# What is the amplitude of a gravitational wave at the source?

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## Main Question or Discussion Point

In the most recent postings on LIGO, it is stated that the amplitude of the signal is less than the diameter of a proton after the propagation of the wave over billions of light years. I am assuming that the wave amplitude will decay as 1/r^2, but perhaps that is an incorrect assumption. So is there a way to calculate how big the magnitude of the wave will be at the source (colliding black holes or neutron stars)? If the wave does decay with distance, I assume it would start with an amplitude in kilometers, what would that mean in the physical space close to the source? Would an object in front of me be suddenly miles away as the wave passes, or would everything just be ripped apart? How far would you have to be before you could survive the passage?

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pervect
Staff Emeritus
In the most recent postings on LIGO, it is stated that the amplitude of the signal is less than the diameter of a proton after the propagation of the wave over billions of light years. I am assuming that the wave amplitude will decay as 1/r^2, but perhaps that is an incorrect assumption. So is there a way to calculate how big the magnitude of the wave will be at the source (colliding black holes or neutron stars)? If the wave does decay with distance, I assume it would start with an amplitude in kilometers, what would that mean in the physical space close to the source? Would an object in front of me be suddenly miles away as the wave passes, or would everything just be ripped apart? How far would you have to be before you could survive the passage?
In the simplest case, for a + polarized gravity wave, if h(t) is the strain function, the effective stress-energy tensor, which you can regard as the energy / unit volume averaged over a cycle of the radiation, is proportional to $(dh/dt)^2$. See http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec14.pdf. Things are more complicated if you have both + and x polarizations (which Ligo almost surely does) - see the paper for more details for the formula where you have both polarizations.

Before I get into explaining some of the fine print, let me say that this means that assuming the wave-packet keeps the same shape (which should be true), then the strain amplitude dies off as 1/r, not 1/r^2. This follows from the fact that if the waveshape is the same, $d/dt \, [a h(t)] = a \, dh/dt$.

Now for the fine print from the reference I cited earlier.

Now if we are interested in the average effective energy flux carried by gravitational waves, we may average this over a wave period. In fact, it is not even meaningful to average this over a fraction of a wave period, since the result is then gauge-dependent.
I forget the name of the lecturer who illustrated translation symmetry by translating (i.e. moving) his hands, rotation symmetry by rotating and twisting them, and gauge symmetry by waving his hands and making mystical noises.. But informally, a dependence on gauge means that something just isn't well defined, because equally plausible assumptions give you different answers.

Energy in GR is just not as well defined as it is in Newtonian mechanics. I can't begin to offer a full explanation in this short post, but I can at least warn readers of this (getting them to believe it seems to be harder).

Just some simple arithmetic: The wavelength reaching us is on the order of 10-21, the distance in meters is on the order of 1025, so that would put the wavelength at the source on the order 104. Gravitational waves stretch space, but I can't fathom what that size gravity wave would have on matter. It probably wouldn't be a good outcome if you were close to that event.

pervect
Staff Emeritus
Just some simple arithmetic: The wavelength reaching us is on the order of 10-21, the distance in meters is on the order of 1025, so that would put the wavelength at the source on the order 104. Gravitational waves stretch space, but I can't fathom what that size gravity wave would have on matter. It probably wouldn't be a good outcome if you were close to that event.
The $10^{-21}$ figure is not a wavelength, or a length of any kind, but a dimensionless strain. You can think of strain as the change in length divided by the length, so with a 2,000 meter arm, a strain factor of $10^{-21}$ means that the separation between the mirrored test masses changes by $2\,10^{-18}$ meters.

Close to the origin of the event, the idea of a flat background space-time through which a wave travels becomes misleading. One needs to study the geometry of the space-time as a whole, there isn't any flat background to regard the wave as a pertubation of.

Pervect, I must be getting dotty in my old age. 10-21 is the amplitude of the wave when it reaches us, not its wavelength.

Consider the Schwarzschild Radius of the two black holes before they merged. I get 86 an 106 km. Add them together to get 192 km. Now divide this by the distance from LIGO to the merging event, $1.256 * 10^{22} km$. Now take the ratio of the black hole sizes and their distance to get:
$\frac{106 km} {1.256 * 10^{22} km} = 1.5 * 10^{-20}$.

This is about the same order of magnitude for the measured LIGO strain.

What would a gravitational wave with that amplitude do to matter in its path of propagation?

mfb
Mentor
Strain of order 1? Rip it apart, in the same way strong tidal gravity does.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Strain of order 1? Rip it apart, in the same way strong tidal gravity does.
This would depend on the frequency of the wave. A very low frequency wave and other forces will very likely be able to keep you whole. Unlike the LIGO mirrors, matter is usually held together by EM interactions.

mfb
Mentor
It also depends on the size of the object. Sure, you won't rip molecules apart with some stellar black holes.
The LIGO mirrors also have a connection via complicated suspension systems and the ground, but that is too slow for 100 Hz waves.