13.10.1
Problem 13.10.1: Mass scales
Show that effectiveness of gravitational wave production increases with the increase of mass.
solution
Intensity of gravitational radiation, as well as the electromagnetic one, is quadratic in charge of the source, e.g. for gravitation it is quadratic in mass $M^2$, while the total energy reserve ($Mc^2$) is proportional to mass in the first power. This means that with increase of mass with the same accelerations the effectiveness of gravitational wave generation should grow.
13.10.2
Problem 13.10.2: The intangibility
What is the reason for the very low efficiency of gravitational waves' production, i.e. conversion of mechanical energy into that of gravitational waves?
solution
The two main reasons for the low effectiveness of energy conversion from mechanical into gravitational waves are the following. First, weakness of gravitational interaction -- the gravitational constant $G$ is small. Second, gravitational radiation in the first non-vanishing order is quadrupole, as opposed to electromagnetic radiation, which can be produced already in the dipole approximation.
13.10.3
Problem 13.10.3: Single particle effects
What happens to a single particle as a gravitational wave passes through?
solution
Nothing. Nothing at all, and this actually follows from the principle of equivalence. Also in the first (linear) order by perturbation the geodesic equation in the TT gauge is $du^\mu /dt =0$. The wave only reveals itself when acting on a system of particles, by varying distances between them; the geodesic deviation is non-zero in the linear theory.
13.10.4
Problem 13.10.4: Waves vs static field
One may wonder, how it is possible to infer the presence of an astronomical body by the gravitational waves that it emits, when it is clearly not possible to sense its much larger stationary (essentially Newtonian) gravitational potential. What gives us hope to overcome this problem?
solution
In General Relativity, the effects of both the stationary field and gravitational radiation are described by the tidal forces they produce on free test masses. In other words, single geodesics alone cannot detect gravity or gravitational radiation; we need at least a pair of geodesics.
While the stationary tidal force due to the Newtonian potential $\phi$ of a self-gravitating source at a distance $r$ falls off as $\nabla \nabla \phi \sim {{r}^{-3}}$, the tidal force due to the gravitational wave of amplitude $h$ that it emits at wavelength $\lambda $ decreases as $\nabla \nabla h\sim r^{-1}\lambda^{-2}$. Therefore, the stationary Coulomb gravitational potential is the dominant tidal force close to the gravitating body (in the near zone, where $r\le \lambda $). However, in the far zone ($r\gg \lambda $) the tidal effect of the waves is much stronger.
The stationary part of the tidal field is a DC effect, and simply adds to the stationary tidal forces of all other objects in the universe. It is not possible to discriminate one source from another. Gravitational waves carry time-dependent tidal forces, and so they can be discriminated from the stationary field if one knows what kind of time dependence to look for. Interferometers are ideal detectors in this respect because they sense only changes in the position of an interference fringe, which makes them insensitive to the DC part of the tidal field.
The next three problems are inspired by S.Hughes, Listening to the Universe with gravitational-wave astronomy, arXiv:astro-ph/0210481
13.10.5
Problem 13.10.5: Listening to the Universe
What could be the origin of this poetic terminology?
solution
The high frequency band of gravitational wave, $1Hz\le f\le {10}^{4}Hz$, $M_{\odot }\le M\le {10}^{4}M_{\odot}$ , is the band targeted by the modern generation of ground-based laser interferometric detectors, such as LIGO. It also corresponds roughly to the audio band of the human ear: when converted to sound, LIGO sources are human audible without any frequency scaling.
The low frequency end of this band is set by the fact that it is extremely difficult to isolate against ground vibrations at low frequencies, and probably impossible to isolate against gravitational coupling to ground vibrations, human activity, and atmospheric motions. The high end of the band is set by the fact that it is unlikely any interesting gravitational-wave source radiate at frequencies higher than a few kilohertz.
13.10.6
Problem 13.10.6: Measuring infinitesimal distances
The most promising sources (neutron binaries, supernovae) of gravitational waves should give the amplitudes of the order of $h\sim 10^{-21}$. For every kilometer of baseline $L$ we need to be able to measure a distance shift of $\Delta L$ better than $10^{-16}cm$. How can we possibly hope to measure an effect that is $\sim 10^{12}$ times smaller than the wavelength of visible light (all interferometers use optical lasers)?
solution
In a laser interferometer like LIGO a carefully prepared laser state is split at the beamsplitter and sent into the Fabry-Perot arm cavities of the detector. The reflectivities of the mirrors in these cavities are chosen such that the light bounces roughly 100 times before exiting the arm cavity (that is, the finesse $F$ of the cavity is roughly 100). This corresponds to about half a cycle of a 100 Hz gravitational wave. The phase shift acquired by the light during those 100 round trips is
\[\Delta \varphi_{GW}
\sim 100\times 2\times \Delta L\times 2\pi /\lambda
\sim 10^{-9}\]
This phase shift can be measured provided that the shot noise at the photodiode, $\Delta {{\varphi }_{shot}}\sim 1/\sqrt{N}$ is less than $\Delta {{\varphi }_{GW}}$. $N$ is the number of photons accumulated over the measurement; $1/\sqrt{N}$ is the phase fluctuation in a quantum mechanical coherent state that describes a laser. We therefore must accumulate $\sim {10}^{18}$ photons over the roughly 0.01 second measurement, translating to a laser power of about 100 watts.
13.10.7
Problem 13.10.7: Surface effects
The atoms on the surface of the interferometers’ test mass mirrors oscillate with an amplitude
\[\delta l_{atom}
=\sqrt{\frac{kT}{m\omega^{2}}}
\sim {10}^{-10}cm\]
at room temperature $T$, with $m$ the atomic mass, and with a vibrational frequency $\omega \sim {10}^{14}s^{-1}$. This amplitude is huge relative to the effect of the gravitational waves. Why doesn't it wash out the gravitational wave?
solution
Gravitational waves are detectable because the atomic vibrations are random and incoherent. The $\sim 7cm$ wide laser beam averages over about ${10}^{17}$ atoms and at least ${10}^{11}$ vibrations in a typical measurement. These atomic vibrations cancel out, becoming irrelevant compared to the coherent effect of a gravitational waves.
Other thermal vibrations, however, end up dominating the detectors' noise spectra in certain frequency bands. For example, the test masses' normal modes are thermally excited. The typical frequency of these modes is $\omega \sim {10}^{5}s^{-1}$, and they have mass $m\sim 10kg$, so $\delta l_{mass}\sim {10}^{-14}cm$. This, again, is much larger than the effect we wish to observe. However, the modes are very high frequency, and so can be averaged away provided the test mass mode is confined to a very narrow band near $\omega$, and thus doesn't leak into the band we want to use for measurements.