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annaandherdad
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Hi careysub, the power flux falls off as the inverse square, but the amplitude of the wave falls off as the inverse power of the distance. The
10^{-21} quoted is the amplitude of the wave. So you'd have to be 10^{20} times closer to get an amplitude of 10%. So that would be 10^{-11} light
year or 3 x 10^{-3} light second or 1000 km, if my rough calculation is right. This sounds about right, the amplitude is basically 1 at the
coalescence of the black holes.
Any other SF Bay chemists?
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hissingnoise
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Hawking says in this article that he would like to use gravitational waves to test his area theorem: that “the area of the final black hole is greater than the
sum of the areas of the internal black holes.” He adds: “This is satisfied by the observations.”
The comments are a read, too . . .
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phlogiston
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Quote: Originally posted by careysub  | | Quote: Originally posted by phlogiston |
Given the extreme distances between cosmic events and earth and the resulting 'dilution' of the signal with distance, I am still a little puzzled that
local signals of low absolute intensity are not far more easily detected.
[Edited on 12-2-2016 by phlogiston] |
The frequency issue is part of it. Another part is how often these events occur.
If by "local" you mean the Local Group of galaxies for example (Milky Way, Andromeda, Triangulum, plus dwarf galaxies) then the volume of the whole
observable Universe is 100 billion times bigger. We are far more likely to see very distant events.
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Good point, but no, I meant way more local. Consider the adjacent room. Sorry for being vague.
I understand that in the much smaller volume of 'local space' there will be fewer detectable events.
What puzzles me is why it is not possible to detect the gravitational effect of accelerating masses moving very near LIGO, ie meters away instead of
billions of lightyears.
For electromagnetic signals, it is trivial to produce an artificial signal many times stronger than any 'natural source', including the sun, using
simple means (big lamp, laser), if measured near enough to the source.
This should hold for gravitational waves too. Would smashing two large blocks of osmium together meters away from one of the detector arms not yield a
stronger signal than a black hole merger billions of light years away?
[Edited on 14-2-2016 by phlogiston]
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"If a rocket goes up, who cares where it comes down, that's not my concern said Wernher von Braun" - Tom Lehrer |
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Oscilllator
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phlogiston if you read careysub's answer he mentioned that ~10% of the mass of the black hole is converted to gravitational wave energy. When you
smash two blocks of osmium together, 0% of the energy is converted to gravitational waves. That's probably why.
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j_sum1
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two blocks of osmium sounds like an expensive way to go. W or even Pb is a more likely option for gravity experiments.
But yes, Oscillator is right. You need something that will make some sort of gravitational wave. Playing with normal sized masses might not do it.
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annaandherdad
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Here are some contributions to the discussion of the comparison between electromagnetic radiation and gravitational radiation. I'll start with some
atomic physics.
Electromagnetic radiation arises when charges are accelerated. An electron orbiting in an atom is accelerated, so the atom radiates. As it does
so, the electron moves closer to the atom (a lower energy state). As the size of the electron orbit shrinks, the orbital frequency increases, as
does the velocity of the electron. However, the total energy of the atom, kinetic plus potential, decreases because of the emitted radiation.
As far as this goes, this is very similar to what happens with the orbiting black holes. The main difference is that the energy states of the atom
are quantized, which prevents the electron from spiraling ever closer to the nucleus and emitting ever higher frequency radiation. This is a
"catastrophe" predicted by classical physics which is prevented in reality by quantum mechanics. Since the atom has a ground state, the inspiral of
the electron stops there. In the case of the orbiting black holes, what prevents this catastrophe is that when the two black holes are close enough
together that their event horizons merge, then they combine to form a single, larger black hole, and radiation ceases.
Another important difference between the electromagnetic and gravitational cases is that electromagnetic radiation is dipole, while gravitational is
quadrupole. More exactly, the electromagnetic radiation from a localized source can be decomposed into a sum, dipole+quadrupole+octupole+ ... but if
the source (the "antenna") is "small" (in a manner described precisely below) then the quadrupole contribution is much smaller than the dipole, etc.
In the case of gravity, the radiation sum is quadrupole+octupole+ ..., that is, without the dipole term. Again, if the "antenna" is "small", then
the leading term (the quadrupole) dominates. Here I am glossing over the difference between electric and magnetic poles, and the analogous
distinction in gravitational waves.
In atoms, sometimes the dipole term is forbidden by selection rules, so one only sees the quadrupole term. In this case, the intensity of the
spectral line is much weaker than other transitions of a comparable frequency, which are allowed as a dipole transition. For example, in hydrogen,
the 3p->1s is allowed as an electric dipole transition, but 3d->1s only as a quadrupole. The quadrupole line is roughly 10^4 times weaker than
the dipole line; both have almost exactly the same frequency. The factor of 10^{-4} is physically (v/c)^2, where v is the electron velocity,
multiplied by various factors of order unity.
It is shown in atomic physics that the transition rate (in sec^{-1}) for an electric dipole transition between states A and B in a single-electron
atom is
$$w=\frac{4}{3} \frac{\omega^3 e^2}{\hbar c^3}
|\langle A \vert {\bf x} \vert B \rangle|^2. $$
These formulas are in Gaussian units; if you prefer SI, replace e^2 by e^2/4 pi epsilon_0. Here omega is the frequency of the photon emitted. If
you multiply this by hbar*omega, the energy of the photon, you get the power emitted by the atom:
$$P=\frac{4}{3} \frac{e^2 \omega^4}{c^3}
|\langle A \vert {\bf x} \vert B \rangle|^2. $$
This formula is derived via quantum mechanics, but it is closely related to the classical Larmor formula for the power radiated by a charge e
undergoing harmonic oscillation of frequency omega and amplitude l:
$$P=\frac{2}{3} \frac{e^2 \omega^4 \ell^2}{c^3}$$
Although this formula applies to a charged particle undergoing harmonic oscillation, apart from the prefactor 2/3 the same formula applies for a
charged particle in a circular orbit, where l is the radius and omega is the frequency of the orbit.
The atom is a little antenna, for radiating electromagnetic waves. Now there are two kinds of antennas, small ones and big ones. An antenna is
small if its length is small compared to the wavelength of the radiation, otherwise it is large. An atom is a small antenna, since its size is about
a thousand times smaller than the wavelength of the light emitted (in simple electronic transitions). And the Larmor formula also only applies to a
small antenna, where
$$\ell \ll \lambda$$
Here lambda is the wave length,
$$\lambda = 2\pi \frac{c}{\omega}$$
To bring out the dimensionless (and small) ratio
$$\frac{\ell}{\lambda}$$
we can rewrite the Larmor formula as
$$P= \frac{e^2\omega^2}{c} \Bigl(\frac{\ell}{\lambda}\Bigr)^2$$
where I'm dropping all numerical factors like 2/3 and 2*pi.
An important point is that the dimensionless ratio (l/\lambda) is the same as (v/c) for the charged particle undergoing harmonic or circular motion,
apart from factors of 2, 2*pi, etc. So if you have a single charged particle radiating by undergoing some motion, circular or harmonic for example,
the antenna will be "small" if the velocity of the particle is much less than the speed of light (the particle is nonrelativistic). Then these
simple formulas for the power radiated apply.
Antennas that are not "small" are much harder to analyze, because at a fixed time the phase of the wave is not constant over the extent of the
antenna.
Now this is electric dipole radiation. The power radiated by electric quadrupole radiation is similar, except the dimensionless ratio (l/lambda) is
raised to the 4th power (and the numerical factors in front are different). That is, there is another factor of (v/c)^2 in the formula for the power
radiated. So for electric quadrupole radiation, the power is
$$P= \frac{e^2\omega^2}{c} \Bigl(\frac{\ell}{\lambda}\Bigr)^4$$
The only thing that has changed is the exponent of (l/lambda) which is essentially (v/c).
Gravitational radiation does not exist in dipole form, the leading order multipole is the quadrupole. And if you take the formula above for the
electromagnetic power emitted in quadrupole radiation and just replace e^2 by GM^2, for two equal masses M in circular orbit around each other, where
G is Newton's constant, then you get the power radiated by gravitational waves:
$$P=\frac{GM^2\omega^2}{c} \Bigl(\frac{\ell}{\lambda}\Bigr)^4$$
Here l is the radius of the orbit, and if you reexpress lambda in terms of omega and c, this formula becomes
$$P = \frac{GM^2\omega^6 \ell^4}{c^5}$$
Here omega is the orbital frequency, and the antenna is "small" if the radius is small compared to the wavelength. But since gravitational waves
travel at the speed of light, this means that the antenna is small as long as the orbital velocity is small compared to the speed of light.
For the orbiting black holes in the recent event, their orbital velocities are only a fraction of the speed of light up until the final stages of the
merger. When the merger finally starts to take place and the event horizons are starting to merge, then the orbital velocities are becoming
comparable to the speed of light and the gravitational quadrupole version of the Larmor formula, quoted above, ceases to be valid. But it works ok
up until the final moments.
More later.
[Edited on 14-2-2016 by annaandherdad]
[Edited on 14-2-2016 by annaandherdad]
Any other SF Bay chemists?
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careysub
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| Quote: Originally posted by phlogiston |
Good point, but no, I meant way more local. Consider the adjacent room. Sorry for being vague.
I understand that in the much smaller volume of 'local space' there will be fewer detectable events.
What puzzles me is why it is not possible to detect the gravitational effect of accelerating masses moving very near LIGO, ie meters away instead of
billions of lightyears.
For electromagnetic signals, it is trivial to produce an artificial signal many times stronger than any 'natural source', including the sun, using
simple means (big lamp, laser), if measured near enough to the source.
This should hold for gravitational waves too. Would smashing two large blocks of osmium together meters away from one of the detector arms not yield a
stronger signal than a black hole merger billions of light years away?
[Edited on 14-2-2016 by phlogiston] |
Refer to my posts above, and check out the Caltech link I included. An ordinary oscillating mass produces gravitational wave energy in the order of
10^-43 watts. But the LIGO detection detected a pulse with the intensity of 0.1 watts/M^2.
That is why, Detection takes a signal with a lot of energy, Ordinary oscillating masses produce incredibly small amounts of energy.
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j_sum1
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I can think of another issue with an experimental setup involving conventional masses.
Imagine you had a large dumbell mass and were spinning it at a few thousand rpm to try to produce a gravitational wave. Your detector would
necessarily have to be close. It would be near impossible to filter out the comparatively enormous effect of mechanical shaking that would be detected
along with any gravitational signal generated.
[Edited on 14-2-2016 by j_sum1]
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Marvin
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The practical problem there is that you could potentially measure gravity, but unless you can spin the masses at a large fraction of the speed of
light you'd never measure the delay caused by the distance.
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careysub
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Thanks annaandherdad, keep it coming!
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annaandherdad
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You're welcome. I edited the post above to make it more comprehensible. I'll add more later.
Any other SF Bay chemists?
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phlogiston
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The Fermi gamma-ray space telescope recorded a gamma ray burst that arrived just 0.4 seconds after GW150914.
http://gammaray.nsstc.nasa.gov/gbm/publications/preprints/gb...
The region of sky where it happened also matches with that of the source of GW140914. It is very likely that the photons and the gravity wave were
produced in the same event.
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"If a rocket goes up, who cares where it comes down, that's not my concern said Wernher von Braun" - Tom Lehrer |
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careysub
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Holy Cow! We really are in a new era of high energy astronomy!
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annaandherdad
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This is fantastic and incredible news. The article is a bit disappointing, because it doesn't speculate much on how there could be an
electromagnetic signal associated with the black hole merger. But that's the point---there was no reason to expect one, and that group just made the
observation. It's hard to believe that the gamma ray signal is not associated with the black hole merger.
Black holes often emit x-rays, but this comes from the accretion disk, matter that is spiraling inward toward the black hole and heating up in the
process so hot that it emits x-rays. As the article said, no one expected there to be accretion disks around these small black holes. (Small in
the sense that they are not supermassive.)
Now there's this very interesting question that every theorist in the world has already jumped on, what could give rise to an electromagnetic signal
when black holes merge?
Any other SF Bay chemists?
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annaandherdad
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You can take the formula for the gravitational power radiated and play some games. You can plug in numbers, for example, for two pyramids of Giza
orbiting around each other at a distance of 500m and a velocity of 1 km/sec.
Here is a slightly different approach. Suppose you have two lead balls of some mass M connected by a cable. The cable has some tensile strength,
say, it's enough so that when the force on the cable exceeds 100 Mg, where g=9.8 m/sec^2 is the acceleration of gravity, the cable will break. In
other words, the cable can support 100 times the weight of one of the balls before breaking. We'll ignore the mass of the cable. Then
$$\ell \omega^2 = 100g$$
Now with this constraint, do we get more gravitational power radiated if the cable is long or short? (The velocity of the masses is adjusted to make
the tension on the cable just short of the breaking tension.) To answer this, we use the relation above to eliminate omega from the power formula,
which gives,
$$P=\frac{GM^2}{c^5}(100g)^3 \ell$$
So longer cables, at constant tensile strength, generate more power. Suppose we are limited by the velocity of the lead balls, at v_max. Then
$$\omega\ell = v_{\rm max}$$
or, with the constraint on the tension,
$$\ell = \frac{v_{\rm max}^2}{100g}$$
so
$$P=\frac{GM^2}{c^5} (100g)^2 v_{\rm max}^2$$
When I plug in M=1000kg and v_max =100m/sec, I get a power of about 3 10^{-15} watts. Not encouraging for generating gravitational radiation via
terrestrial sources.
By the way, when the gravitational wave passes through the earth, it causes small motions in the body of the earth. These in turn radiate their
own gravitational waves, which are the scattered wave. Although the earth is very massive, the numerical factors are very small, and the scattered
radiation is very small. In fact, gravitational waves pass through ordinary matter almost completely unscathed. This is part of what may make them
useful in astronomy, they can be detected from places where light cannot get out.
One way to see why these numbers are so small is to express the power in terms of (v/c) of the mass, which is the same as (l/lambda). The quantity
P/omega is the energy emitter per cycle; it is
$$\frac{P}{\omega} = \frac{GM^2}{\ell} \Bigl(\frac{v}{c}\Bigr)^5$$
It is the factor (v/c)^5 that makes this number so small for terrestrial radiators. The prefactor, GM^2/l, is the Newtonian gravitational potential
of the two bodies at a distance of l.
By arranging a collection of radiators in a phased array, you can make the power radiated grow as the square of the number of radiators, but you still
won't get much power in ordinary terms.
[Edited on 15-2-2016 by annaandherdad]
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neptunium
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No detection of GR would have been a major blow to general relativity and physics . this is indeed welcome news. The emission of light in the form of
GRB is a curious phenomenon .
Maybe have its source in the quantum world. Large release of energy are often accompanied or generate light.I cannot wait to see what LIGO and VIRGO
have in store waiting for the next generation of GW detectors! I've just listen to an interview with the LIGO team who said they could also "see" the
ringing of the final black hole dissipating ... Isn't that amazing?
[Edited on 16-2-2016 by neptunium]
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