Phillip Helbig (undress to reply) wrote:
David Staup writes:=20
This surprises me, the equivalent of 3 solar masses radiated away in
less than a second from 96 million miles away and we wouldn't notice?

It's not just the energy, but rather the effect it produces on
whatever it interacts with (or not).

Right. I just ran some numbers, and I came to the remarkable
conclusion that the total solar power output (4E+26 watts) could
harmlessly pass through you if it was in the form of gravitational
waves rather than heat and light.

I used the formula c^3 h^2 f^2 pi / (8 G) to convert strain to flux.
G is the gravitational constant, h is the strain, c is the speed of
light, and f is the frequency in Hz (250 in this case). The peak
strain of the recent event was 1E-21, so I get a peak flux of 10
milliwatts per square meter.

No wonder they always list the sensitivity of LIGO in terms of strain
rather than in terms of watts per square meter. The latter doesn't
sound nearly as impressive! Indeed, if the event had given off light
rather than gravitation waves, it would have been not only bright
enough to see from here, but bright enough to read by!

As a sanity check, I divided the reported peak power output of the
event, 3.6E49 watts, i.e. 200 solar masses per second annihilated, by
the area of a sphere 1.3 billion light years in radius. I get about
20 milliwatts per square meter. What accounts for the factor of two
discrepancy? Probably polarization. LIGO, if I understand correctly,
is sensitive to only one of the two polarizations.

("Only" 3 solar masses were annihilated, because the event lasted less
than a second.)

Lets get closer to the event and see what happens. I hope you're
reading this with a fixed font.

Distance flux (W/m^2) strain N

1.3E25 m (1.3E9 ly) 1E-2 1E-21 4E33
1.3E22 m (1.3E6 ly) 1E+4 1E-18 4E39
1.3E19 m (1.3E3 ly) 1E+10 1E-15 4E45
1.3E16 m (1.3 ly) 1E+16 1E-12 4E51
1.3E13 m (66 AU) 1E+22 1E-9 4E57
1.3E10 m (8 M miles) 1E+28 1E-6 4E63

The last column is the number of gravitons per square meter per
second. I get that by multiplying the flux by the frequency and
dividing by Plank's constant.

In each case, I assume you're floating in space, in a good spacesuit,
facing toward the event.

I assume that a strain of one part in a million isn't going to hurt
you, especially if it's front-to-back rather than head-to-toes. Note
that that last distance is much less than 1 AU. 1E+28 watts per
square meter -- your cross-sectional area is probably roughly one
square meter -- means 25 times the sun's total power output is going
through you. I wonder what it would feel like.

Of course I'm also assuming it was a "clean" event, i.e. nothing but
gravitational waves was given off. If it consisted of nothing but two
black holes, that's pretty much certain. But if there was other stuff
in the area, all bets are off. Indeed, there was a weak gamma ray
burst half a second after the event, which may or may not be a
coincidence. We don't know the direction of either the event or
the gamma ray burst, except very roughly.

Supernovae radiate a huge amount of energy in neutrinos, but these
hardly affect anything else.

Neutrinos aren't nearly as stealthy as gravitons. According to
Randall Munroe, a typical supernova will emit 1E57 neutrinos, and
they will be lethal at about 2 AU. During the peak tenth of a second
of the event at the closest distance I list, 100,000 times as many
gravitons will harmlessly pass through you as the *total* number
of neutrinos given off by a supernova!

--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.