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Gravitational wave

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In physics, a gravitational wave is a fluctuation in the curvature of spacetime which propagates as a wave, traveling outward from a moving object or system of objects. Gravitational radiation is the energy transported by these waves. Important examples of systems which emit gravitational waves are binary star systems, where the two stars in the binary are white dwarfs, neutron stars, or black holes.

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.

(Gravitational waves are sometimes called gravity waves, but this term should be reserved for a completely different kind of wave encountered in hydrodynamics.)

Contents

[edit] Introduction

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.


The simplest example of a strong source of gravitational waves is a spinning neutron star with a small mountain on its surface. The mountain's mass will cause curvature of the spacetime. Its movement will "stir up" spacetime, much like a paddle stirring up water. The waves will spread out through the Universe at the speed of light, never stopping or slowing down.


As these waves pass a distant observer, that observer will find spacetime distorted in a very particular way. Distances between objects will increase and decrease rhythmically as the wave passes. The size of this effect will go down the farther the observer is from the source. Any gravitational waves expected to be seen on Earth will be quite small; the change in size of any object will never be much more than <math>10^{-18}</math> percent. Still, scientists are attempting to measure the effects of these waves using extraordinarily precise experiments.


By measuring these waves, astrophysicists hope to learn about systems they could not observe with more traditional means such as optical telescopes, radio telescopes, etc. Gravitational waves can penetrate regions that the more familiar waves cannot, providing us with a view of black holes and other mysterious objects in the distant Universe. Using precise measurements of gravitational waves in this way will also allow us to test the theory of general relativity more thoroughly.

[edit] Effects of a passing gravitational wave

Image:GravitationalWave PlusPolarization.gif Image:GravitationalWave CrossPolarization.gif Imagine a perfectly flat region of spacetime, with a group of motionless test particles lying in a plane. Then, a weak gravitational wave arrives, passing through the particles along a line perpendicular to the plane of the particles. What happens to the test particles? Roughly speaking, they will oscillate in a "cruciform" manner, as shown in the animations. The area enclosed by the test particles does not change, and there is no motion along the direction of propagation. In the animation at the right, the wave would be passing from you, through the screen, and out the back.


Like other waves, there are a few useful numbers describing a gravitational wave:

  • Amplitude: Usually denoted <math>h</math>, this is the size of the wave — the fraction of stretching or squeezing in the animation. The amplitude shown here is roughly <math>h=0.5</math> (or 50%). Gravitational waves passing through the Earth are many billion billion times weaker than this.
  • Frequency: Usually denoted <math>\nu</math>, this is the frequency with which the wave oscillates (1 divided by the amount of time between maximum stretch or squeeze)
  • Wavelength: Usually denoted <math>\lambda</math>, this is the distance along the wave between points of maximum stretch or squeeze.
  • Speed: This is the speed at which a point on the wave (for example, a point of maximum stretch or squeeze) travels. For gravitational waves with small amplitudes, this happens to be the speed of light, <math>c</math>


The frequency, wavelength, and speed of a gravitational wave are related by the equation

<math>c = \lambda\, \nu\ ,</math>

just like the equation for a light wave. The animations shown here oscillate roughly once every two seconds. This means that <math>\nu = \frac{1}{2\mathrm{\ sec}} = \frac{1}{2}</math> Hertz. Now, using the speed of light (c = 3 x 108 meters per second) the wavelength of the waves would be roughly

<math>\lambda = \frac{c}{\nu} = \frac{3\times 10^8 \textrm{meters\ per\ second}} {1/2\textrm{sec}} = 6 \times 10^8\textrm{\ meters}\ ,</math>

or about 50 times the width of the Earth.


In the example just discussed, we actually assume something special about the wave. We have assumed that the wave is linearly polarized, with a "plus" polarization, written <math>h_{\,+}</math>. Polarization of a gravitational wave is just like polarization of a light wave, except that the polarizations of a gravitational wave are at 45 degrees, as opposed to 90 degrees. In particular, if we had a "cross"-polarized gravitational wave, <math>h_{\,\times}</math>, the effect on the test particles would be basically the same, but rotated by 45 degrees, as shown in the second animation. Just as with light polarization, the polarizations of gravitational waves may also be expressed in terms of circularly polarized waves. Gravitational waves are polarized because of the nature of their sources. The polarization of a wave actually depends on the angle from the source, as we will see in the next section.

[edit] Sources of gravitational waves

In general terms, gravitational waves are radiated by objects whose motion involves acceleration, provided that the motion is not perfectly spherically symmetric (like a spinning, expanding or contracting sphere) or cylindrically symmetric (like a spinning disk).


The simplest example is the spinning dumbbell. The dumbbell will not radiate if it spins around its vertical axis. It will radiate if it tumbles end-over-end. The heavier the dumbbell, and the faster it tumbles, the greater is the gravitational radiation it will give off. If we imagine an extreme case in which the two weights of the dumbbell are 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 in a quasi-Keplerian planar orbit (basically, as a planet would orbit the Sun) will radiate.
  • A spinning non-axisymmetric planetoid — say with a large bump or dimple on the equator — will radiate.
  • A supernova will radiate except in the unlikely event that it is perfectly symmetric.
  • An isolated non-spinning solid object moving at a constant speed will not radiate. This can be regarded as a consequence of the principle of conservation of linear momentum.
  • A spinning disk 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.
  • A spherically pulsating spherical star (non-zero monopole moment or mass, but zero quadrupole moment) will not radiate, in agreement with Birkhoff's theorem.


More technically, the second time derivative of the quadrupole moment (or the l-th time derivative of the l-th multipole moment) of an isolated system's stress-energy tensor must be nonzero in order for it to emit gravitational radiation. This is analogous to the changing dipole moment of charge or current necessary for electromagnetic radiation.


[edit] Power radiated by the Earth-Sun system

We imagine a simple system of two masses — such as the Earth-Sun system — moving slowly compared to the speed of light. Assume that these two masses orbit each other in a circular orbit in the <math>x</math>-<math>y</math> plane. To a good approximation, the masses follow simple Keplerian orbits. However, such an orbit represents a changing quadrupole moment. That is, the system will give off gravitational waves.


Suppose that the two masses are <math>M_1</math> and <math>M_2</math>, and they are separated by a distance <math>R</math>. The power given off (radiated) by this system is:


<math>P = \frac{dE}{dt} = - \frac{32}{\pi}\, \frac{G^4}{c^5}\, \frac{(M_1M_2)^2 (M_1+M_2)}{R^5}\ ,</math>


where the negative sign means that power is being given off by the system, rather than received. For a system like the Earth and the Sun, <math>R</math> is very large, and <math>M_1</math> and <math>M_2</math> are relatively very small. In this case, the power is:


<math>P = - \frac{32}{\pi}\, \frac{(6.7 \times 10^{-11}\, \frac{\mathrm{m}^3} {\mathrm{kg}\, \mathrm{s}^{2}})^4}{(3\times10^8\, \mathrm{m/s})^5}\, \frac{(6\times 10^{24}\,\mathrm{kg}\cdot 2\times 10^{30}\,\mathrm{kg})^2 (6\times 10^{24}\,\mathrm{kg} + 2\times 10^{30}\,\mathrm{kg})} {(1.5\times10^{11} \mathrm{m})^5} = 313\, \mathrm{Watts}\ .</math>


Thus the total power radiated by the Earth-Sun system in the form of gravitational waves is about 300 Watts (i.e. about five 60 Watt light bulbs). This is truly tiny compared to the total electromagnetic radiation given off by the Sun (roughly 3.86 x 1026 Watts).


The energy of the gravitational waves comes out of the kinetic energy of the Earth's orbit. Theoretically, this slow radiation from the Earth-Sun system could steal enough energy to drop the Earth into the Sun. However, the kinetic energy of the Earth orbiting the Sun is about 2.7 x 1033 Joules. As the gravitational radiation is given off, it takes about 300 Joules per second away from the orbit. At this rate, it would take many billion times more than the current age of the Universe for the Earth to fall into the Sun. (Though, worse things are expected to happen before such a time.)

[edit] Wave amplitudes from the Earth-Sun system

We can also think in terms of the actual amplitude of the wave. Suppose that an observer is positioned at a distance <math>r</math> from the center of mass of the system, at spherical coordinates <math>(r, \theta, \phi)</math>. If the observer is well outside the system (in fact, we need <math>r > c/\Omega</math>), the two polarizations of the wave will be


<math>h_{+} = -\frac{1}{r}\, \frac{G^2}{c^4}\, \frac{2 M_1 M_2}{R} (1+\cos^2\theta) \cos\left[2\Omega (t - r) - 2\phi\right]\ ,</math>
<math>h_{\times} = -\frac{1}{r}\, \frac{G^2}{c^4}\, \frac{4 M_1 M_2}{R}\, \cos{\theta} \sin \left[2 \Omega (t - r) - 2\phi \right]\ .</math>


Here, we use the angular velocity <math>\Omega=\sqrt{G(M_1+M_2)/R^3}</math> of a simple Keplerian system. Note that the polarization depends on the angle to the system. For example, if the observer is in the <math>x</math>-<math>y</math> plane, <math>\theta=\pi/2</math>, and <math>\cos\theta = 0</math>, so the <math>h_\times</math> polarization is always zero. We also see that the frequency of the wave given off is <math>\nu = 2\Omega/2\pi = \Omega / \pi</math>. If we put in numbers for the Earth-Sun system, we find


<math>-\frac{1}{r}\, \frac{G^2}{c^4}\, \frac{4M_1 M_2}{R} = -\frac{1}{r}\, 1.7\times 10^{-10}\, \mathrm{meters}\ .</math>


In this case, the minimum distance to find waves is <math>r =1</math>light year, so typical amplitudes will be <math>h \approx 10^{-26}</math>. That is, a ring of particles would stretch or squeeze by just <math>10^{-24}</math> percent.

[edit] Radiation from other sources

Although the waves from the Earth-Sun system are minuscule, astronomers can point to other sources for which the radiation should be substantial. One important example is the Hulse-Taylor binary — a pair of stars, one of which is a pulsar. The characteristics of their orbit can be deduced from the Doppler shifting of radio signals given off by the pulsar. Each of the stars is far more massive than the Earth, and about 1.4 times heavier than the Sun. Also, their orbit is about 75 times smaller than the distance between the Earth and Sun — which means the distance between the two stars is just a few times larger than the diameter of our own Sun. This combination of greater masses and smaller separation means that the energy given off by the Hulse-Taylor binary will be far greater than the energy given off by the Earth-Sun system — roughly <math>10^{22}</math> times as much.


The information about the orbit can be used to predict just how much energy (and angular momentum) should be given off in the form of gravitational waves. As the energy is carried off, the orbit will change; the stars will draw closer to each other. This effect of drawing closer is called an inspiral, and it can be observed in the pulsar's signals. The measurements on this system were carried out over several decades, and it was shown that the changes predicted by gravitational radiation in general relativity matched the observations very well. In 1993, Russell Hulse and Joe Taylor were awarded the Nobel Prize in Physics for this experiment, which was (and remains) the first experimental evidence for gravitational waves.


Inspirals are very important sources of gravitational waves. Any time two compact objects (white dwarfs, neutron stars, or black holes) come close to each other, they send out intense gravitational waves. As the objects come closer and closer to each other (that is, as <math>R</math> becomes smaller and smaller), the gravitational waves become more and more intense. At some point these waves should become so intense that they can be directly detected by their effect on objects on the Earth. This direct detection is the goal of several large experiments around the world.


The only difficulty is that systems like the Hulse-Taylor binary are so far away. The amplitude of waves given off by the Hulse-Taylor binary as seen on Earth would be roughly <math>h \approx 10^{-26}</math>. There are some sources, however, that astrophysicists expect to find with the somewhat larger amplitudes of <math>h \approx 10^{-20}</math>.

[edit] Astrophysics and gravitational waves

Unsolved problems in physics: Is our universe filled with gravitational radiation from the big bang? From astrophysical sources, such as inspiraling neutron stars? What can this tell us about quantum gravity and general relativity?

During the past century, astronomy has been revolutionized by the use of new methods for observing the universe. Astronomical observations were originally made using visible light. Galileo Galilei pioneered the use of telescopes to enhance these observations. However, visible light is only a small portion of the electromagnetic spectrum, and not all objects in the distant universe shine strongly in this particular band. More useful information may be found, for example, in radio wavelengths. Using radio telescopes, astronomers have found pulsars, quasars, and other extreme objects which push the limits of our understanding of physics. Observations in the microwave band have opened our eyes to the faint imprints of the Big Bang a discovery Stephen Hawking called the "greatest discovery of the century, if not all time". Similar advances in observations using gamma rays, x-rays, ultraviolet light, and infrared light have also brought new insights to astronomy. As each of these regions of the spectrum has opened, new discoveries have been made that could not have been made otherwise. Astronomers hope that the same holds true of gravitational waves.


Gravitational waves have two important and unique properties. First, there is no need for any type of matter to be present in order for the waves to be created. For example, two black holes crashing into each other could generate powerful gravitational waves, while being completely black in electromagnetic radiation. Second, gravitational waves can pass through any intervening matter without being scattered. Whereas light from distant stars may be blocked out by interstellar dust, for example, gravitational waves will pass through unimpeded. These two features allow gravitational waves to carry information about astronomical phenomena never before observed by humans.

[edit] Energy, momentum, and angular momentum carried by gravitational waves

Waves familiar from other areas of physics such as water waves, sound waves, and electromagnetic waves are able to carry energy, momentum, and angular momentum. By carrying these away from a source, waves are able to rob that source of its energy, momentum, or angular momentum.


In much the same way, a gravitational wave can carry off energy, momentum, and angular momentum, robbing them from its source. 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. This could decrease the number of black holes found in star clusters, which might be observable.

[edit] Gravitational wave detectors

Though the Hulse-Taylor observations were very important, they were only indirect evidence for gravitational waves. A more interesting observation would be a direct measurement of the effect of a passing gravitational wave. Not only would a direct measurement of gravitational waves rule out other possible (however unlikely) reasons for changes to the orbit of an inspiraling system, it would also provide us more information on the system. Perhaps more importantly, such a detection could give us information about things we can't see with radio or light waves — such as black holes. This would provide us with a rigorous test for Einstein's theory of general relativity.


The great challenge of this type of detection, though, is the extraordinarily small effect the waves would produce on a detector. The amplitude of any wave will fall off as the inverse of the distance from the source (the <math>1/r</math> term in the formulas for <math>h</math> above). Thus, even waves from extreme systems like merging binary black holes die out to very small amplitude by the time they reach the Earth. Astrophysicists expect that some gravitational waves passing the Earth may be as large <math>h\approx 10^{-20}</math>, but generally no bigger. For an object 1 meter in length, this means that its ends would move by <math>10^{-20}</math> meters relative to each other. This distance is about 1 billionth of the width of a typical atom, and roughly one one-hundred-thousandth the width of a proton.


A simple device to detect this motion is called a Weber bar — a large, solid piece of metal with electronics attached to detect any vibrations. This type of instrument was the first type of gravitational wave detector. The idea is to wait for a passing gravitational wave to "ring up" a bar at its resonant frequency, which would basically amplify the wave naturally. Alternatively, a nearby supernova might be strong enough to be seen without resonant amplification. Modern forms of the Weber bar are still operated, cryogenically cooled, with superconducting quantum interference devices to detect the motion. Unfortunately, Weber bars are not sensitive enough to detect anything but extremely powerful gravitational waves.


Image:Ligo.gif

A more sensitive version is the laser interferometer, with separate masses placed many hundreds of meters to several kilometers apart acting as two ends of a bar. Ground-based interferometers are now operating, and taking data. The most sensitive is LIGO — the Laser Interferometer Gravitational Wave Observatory. This is actually a set of three devices: one in Livingston, Louisiana; the other two (essentially on top of each other) in Hanford, Washington. Each consists of two light storage arms which are 2 to 4 kilometers in length. These are at 90 degree angles to each other, and consist of large vacuum tubes running the entire 4 kilometers. A passing gravitational wave will then slightly stretch one arm as it shortens the other. This is precisely the motion to which an interferometer is most sensitive.


Even with such long arms, a gravitational wave will only change the distance between the ends of the arms by about <math>10^{-17}</math> meters at most. This is still only a fraction of the width of a proton. Nonetheless, LIGO's interferometers are now running routinely at an even better sensitivity level. LIGO's should be able to detect gravitational waves as small as <math>h \approx 5\times 10^{-22}</math>, but needs to wait until a gravitational wave with at least that amplitude passes. Upgrades to LIGO and other detectors such as VIRGO, GEO, and TAMA should increase the sensitivity still further — by a factor of up to 100.


All detectors are limited at high frequencies by shot noise, which occurs because the laser beams are made up of photons. If there are not enough photons arriving in a given time interval (that is, if the laser is not intense enough), it will be impossible to tell whether a measurement is due to real data, or just random fluctuations in the number of photons.


All ground-based detectors are also limited at low frequencies by seismic noise, and must be very well isolated from seismic disturbances. Passing cars and trains, falling logs, and even waves crashing on the shore hundreds of miles away are all very significant sources of noise in real interferometers.


Space-based interferometers, such as LISA, are also being developed. LISA's design calls for test masses to be placed five million kilometers apart, in separate spacecraft with lasers running between them. Because of the distance between the spacecraft, it will be impossible to create light storage arms. Also, the arms will be at 60 degree angles to each other, rather than 90 degree angles. Still, the principle will be the same. Although LISA will not be affected by seismic noise, it will be affected by other noise sources, including noise from cosmic rays and solar wind and — of course — shot noise.


There are other prospects such as MiniGRAIL, a spherical gravitational wave antenna based at Leiden University. Some scientists even want to use the moon as a giant gravitational wave detector. The moon should be somewhat pliable to the contortions caused by gravitational waves, and the hope is that the motion of the moon caused by these waves will be detectable, much like the motion of a Weber bar.

[edit] Einstein@Home

Main article: Einstein@Home

In some sense, the easiest signals to detect should be constant sources. Supernovae and neutron star or black hole mergers should have larger amplitudes and be more interesting, but their waves will be more complicated. The waves given off by a spinning, bumpy neutron star would be "monochromatic" — like a pure tone in music. It wouldn't change (much) in amplitude or frequency.


The Einstein@Home project is a distributed computing project similar to SETI@home intended to detect this type of simple gravitational wave. By taking data from LIGO and GEO, and sending it out in little pieces to thousands of volunteers for analysis on their home computers, Einstein@Home can sift through the data far more quickly than would be possible otherwise. Searches for gravitational waves from other types of systems require large supercomputers running for long periods.


A simple computer program can be downloaded to any home computer, and acts as a screen saver to use computer time that would otherwise be wasted. The program automatically downloads the data, analyzes it while the screen saver is running, and sends the final results back to a central computer.

[edit] Mathematics

Einstein's equations form the fundamental law of general relativity. The curvature of spacetime can be expressed mathematically using the metric tensor — denoted <math>g_{\mu \nu}</math> — and with respect to a covariant derivative, <math>\nabla</math>, in the form of the Einstein tensor — <math>G_{\mu \nu}</math>. This curvature is related to the stress-energy tensor — <math>T_{\mu\nu}</math> — by the key equation

<math>G_{\mu \nu} = \frac{8\pi G_N}{c^4} T_{\mu \nu}\ ,</math>

where <math>G_N</math> is Newton's gravitational constant, and <math>c</math> is the speed of light. We assume geometrized units, so <math>G_N = 1 = c</math>.


With some simple assumptions, Einstein's equations can be rewritten to show explicitly that they are just wave equations. To begin with, we adopt some coordinate system, like <math>(t,r,\theta,\phi)</math>. We define the "flat-space metric" <math>\eta_{\mu\nu}</math> to be the quantity which — in this coordinate system — has the components we would expect for the flat space metric. For example, in these spherical coordinates, we have

<math>

\eta_{\mu \nu} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 \sin^2\theta \end{bmatrix}\ . </math>

This object has no physical significance; it is a purely mathematical device necessary for the analysis. Tensor indices are raised and lowered using this "metric".


Now, we can also think of the physical metric <math>g_{\mu \nu}</math> as a matrix, and find its determinant, <math>\det\ g</math>. Finally, we define a quantity

<math>\bar{h}^{\alpha \beta} \equiv \eta^{\alpha \beta} - \sqrt{|\det g|} g^{\alpha \beta}\ .</math>

This is the crucial field, which will represent the radiation. It is possible (at least in an asymptotically flat spacetime) to choose the coordinates in such a way that this quantity satisfies the "de Donder" gauge conditions (conditions on the coordinates):

<math>\nabla_\beta\, \bar{h}^{\alpha \beta} = 0\ ,</math>

where <math>\nabla</math> represents the flat-space derivative operator. These equations say that the divergence of the field is zero. The full, nonlinear Einstein equations can now be written<ref name="RMP80">Thorne, Kip (April 1980). "Multipole expansions of gravitational radiation". Reviews of Modern Physics 52.</ref> as

<math>\Box \bar{h}^{\alpha \beta} = -16\pi \tau^{\alpha \beta}\ ,</math>

where <math>\Box = -\partial_t^2 + \Delta</math> represents the flat-space d'Alembertian operator, and <math>\tau^{\alpha \beta}</math> represents the stress-energy tensor plus quadratic terms involving <math>\bar{h}^{\alpha \beta}</math>. This is just a wave equation for the field with a source, despite the fact that the source involves terms quadratic in the field itself. That is, it can be shown that solutions to this equation are waves traveling with velocity 1 in these coordinates.

[edit] Linear approximation

The equations above are valid everywhere — near a black hole, for instance. However, because of the complicated source term, the solution is generally too difficult to find analytically. We can often assume that space is nearly flat, so the metric is nearly equal to the <math>\eta^{\alpha \beta}</math> tensor. In this case, we can neglect terms quadratic in <math>\bar{h}^{\alpha \beta}</math>, which means that the <math>\tau^{\alpha \beta}</math> field reduces to the usual stress-energy tensor <math>T^{\alpha \beta}</math>. That is, Einstein's equations become

<math>\Box \bar{h}^{\alpha \beta} = -16\pi T^{\alpha \beta}\ .</math>

If we are interested in the field far from a source, however, we can treat the source as a point source; everywhere else, the stress-energy tensor would be zero, so

<math>\Box \bar{h}^{\alpha \beta} = 0\ .</math>

Now, this is the usual homogeneous wave equation — one for each component of <math>\bar{h}^{\alpha \beta}</math>. Solutions to this equation are well known. For a wave moving away from a point source, the radiated part (meaning the part that dies off as <math>1/r</math> far from the source) can always be written in the form <math>A(t-r,\theta,\phi)/r</math>, where <math>A</math> is just some function. It can be shown<ref>C.W. Misner, K.S. Thorne, and J.A. Wheeler (1973). Gravitation. W.H. Freeman and Co..</ref> that — to a linear approximation — it is always possible to make the field traceless. Now, if we further assume that the source is positioned at <math>r=0</math>, the general solution to the wave equation in spherical coordinates is

<math>

\bar{h}^{\alpha \beta} = \frac{1}{r}\, \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & A_{+}(t-r,\theta,\phi) & A_{\times}(t-r,\theta,\phi) \\ 0 & 0 & A_{\times}(t-r,\theta,\phi) & -A_{+}(t-r,\theta,\phi) \end{bmatrix} \equiv \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & h_{+}(t-r,r,\theta,\phi) & h_{\times}(t-r,r,\theta,\phi) \\ 0 & 0 & h_{\times}(t-r,r,\theta,\phi) & -h_{+}(t-r,r,\theta,\phi) \end{bmatrix}\ , </math>

where we now see the origin of the two polarizations.


[edit] Relation to the source

If we know the details of a source — for instance, the parameters of the orbit of a binary — we can relate the source's motion to the gravitational radiation observed far away. With the relation

<math>\Box \bar{h}^{\alpha \beta} = -16\pi \tau^{\alpha \beta}\ ,</math>

we can write the solution in terms of the tensorial Green's function for the d'Alembertian operator<ref name="RMP80"/>:

<math>

\bar{h}^{\alpha \beta} (t,\vec{x}) = -16\pi \int\, G^{\alpha \beta}_{\gamma \delta} (t,\vec{x};t',\vec{x}')\, \tau^{\gamma \delta}(t',\vec{x}')\, dt'\, d^3x'\ . </math>

Though it is possible to expand the Green's function in tensor spherical harmonics, it is easier to simply use the form

<math>G^{\alpha \beta}_{\gamma \delta} (t,\vec{x};t',\vec{x}') = \frac{1}{4\pi} \delta_{\gamma}^\alpha\, \delta_{\delta}^\beta\, \frac{\delta(t\pm|\vec{x}-\vec{x}'|-t')} {|\vec{x}-\vec{x}'|}\ ,</math>

where the positive and negative signs correspond to ingoing and outgoing solutions, respectively. Generally, we are interested in the outgoing solutions, so

<math>

\bar{h}^{\alpha \beta} (t,\vec{x}) = -4 \int\, \frac{\tau^{\alpha \beta}(t-|\vec{x}-\vec{x}'|,\vec{x}')}{|\vec{x}-\vec{x}'|}\, d^3x'\ . </math>

If the source is confined to a small region very far away, to an excellent approximation we have:

<math>

\bar{h}^{\alpha \beta} (t,\vec{x}) \approx -\frac{4}{r}\, \int\, \tau^{\alpha \beta}(t-r,\vec{x}')\, d^3x'\ , </math>

where <math>r=|\vec{x}|</math>.


Now, because we will eventually only be interested in the spatial components of this equation (time components can be set to zero with a coordinate transformation), and we are integrating this quantity — presumably over a region of which there is no boundary — we can put this in a different form. Ignoring divergences with the help of Stokes' theorem and an empty boundary, we can see that

<math>

\int\, \tau^{i j}(t-r,\vec{x}')\, d^3x' = \int\, x'^i x'^j \nabla_k \nabla_l \tau^{k l} (t-r,\vec{x}')\, d^3x'\ , </math>

Inserting this into the above equation, we arrive at

<math>

\bar{h}^{i j} (t,\vec{x}) \approx -\frac{4}{r}\, \int\, x'^i x'^j \nabla_k \nabla_l \tau^{k l} (t-r,\vec{x}')\, d^3x'\ , </math>

Finally, because we have chosen to work in coordinates for which <math>\nabla_\beta\, \bar{h}^{\alpha \beta} = 0</math>, we know that <math>\nabla_\beta\, \tau^{\alpha \beta} = 0</math>. With a few simple manipulations, we can use this to prove that

<math>\nabla_0 \nabla_0 \tau^{00} = \nabla_j \nabla_k \tau^{jk}\ .</math>

With this relation, the expression for the radiated field is

<math>

\bar{h}^{i j} (t,\vec{x}) \approx -\frac{4}{r}\, \frac{d^2}{dt^2}\, \int\, x'^i x'^j \tau^{0 0} (t-r,\vec{x}')\, d^3x'\ . </math>

In the linear case, <math>\tau^{00} = \rho</math>, the density of mass-energy.


To a very good approximation, the density of a simple binary can be described by a pair of delta-functions, which eliminates the integral. Explicitly, if the masses of the two objects are <math>M_1</math> and <math>M_2</math>, and the positions are <math>\vec{x}_1</math> and <math>\vec{x}_2</math>, then

<math>\rho(t-r,\vec{x}') = M_1 \delta^3(\vec{x}'-\vec{x}_1(t-r)) + M_2 \delta^3(\vec{x}'-\vec{x}_2(t-r))\ .</math>

We can use this expression to do the integral above:

<math>

\bar{h}^{i j} (t,\vec{x}) \approx -\frac{4}{r}\, \frac{d^2}{dt^2}\, \left\{ M_1 x_1^i(t-r) x_1^j(t-r) + M_2 x_2^i(t-r) x_2^j(t-r) \right\}\ . </math>

Using mass-centered coordinates, and assuming a circular binary, this is

<math>

\bar{h}^{i j} (t,\vec{x}) \approx -\frac{4}{r}\, \frac{M_1 M_2}{R}\, n^i(t-r) n^j(t-r)\ , </math>

where <math>\vec{n} = \vec{x}_1 / |\vec{x}_1|</math>. Plugging in the known values of <math>\vec{x}_1(t-r)</math>, we obtain the expressions given above for the radiation from a simple binary.


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[edit] References

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