4 ms·

This event was 1.8 billion (10^9) light years away. The so called "strain" (i.e. the extent to which the wave stretches / compresses things) is around 10^-20 or

This event was 1.8 billion (10^9) light years away. The so called "strain" (i.e. the extent to which the wave stretches / compresses things) is around 10^-20 or somewhere around that order of magnitude. To make things dangerous I reckon you'll need to have a strain on the order of 1, and since strain is inversely proportional to distance that would give a distance of 10^-11 ly, or around 100 km. Which is not a 'safe' distance to a black hole in any scenario.

eponeponepon 9y ago
Would that be 100km from the event horizon, or from the singularity?
- contravariant 9y ago
  Well the estimate kind of assumes a point source, but at 100km from the merger that assumption is probably no longer accurate. You'll need to simulate the black hole merger to figure out where exactly it would be unsafe, but it looks like it would be distance that's pretty unsafe even without the merger.
  - raattgift 9y ago
    It's possibly more interesting to think in terms of $\nabla_\mu T^{\mu\nu} = 0$ which is the generalized conservation of stress-energy-momentum, and to think in terms of energy-momentum entering and exiting regions of spacetime. The gravitational radiation carries momentum out of a (comfortably flat-ish at the boundary) region of spacetime tightly enclosing the inspirallers and merger, and delivers momentum elsewhere in the inspiraller's forward light cone. In an otherwise expanding vacuum solution, the radiated momentum just propagates to future null infinity. Since we have other matter in our universe, the gravitational radiation can donate some of the original momentum to matter outside our chosen region, causing bits of matter along a lightlike path to move a small amount (or if you want to switch to an energy picture, ordinary matter outside our chosen region will become a tiny bit hotter upon feeling the gravitational radiation). Although the amount of stress-energy-momentum radiated out of our region comprising the inspirallers is enormous (and at peak is so brief that the power is almost unimaginably enormous), the coupling of gravitational radiation to matter is extremely weak, so there will be very little scattering or absorption by matter outside the region. Consequently, most of the momentum crossing to the outside of our Schwarzschild-like boundary around the merger will still go to future null infinity. It's this weak coupling that leads to your intuition about a human having to be very close to the merger itself in order to feel anything noticeable from the gravitational radiation. By comparison one would expect balding to happen so that we recover a no-hair state from the merged BH, and there are decent reasons to expect more than gravitational radiation to be shed during balding. Additionally the individual BHs may have a lot of energetic particles entrained around them, and much of that will be moving on geodesics conducive to extremely high energy interactions (think of a pair of BHs whose rotational axes are parallel but with each BH dragging its accretion disc in the opposite CW/CCW direction, and we can add in extreme inverse Compton scattering just prior to contact as the photon rings around each BH interact relativistically with charged particles around the other). And of course this picture is likely to be much more dramatic in the case of an NS-BH merger. Another way of thinking about this is that the build-up of momentum due to gravitational interactions is typically slow (and steady) but may have no upper limit, and that's the source of danger rather than a quick (and briefly high amplitude) gravitational interaction that doesn't take away more than a few percent of the built-up momentum of the matter at and near the source of the burst of gravitational radiation. A human being close enough to the merger to feel the effects of gravitational radiation is thus likely to be feeling an awful lot more from excitations in the non-gravitational fields, or alternatively, the nearby human being is at risk of collision with ultrarelativistic (i.e., ultra-high-momentum) matter.
    - pdonis 9y ago
      > $\nabla_\mu T^{\mu\nu} = 0$ which is the generalized conservation of stress-energy-momentum Unfortunately, for gravitational waves this is trivial, since they are vacuum--so T^{\mu\nu} itself is zero, and its covariant divergence is trivially zero. There are ways to mathematically compute the "energy carried away by gravitational waves" from an isolated system like a black hole merger, but you have to use global methods to do it. For example, you can compare the ADM mass with the Bondi mass of the system; the latter decreases over time as GWs carry energy away, while the former stays the same.
      - raattgift 9y ago
        Sure, I just didn't want to go as deep as different slicings of spacetime and comparing position-dependent dynamical quantities on a slice of scri+ with quantities on a spacelike slice that you expect to always appear in the same form somewhere in every successor slice. On that last point, the latter (ADM) always stays the same, doesn't it? In early spacelike slices when the binary system is radiating basically nothing, the quantities you will find in the GWs in later slices are just clustered around the black holes. Mass in General Relativity will give anyone a headache; I was trying to side-step that without being caught shuffling gravitational radiation off to the source side using pseudotensors. :-)
        pdonis 9y ago
        > the latter (ADM) always stays the same, doesn't it? On spacetimes where it is defined, yes, I believe it does.
- pdonis 9y ago
  > Would that be 100km from the event horizon, or from the singularity? 100km from the horizon. "Distance" from the singularity does not actually make sense; the singularity is best thought of as a moment of time, not a place in space.
JohnBooty 9y ago
> Which is not a 'safe' distance to a black hole in any scenario. Sure, if you're a pansy. HOLD MY BEER!
- raattgift 9y ago
  As you take the size of a black hole to extreme mass, just outside the event horizon is as safe as any other place in outer space, as curvature (in the sense of tidal forces) at and near the horizon falls off dramatically as the horizon radius increases. The increase in the horizon area also reduces your chances of colliding with something on a different orbit around the black hole, so you're less likely to be vaporized by matter in the accretion disc moving relativistically with respect to you. Your buddy could be holding your beer for a very long time though, because gravitational time dilation at a given small distance outside the horizon increases with the mass of the black hole, and it will take you more of your own proper time to move through the region near a very large black hole than through the region near a small one.
yiyus 9y ago
Normally, I would say that a strain of 1 would be way too much. A strain of 0.1 would already mean a change of length of 10%, which is probably more than enough to destroy your body. But we are talking about orders of magnitude here, so the comparison of 10^-20 with 1 seems precise enough.

eponeponepon 9y ago

Would that be 100km from the event horizon, or from the singularity?

contravariant 9y ago
Well the estimate kind of assumes a point source, but at 100km from the merger that assumption is probably no longer accurate. You'll need to simulate the black hole merger to figure out where exactly it would be unsafe, but it looks like it would be distance that's pretty unsafe even without the merger.
- raattgift 9y ago
  It's possibly more interesting to think in terms of $\nabla_\mu T^{\mu\nu} = 0$ which is the generalized conservation of stress-energy-momentum, and to think in terms of energy-momentum entering and exiting regions of spacetime. The gravitational radiation carries momentum out of a (comfortably flat-ish at the boundary) region of spacetime tightly enclosing the inspirallers and merger, and delivers momentum elsewhere in the inspiraller's forward light cone. In an otherwise expanding vacuum solution, the radiated momentum just propagates to future null infinity. Since we have other matter in our universe, the gravitational radiation can donate some of the original momentum to matter outside our chosen region, causing bits of matter along a lightlike path to move a small amount (or if you want to switch to an energy picture, ordinary matter outside our chosen region will become a tiny bit hotter upon feeling the gravitational radiation). Although the amount of stress-energy-momentum radiated out of our region comprising the inspirallers is enormous (and at peak is so brief that the power is almost unimaginably enormous), the coupling of gravitational radiation to matter is extremely weak, so there will be very little scattering or absorption by matter outside the region. Consequently, most of the momentum crossing to the outside of our Schwarzschild-like boundary around the merger will still go to future null infinity. It's this weak coupling that leads to your intuition about a human having to be very close to the merger itself in order to feel anything noticeable from the gravitational radiation. By comparison one would expect balding to happen so that we recover a no-hair state from the merged BH, and there are decent reasons to expect more than gravitational radiation to be shed during balding. Additionally the individual BHs may have a lot of energetic particles entrained around them, and much of that will be moving on geodesics conducive to extremely high energy interactions (think of a pair of BHs whose rotational axes are parallel but with each BH dragging its accretion disc in the opposite CW/CCW direction, and we can add in extreme inverse Compton scattering just prior to contact as the photon rings around each BH interact relativistically with charged particles around the other). And of course this picture is likely to be much more dramatic in the case of an NS-BH merger. Another way of thinking about this is that the build-up of momentum due to gravitational interactions is typically slow (and steady) but may have no upper limit, and that's the source of danger rather than a quick (and briefly high amplitude) gravitational interaction that doesn't take away more than a few percent of the built-up momentum of the matter at and near the source of the burst of gravitational radiation. A human being close enough to the merger to feel the effects of gravitational radiation is thus likely to be feeling an awful lot more from excitations in the non-gravitational fields, or alternatively, the nearby human being is at risk of collision with ultrarelativistic (i.e., ultra-high-momentum) matter.
  - pdonis 9y ago
    > $\nabla_\mu T^{\mu\nu} = 0$ which is the generalized conservation of stress-energy-momentum Unfortunately, for gravitational waves this is trivial, since they are vacuum--so T^{\mu\nu} itself is zero, and its covariant divergence is trivially zero. There are ways to mathematically compute the "energy carried away by gravitational waves" from an isolated system like a black hole merger, but you have to use global methods to do it. For example, you can compare the ADM mass with the Bondi mass of the system; the latter decreases over time as GWs carry energy away, while the former stays the same.
    - raattgift 9y ago
      Sure, I just didn't want to go as deep as different slicings of spacetime and comparing position-dependent dynamical quantities on a slice of scri+ with quantities on a spacelike slice that you expect to always appear in the same form somewhere in every successor slice. On that last point, the latter (ADM) always stays the same, doesn't it? In early spacelike slices when the binary system is radiating basically nothing, the quantities you will find in the GWs in later slices are just clustered around the black holes. Mass in General Relativity will give anyone a headache; I was trying to side-step that without being caught shuffling gravitational radiation off to the source side using pseudotensors. :-)
      - pdonis 9y ago
        > the latter (ADM) always stays the same, doesn't it? On spacetimes where it is defined, yes, I believe it does.
pdonis 9y ago
> Would that be 100km from the event horizon, or from the singularity? 100km from the horizon. "Distance" from the singularity does not actually make sense; the singularity is best thought of as a moment of time, not a place in space.

JohnBooty 9y ago

> Which is not a 'safe' distance to a black hole in any scenario. Sure, if you're a pansy. HOLD MY BEER!

raattgift 9y ago
As you take the size of a black hole to extreme mass, just outside the event horizon is as safe as any other place in outer space, as curvature (in the sense of tidal forces) at and near the horizon falls off dramatically as the horizon radius increases. The increase in the horizon area also reduces your chances of colliding with something on a different orbit around the black hole, so you're less likely to be vaporized by matter in the accretion disc moving relativistically with respect to you. Your buddy could be holding your beer for a very long time though, because gravitational time dilation at a given small distance outside the horizon increases with the mass of the black hole, and it will take you more of your own proper time to move through the region near a very large black hole than through the region near a small one.

yiyus 9y ago

Normally, I would say that a strain of 1 would be way too much. A strain of 0.1 would already mean a change of length of 10%, which is probably more than enough to destroy your body. But we are talking about orders of magnitude here, so the comparison of 10^-20 with 1 seems precise enough.