Here's a really cool and exciting section from the ET report ( 32 of 452, or 34 of 455 ) :
Quote:
Standard Sirens of Gravity Cosmologists have long sought for standard candles that can work on large distance scales without being dependent on the lower rungs of the cosmic distance ladder. In 1986, Schutz [43] pointed out that gravitational astronomy can provide such a candle, or, more appropriately, a standard siren, in the form of a chirping signal from the coalescence of compact stars in a binary. The basic reason for this is that the gravitational-wave amplitude depends only on the ratio of a certain combination of the binary masses and the luminosity distance. For chirping signals observations can measure both the amplitude of the signal and the masses very accurately and hence infer the luminosity distance.
Wow! A good chirp observation weighs the entire system giving an absolute luminosity, the deflection at the IFO gives the apparent luminosity and bing/badda/boom you have the distance. Amazing! :-)
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
Yes, but why a 60 degrees angle? 90 degrees seems more likely.
Tullio
From the design study ( section 1.3.1 ) :
Quote:
In contrast to the traditional L-shaped geometry of the first and second generations of gravitational wave detectors this arrangement is equally sensitive for both polarisations of the gravitational wave. Additionally it shows a more isotropic antenna pattern compared to the L-shaped detectors, as shown in figure 16. The overall frequency range covered will reach from a few Hertz to about 10 kHz.
The quoted fig. 16 shows that not only is the ( overall, sky position and polarization averaged ) sensitivity up by 50% compared to a similiar dimensioned IFO with perpendicular arms but there are no nulls - that is : no choice of source direction and polarisation for which the triangular detector geometry causes nil differential phase delay for photon round trips. The essential reason is the two base gravitational wave polarisations are rotated by 45 degrees one from the other, and 60 degrees won't neatly catch that, whereas 90 degrees might.
After some long thought I think I have found a useful analogy for this : airports with multiple runways. You want to operate ( take off and land ) :
(a) nett into the wind.
Quote:
[aside] Nearly all planes virtually double their runway performance distances with minimal tailwind - you'd think it would scale linearly upon wind direction reversal but kinetic energy is quadratic and that's what you have to bleed off with linear mechanisms to bring the damn thing to a stop ( or gain to lift off ). This is also why car drivers persistently run vehicles into the rear end of the one in front - the police & driving schools have taught this for decades but no one listens. [/aside]
(b) with minimum crosswind component for stability and other reasons.
So what's the best choice on a given day ( you may choose not to take off, but you can't stay aloft forever if the wind won't change ), with a given wind direction and certain fixed runway geometry? Note that any specific runway actually represents two choices of landing direction !
If you have two perpendicular runways, the worst crosswind you ought have to suffer is at 45 degrees to both.
If you have three runways at a mutual 60 degree orientation ( Boston Logan say ) then the 'best of the worst' crosswind choices is going to be at 30 degrees ( one runway intersection to the midpoint of the opposite one ).
Interestingly, as the sin[30 degrees] = 0.5 then you get half the total windspeed magnitude as a component across the line of flight. Which is a doddle to calculate vs say 0.7071 for 45 degrees. Anyway for GW IFO's in a triangular pattern no wave can 'sneak past' undetected on account of that geometry. It may sneak past on other grounds though.
Cheers, Mike.
( edit ) And also why going "just a couple of k's over the speed limit" is not equivalent to going from zero to the same "couple of k's" in your driveway. Ever wondered if/why that hot smell from the brakes and tyres might be real important ? { Answer : your energy bleed mechanisms are failing because the materials are falling apart at a microscopic level, and such structure doesn't spontaneously reform on cooling .... so you'd have to go back up the road and pick up all the little bits yourself. :-) }
( edit ) The wide frequency range isn't due to the triangular geometry alone : each vertex of the triangle is a corner station for two IFO's, a lower and a higher frequency one, that thus have a common geometry with respect to a wave. This is dubbed the 'xylophone configuration'. The lower frequency one will be at cryogenic temperatures and uses really heavy mirror masses ( ~200kg ) and low laser power to keep radiation pressure down ( which otherwise would displace the end mirrors away from nul/lock ), plus being well underground this reduces the seismic and gravity gradient noises ( one hopes ) too. The higher frequency one uses high beam power to reduce the effect of shot noise : fluctuation in photon arrival times ( which is a quantum thingy ) that would blur the phase comparisons b/w arms. They'll be in separate vacuum cavities etc ...
The other idea is 'squeezed light' which doesn't actually abrogate the uncertainty principle but allows one to trade off the components that multiply within it. That'll go into the low power one. I don't quite understand the detailed logic here - I will inquire - but appears similiar to a NASCAR race start where you bunch up the cars ready for the 'go flag'. What isn't clear to me is precisely what you lose by doing that, as Heisenberg says you must not allow the product to go under Planck's constant ( roughly ). I need to grasp this feature better. :-)
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
Here's a really cool and
)
Here's a really cool and exciting section from the ET report ( 32 of 452, or 34 of 455 ) :
Wow! A good chirp observation weighs the entire system giving an absolute luminosity, the deflection at the IFO gives the apparent luminosity and bing/badda/boom you have the distance. Amazing! :-)
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
A short article in IEEE
)
A short article in IEEE Spectrum on the Einstein Telescope
Einstein Telescope
There are some who can live without wild things and some who cannot. - Aldo Leopold
RE: A short article in IEEE
)
Yes, but why a 60 degrees angle? 90 degrees seems more likely.
Tullio
RE: RE: A short article
)
From the design study ( section 1.3.1 ) :
The quoted fig. 16 shows that not only is the ( overall, sky position and polarization averaged ) sensitivity up by 50% compared to a similiar dimensioned IFO with perpendicular arms but there are no nulls - that is : no choice of source direction and polarisation for which the triangular detector geometry causes nil differential phase delay for photon round trips. The essential reason is the two base gravitational wave polarisations are rotated by 45 degrees one from the other, and 60 degrees won't neatly catch that, whereas 90 degrees might.
After some long thought I think I have found a useful analogy for this : airports with multiple runways. You want to operate ( take off and land ) :
(a) nett into the wind.
(b) with minimum crosswind component for stability and other reasons.
So what's the best choice on a given day ( you may choose not to take off, but you can't stay aloft forever if the wind won't change ), with a given wind direction and certain fixed runway geometry? Note that any specific runway actually represents two choices of landing direction !
If you have two perpendicular runways, the worst crosswind you ought have to suffer is at 45 degrees to both.
If you have three runways at a mutual 60 degree orientation ( Boston Logan say ) then the 'best of the worst' crosswind choices is going to be at 30 degrees ( one runway intersection to the midpoint of the opposite one ).
Interestingly, as the sin[30 degrees] = 0.5 then you get half the total windspeed magnitude as a component across the line of flight. Which is a doddle to calculate vs say 0.7071 for 45 degrees. Anyway for GW IFO's in a triangular pattern no wave can 'sneak past' undetected on account of that geometry. It may sneak past on other grounds though.
Cheers, Mike.
( edit ) And also why going "just a couple of k's over the speed limit" is not equivalent to going from zero to the same "couple of k's" in your driveway. Ever wondered if/why that hot smell from the brakes and tyres might be real important ? { Answer : your energy bleed mechanisms are failing because the materials are falling apart at a microscopic level, and such structure doesn't spontaneously reform on cooling .... so you'd have to go back up the road and pick up all the little bits yourself. :-) }
( edit ) The wide frequency range isn't due to the triangular geometry alone : each vertex of the triangle is a corner station for two IFO's, a lower and a higher frequency one, that thus have a common geometry with respect to a wave. This is dubbed the 'xylophone configuration'. The lower frequency one will be at cryogenic temperatures and uses really heavy mirror masses ( ~200kg ) and low laser power to keep radiation pressure down ( which otherwise would displace the end mirrors away from nul/lock ), plus being well underground this reduces the seismic and gravity gradient noises ( one hopes ) too. The higher frequency one uses high beam power to reduce the effect of shot noise : fluctuation in photon arrival times ( which is a quantum thingy ) that would blur the phase comparisons b/w arms. They'll be in separate vacuum cavities etc ...
The other idea is 'squeezed light' which doesn't actually abrogate the uncertainty principle but allows one to trade off the components that multiply within it. That'll go into the low power one. I don't quite understand the detailed logic here - I will inquire - but appears similiar to a NASCAR race start where you bunch up the cars ready for the 'go flag'. What isn't clear to me is precisely what you lose by doing that, as Heisenberg says you must not allow the product to go under Planck's constant ( roughly ). I need to grasp this feature better. :-)
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