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Congrats to the LIGO team on the announcement of their discovery of gravity waves!

The articles I've read say that the distortion we see here is much smaller than a proton. What about at the source? Would these waves have been strong enough to see macroscopic effects near the binary black holes themselves? Could you orbit the system at a "safe" distance, observe the merger, and "feel" the waves?

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We are "orbiting" at a safe distance, aren't we? What's a macroscopic effect for you? Obviously the effect here was still macroscopic enough to see it, wasn't it? Is there a distance at which you could feel the wave in your stomach? Probably... not sure it would be safe for other reasons, though. At that distance the gravity would certainly not do more to you than the acceleration in a roller coaster. Having said that, the frequency at which LIGO made the detection will probably not be near the peak sensitivity of your stomach. – CuriousOne Feb 11 at 17:01
the press conference has nice videos – anna v Feb 11 at 20:13
up vote 7 down vote accepted

According to the press announcement, the merger of the black holes released an amount of energy equivalent to three solar masses (an energy given by $E=mc^{2}$ with $m$ being three times the mass of the sun), which is an absolutely mind-boggling amount of energy. If you were too close, the rapidly fluctuating tidal forces on your body would cause alternating compressive and tensile stresses on your body, tearing your body to bits, which would certainly count as a "macroscopic effect".

Sure, at an appropriate distance, you'd still be able to feel the gravitational wave without being killed. However, with such an enormous amount of energy being released from two objects that are quite small and close together by astronomical standards, for the experience to be survivable you'd have to be so far away that you wouldn't really be able to "observe the merger" with the naked eye. I.e., you wouldn't be able to see two black blobs in the sky that merge into one.

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Why would you be vaporized? Wasn't this energy released as gravitational waves? – user20936 Feb 11 at 21:52
@MichaelT Good point. After the rapidly fluctuating tidal forces tore your body to pieces, the pieces would perhaps then be too small for the tidal forces on them to generate enough heat from friction to cause vaporization. I didn't do a detailed calculation. I have change my answer's cause of death accordingly. – Red Act Feb 11 at 22:28
If you only have distortions of ≈1% in a distance of ≈1000km I don't think that you would have to be so far away that you couldn't see anything with the naked eye to survive the event. It also does not matter how much energy is involved, what matters is the power (energy per time). – Симон Тыран Feb 12 at 0:26
@СимонТыран If you try to stretch a bone in a few milliseconds such that the bone is "only" 1% longer than it normally is, the bone is going to snap. If you try to squeeze a skull in a few milliseconds to be "only" 1% narrower than it normally is, that's likely to be fatal. A uniform 1% strain on a human body is a lot. – Red Act Feb 12 at 2:13
This sounds like a good subject for XKCD's "What If?" feature. – Barmar Feb 12 at 8:38

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 $1300\ \mathrm{km}$. For reference, the observed signal is from black holes that were about $100\ \mathrm{km}$ in radius initially.

Much closer than this, and the linearized theory of GR breaks down. Gravitational waves are only well-defined in the small-amplitude limit. Close-in, we have the near-field regime where nonlinear effects that can't really be described as simple waves dominate. The only simple statement that really can be made here is that distortions are even stronger closer in.

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Is this how they estimated the distance of the event? By comparing the observed strain to what computer models using GR predict by looking at waveform? – dualredlaugh Feb 11 at 18:47
@dualredlaugh Yes, exactly. – Chris White Feb 11 at 18:47
So at 1300km, I'd be shrinking and growing + or - 1% of my height, basically? – John Feb 11 at 20:04
@John : You're slightly more squishy. – MSalters Feb 11 at 23:25
@John Freely floating objects would move that way, if you factor in the frequency, you can find the effective force (calculate the acceleration), which is then proportional to the mass. Your body will experience this effective force. – Count Iblis Feb 12 at 0:12

Knowing that the emitted energy is of about 3 solar masses, and if we use the famous Einstein energy formula $E=mc^2$ we have then :

$\text{A.N.:}\ \ E=1,989\cdot10^{30} \times 3 \times c^2\approx2\cdot10^{43} \ \text{J}$

which is a huge amount of energy. This is a bigger emission of energy than a supernova (!) and similar to gamma-ray bursts (biggest emission of light after the Big-Bang)...

enter image description here

Making "popular science", this is the caloric intake (energy) that $5\cdot10^{34}$ Nutella pots of 400kg would give you.

If that amount of energy would be inside the gravitational waves passing threw us, we would be looking like spaghettis. Our body and Earth would be visibly distorted or elongated. Maybe the word "feel" is not appropriate because it would be more "die". Orbiting the system at a safe distance is plausible, knowing that we have to take care the combined ISCO (innermost stable circular orbit or last stable orbit) of both black holes before they merge. We can determine it by :

$R_{ISCO} = \dfrac{6GM}{c^2}$

where $M$ is the mass of the black hole in $kg$ and $G$ the universal gravitational constant in $N\cdot m^2\cdot kg^{-2}$ and $c$ the velocity of light in $m\cdot s^{-1}$. We then have, for the most massive black hole (here $36$ solar masses) :

$R_{ISCO} = \dfrac{6\times6,67384\cdot10^{-11}\times36\times1,989\cdot10^{30}}{299 792 458^2}=319,0\cdot10^3m=319,0km$

For the ring-down black hole of $62$ solar masses, we have ${{R}_{rd}}_{ISCO}=5,494\cdot10^5m=549,4km$

enter image description here

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Great visual aids, thanks. – John Feb 15 at 16:09
My back-of-envelope calculation has the energy of the black hole event roughly matching the nutritional energy of enough Nutella to fill the orbit of Venus. – Richard Gadsden Feb 17 at 12:02

protected by Qmechanic Feb 13 at 16:03

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