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0 pointsswombat10y ago0 comments

On the other hand, we may well detect the 3 solar masses radiated away as energy. That decreases as an inverse square law, so as one solar mass is about 10^30kg, and 1kg gives off about 10^17 J, we're talking about an explosion releasing something around 10^47 J. For comparison, a 1 kiloton nuclear bomb gives off about 10^15 J.

So, inverse square that explosion... 1 light year is about 10^16m, so we square that and get 10^32m, so we're now talking about ... 10^15 J.

So, unless my maths is all off (which is possible), if this happened about a light year away, whoever's on the side facing towards the blast wouldn't get to observe very much because they'd feel as if a 1kt nuke just went off above their head. Not a great way to start the day.

Chances are it would wipe out life on Earth too, through the ensuing side-effects like lighting the atmosphere on fire, sterilising half the planet, significantly heating up the oceans, possibly even stripping part of the atmosphere away, etc.

For a great novel based around a strikingly similar premise to what was just observed (and the main reason I even bothered to calculate this), Diaspora by Greg Egan is a fantastic book.

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bzbarsky10y ago

The big question is how much of the energy would get transferred in practice.

I agree that 3 solar masses worth of electromagnetic radiation at 1 light year distance would feel like a nuke going off. What I don't know is to what extent the energy of the equivalent gravitational waves (which _would_ have a lot of energy I agree) would actually get transferred to things we care about, like the atmosphere and us. If it's a few percent, say, we'd clearly be in trouble. If it's more like what neutrinos do, it would probably be detectable but probably not by unaided human senses.

I tried doing some quick looking around for estimates of gravitational wave coupling and energy transfer and didn't find anything so far...

platz10y ago

I would like to understand why a gravitational wave distorts length in relation to normal gravity wells; specifically is this particular to waves? Why don't lengths get distorted in a normal gravity well, or do they? In essence, what is different between a gravity wave and a gravity well, which i understand both distort space, but only the wave distorts it in a way we can measure? Does the gravity well change lengths proportionally in all directions and thus isnt measurable?

bzbarsky10y ago

A gravity well also distorts lengths, as best I understand (which is not very well, to be honest; take everything I'm saying here with a big grain of salt).

The difference in terms of detection is that the wave does this in a time-varying, periodic fashion.

For something like LIGO, we're trying to measure length changes on the order of 1e-18 meters. We're not actually measuring the lengths of LIGO's arms to that accuracy, though. What we're measuring is the difference between the times light takes to travel down those arms. And even that's hard to measure on an absolute scale, so what we really measure is how that difference changes in time.

Or put another way, the effect of Earth's gravitational well is not really distinguishable from inaccuracies in making the two legs of the interferometer equal length to start with, and is a much smaller effect than those inaccuracies. Again, if I understand this right...

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swombatOP10y ago

That is a good point - perhaps it would all come out as neutrinos or gravitons... then we'd be fine. I doubt such an event would result in no electromagnetic radiation at all however... Why would it? The creation of a new black hole typically releases enormous amounts of all kinds of energy, electromagnetic as well as in the form of neutrinos.

bzbarsky10y ago

I really don't have any good answers here; this just isn't something I know very much about....

Florin_Andrei10y ago

If the gradient is small (as it should be 1 ly away) then the coupling ought to be very weak.

pja10y ago

The energy was all dumped into the gravitational waves we detected, not into electromagnetic radiation: Gravity waves don’t interact with matter very much (the cross section of the graviton is believed to be extremely small) so the quantity of energy transferred to matter as the wave passes through is likewise extremely small. I haven’t run the numbers, but I’m not sure you’d notice this even from a light year away without fairly sensitive detectors.

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0 pointsswombat10y ago0 comments

So, inverse square that explosion... 1 light year is about 10^16m, so we square that and get 10^32m, so we're now talking about ... 10^15 J.

For a great novel based around a strikingly similar premise to what was just observed (and the main reason I even bothered to calculate this), Diaspora by Greg Egan is a fantastic book.

0 comments

bzbarsky10y ago

The big question is how much of the energy would get transferred in practice.

I tried doing some quick looking around for estimates of gravitational wave coupling and energy transfer and didn't find anything so far...

platz10y ago

bzbarsky10y ago

A gravity well also distorts lengths, as best I understand (which is not very well, to be honest; take everything I'm saying here with a big grain of salt).

The difference in terms of detection is that the wave does this in a time-varying, periodic fashion.

1 more reply

swombatOP10y ago

bzbarsky10y ago

I really don't have any good answers here; this just isn't something I know very much about....

Florin_Andrei10y ago

If the gradient is small (as it should be 1 ly away) then the coupling ought to be very weak.

pja10y ago

j / k navigate · click thread line to collapse