + m2
a is the semi-major axis

There may be some largish (30x) discrepancies between the two formula - the second formula is more accurate as well as being easier to understand (though less general).

I've ignored some correction terms for the eccentricity of the oribt.

Note that all of this has very litttle to do with "gravitational potential energy" which is a concept that's Newtonian in nature anyway.

For human sized source, it might be useful to know that a 490 ton steel bar of radius 1 meter, length 20 meters, spinning at 28 radians / second (placing it close to its yield point) radiates gravitational waves at approximately 10^-23 ergs/second (an example from MTW).

For really precise answers for things like colliding black holes, etc. more sophisticated methods numerical simulations need to be done.
 
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Related to Gravitation waves from two objects

1. What are gravitational waves?

Gravitational waves are ripples in the fabric of space-time that are created when two massive objects, such as black holes or neutron stars, orbit each other. These waves are similar to the ripples that form on the surface of a pond when a stone is thrown into it.

2. How are gravitational waves produced?

Gravitational waves are produced when two massive objects orbit each other, causing a distortion in the fabric of space-time. This distortion creates waves that radiate outwards at the speed of light.

3. How can we detect gravitational waves?

Gravitational waves can be detected using specialized instruments called interferometers. These instruments measure tiny changes in the length of two perpendicular arms caused by passing gravitational waves.

4. What are the implications of detecting gravitational waves?

The detection of gravitational waves provides evidence for the theory of general relativity and gives us a new way to study the universe. It also allows us to observe events, such as the collision of black holes, that were previously undetectable.

5. How do gravitational waves affect us?

Gravitational waves have a very small effect on us and our surroundings, as they are incredibly weak by the time they reach Earth. However, their detection has opened up a new field of study in astrophysics and has the potential to advance our understanding of the universe.

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