Masses of both black holes exceed 10 solar masses
Because many of us eagerly expect the announcement of a hugely exciting discovery (direct detection) of the gravitational waves, the background image shows three rural experiments: LIGO in Hanford WA (the brown "desert"), LIGO in Livingston LA (the forest, an identical gadget thousands of miles away), and VIRGO in Italy (the green fields beneath mountains). All these experiments are Lshaped pairs of orthogonal tunnels. The length of each LIGO bitunnel is 2x 4 kilometers; VIRGO has 2x 3 kilometers. Their mechanism is the same and they share the data.
The tunnels contain vacuum with laser beams. Interference is used to measure the difference between the length of both arms with the accuracy around \(10^{20}\) meters, an error margin ten billion times shorter than the atomic radius.
The gravitational waves are analogous to electromagnetic waves. They basically propagate as periodic functions\[
\cos(\omega t  \vec k \cdot \vec r)
\] in space. What is the field that has this cosine form?
In electromagnetism, we may say that it is the electromagnetic potential \(A_\mu(\vec r, t)\) that has this form, or its derivatives, the electric and magnetic field strengths \(\vec E,\vec B\). In the case of gravity, the field has two indices, \(h_{\mu\nu}\), and we may define it as\[
h_{\mu\nu} = g_{\mu\nu}  \eta_{\mu\nu}
\] where \(g_{\mu\nu}\) is the actual oscillating metric tensor determining the curved spacetime geometry and \(\eta_{\mu\nu} = {\rm diag} (+1,1,1,1)\) is the special relativistic Minkowskian metric tensor.
Except for the number of indices, the Einsteinian equations of gravity may be viewed as analogous to the equations for electromagnetism. Schematically, electromagnetism says\[
\square A^\mu = \left(\partial_\lambda F^{\lambda\mu} =\right) j^\mu.
\] The box operator \(\square=\partial^2/\partial t^2  \Delta\) acting on the electromagnetic potential is equal to the electric charge density plus current. There are actually some numerical coefficients and different contractions of the indices as well but the equation above captures the spirit of the most important terms. Similarly, Einstein's equations, when rewritten in terms of \(h_{\mu\nu}\), say something like\[
\square h_{\mu\nu} = G\cdot T_{\mu\nu}
\] The box operator (coming from the most important secondderivative terms in the Einstein tensor on the left hand side of Einstein's equations) acting on the "fluctuations of the metric tensor" is governed by the sources, the stressenergy tensor multiplied by Newton's gravitational constant. In the cases of both fields, we try to vibrate with the sources on the right hand side (the electric charges or magnetic dipoles; or the distribution of mass, energy and momentum) and the wave equations guarantee that we create waves in the fields \(A_\mu\) or \(h_{\mu\nu}\) that may be detected.
All of our cell phones, WiFi, Bluetooth, and even ordinary light detected by our eyes, if you allow me to mention an oldfashioned technology that is getting superseded ;), are based on the transmission of the electromagnetic waves. On the other hand, the gravitational waves have never been directly detected and heard, except for the notyetannounced LIGO rumors.
If you want to create the simplest and strongest electromagnetic waves, you use a dipole antenna. The electric charge is sent back and forth in a linear piece of a metal. This motion of the charge creates an electromagnetic wave that may be detected by another, distant dipole antenna. Courses on electromagnetism show you how to derive the total power (energy per second) emitted by such an antenna given a maximum dipole \(\vec p\sim Q\cdot \Delta \vec r\) and the frequency \(\omega\). Aside from the electric dipole radiation, there also exists the magnetic dipole radiation, the electric quadrupole radiation, and so forth, and they're getting weaker.
Because the gravitational field has two indices, the strongest analogous radiation comes from the quadrupole of the distribution of matter. Why? The monopole radiation exists neither in the electromagnetic nor gravitational setup because the total charge and total mass are conserved. In the gravitational case, the location of the centerofmass (and also the momentum) is conserved as well which is why we have no dipole radiation, either. So the leading terms contributing to the gravitational waves come from the quadrupoles.
The quadrupole is a tensor of the type \[
T_{ij} = \int dV\,\rho\,\,r_i r_j.
\] You write down the integral expressing the total mass from the volume integral of its density \(\rho\) but you add two factors of the location \(r_i\) and \(r_j\). This tensor fluctuates whenever two objects orbit each other because during "some seasons", the masses are separated in the \(x\) direction, and they're separated in the \(y\) direction later – which makes different components of the tensor above nonzero. You may substitute these oscillating sources to some multipole expansion for \(T_{\mu\nu}\) as well as the field \(h_{\mu\nu}\) and calculate the power and the waves, too. If two objects of masses \(m_1,m_2\) orbit each other, the total energy emitted by the gravitational waves (an analogy of the dipole radiation) ends up being\[
P = \frac{\mathrm{d}E}{\mathrm{d}t} =  \frac{32}{5}\, \frac{G^4}{c^5}\, \frac{(m_1m_2)^2 (m_1+m_2)}{r^5}.
\] It's calculable by similar methods as the dipole radiation. One only deals with a greater number of indices. But you're invited to ignore the fact that the waves come from some beautiful full tensor describing a nicely curved spacetime. The equations for \(h_{\mu\nu}\) are analogous to those for \(A_\mu\) – just another field in the spacetime – and linearization, multipole expansions, and other methods in mathematical physics are equally powerful.
If the gravitational wave is moving in the direction \(z=x^3\), the vertical dimension, only two components of the metric tensor matter, basically \(h_{11}h_{22}\) and \(2h_{12}\). When these two components periodically oscillate, the horizontal plane (surface of the Earth) gets stretched in these two patterns:
The first one just stretches and shrinks the two axes in the plane in the opposite way; the second one does the same with the tilted axes rotated by 45 degrees. Note that if you rotate any of the pictures by a multiple of 90 degrees, you get the same picture, perhaps with some phase shift in time (a delay). The gravitational wave may only have these two independent polarizations because the tensor \(h\) is symmetric; because all the vertical and timelike components may be set to zero by the diffeomorphism symmetry or are required to vanish because of the nondynamical part of Einstein's equations (constraints) – those eliminations are totally analogous to the elimination of the timelike and longitudinal polarizations of photons due to the \(U(1)\) gauge symmetry and the Gauss' law; and because the trace of the tensor may be set to zero, too (that's a new technicality that arises for tensors and at this level, there are no good electromagnetic analogies).
These two polarizations, "Plus" and "Cross", are analogous to the \(x\) and \(y\) linear polarizations of the electromagnetic wave. But the "directions" of the two linear electromagnetic polarizations differ by 90 degrees. In the case of the gravitational wave, they differ by 45 degrees – a consequence of the gravitational tensors' having two indices instead of one (equivalently, the spin of the graviton is two and not one like photon's).
You may also pick a different basis in this 2D space of polarizations, the circular polarizations. Just like in the electromagnetic case, the righthanded or lefthanded circular polarization may be obtained by taking "Plus" plus minus \(i\) times "Cross". The animated GIFs for the two circular polarizations would simply show a fixedeccentricity ellipse rotating clockwise or counterclockwise. Because all the gravitational waves are even under rotations by 180 degrees, and not just 360 degrees (as in the case of electromagnetic vectors), the period of the rotation is 2 times shorter which is why the spin is doubled (gravitons' spin is two): the period \(\Delta\phi\) of \(\exp(ij_z \phi)\) halves when \(j_z\) doubles from one to two.
Great. When a wave vertically reaches a LIGO detector, the two horizontal arms' length will start to oscillate exactly as dictated by the full metric tensor \(\eta+h\). The oscillating animated elliptical GIFs above do capture the plane of the LIGO detector (the horizontal surface of the Earth); the direction of the wave is vertical (or that's the arrangement for which the measured signal is strongest). So if you measure the lengths, especially the differences between the two arms' length, they will show exactly the same oscillations as the underlying component of the \(h_{\mu\nu}\) tensor.
You might object that the normal stresses in the material of the Earth and the detectors always want to return the proper length of the arms to the equilibrium value. Right. But they don't have enough time for that. These stresses are propagating by the speed of sound and the sound needs a second or several seconds to fly over those 4 kilometers of the arm (in solids, the sound is faster than in the air but not too much faster). On the other hand, the waves that LIGO is supposed to detect have frequencies comparable to 100 hertz, as we will compute momentarily. The period is much shorter. So you may assume that the atoms of LIGO keep on moving along the geodesics when the wave is going through. And when \(h\) abruptly changes, and it's changing rather abruptly relatively to the "sound frequencies" from the stress in the Earth, it just means that the proper length measured by the laser beams changes, too. The laser beams need an even (much) shorter time to get back and forth in the arms than the period of the measured gravitational waves so they basically measure the proper length of the arms in "real time".
There are several "traps" that may make you think that LIGO shouldn't work at all (I was tempted to be confused by several such traps) and for decades after the discovery of GR, people felt uncertain whether the gravitational waves could have been physical at all (Mach's principle was the primary misconception that drove those who wanted to say that they were unphysical) but at the end, all of them are wrong. Gravitational waves do exist and LIGOlike detectors may detect them. Note that the lengths of the 4kilometer arms are measured with the accuracy 100,000 times better than the radius of the atomic nucleus. Because it's so, a LIGO discovery will eliminate all conceivable theories that claim that Nature has an unavoidable error margin in positions that would be longer than \(10^{20}\) meters (e.g. the nuclear radius). Nature only prevents you from measuring the positions and momenta simultaneously; but it surely keeps track of the position separately and the position makes sense with an arbitrary accuracy – at most, at the Planck scale, there may be some issues (but not the issues that nonstringy quantum gravity babblers are sometimes imagining).
Fine, back to the rumor.
Ask Ethan: Is Zero Gravity Really A Thing? (Synopsis)

“It was a strange lightness, a drifting feeling. Zero gravity. I understood
that everything that once seemed solid and immovable might just float
away.” L...
1 day ago
snail feedback (0) :
Post a Comment