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chornedsnorkack
2016-Jun-18, 06:23 AM
When a binary black hole spirals in due to gravity wave emission - are there any directions of space into which gravity waves are not emitted for reasons of symmetry?

korjik
2016-Jun-18, 10:54 AM
If there is any direction with no gravity waves, it would be along the axis of rotation. I dont know my GR well enough to know if that is true tho.

ShinAce
2016-Jun-18, 11:54 AM
My understanding is that you are both correct. We could say it is due to reasons of symmetry, but the truth is GR was built upon a specific symmetry. So no surprise there.

I believe the multipole expansion of the stress-energy tensor gives us a quadrupole for gravitational radiation when the source has v << c. Which means the only place you won't measure any radiation is along the axis of rotation.

chornedsnorkack
2016-Jun-18, 12:58 PM
For the gravity wave emissions in September and December 2015, is it possible to measure from the arriving gravity waves what the angle was between the axis of binary hole and the direction to Earth?

Crawtator
2016-Jun-18, 01:08 PM
just a layman, so pardon this question if it's stupid...

Is there any attenuation of waves towards the axis of rotation? Wouldn't that affect the measurements concerning the size of the originator bhs? Or is it predicted to be a firm line where detection of radiation stops?

ShinAce
2016-Jun-18, 01:55 PM
To measure; I don't think so.
To infer; I don't see why not.

JCoyote
2016-Jun-18, 02:15 PM
There is a third gravity wave detector being brought online for the sake of gaining better directional information on them.

swampyankee
2016-Jun-18, 02:24 PM
Just looking at the geometry -- two massive points separated by a relatively small distance -- I'd think that the radiation pattern would be that of a dipole emitter.

Strange
2016-Jun-18, 03:40 PM
For the gravity wave emissions in September and December 2015, is it possible to measure from the arriving gravity waves what the angle was between the axis of binary hole and the direction to Earth?

The parameter estimation paper has this (about 150): https://cplberry.com/2016/02/23/gw150914-the-papers/#parameter-estimation

chornedsnorkack
2016-Jun-18, 07:49 PM
Just looking at the geometry -- two massive points separated by a relatively small distance -- I'd think that the radiation pattern would be that of a dipole emitter.

But unlike electromagnetic radiation, which is dipole radiation, gravity waves are quadrupole waves.

Ken G
2016-Jun-18, 07:52 PM
But unlike electromagnetic radiation, which is dipole radiation, gravity waves are quadrupole waves.
The reason is straightforward-- if you look at the center of mass of a closed system, it's mass dipole is always zero. So the dipole of the mass cannot change, so there's no dipole term to the radiation.

swampyankee
2016-Jun-20, 12:34 PM
The reason is straightforward-- if you look at the center of mass of a closed system, it's mass dipole is always zero. So the dipole of the mass cannot change, so there's no dipole term to the radiation.

That makes them even harder to detect, then. If I recall, quadrupoles result in intensity proportional to distance to the negative fifth power, it's to the negative third with dipoles.

Ken G
2016-Jun-20, 05:48 PM
That makes them even harder to detect, then. If I recall, quadrupoles result in intensity proportional to distance to the negative fifth power, it's to the negative third with dipoles.There's a difference between the field and the waves it carries. Masses have no dipole moment in the center-of-mass frame, but they do have a monopole moment! So the gravitational field itself falls off like 1/r2. The quadrupole moment variations that produce waves must produce waves whose energy fans out into space like any spherically symmetric energy-carrying flux, to the energy flux density must also fall off like 1/r2. I'm not positive how the field that carries the waves (i.e., the field perturbations) fall off, but normally a field perturbation falls off like 1/r to get an energy flux density that falls off like 1/r2, even if the waves come from a perturbation in the quadrupole moment. This means that eventually the field perturbation becomes much stronger than the background field that carries it. That would happen for light and the electromagnetic field, for example, even if it was quadrupole radiation. This would be true of any field in the linear limit, and I'm pretty sure gravitational radiation is a linear limit of GR.

Grey
2016-Jun-20, 07:29 PM
The parameter estimation paper has this (about 150): https://cplberry.com/2016/02/23/gw150914-the-papers/#parameter-estimationLooking at this summary, it appears that the signal would actually be loudest if we're along the axis of rotation, and quietest if we were somewhere on the "equator".


That makes them even harder to detect, then. If I recall, quadrupoles result in intensity proportional to distance to the negative fifth power, it's to the negative third with dipoles.You'd think so, but apparently the size of the strain in the detectors is proportional to the amplitude (not the power) of the gravitational waves, which decreases linearly with distance. So things that are far away are easier to detect than if we were looking at electromagnetic radiation. Which is kind of cool: with additional detectors to provide better localization of signals, we have a new method of probing deeply into the universe.

Ken G
2016-Jun-20, 08:28 PM
So things that are far away are easier to detect than if we were looking at electromagnetic radiation. It depends on the type of electromagnetic radiation. Your comment applies to whenever we detect either power flux (typical in the optical) or photon flux (typical at very high energies, like gamma rays). But radio receivers typically measure the electric field, not the power, so radio detection is a lot like gravitational wave detection-- it falls off like 1/r not 1/r2.