Tag Info

Hot answers tagged

39

The actual paper (pdf) is very heavy in error quantification - and rightly so. They presented an experiment result that is statistically extremely difficult to obtain. But for the rest of us, conclusion is the most important part. The abstract says: The observed B-mode power spectrum is well fit by a lensed-$\lambda$CDM + tensor theoretical model with ...


15

They announced that through observation of the Cosmic Microwave Background, via the BICEP2 experiment in Antarctica, particularly the polarization on a 2-4 degree angular scale, gravitational waves from inflation during the early universe are being indirectly observed. Link to FAQs about the release: http://bicepkeck.org/faq.html Link to pre-print: ...


13

Gravitational waves are transverse waves but they are not dipole transverse waves like most electromagnetic waves, they are quadrupole waves. They simultaneously squeeze and stretch matter in two perpendicular directions. Gravitational waves definitely propagate in a given direction but the effect that they have on matter is completely perpendicular to the ...


8

A gravitational wave will distort space-time and the light that is on a path affected by such a wave will be similarly affected, but it will still take longer (or shorter) to travel that path. Imagine a car travelling along the surface of a trampoline, a wave on the trampoline could cause its path to become longer, but it won't be impossible to detect ...


7

Nope. Gravitational radiation is a kind of radiation and it has a completely different equation of state than the cosmological constant. The cosmological constant has pressure equal to the energy density with a minus sign, $p=-\rho$: the stress-energy tensor is proportional to the metric tensor so the spatial and temporal diagonal components only differ by ...


7

An addendum to the answers of Daniel Grumiller and sb1: The major difference of the gravitational field and other fields is that according to general relativity the gravitational field defines space and time and therefore defines the relation of events. It is true that it is possible to do an "arbitrary" split of a certain linear approximation of the ...


7

The amplitudes do become arbitrarily small, and there's nothing at all wrong with this. In fact the exact same thing happens with electromagnetic waves. Sure we have a quantum theory with photons that places limits on how small a packet of energy can be detected, but light can travel across the universe just fine and become as dim as it wants. The intensity ...


7

Photons or cosmic rays don't (normally) emit gravity waves. Consider the comparison with radio waves. A moving electron doesn't emit radio waves. It has to be accelerating to emit EM radiation. Specifically radio waves are only emitted when there is a changing dipole moment. So you wouldn't expect a particle moving at constant velocity (photon or ...


7

Dr. Matt Strassler has some great info on his site, see here: http://profmattstrassler.com/2014/03/17/a-primer-on-todays-events/ http://profmattstrassler.com/2014/03/17/bicep2-new-evidence-of-cosmic-inflation/ http://profmattstrassler.com/2014/03/18/if-its-holds-up-what-might-bicep2s-discovery-mean/ Here's a summary of some key points in my own words (any ...


7

A common procedure to determine the spin of the excitations of a quantum field is to first determine the conserved currents arising from quasi-symmetries via Noether's theorem. For example, in the case of the Dirac field, described by the Lagrangian, $$\mathcal{L}=\bar{\psi}(i\gamma^\mu \partial_\mu -m)\psi $$ the associated conserved currents under a ...


6

There is no gravitational waves for a uniformly rotating axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fields, §88 The constant gravitational field: However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the ...


5

First it's important to note that gravitational waves do require energy to produce. A good example of this is a binary pulsar, where the emission of gravity waves carries energy away so the two pulsars spiral in towards each other and will eventually merge. Having said this, it is theoretically possible to modulate a gravitational wave and use it to ...


5

yes, the waves fall off in intensity as they get farther from the source. This does not violate conservation of energy, because you'll just be spreading the same amount of energy out over an ever larger volume, but the (energy density)*volume will be constant, minus energy transferred from the waves to matter.


5

As mentioned, this is not the first evidence for gravitational waves. The data from BICEP2 shows that there is a much higher amount of B-mode polarization than what is predicted by gravitational lensing alone. According to theory, this could only be due to higher amplitude tensor modes in the CMB than previously observed (or rather, lack of observed). These ...


4

It was not showed experimentally - the only experimental evidence of gravitational waves is that some binary pulsars change their orbiting frequency exactly as expected from their losing energy by gravitational waves of the GR-predicted intensity. However, it is surely established theoretically. The answer is Yes, gravitational waves that carry enough ...


4

in the linearized limit of General relativity, as FrankH said, all propagating perturbations of the metric are transverse. However, it must be noted that the full theory does allow for nonlinear longitudinal modes of propagation. According to the Petrov classification, such regions of longitudinal propagation are region III. There are usually not taken as ...


4

In short, if there is nothing to interact with the wave, it can't lose energy. EM and gravity waves do not experience friction with vacuum, so they just keep going. Of course, as they spread out, their energy becomes spread out as well. The power per unit area, or flux, is (somewhat trivially) inversely proportional to the area of the wavefront, so as long ...


4

At the level of understanding the data and observations we have up to now, General Relativity describes well what we perceive of the Cosmos and Quantum Field Theory what we observe in the microcosm of elementary particles and their interactions. The two have not been joined up to now, i.e. there is no accepted unified theory that joins smoothly these two ...


4

This is a very hypothetical question. a) we do not have a quantized theory of gravity working in conjunction with elementary particles ( the photon is an elementary particle). b) we have not detected a graviton in the way we detect photons If we forget a) and assume for b) that if gravitons exist they are the 0 mass carriers of the gravitational ...


3

You probably should look here first for some relatively up-to-date predictions of signal-to-noise from cosmological sources. Astrophysicists are very confident that advanced LIGO will see signals from merging compact objects like neutron star binaries and black hole binaries, on the order of a few tens per year (see first link). It is far less certain ...


3

I don't think it possible to easily tell, but here is how you might approach this. The multipole moments $T_{\ell, m}(t)$ of the energy density $T^{00}(t, \mathbf x)$ of a source can be written on surfaces of constant time in terms of spherical harmonics as follows: $$ T_{\ell m}(t) = \int d^3x \, Y^*_{\ell, m}(\theta, \phi)r^\ell\,T^{00}(t, \mathbf x) $$ ...


3

There are experimental projects (LIGO, and friends) to detect large gravitational waves from the collisions of neutron stars and black holes with other dense massive objects, but the mechanism of such waves in no way resembles waves breaking on the beach: neither in mechanism nor in mathematics. If you have some space cycles, Einstein@Home is a BOINC ...


3

Given boundary conditions allow for a unique solution of the wave equation. If you do it your way and immediately guess the retarded Green's function you are fine, but in principle there is an infinite amount of solutions which have to be fixed by boundary conditions. By imposing Sommerfeld conditions, you make sure that only the retarded solution survives, ...


3

I, also a layman in physics, after watching one of Steinhardt's lectures, became curious as to the results he mentioned as well. I found this article in my search, which seems to state that the data which he is referring to has not yet been released, and is slated for a 2014/2015 release. In particular he refers to the polarized view of the cosmic microwave ...


3

Read A.Zee, Quantum Field Theory In a Nutshell, Princeton, Chapter I.5, p 30 (first edition) In Quantum Field Theory, "forces" between 2 "charged" particles correspond to an exchange of "virtual gauge bosons". For instance, the repulsive force between 2 electrons, corresponds to an "exchange" of a "virtual photon" (a perturbation of the photon field). Here ...


3

In the nonrelativistic limit the energy lost by the system due to gravitational radiation is defined by the third time derivative of quadrupole moment: $$- \frac{d E}{dt} = \frac{G}{45 c^5}\dddot{D}^2_{ij}.$$ Where indices $i$, $j$ correspond to (flat) 3D space, and dot denotes time derivative. This equation is taken from Landau & Lifshitz' 'Classical ...


3

The Weyl tensor is the trace-free part of the Riemann tensor. The latter describes the curvature of spacetime. In the absence of sources, the trace part of the Riemann tensor will vanish due to the Einstein equations, but the Weyl tensor can still be non-zero. This is the case for gravitational waves propagating in vacuum. The physical reason is that even ...


3

Just test the invariance of the action (so using integrations by part and neglecting surface terms) under "linear diffeomorphism" $h_{\mu\nu} \to h_{\mu\nu} + \partial_\mu \epsilon_\nu+\partial_\nu \epsilon_\mu$ [EDIT] More precisions : The linear diffeomorphism is only the linearization of the standard diffeomorphism : $g^{'\mu\nu} = \frac{\partial ...



Only top voted, non community-wiki answers of a minimum length are eligible