HN Gopher Feed (2017-09-27) - page 1 of 10 ___________________________________________________________________
Gravitational waves from a binary black hole merger observed by
LIGO and Virgo
299 points by nickcw
https://www.ligo.caltech.edu/news/ligo20170927___________________________________________________________________
ricardobeat - 4 hours ago
What are the practical applications of technology that could be
derived from these discoveries? Any chance of us harvesting this
energy?
sleavey - 2 hours ago
(LIGO collaboration postdoc)There have been plenty of
technologies pioneered by the collaboration in the development of
the detectors - control of optical cavities via radio frequency
signals being one big one, and low thermal noise monolithic
silica fibre suspensions being another. The discovery of
gravitational waves, however, simply improves our understanding
of the universe and has no known practical applications.It's
great to live in a time when countries are willing to fund such
bold leaps into the unknown, without obvious monetary pay-offs.
throwaway613834 - 4 hours ago
You mean from this distance? Or do you mean if a closer one
occurred?
stefco_ - 3 hours ago
Hi, I'm a grad student working on LIGO.There is no practical way
to harvest this energy. Gravitational waves interact very weakly
with matter. This is part of why they are useful for
observations: GWs can travel from the core of violent events
unmodified, unlike light, which gets blocked by surrounding
matter. You can't see the center of a core collapse supernova
with light, but you can see it with GWs and neutrinos. They also
won't get bent by EM or gravitational fields, meaning that the
direction they come from in the sky points straight back to the
source. And unlike very high energy gamma rays, they don't pair
produce into electrons.This all makes them great for astronomy
and totally useless for extracting energy.
chebucto - 1 hours ago
What happens to that energy, though? Energy is conserved, even
with gravitational waves, correct?So, what happens to the
energy - does some of it get converted into 'normal' potential
energy by changing the (amount of gravity? gravitational
fields?) of the stuff that it passes through - the weak
interaction that you refer to?And, if it interacts very weakly,
does all of the energy eventually get used up this way - do GWs
get 'used up' before they travel across the universe?(Probably
a meaningless question): can GWs hit the edge of the universe,
and if so, what happens then?
vorotato - 1 hours ago
Finding aliens using warp drives! :D ....
dpcx - 4 hours ago
A coworker earlier mentioned a paper questioning the validity of
the findings around gravitational waves. Can anyone point me to
them and (as a non-scientist) explain why this paper would think
those findings are inaccurate?
ISL - 4 hours ago
Context (too busy getting physics done today to expound further,
except to state that I presently agree with
LIGO):https://www.wired.com/story/strange-noise-in-
gravitational-w...
InclinedPlane - 42 minutes ago
Your coworker might be confusing the LIGO observations (widely
considered to be some of the most meticulously done research in
science) with the BICEP2 early big bang gravitational wave
research, which has since been called into question
substantially.
batmansmk - 4 hours ago
Can you coworker provide some info instead?
dpcx - 4 hours ago
He could provide information about the paper, sure. The
explanation, I'm fairly certain, he could not.
sleavey - 2 hours ago
LIGO made an unofficial response via the blog of Sean Carroll
[1]. The problem with the analysis in the paper that questioned
LIGO's results was that they misinterpreted the filtering that
had been applied to the data shown in the original GW150914
paper, which led them to think there was noise produced by the
detectors themselves.[1]
http://www.preposterousuniverse.com/blog/2017/06/18/a-respon...
autocorr - 3 hours ago
From an observational astronomy point of view, perhaps the most
interesting thing about the report is that by including the Virgo
detector with both LIGO detectors, the uncertainty in the direction
the gravitational wave event was reduced by a factor of 10! The
"error ellipse" on the sky is now ~60 square degrees as opposed to
~600. This makes it much more feasible to do rapid observational
follow-up with other observatories at all wavelengths (radio,
optical, X-ray, etc.), to search for the counterparts.Many
observatories like the VLA, Chandra, Fermi, Gemini, have a rapid
response trigger for the detection events, where within minutes
they will stop their current operations. Observations in the EM
spectrum are crucial for learning about a huge number of questions
in astrophysics, like what kinds of galaxies do the merger events
originate from, where are they located in those galaxies, how does
the luminosity decay at different wavelengths, etc. It will be
seriously exciting when the first counterpart is discovered!
c3534l - 2 hours ago
A factor of 10 would be impressive enough, but a factor of
3,628,800 is incomprehensible.
vorotato - 1 hours ago
you could say it's astronomical
viraptor - 21 minutes ago
> a rapid response trigger for the detection events, where within
minutes they will stop their current operationsIs that a large
enough margin? What would be typical delay between noticing
gravitational wave event and seeing the related EM waves?
sleavey - 3 hours ago
Here's a skymap [1] that really drives home the point about how
much having a third detector helps.[1] http://www.virgo-
gw.eu/skymap.html
stefco_ - 2 hours ago
Having extra detectors also helps hugely with uptime.We're
currently only active and collecting clean data ~50% of the
time for each detector. We need at least two detectors active
for a detection, and need three detectors for this kind of
beautiful directional isolation. So having extra detectors
decreases our skymap areas significantly, which helps in EM
followup, but it also significantly reduces GW detector network
downtime. And it allows for graceful degradation; with 5
detectors, you can afford to have a couple detectors down and
still detect an event with good direction reconstruction.We
also have tons of downtime for upgrades. Now that our second
observing run is over, we have a ~1.5yr upgrade cycle. There's
no point in running VIRGO without LIGO (except for engineering
reasons); having more detectors will increase flexibility in
this regard, too.When KAGRA and LIGO India come online, we
should see much better skymaps, but we will also see much
higher uptimes (and hence higher detection rates).It's
especially exciting to see improved direction reconstruction
given that VIRGO has faced setbacks this year with their
mirrors. I wasn't expecting to see such beautiful skymaps until
our third observing run! Needless to say it was an exciting
summer :)- grad student working on LIGO
saganus - 1 hours ago
1.5 years for an upgrade?Can you expand on why it takes so
much time to do an upgrade?I'm guessing it's due to the
complexity of the system, kinda like with the LHC? or is it
something different?I mean, having equipment that expensive
on downtime for more than a year seems like a lot!
sleavey - 56 minutes ago
The required time is due to how complicated the instruments
are, and how hard it is to tune them such that they achieve
the sensitivities they are designed to achieve across such
a wide frequency band.In the downtime until O3, new optics
are to be installed. As these are 40kg each and suspended
from monolithic fused silica fibres (each a couple hundred
microns thick), this process is complicated and requires
great care to avoid damaging either the fibres or the
optics during installation. In addition, both LIGO and
Virgo are going to get squeezed light sources installed,
which allow for reduction of limiting quantum noise at high
frequencies. These are immensely sensitive to light loss in
the optical path, so commissioners need to be very careful
to minimised scattered light and ensure good
alignment.After making all of the planned upgrades, the
interferometers then need to be recommissioned: all of the
control loops (of which there are thousands) need to be re-
tuned from their settings as of now, given the changes in
the behaviour of the interferometer due to the upgrades.
This takes a lot of person power and a lot of time to get
right - for the first observing run this process was
basically also 1.5 years.
saganus - 37 minutes ago
Wow!Great explanation. Thanks!I just can't start to
fathom the complexity of these instruments.Also... how
cool that you get to work on such interesting field and
with such cool "lab" devices!
rcthompson - 4 hours ago
> about 3 solar masses were converted into gravitational-wave
energy during the coalescenceThis is such a staggering amount of
energy to be released in such a short time.
leeoniya - 4 hours ago
more or less than https://en.wikipedia.org/wiki/Gamma-ray_burst ?
throwaway613834 - 4 hours ago
I think less? When a gamma-ray burst is pointed towards Earth,
the focusing of its energy along a relatively narrow beam
causes the burst to appear much brighter than it would have
been were its energy emitted spherically. When this effect is
taken into account, typical gamma-ray bursts are observed to
have a true energy release of about 10^44 J, or about 1/2000 of
a Solar mass (M) energy equivalent. [1][1]
https://en.wikipedia.org/wiki/Gamma-ray_burst#Energetics_and...
rcthompson - 4 hours ago
I think another sentence in that article might need
revision:> No known process in the universe can produce this
much energy [1 solar mass equivalent] in such a short time.
danharaj - 4 hours ago
The wikipedia article says that if a GRB were a spherical
release of energy, they would be on the same order. However,
GRB's are highly anisotropic; most of the energy is expelled in
a smaller cone. One of the estimates on the page is that a
typical GRB expels 1/2000th a solar mass of energy. So these
gravity events are thousands of times more powerful.These
scales are incomprehensible, at least to me.
ntumlin - 4 hours ago
From the wiki page:> When this effect is taken into account,
typical gamma-ray bursts are observed to have a true energy
release of about 1044 J, or about 1/2000 of a Solar mass (M)
energy equivalentSo more, by about 6000 times.This is
consistant with WolframAlpha's output for amount of energy:
5.361e47 J. I tried to convert this to "number of years we
could power the Earth", but it just changed it from one mind
boggling number into another.
dragonwriter - 4 hours ago
> > When this effect is taken into account, typical gamma-ray
bursts are observed to have a true energy release of about
1044 J,Copy-paste error here from losing sup/sub formatting,
that's 10^44 J, not 1044 J.
[deleted]
jobu - 4 hours ago
Didn't realize that energy could be converted into gravitational
waves. It does make sense I suppose, but I always thought gravity
and gravitational waves were just a side-effect of the way mass
interacted with space-time.
SomeStupidPoint - 4 hours ago
Can't you radiate potential gravitational energy as
gravitational waves without any sort of particle changes,
because potential gravitational energy is stored in the
configuration of the system?If I drop my bowling ball, it has
lower potential energy, but can't some of that be compensated
for by the Earth-ball system radiating a gravitational wave
(rather than all being converted into kinetic
energy)?Similarly, don't orbiting bodies emit gravitational
waves? I forget the exact dynamics, but the way that they drag
just a little bit as they orbit creates waves that radiate
power out of the system -- I think the Sun and Jupiter radiate
something like 55 watts of gravitational waves.If you take two
things that are each tens of times larger than the Sun and set
them spinning an appreciable amount of the speed of light very
nearby -- to the point they eventually collide -- it's not
surprising they emit quite a bit more than our solar
system.Think of it this way:Two large objects start orbiting
slowly far apart. But because they're large and the system is
big, they have a lot of orbital momentum.They pull each other
close, but as they get closer, that orbital momentum has
nowhere to dissipate, so they orbit faster and
faster!......Except that because they're so massive, they
slightly tug space around with them as they orbit. And because
they're orbiting so fast, space doesn't have a good way to
smooth itself out. So some of the momentum from the heavy
things spinning far apart gets stored in wrinkles in space as
they come together.Which then radiates away from there to us,
...and slightly stretches one direction of reality as it passes
by. Which we detect by measuring how long two perpendicular
lines are very accurately a few different places on a nearly
spherical object....So you can model it as just the way that
mass interacts with spacetime if you remember to include things
like the distribution of your mass as a source of potential
energy that can itself be used to generate gravitational waves.
However, this does not make the whole situation any less weird
to explain.Reports just tend to convert the amount of energy
released through mechanisms like that in terms of mass, because
it's the only thing vaguely comprehensible. (eg, 3 suns)
teraflop - 3 hours ago
> If I drop my bowling ball, it has lower potential energy,
but can't some of that be compensated for by the Earth-ball
system radiating a gravitational wave (rather than all being
converted into kinetic energy)?Nothing wrong with your
comment, but I just want to emphasize that for objects
smaller than stars, the amount of energy carried by
gravitational waves ranges from tiny to unimaginably tiny.
The Earth-Moon system emits about 7 microwatts of
gravitational waves. A bowling ball in low earth orbit would
radiate about 5*10^-41 W, which is roughly equivalent to one
photon of visible light every quadrillion years.
WhoBeI - 3 hours ago
According to Einstein there is no gravity that pulls us to the
center of Earth. Instead mass curves spacetime in such a way to
make you experience traveling along the curvature as gravity.
If the mass is rotating the curvature will also drag on the
surrounding spacetime. Imagine something like a whirlpool and
then consider what happens if two of those merge. While merging
some of their energy will cause waves in the surrounding water
that eventually fades away with range. Those waves would be the
gravitational waves - ripples in spacetime.
lomnakkus - 2 hours ago
Incidentally, I also find the "geometrical" approach a great
way to look at the whole "light vs. black holes" thing: It's
not that black holes directly "trap" light or "prevent light
from escaping" per se, it's more that spacetime gets curved
so much that regardless of which direction a ray of light is
traveling it'll always find itself back inside the event
horizon.(Not sure where I first heard this explanation, but I
can definitely say that I didn't originate it!)
Florin_Andrei - 3 hours ago
Well, if it's a wave, it carries energy. That energy has to
come from somewhere. And relativity teaches us that energy and
mass are the two sides of the same coin.
InclinedPlane - 45 minutes ago
In a system of binary black holes where one is 31 solar masses
and the other is 25 they have event horizon radii of 93 and 75
km, respectively. At the point their event horizons start to
touch they have a relative distance of 168 km. Each of them
have a gravitational potential energy in the other's
gravitational field of G m1 m2 / r or 1.22e48 Joules. Both of
them together have a total gravitational potential energy of
2.44e48 Joules or 1.36e31 kg (6.8 solar masses).(In reality the
mass contribution due to the gravitational potential energy is
included in the "mass", this is just illustrative.)
wlesieutre - 4 hours ago
EDIT - Thanks to bonzini for catching a mistake, I'd typod 47 as
74 on the first line and drastically inflated everything after
that. It's still huge though.Wolfram Alpha gives a solar mass as
1.988435?10^30 kg. Multiply that by (299,292,458 m/s)^2, you get
~1.78?10^47 Joules.The most powerful manmade explosion in history
was Tsar Bomba, a Soviet hydrogen bomb with 210 Petajoule
(210?10^15 J) yeild.Amount of energy released by the black holes
merging is 8.48?10^29 times greater, so you'd need to set off
that many Tsar Bombas to get equal total yeild.The universe is
4.3?10^17 seconds old, meaning if you set off a Tsar Bomba every
second since the beginning of time, you'd still be off by a
factor of 1.953?10^12.You need about 2,000,000,000,000,000,000
Tsar Bombas every second since the big bang to match it.The mind
boggles.
bonzini - 4 hours ago
That 10^74 should have been 10^47, shouldn't it?So it's "only"
a million billions (10^15) tsar bombas (10^15) every second
(10^17, 15+15+17=47).
wlesieutre - 4 hours ago
Good catch, can't believe I flipped that! Certainly changes
things, one second while I revise.
wlesieutre - 4 hours ago
EDIT - originally compared the last factor to number of grains
of sand on Earth squared, but after catching the error above
that's much too large.So how big is the remaining 2x10^12 bombs
per seconds?Wikipedia has some other suggestions: ~10^12
stars in the Andromeda galaxy 1.98x10^12 links on Wikipedia
3.04?10^12 trees on Earth in 2015 3.5?10^12 estimated fish
in the ocean So take every link on Wikipedia (or two thirds
of the trees on Earth), turn each one into the largest hydrogen
bomb ever detonated, and blow them all up every second since
the beginning of the universe, and that's how much energy we're
talking about.Thanks again to bonzini for spotting the rather
large error!
rcthompson - 4 hours ago
Funnily enough, the past two XKCD comics are both relevant
here: https://xkcd.com/1894/ https://xkcd.com/1895/
bas - 4 hours ago
Nicely (back of) napkinned!
wlesieutre - 3 hours ago
Originally typo'd pretty hard and had numbers much too
high! It's still a larger amount of energy than I can even
imagine.That's what I get for trying to do math in a