The Gravitational-Wave “Revolution” Is Underway

Exactly. As soon as you magically remove the gravitational body, you are magically removing the waves too according to GR. There is no such thing as curved spacetime without mass. (Except for the cosmological constant, but that's different.)

It is a thought experiment, you have to engage with it charitably.

Edited for pedantry. :)

A blackhole traveling at near light speed is pretty darned close to the analogy of a massive object instantaneously disappearing, similar to fictional spaceships engaging their warp engines.

Of course, it's not actually disappearing, just moving, but the original point was about detecting sharp changes in the gravitational waves. A quick Google search tells me that gravitational wave red shifting is a thing, and I imagine that with blackholes it's a very important phenomenon and area of study. And I would guess that there can also be interesting second-order effects that such a blackhole's movements have on the propagation of gravitational and electromagnetic waves from other objects.

Eh, it's a gedanken experiment, it's done all the time in physics. The cautionary advice ought to apply without saying.

An object moving and an object vanishing are the same from the perspective of wave propagation, the only difference is that one event will have a more dramatic (therefore easier to visualize) effect.

If the sun instantaneously vanished, we would see it disappear at the same instant as its gravitational effect stops, 8 minutes after the actual event occurred. For those 8 minutes while the light and gravitational information are in transit, the Earth will continue to revolve around a visible (though now nonexistant) sun.

In the same way as if the sun suddenly jerked ten million miles to the south, we would see it move at the same instant as its gravitational force vector changed, 8 minutes after the actual event occurred, but that's harder to keep in your head.

They do, it's just of very-very tiny amplitude

Source: Bachelor at GTR

A ripple of water traveling along the surface of a vortex still experiences the curvature of the vortex, even though both are disturbances. (this is a terrible analogy, please take with a ladle of salt)

Firstly, somewhat nontechnically: if you think of a binary as being two ends of a barbell with an infinitesimally thin bar and heavy weights, and put the observer in the plane of rotation and far from the binary, sometimes one of the weights on the end of the "bar" is closer, sometimes one eclipses the other so the weights line up, sometimes both are equidistant. What a pair of accelerometers at the observer reports depends on the geometry, and in particular the orbital phase. We have, in effect, a giant Cavendish experiment. If we put a large compact mass between the binary and the observer, then the light image of the binary forms a https://en.wikipedia.org/wiki/Einstein_ring . Since experimentally we know that accelerometers point to the visible (and radio and so on) image of a massive system, we would expect that gravitational radiation of all sorts gravitationally lensed just like light. We usually just make a formal analogy on that, which I'll deal with a few paragraphs further down.

Slightly technically: the "curvature of spacetime" in your question is the metric tensor field ("the metric") which fills the whole of spacetime. Yes, when we have a binary like above, the metric is dynamical. We can deliberately fix a background metric chosen for calculational or conceptual ease, and capture much of the metric dynamics as small perturbations on the background.

In this picture, we usually get gravitational waves by imposing a coordinate time on a relevant part of the spacetime, and then finding which small perturbations obey an appropriate time-dependent massless wave equation.

More physically, what this picture is saying is that in the case of a single binary, if an observer stays at one spatial location and from time to time checks its accelerometers, and they will point to the retarded positions of the objects in the binary. If you put a gravitational lense between the binary and the observer, one is free to encode the lense as a perturbation, or one can add it into the background. (The former is more popular for reasons I'll explain further below.)

Alternatively, it might help to think of gravitation as a gauge theory, wherein one can only measure potential differences between two points in spacetime rather than some absolute potential.

Let's start by defining a background wherein the potential is everywhere identical and calling that the vacuum expectation value (vev). If our background has some static spherical mass on it, we can measure a potential difference between fairly-close-to-the-mass and far-from-the-mass[1]. The potential difference is not time-dependent.

As we add any such mass we may update our vev a little, but at enormous distances from all the masses, it's only a very little, so we can largely take a value much closer to our set of masses.

But when we add in more than one mass, we lose time-independence.

If the initial conditions are picked so the masses do not orbit or twist around each other, our masses will obey Raychaudhuri's focusing theorem, and we can reason using shell theorem or Birkhoff's theorem, and expect no gravitational waves detectable outside the collapsing system. On the other hand, if we let our masses fall into orbits, we get gravitational waves.

In the focusing-only case we end up seeing vev - potential_near increasing during the collapse, where near is at a large finite radial distance (in Schwarzschild coordinates) from the collapsing mass, and we measure vev at radial infinity. In the orbiting case, vev - potential_near has a long term tendency to increase, but will vary slightly depending on the orientation of the measurement points to the binary.

A large mass -- like a galaxy cluster -- far from the source is just a region where the potential departs from vev. Helpfully, when the lense mass is that large and the GW frequency is high (e.g. near the end of a binary inspiral, or quantitatively greater than ~ 1 Hz), we can draw a direct formal analogy with electromagnetism: the geometrical optics approximation [2]. Following gravitational wave analogy with https://en.wikipedia.org/wiki/Geometrical_optics we map the difference in potential from the vev to a difference in index of refraction and let the GW plane wave refract. It's useful to remember here that GWs are modelled in a linearization of the Einstein Field Theory and so treating galaxy clusters as linear media or galaxies as nested shells of linear media is a reasonable approach. Unfortunately, in general wave optical solutions are only obtainable numerically.

Using the full GR, you have a metric with pretty much no symmetries and good luck with the calculations. Your starting point for extragalactic GW sources would be a "swiss cheese" like Einsten-Straus 1931 with multiple vacuoles and then combine that with a pp-wave, probably perturbatively, ideas which have been explored in the literature although usually in cosmological/primordial GW contexts [3]. (Compare with a more formal statement of the approach in previous paragraphs, e.g. in §1.2 at http://aether.lbl.gov/www/classes/p139/homework/hw12.pdf )

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[1] If we are sufficiently close to Newton, by splitting spacetime according to a slow-moving-compared-to-light observer not very close to any compact massive objects then we can can use : https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity#Pois...

[2] Thorne, K. S. 1987, in Three Hundred Years of Gravitation, ed. S. W. Hawking & W. Israel (Cambridge: Cambridge Univ. Press), p. 361, and Isaacson [1967] https://doi.org/10.1103/PhysRev.166.1263 (PDF version https://drum.lib.umd.edu/bitstream/handle/1903/17447/Isaacso... )

[3] for some pointers, see the Introduction section of https://arxiv.org/abs/1307.4371

Yes. We actually detect gravitational waves because photons we created did not behave the way we expected.

I see what you're aiming at, but technically it's not true. It can be similar though.

The point about how we can only infer their existence from a mass of data is similar to how we talk about deep learning. The pretrained neural net works, but we can't pinpoint why, we can only test it on various data and receive results.

There is no "dog that didn't bark", because we clearly can see the graphs and the correlations.

The way we detect gravitational waves is not "the photon haven't arrived", but "the photon has arrived a few nanoseconds earlier than we expected it". (But multiplied by a few thousands of reflections and detected only by interference pattern, not the photon arrival itself).

So, in a way, it's similar to deviations from paths by which we presume dark matter. But also, it's not that: we are almost certain here that these exact deviations are caused by the gravitational waves, because calculations and because multiple unconnected detectors, and because the form of the graph matches.

Oh come on, just collapsing some million black hole pairs to get the latest JavaScript based ad across the whole Galaxy. Let's not pretend there are still people with a 56k gravitation modem out there.

We will send you thin information even if you don't want it! Otherwise, there isn't much obstacles in empty space.

So do neutrinos. They're a lot easier to make.

No, both gravitational waves and electromagnetic waves decrease in intensity with the square of distance.

(This is true of any wave in 3D space, because the area covered by a wavefront increases with the square of distance. The inverse square law is a consequence of conservation of energy.)