14
$\begingroup$

In a hypothetical situation I'm still sitting in a coffee shop but a gravitational wave similar to the three reported by LIGO passes through me from behind. The source is much closer though, so this one is perceptible but (let's hope) not yet destructive. (If a perceptible event would destroy the Earth, please put my coffee shop in a large spacecraft if necessary)

Let's say the orbital plane of the black holes is coplanar to the floor of my coffee shop so that the wave is aligned vertically/horizontally. If I stand up and extend my arms, will I sense alternating compression and pulling as if I were in an oscillating quadrupole field (pulling on my arms while compressing my height, and vice versa)?

The term "strain" is used do describe the measurement, but would I feel the effect of this strain as a distributed force gradient, so that my fingers would feel more pulling than my elbows?

If I had a ball on the end of a stretchy rubber band, would it respond to this strain (especially if $k/m$ were tuned to the frequency of the wave)? Would I see it start oscillating?

There is an interesting and somewhat related question and answer; How close would you have to be to the merger of two black holes, for the effects of gravitational waves to be detected without instruments? but I'm really trying to get at understanding what the experience would be like hypothetically.

This answer seems to touch on this question but the conclusion "...if the ripples in space-time were of very long wavelength and small amplitude, they might pass through us without distorting our individual shapes very much at all." isn't enough. If it were strong enough to notice, how would a passing gravitational wave look or feel?

$\endgroup$
  • 3
    $\begingroup$ look at this youtube.com/watch?v=f9dhj3mP6F0 $\endgroup$ – anna v Jun 12 '17 at 12:24
  • 3
    $\begingroup$ @annav I've put some careful thought into this question as asked. That's a cartoon without equations or explanation. My question is about sensation/perception, those are dots. If you can answer the question as asked, please do! $\endgroup$ – uhoh Jun 12 '17 at 12:27
15
$\begingroup$

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.)

What is the effect of a gravitational wave on a physical object?

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.

How would a human experience this?

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.

Is it physically plausible that you could be exposed to high enough GW amplitudes?

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!

$\endgroup$
  • 1
    $\begingroup$ Thanks for the great answer! This is literally the visceral insight I was looking for. The idea that flesh would move relative to bone really adds a level of basic understanding. It seems that the wave "feels like" gravity, except that the direction is a function of position and time, better described in the answer of @AGML. The stiff skeleton provides the reference and the flesh is the detector, roughly speaking. It's possible at 1 million kilometers there would be a substantial dose of X-rays unrelated to the GWs; c'est la vie. $\endgroup$ – uhoh Jul 9 '17 at 2:00
  • $\begingroup$ Is it possible to add a link to a definition of proper distance that would best apply here? $\endgroup$ – uhoh Jul 9 '17 at 2:04
  • 1
    $\begingroup$ @uhoh: I did not immediately find a good reference that nicely explains the concept of proper distance in this context. Proper distance along a path is given by integrating the spacetime metric along that path (see e.g. the Wikipedia article). For two lightlike separated events the proper distance measures the light-travel time between them, and it is also what determines the propagation of electro-magnetic forces. In a curved spacetime it's important to distinguish the proper distance from the coordinate distance. $\endgroup$ – Emil Jul 9 '17 at 13:26
  • 1
    $\begingroup$ Hmm. So gravitational waves may be the closest thing there is to "sound in space", as in a mechanical distortion that propagates without a material medium - to the point you could even hear them under the right circumstances :) $\endgroup$ – The_Sympathizer Apr 6 '19 at 4:05
2
$\begingroup$

The "plus and cross" mode youtube video Anna V linked to in her comment is actually pretty much the answer.

A gravitational wave can, at large distances, be thought of as some pretty much arbitrary combination of these two modes: a plus mode, which oscillates in a plus-sign-shaped pattern compared to some reference frame, and a cross mode, which oscillates in a similar pattern rotated by a 45 degree angle.

What the wave would "feel" like depends on the specific mixture of modes you have in mind. The plus mode will alternate between stretching and squeezing you out from toe to tip while squeezing and stretching you from arm to arm (we are always free to decompose the waves into modes aligned this way). The cross mode will do the same thing, except roughly from hip to shoulder instead of toe to tip. The two modes need not have the same frequencies, so the combined effect can in principle be an almost-arbitrary three-dimensional distortion that oscillates in time. This could be lethal, agonizing, tickly, or imperceptible, depending on the amplitude, frequency, and respective importance of the modes.

At small (i.e. strong-field) distance from the source, where nonlinearities might be important, we can't understand time-dependent gravitational effects as being linear combinations of tensor modes like those above. In this regime basically anything is possible, but one would also not describe the behaviour as "gravitational radiation". By analogy with light, that term is reserved for the long-distance behaviour where the theory can be linearized.

$\endgroup$
  • $\begingroup$ Are you sure I would feel the stretching and squeezing? My arms would not be moving in space as in normal stretching, it's the space that is stretching; wouldn't my body parts be fixed in that space, and so not appear or feel to be stretching? And the ball on the rubber band, would appear to me as though someone was pulling on it? Would it start getting longer and shorter compared to a ruler? Can you find some link or source to something that addresses this? I'd like to read further. $\endgroup$ – uhoh Jul 7 '17 at 16:46
  • 2
    $\begingroup$ I don't know what "moving in space" means. But the wave would cause a relative distortion between the parts of your body, which you would certainly feel. This is the substance of Feynman's sticky bead argument: physicsbuzz.physicscentral.com/2016/06/… $\endgroup$ – AGML Jul 7 '17 at 16:49
  • 3
    $\begingroup$ I don't mean to be dismissive, as the complaint you make above was essentially the substance of the original controversy about whether gravitational waves "really exist and carry energy", which was not settled until the late 50's or so. Your objection is essentially why modern treatments describe the radiation in terms of strain, i.e. tidal forces. Tidal forces are relative distortion between physical bodies and are manifestly not coordinate bound. $\endgroup$ – AGML Jul 7 '17 at 16:57
  • $\begingroup$ Not dismissive at all, thanks for the discusion! I have to admit I don't know what "moving in space" really means either. But I'm aware of the sticky bead. In the scenario in my question the wave is coming from directly behind me and the orbital plane is parallel to my floor. Haven't I already narrowed it down to "plus mode"? Or would there still be "cross mode" in this geometry? Perhaps only if the masses were also unequal? $\endgroup$ – uhoh Jul 7 '17 at 16:59
  • 1
    $\begingroup$ In general, you need both modes, and no reorientation of coordinates can make one vanish. In coordinates that oscillate with the cross mode your body will oscillate correspondingly. Gravity is a rank-2 tensor field, and its radiation requires two orthogonal rank-2 modes: the plus and the cross. EM is a vector field and its waves require two orthogonal rank-1 modes: the E and B vectors. If what you suggest were possible you would only need two vectors to describe the gravitational wave. So gravity would be a vector theory. $\endgroup$ – AGML Jul 7 '17 at 17:09
2
$\begingroup$

I would like to approach the question from a different side. I'm not satisfied with phrases like "a GW distorts spacetime" or "a GW changes proper distance" or other more or less alike. The defect I find is that all try to explain in common words something entirely out of common experience. Misunderstanding is almost unavoidable.

Let me begin with something which has nothing to do with GW, but should approach the reader to "curvature of spacetime". We are in a spaceship, motors shut, no star or planet within several light-years. The spaceship floats free - it is standing or moving at a costant speed according to which (inertial) reference frame we use to measure its position. We may also say that the spaceship itself defines an inertial frame: its own resting frame.

Physicists within the spaceship agree to adopt a (cartesian) coordinate system based on floor and walls of laboratory. They also have high quality clocks. They engage in a complex experiment: take two balls, put in one's hands, then leave them go. No one will be surprised with the result: the balls remain where they were, all coordinates staying constant in time.

Now leave the spaceship for a while and come back to Earth. Let's do the same experiment in an earthbound lab. (Do not think I'm fooling you, please, all this is necessary.) Of course, the balls fall to ground.

But our physicists are very finicky guys and wish to measure the fall with extreme accuracy. They find this: distance between the balls decreases as the balls fall. It decreases with accelerted motion, i.e. there is a negative horizontal relative acceleration. If the initial distance was 1 meter, horizontal relative acceleration is $-1.5\cdot10^{-6}\,\rm m/s^2$. They also observe that this acceleration is proportional to initial distance, so that it is more correct to write $$a = k\,d \qquad k = -1.5\cdot10^{-6}\,\rm s^{-2}.$$

Surely all of you have understood the trivial explanation. Each ball falls with acceleration $g$ towards Earth's center, and their acceleration vectors are not parallel: they form a very small angle of $d/R$ radians ($R$ is Earth's radius). Then relative acceleration amounts to $a=-g\,d/R$. The mysterious $k$ I wrote above is nothing but $-g/R$.

There is however a slightly more sophisticated version of the last experiment: to do it in a free falling elevator (Einstein's elevator, you know). What happens now? There is no free fall of the balls wrt elevator, but the horizontal negative acceleration remains unchanged. From the viewpoint of GR a free falling elevator is an inertial frame, like the spaceship in deep space. But a difference shows between these frames, since in the spaceship frame the balls stay put, whereas in the elevator they approach each other.

According the geodesic principle (GP) of GR every free falling body follows a geodesic of spacetime, and we see a difference between geodesics of spacetime in spaceship's neigbourhood and in Earth's one. Someone could ask "where is spacetime in our experiment?" Answer: spacetime is always everywhere, and we are allowed to draw maps of regions of it.

As to spaceship's frame, we had already prepared space coordinates and we had clocks. So it's an easy matter to draw a map. On a paper leaf we draw two cartesian axes: the horizontal one we label $x$ and represents the balls' space positions. The vertical axis we label $t$ and use it to mark instants of time. In this drawing a stationary ball is represented by a vertical line: $x$ stays constant as time $t$ goes by.

What about the same map for the experiment in Einstein's elevator? We'll draw $x$ and $t$ axes as before, but balls don't stay put. They move of accelerated motion, starting from rest. So their worldlines are curved (more exactly, are parabolas with axis parallel to $x$). Very very near to vertical lines, but in a drawing we are free to choose the scales on each axis in order to make curvature visible.

Here is the difference: in spaceship's frame the balls' geodesics are parallel to each other. In Einstein's frame near Earth, they are not: they start parallel but then curve to exhibit the lessening distance of balls. Note that this distance is measurable: there is nothing conventional or arbitrary in our maps!

It is precisely such behavior which defines a curved spacetime: geodesics starting parallel and then deviating, approaching each other or getting away. So we say that near the spaceship spacetime is flat (not curved) whereas near Earth it is curved. We could also, with little effort, get a definition of curvature, a measurable quantity - but I can't allow myself that luxury. I've already been writing too much...


Up to now we have been talking about a static spacetime: loosely speaking, one whose properties remain the same at different times. This is not the case, however, of a GW, which on the contrary comes and passes by in an otherwise static spacetime. We have now a spacetime curvature which varies with time.

But the effect of a GW on our balls is not different from what it is in a static spacetime - it's only that the balls' acceleration varies with time, and lasts until the GW is present. For instance, if physicists in the spaceship have arranged things to measure the balls' distance in time, they will observe a temporary variation, maybe an oscillation, of this distance. It's exactly what GW interferometers like LIGO or VIRGO are designed to do. (A private note, if I'm allowed: VIRGO is placed at less than 10 km from my home.)


I must pause discussing how distance can be measured. The most naive way would be to use a measuring rod. (A rod with unbelievably tight marks would be needed, but don't care of it.) The real problem is another: wouldn't the rod experience the same lengthening or shortening as the distance it had to measure? If you believe that a GW changes proper distance, why wouldn't you think it happens to everything, rod included?

The answer brings good news: it doesn't happen. The reason is in the very GP. The balls get closer or away from each other because they, being free, are obliged to follow geodesics of spacetime. But the ends of the measuring rod are in a different situation: they are part of a body, approximately rigid, which will not change easily its length. There are interatomic forces that oppose it.

We can see the same fact in another setting. Suppose the laboratory's walls, floor and ceiling, which are part of the spaceship, were disassembled and left floating in their initial places. What would happen if a GW passes by? Obviously they would move like the balls (I'm neglecting complications ensuing from the peculiar character of GW, transverse tensor waves). Then it would be difficult to ascertain the motion of our balls by simply referring their positions to walls etc. But if laboratory's cabinet is kept assembled their parts are not free to move one relative to another and we will see that distances between balls and walls are changing, oscillating. The same happens wrt the ruler. (Needless to say, GW interferometers use a much cleverer method to measure distance variations, based on time employed by light to go back and forth between mirrors kms away. I can't delve on this.)


Before closing this long, too long post I have to touch another subject. How could energy be drawn from a GW? The idea is to start with our two balls, but not letting them free. We should instead connect them to a spring or something else, that be capable of doing work thanks to the balls' motion a GW causes.

Let's think a moment. Free balls oscillate, but being free do not transmit energy to anything else. If on the contrary the balls are fastened to a rigid stick, they do not move, so again no work can be obtained. Clearly something intermediate is needed: a mechanism (a spring?) which lets the balls partly free to move, and because of this motion absorbs energy from them.

A simple spring will not work as it is a conservative system: during an oscillation it returns as much energy as it had previously received. A conceptually working if absolutely impractical solution would be a double linear ratchet-and-pawl mechanism. I drew the idea from Feynman's lectures (vol. I, Ch. 46), where a rotating ratchet-and-pawl is employed to extract energy from thermal agitation.

I have no time to draw a decent figure now - I hope to be able to add it later. Let me explain the mechanism's action in words. One ball is needed, the other being replaced by the lab's wall. The ball can slide horizontally and brings a pawl which engages on a horizontal ratchet. The ensemble allows the ball to move freely to the right, whereas its left movement forces the ratchet to move too. The bar housing the ratchet is connected at its left end to a spring, whose other extremity is blocked at the left wall. The same bar also houses a second ratchet below, the pawl's pivot being fixed to the wall. This second ratchet also allows left movement of the bar.

Operation is as follows. When a GW pushes the ball to right, it moves freely. When the GW pulls the ball left it moves, training with itself the ratchet and the bar, thus compressing the spring. When the ball returns to right the lower ratchet forbids the spring to expand. The spring gets progressively compressed, accumulating elastic energy at the expense of the GW.

Of course this primitive mechanism couldn't go on working forever, even if GW's arrived continuously. Once the full length of ratchets is reached, the system has to be reset, detaching the compressed spring and letting it to do useful work on a load. But the aim of getting work from a GW is attained.

$\endgroup$
  • 3
    $\begingroup$ Could you highlight or otherwise indicate which part addresses How would a passing gravitational wave look or feel? and/or Would I see it (ball on end of stretchy rubber band) start oscillating? I'm sure it is here but for me an future readers, maybe some headings, or a tl;dr or a summary might make the answer to the question more accessible. Thank you! $\endgroup$ – uhoh Oct 14 '18 at 14:53
  • $\begingroup$ I think this is a great answer. What I still miss to understand is, that if the GW changes the geometry of space(time) how is that the atoms (e.g. of the rod) can feel that they are pulled apart. The electromagnetic force between the atoms does not have to travel longer/shorter than without the GW, since only "notion" of distance is changing, or? In other words, 1 meter rod will still (locally) think it is 1 meter long, even if an outside observer sees that it shrinks to 1 cm as the GW is traveling through the rod. $\endgroup$ – pisoir May 26 '19 at 18:01
  • $\begingroup$ @pisoir Thanks for your comment. I'm afraid I'm not able to answer your question in the limited space of a comment. Would you like to turn your comment into a new question? I'll do my best to write a satisfactory answer. $\endgroup$ – Elio Fabri May 28 '19 at 14:58
-1
$\begingroup$

The stresses and strains of a gravitational wave as seen here are those of a quadrupole source. They generate a wave, but the wavelengths are enormous.

gravwavelen

Humans being of orders of meter, and cells of microns and smaller, the energy/amplitude of the classical wave should be enormous in order for a body to be affected as a gravitational wave passes.

An analogy would be the effect on cm stick of wood, floating on a tsounami kilometre wavelength wave.

Gravitational waves should not be confused with gravity waves. Gravitational waves are expected to emerge from an enormou number of gravitons in analogy with electromagnetic waves emerging from an enormous number of photons. Gravity waves are the motion of matter by changes in gravitational interactions as are tides and earthquakes.

The effect on the body of extreme gravity waves will be seen if a body falls in a black hole, but that is a different effect than the passing of a strong gravitational wave. The gravitational wave is characterized by its wavelength and human sized "feel" and "look" detection would not notice it, imo of course. If one were close to the LIGO black hole merger, it would not be the gravitational waves a body would feel, but the gravity waves and spaghetification.

$\endgroup$
  • 3
    $\begingroup$ "enormous" is not a useful word. At some distance from the source of the wave, it will be strong enough to feel. At that distance, what will it feel like? In the 1 to 100 Hz range, it's likely to have a tactile component. If I stretch my arm by 1 or 10 microns at 100 Hz, I'm going to feel it. You shouldn't try to dismiss the question with qualitative hand-waving. There will be a distance from the source such that one will feel the effects. This is actually a good question, but it's also a hard question. $\endgroup$ – uhoh Jul 7 '17 at 15:29
  • 1
    $\begingroup$ Consider atomic or molecular absorption of IR or microwaves. There the wavelength is 10${}^4$ to 10${}^8$ times longer than the atom. As long as the frequency is right, there is an interaction.. $\endgroup$ – uhoh Jul 7 '17 at 15:37
  • 2
    $\begingroup$ @uhoh It does not hold. The atomic states are quantized states of the electromagnetic field, and of course wavelengths can be as small as possible. The human body is not in eigenstates of the gravitational interaction ( assuming it is quantized) so as to exchange energies in the way it does with microwaves. $\endgroup$ – anna v Jul 7 '17 at 15:43
  • 2
    $\begingroup$ @Rococo the gravitational interaction is so weak that any strong enough gravitational waves in amplitude s have to come from large masses. (in addition to the quadrupole form of them). Going close to the source of this wave is what I am considering, and the wavelengths are such that , imo, a body would not feel them, before getting caught by the gravity waves of the annihilation of the black holes. $\endgroup$ – anna v Jul 7 '17 at 15:47
  • 1
    $\begingroup$ @Emil this is not a motion of a body in one direction, a gravitational wave is a stretching and compressing of space itself, each cell will be distorted by its passing. I am saying that the wavelength plays a role because the cell is 10 orders of magnitude of the wavelength and will not be perceptibly stretched unless in a very strong field where gravity waves will take over before realizing the hit by the gravitational wave. I have given a link for the gravity wave. $\endgroup$ – anna v Jul 8 '17 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.