# Tag Info

## Hot answers tagged gravitational-waves

135

Gravitational waves are qualitatively different from other detections. As much as we have tested GR before, it's still reassuring to find a completely different test that works just as well. The most notable tests so far have been the shifting of Mercury's orbit, the correct deflection of light by massive objects, and the redshifting of light moving against ...

122

This represents a major misunderstanding of what a gravitational wave is. The effect presented is simply the semi-static gravitational field at earth due to the earth, moon and sun. It is predicted by Newtonian gravity. There is no 'wave' that propagated, it's the instant positions of the 3 bodies that change over 1 day (and over 1 year also). It does not ...

74

It's an incredibly misleading statement, so it's not you. Gravitational waves propagate at the speed of light, so their detection by Earth-bound detectors is expected to correlate with the arrival of light from distant events assuming the source of light generation is identical (not spatially or temporally separated) to the source of the gravitational ...

69

A wave is a traveling distortion. This goes for any type of wave. An ocean wave is a distortion of the water surface. A sound wave is a distortion in air pressure. A light wave is a distortion in electromagnetic fields. A wave is made of the thing that is vibrating--ocean waves are made of water, etc. So, a gravitational wave is made of space and time.

61

The time delay depends on the direction the wave is travelling. If it is travelling along the line connecting Livingston and Hanford then the delay time would indeed be the Livingston-Hanford distance divided by $c$: However suppose the wave was travelling normal to the line connecting the two detectors. In that case the wave would arrive at both of them ...

59

Chris' answer provides an excellent explanation as to why gravitational waves are useful to detect in general. Here's my take (as someone who works in the theory of black holes) on what is particularly interesting about the signal that was announced yesterday. Many of my thoughts are taken from the official NSF press conference and from colloquia at my ...

55

[Note: I work on gravitational waves, and am an author on several of the recent LIGO papers on GW150914 — though I am not a member of the LIGO collaboration. So if you're looking for conspiracy theories, that can be your reason to ignore me.] Zumberge, Rinker, and Faller did not measure gravitational waves. To explain this, I'll start off with an analogy, ...

47

In additions to what Chris White lists, I'd like to point to the fact that, except for a few meteorites and some dust collected on the plates of satellites and rocks from Mars (and cosmic rays and a handful of neutrinos; thanks Ruslan and Kyle Oman), until now all information reaching us from the Universe — whether it is the Sun, the more distant planets, ...

46

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 ...

46

This is the SAME gravitational wave effect measured by the LIGO researches recently It is not. Bob already gave a nice answer, but I would like to add a couple of layman analogies. Picture a pond. With simple instruments you can measure the water level over seasons and you will get some wavy trend like low in summer, high in winter, however this is just ...

44

The frequency of the recent experiment was in the audible range. The amplitude was off by unspeakable orders of magnitude. But yes, you would hear it (even in vacuum, if you were to survive). Yes, the GW are transverse (quadrupolar). But they do move things (they cause change in distances, that's actually how they detected them: the length of the 4km tube ...

38

We can already manipulate gravity like we manipulate electromagnetic waves. Tie something at the end of a rope and swing it around your head: you're now generating gravitational waves. And yes, you can transmit information with gravitational waves in the same way you can transmit information with any other modulable wave. The problem is not generating waves,...

33

The strain (ratio of displacement from equilibrium to equilibrium separation) of gravitational waves decreases as $1/r$ for a distance $r$ from the source. Since the strain in this case peaked at $10^{-21}$ at a distance of $1.3\times10^9\ \mathrm{ly} = 1.3\times10^{25}\ \mathrm{m}$, you would expect strains on the order of $1\%$ at a distance of $1300\ \... 30 Laser light generation is intimately related to processes that generate single photons. To date, gravitational waves have not been detected, and there are no known processes that produce single gravitons (not to mention there is no direct evidence that the gravitational field is quantized at all -- just logical arguments based on the structure of general ... 24 Yes, gravitational waves will undergo the same red-shift as any wave that propagates at$c$. There were probably very violent gravitational waves in the very early universe. If those waves hadn't been red-shifted, they'd be ripping us apart right now. If so, could observations of them be used like red-shifted electromagnetic waves from distant sources ... 23 Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves. Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its ... 23 From the FAQ of LIGO: LIGO uses two basic strategies to shield the detectors from vibrations of the Earth. They are referred to as “passive” and “active” vibration isolation systems. Basic "quad" pendulum to demonstrate passive damping through suspension. LIGO’s passive vibration isolation system absorbs vibrations before they reach the all-... 22 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 ... 22 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 ... 22 I have a direct quote from the website: The event would not have registered in LIGO's first-generation detectors; the fact that it appeared with striking clarity in both L1 and H1 indicates the leap in detector performance that the Advanced LIGO program has produced. This was a sensitivity issue: in most frequencies, Advanced LIGO is more strain-... 21 Suppose you have two black holes of the same mass$M$and$m = GM/c^2$. The radius of each black hole is then$r = 2m$, and the horizon area is$A = 4\pi r^2 = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational ... 20 Adding briefly to Chris' answer. Gravitational waves are not obscured by anything. If detectors are made to work at lower frequencies (in space) then they can "see" gravitational waves originating from beyond the cosmic microwave background right back to the inflationary epoch. Another thing that has become clear today is that binary mergers give a chirp ... 20 To expand on HDE's answer, initial LIGO indeed wouldn't have detected GW150914, but it's not quite as simple as the peak strain being below the curve in the sensitivity plot: the integration time also matters. These plots can be misleading; the curves they show don't represent a minimum detectable strain. Indeed, the units on the y-axis of these plots are$...

20

In General Relativity the effect of gravitation is represented by the stress energy tensor, $T_{μν}$: $$R_{μν} - \frac{1}{2}Rg_{μν} + \Lambda g_{μν} = \frac{8\pi G}{c^4}T_{μν}$$ where $R_{μν}$ is the Ricci curvature tensor, $R$ is the scalar curvature, $g_{μν}$ is the metric tensor, $Λ$ is the cosmological constant, $G$ is Newton's gravitational ...

19

Have a look at an announcement from LIGO where they describe the experiment. The first plot shows the frequencies detected. The original waves are redshifted. Estimated source parameters for GW150914. We report the median value as well as the range of the 90% credible interval. Masses are measured in the source frame; to convert masses to detector ...

18

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: http:/...

18

Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately: $$P = \frac{32}{... 18 The simple Newton-like explanation of dipole gravitational radiation unexistence is following. The gravitational analog of electric dipole moment is$$ \mathbf d = \sum_{\text{particles}}m_{p}\mathbf r_{p} $$The first time derivative$$ \dot{\mathbf d} =\sum_{\text{particles}}\mathbf p_{p}, $$while the second one is$$ \ddot{\mathbf d} = \sum_{\text{...

17

It is a misconception that LIGO is a very accurate instrument, it has an uncertainty in calibration which is on the order of 10%. This means that the measured strain amplitude of GW150914 of $1.0 \cdot 10^{-21}$ could easily have been $1.1 \cdot 10^{-21}$. Note that this is just a scaling error. LIGO is however extremely sensitive, it can measure relative ...

16

It didn't. The key is that the wave is not point-like, it has an extended "wavefront". The wavefront/event just arrived at the two locations with that delay, which gives some information about its direction. You might also be interested in: Is it really possible to break the speed of light by flicking your wrist with a laser pointer?, which discusses ...

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