# Energy of an event for a human observer at a distance of 1 astronomical unit to physically feel the gravitational waves using inverse square law?

I'm asking this question to satisfy my curiosity about how much energy (in joules) it would take for a human astronaut to feel a gravitational wave from a distance of 1 astronomical unit (AU) from two black holes radiating off energy using inverse square law? How much energy would they have to give off to be felt from a distance of 500 AU? Would the energy of such an event scale up the farther away the astronaut is from the source to feel it displace them?

My guesswork so far uses these two links:

https://www.forbes.com/sites/briankoberlein/2016/02/13/could-gravitational-waves-ever-be-strong-enough-to-feel/#52f9ebdb02f1

https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves

$$4.5 \times 10^{47}$$ joules radiated as gravitational waves from this event.

Using inverse square law and applying the distance needed to be from this event would be $$10^7$$ meters away to feel the effects of the gravitational waves from the source would mean a whopping $$3.57\times10^{32}{ joules/m}^2$$ is the intensity at that distance and using the surface area of a sphere at that distance which would be $$1.26\times10^{15} \text{m}^2$$.

I'm puzzled at how to go about finding the answer to this question because what I have so far, the intensity at that distance would completely kill the observer not just noticeably affect them with a gravitational wave and shift them by a small distance.

## 1 Answer

$$\let\om=\omega \def\ddh{\ddot h} \def\ddxi{\ddot\xi} \def\half{{\textstyle {1 \over 2}}} \def\qy#1#2{#1\,{\mathrm{#2}}} \def\10#1#2{#1\cdot10^{#2}}$$ Amusing question. Let me anticipate that energy isn't the right quantity to be considered. It's rather the wave amplitude that counts. But to understand it we have to embark on an argument neither easy nor short. It's usually believed that GR is hard because of mathematics. You'll see it's not so: there are physical concepts not easily grasped which we must carefully scrutinize.

First question - How could we feel a gravitational wave?

It's very difficult to answer since we have no term of comparison in our everyday experience. We are used being compressed because of gravity when standing upright, but a stretching experience is much rarer - I can only think of some torture practices, like being suspended by tied hands.

An important difference between the effect of a GW and the compressing or stretching I just spoke about is that the latter are incremental: when standing upright your weight burdens on legs and feet, much less on your spine, little on your neck. The opposite happens when suspended: the strongest burden is on arms, then spine, little on legs and negligibly on feet.

Instead a GW acts uniformly on the whole body - any part of it is stressed in proportion to its length, i.e. stress per unit length is the same everywhere.

Another important difference is that weight and suspension are static actions, whereas a GW is a vibration of a short duration (order of one second) and relatively high frequency ($$100\div200$$ Hz).

One might think a GW could be something like an earthquake, but this too is not correct because of the great difference in frequency. Earthquake vibrations have widely variable periods, but most of them are around some seconds. So much slower than a GW.

Second question - How would our body respond to a GW?

It's better to start from much simpler systems than human body, e.g. free bodies or bodies tied by a spring. The first case is well approximated by the suspended mirrors of GW interferometers (LIGO, VIRGO). They are "pendulums" very cleverly built, so that their suspension doesn't transmit to the mirror the slightest fraction of the environment's motions. That's only true in one horizontal direction, but it's enough: it's the direction wherefrom a laser beam comes and gets reflected thus detecting every displacement of the mirror wrt lab.

But why should a free mirror get displaced? This is a case of geodesic deviation. An incoming GW is an oscillating curvature in spacetime. The free mirror follows a spacetime geodesic, and because of curvature that geodesic doesn't keep a constant distance to a nearby one. Note that from GR viewpoint there are no forces involved: it's a purely geometrical effect. Geodesic deviation is a relative quantity (a pure number): the actual displacement is proportional to the distance of both geodesics being measured. In the approximation of small curvature (weak GW) the symbol $$h$$ is used for relative deviation. For the GW150914 event the peak $$h$$ was estimated to be about $$10^{-21}$$.

Consider now two free bodies, A and B, placed a distance $$d$$ apart. If a GW comes in, their distance will be altered to $$\xi(t) = d\,[1 + h(t)].\tag1$$ But how could such variation be detected? We could believe that all distances will be changed in the same proportion, measuring sticks included. If that were true, GW's detection would be impossible. Happily enough, things are different: solid bodies are practically unaffected by a GW, as we'll presently see.

To understand this, consider a variation of the previous setup: still two bodies A an B, of equal mass $$m$$, but connected to each other through a spring. Because of the spring eq. (1) no longer holds - it must be replaced by something like $$\half\,m\,\ddxi = -k\,(\xi - d) \tag2$$ where $$k$$ is spring's constant. (The factor 1/2 keeps into account that each mass' displacement wrt to com is one half their relative displacement.) But (2) can't be right, as it keeps no trace of the incoming GW. The right equation is $$m\,\ddxi = m\,d\,\ddh - 2\,k\,(\xi - d).\tag3$$ You may verify that in absence of GW ($$h=0$$) eq. (3) reduces to eq. (2) whereas if no spring is present ($$k=0$$) then (3) becomes $$\ddxi = d\,\ddh.\tag4$$ If $$h(t)=0$$ for $$t\le0$$ and assuming $$\xi(0)=d$$, $$\dot\xi(0)=0$$ then (4) integrated twice leads to eq. (1).

Elaborating on eq. (3)

Let's assume a monochromatic GW: $$h(t) = h_1 \cos\om t \qquad (t>0)$$ and put for short $$k = m\,\om_0^2$$. Then (3) becomes $$\ddxi = -\om^2 d\,h - 2\,\om_0^2\,(\xi - d).\tag5$$ Two limit cases are of interest:

Weak spring ($$\om_0\ll\om$$)

Then the last term in (5) may be neglected and the stationary solution is $$\xi = d\,(1 + h)$$ i.e. (1), as expected.

Strong spring ($$\om_0\gg\om$$)

This would be the rigid body limit. We may neglect $$\ddxi$$ and get $$\xi = d\left(\!1 - {\om^2 \over 2\,\om_0^2}\!\right)\!h.\tag6$$ Eq. (6) shows that the masses' oscillation is depressed by a factor $$\om^2\!/(2\,\om_0^2)$$ wrt static solution.

Estimating human strain

It remains to be seen if human body is better approximated by one or the other case (or none?). Given the frequency of incoming GW ($$100\div200$$ Hz) it's unlikely that the "human oscillator" has a proper frequency of that order. Surely much lower. Then weak spring case applies and we may safely use eq. (1). (Incidentally the same holds for a much bigger body like the whole Earth.)

I'll follow OP assumption about distance: $$\qy{10^7}m$$ from source, to be compared with the one given for GW150914, i.e. $$\qy{400}{Mpc} = \qy{\10{1.2}{25}}m$$. Thus we have to multiply $$10^{-21}$$ by $$\10{1.2}{25}/10^7$$ (remember that wave's amplitude goes as $$1/r$$, not as $$1/r^2$$) obtaining $$h = \10{1.2}{-3}$$. Finally multiplying by an average height of 1.75 m we get a strain of $$\qy{2.1}{mm}$$.

Would you feel a strain like that, lasting about 1 second and vibrating at $$100\div200$$ Hz? I don't know. Maybe you'd experience a strange shiver. Nothing dangerous, I'd bet.

Caveat

It's likely that some unwanted errors crept into my argument, especially wrong signs. I urge readers to check and let me know. Thanks.