# Energy released by GW150914

Mr Anderson
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Topic 211283

I was reading that the energy released by GW150914 in gravitational waves consumed 3 solar masses and the peak energy output was greater than the combined output of all the stars in the observable universe. So, I was wondering, how that would have affected anything in the general area. As I understand it, since this event involved black holes, no electromagnetic energy was released and assuming there was no other matter around to produce radiation, a nearby observer is not going to get roasted by X-rays etc but what effect would the gravitational waves have on an observer that would be say 1 AU away? Would there be any effect? If so, what about 1 light year away?

Shawn Kwang
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### incrediballs wrote:I was

incrediballs wrote:
I was reading that the energy released by GW150914 in gravitational waves consumed 3 solar masses and the peak energy output was greater than the combined output of all the stars in the observable universe. So, I was wondering, how that would have affected anything in the general area. As I understand it, since this event involved black holes, no electromagnetic energy was released and assuming there was no other matter around to produce radiation, a nearby observer is not going to get roasted by X-rays etc but what effect would the gravitational waves have on an observer that would be say 1 AU away? Would there be any effect? If so, what about 1 light year away?

I'm not fully equipped to answer your question so I asked professor whom I work with. He gave me some order-of-magnitude figures.

First, you are correct in that no EM energy is released in the the binary black-hole merge event GW150914.

The way to roughly calculate the effect of gravitation waves, over a 1 meter distance, is the Schwarzschild radius divided by the distance ( r_sch / r ).

Let's assume the Schwarzschild radius is for a 30 solar mass black hole. That calculates r_sch to 87 km. Round up to 100 km = 10^5 m. If you are at 1 AU = 10^11 m, that's 10^-6 m of strain over a 1 m 'body'. I micrometer of strain, at a few Hertz isn't really going to be noticeable to a human observer. So you would have to be closer, e.g. 10,000 km (10^7 m), to feel a 1 centimeter strain. For comparison, the radius of our sun is 5.6x10^7 m.

Einstein@Home Project

mmonnin
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### That's hard to fatham a 3

That's hard to fatham a 3 solar masses of energy basically disappearing w/o any change on the outside universe. I know it didn't really disappear but w/o EM it would seem like it did. Just a crap ton of energy to go poof.

Mr Anderson
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### Thanks for the information.

Thanks for the information. That's really interesting. It seems weird that so much energy can have so little effect but I suppose the difference is that unlike EM, the gravitational waves are so poorly (or not at all?) absorbed, which of course is why they are not obscured by other matter in the universe.

mmonnin
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### It makes me wonder where all

It makes me wonder where all the energy ends up. We've detected GW from over 1billion LY away so I guess every single atom within the 1bi LY radius (and beyond) would have moved along with the wave. Does that movement transfer some energy of the GW into momentum like ripples on a lake? Then eventually transferred into heat, etc? Or is it different since space is what's moving and not the atom within space?

Mike Hewson
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### mmonnin wrote:It makes me

mmonnin wrote:
It makes me wonder where all the energy ends up. We've detected GW from over 1billion LY away so I guess every single atom within the 1bi LY radius (and beyond) would have moved along with the wave. Does that movement transfer some energy of the GW into momentum like ripples on a lake? Then eventually transferred into heat, etc?

This can be answered at many levels. I'd go for the 'structure of spacetime' being an alias for 'force of gravitation' as per the theory of Mr Einstein. This makes gravitational waves the propagation of structural alterations of spacetime. In turn masses respond to said changes eg. the test masses at the interferometers. Since our interferometers convert spacetime strain changes to photon count variations then indeed there is a teensie weensie absorption of the GW's energy. Otherwise how else could we notice that something had occurred ? One technical way to say this is "impedance mismatch" or that the mechanism to transfer energy across those interfaces that we call our interferometers is horribly inefficient. FWIW this is why our gravitational wave detectors won't ever be used in reverse to radiate signals.

Now at this point in the thinking, one has to accept that GW's dissipate in strength almost exclusively by expansion into the vast volumes of the universe. A bigger wave can hence cause alterations at greater distances than smaller waves. For the inspirals/collisions as detected the several Sun's worth of mass that 'disappear' has an energy equivalent ( mc2 ) which is diluted across spacetime when all is said and done. I could confidently bet you any wager  you care to name that any randomly chosen cubic metre of the Universe will contain some CMB photons, about as many neutrinos, but precisely nothing else. Doug Adams had it right about the extents of space. Structures like galaxies, solar systems, stars, planets, chatty apes with GW receivers are rare to a degree beyond belief.

Quote:
Or is it different since space is what's moving and not the atom within space?

Choose one or the other really. Both are valid but not in the same reference frame. You can say the atoms are static but the metric has changed ( metric = local structure of spacetime ) OR the atoms have moved with respect to each other because of the tidal effect of gravity at slightly different locations. Again it is a question of which alternate labelling you prefer to use, as mentioned above : 'structure of spacetime' being an alias for 'force of gravitation'.

Side Note : in that regard Special Relativity is a particular case of General Relativity. In SR spacetime is 'flat' ie. not curved and neither is there any mention of gravity.

Side-Side Note : if one takes a sufficiently small chunk of spacetime ie. a tiny spatial volume for even the merest brief moment, then it is flat regardless of any curvatures about overall. This is Einstein's Principle of Equivalence ! :-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal