general-relativitygravitational-wavesgravity
A very layman question as in title. Like every wave having a negative side, can a gravitational wave have anti-gravity.
To put it in different words, a gravitational wave passing through a complete vacuum, if in positive cycle, can create a denser space-time, in it's negative cycle, create a rarer space-time?
Part 1: no, the amplitude decays as 1/r. The power or energy of the wave decays like the square of that, like 1/r^2 (one over r squared). This is the same as in electromagnetic waves, the gravitational wave amplitude (in General Relativity) components obey the same equations as that for electromagnetic waves for the electric or magnetic field. The reason is that although forces go like 1 over r squared, that can be also used to see how the electric field (or Newtonian gravitational field) changes with distance r, but it is true for STATIC or slowly changing fields, not true for radiating fields, eg, propagating waves.
To figure out propagating waves you need to figure out how the time change of the fields (electric, magnetic or gravitational, they give the same results) change. You need to derive the wave equations for those. It then comes out that the fields go like 1/r. The fields are the amplitudes, A. The power or energy in all cases those cases goes like A squared. That holds true for radially propagating electromagnetic fields and gravitational fields. For gravitational fields it is the metric of the spacetime in Einstein's General Relativity equations that define the fields. For gravitational waves, in the linearized case, the waves are represented as small changes, with amplitude A, of the background metric. In both cases it is the square of A that is the power or energy (and actually it is power or energy density, per unit area), which then goes like 1/r squared.
It is easy why it must be so. Realize that those energies are conserved as they propagate. So the total energy at a distance r, summed around the whole spherical wavefront (yes, consider them radiating away radially symmetrically) is then the power or energy density, multiplied by the area of a sphere. Since area = 4 x pi x r^2, the energy or power density must go like 1/r^2, so the total power or energy is the same at any spherical distance r. It's energy conservation that leads to that behavior.
Please be aware that these statements for the gravitational field are true for linearized gravity, i.e. For Einstein's equations, but where the waves or disturbances in the gravitational field, the metric of spacetime, is small. For very strong gravity and gravity changes things get much more complicated because Einstein's equations are nonlinear, gravity interacts with itself, and gravity is also be a source of a gravitational field. Things get complicated and you need some understanding of General Relativity to understand what that really means. If you are interested, there's plenty good online lectures and summaries (and bad ones also, careful).
Part 2: the decay of gravitational wave amplitude or power or energy density. Does it depend on time?
It might (remember that in General Relativity you can change coordinates and mix space and time)@ but in the linearized Einstein equations it does not, if you stick with the flat coordinate system as the reference frame. It'll depend on distance. And yes, then there is so called lensing, which is waves passing close to bodies some of the gravity and distort spacetime (in your terms, all of those together really just create the curvature of spacetime), and thus space near them, and thus a bending of the waves like lenses do. So waves going near them will bend, and typically travel a longer distance (but it all depends on the overall distribution of mass around their overall path), and arrive delayed. We can measure the so called lensing delays for electromagnetic waves and actually used it to measure the mass densities around various galaxies and galaxy clusters. We can even measure the mass densities of dark matter around them. We have not yet observed gravitational wave lensing, but it is expected and will be used to measure more details from bodies or objects that are not visible through electromagnetism.
So, yes gravitational waves will vary their amplitude and power with the longer lensing distances, and be delayed. The effects will be small, and the amplitude changes will be much harder to detect accurately than the timing delays. The space based gravitational interferometer to be launched in the next few years will be able to see the delays, not sure about the distance variations through the amplitude changes, but time delays determines distance anyway so we'll measure those distance changes also.
Let me try to answer in a few separate steps. (I'll try to make it simple and people should correct me where I oversimplify things.)
Let's start with just two atoms, bound to each other by interatomic forces at a certain effective equilibrium distance. A passing gravitational wave will start to change the proper distance between the two atoms. If for example the proper distance gets longer the atoms will start to experience an attractive force, pulling them back to equilibrium. Now, if the change of GW strain happens slow enough (for GW frequencies far below the system's resonance) everything will essentially stay in equilibrium and nothing really happens. Stiff objects will keep their length.
However, for higher GW frequencies, and especially at the mechanical resonance, the system will experience an effective force and will be excited to perform real physical oscillations. It could even keep ringing after the gravitational wave has passed. If they are strong enough, these oscillations are observable as any other mechanical oscillations.
All this stays true for larger systems like your example of a ball on a rubber band or for a human body. It is also how bar detectors work.
So, a gravitational wave exerts forces on your body by periodically stretching and compressing all the intermolecular distances inside it. That means you will basically be shaken from the inside. With reference to the stiffer parts of your body the really soft parts will move by the relative amount that is given by the GW strain $h$. The effect can be enhanced where a mechanical resonance is hit.
I guess you would experience this in many ways just like sound waves, either like a deep rumbling bass that shakes your guts, or picked up directly by your ears. I assume that within the right frequency range the ear is indeed the most sensitive sense for these vibrations.
Lets take the GW150914 event where two black holes, of several solar masses each, coalesced. Here on Earth, an estimated 1.3 billion lightyears away from the event, the maximum GW strain was in the order of $h\approx 10^{-21}$ at a frequency of about $250\,\mathrm{Hz}$. The amplitude of a gravitational wave decreases with $1/r$, so we can calculate what the strain was closer by:
Lets go as close as 1 million kilometres, which is about 1000 wavelengths out and so clearly in the far field (often everything from 2 wavelengths is called far field). Tidal forces from the black holes would be only about 5 times higher than on Earth, so perfectly bearable.
At this distance the strain is roughly $h\approx 10^{-5}$. That means that the structures of the inner ear that are maybe a few millimetres large would move by something in the order of a few tens of nanometres. Not much, but given that apparently our ears can pick up displacements of the ear drum of mere picometres that's probably perfectly audible!
Best Answer
Gravitational waves, though transverse, can be thought of as similar to sound waves:
A sound wave, as it moves through a medium the sound wave creates alternating volumes of greater and lesser particle density.
Gravitational waves do something similar, except the medium is spacetime itself. The result is that as a gravitational wave passes through a region of space, at one crest the spacetime is "stretched" in one direction and contracted in the perpendicular direction, like when you stretch a rubber band and it gets narrower. At the trough of the wave, the same thing happens, except the direction that was contracted is now stretched and the direction that was stretch is now contracted. This is why the good ol' perpendicular lasers and mirrors trick worked for detecting them.