I What happens in the area between black holes before they collide

[timmdeeg]

Yes.
To a first approximation, yes, at least for the case where the holes were large enough and far enough apart that linearly superposing the curvatures from the two holes was a good first approximation. But no exact solution is known for this case, and the EFE is nonlinear so as the holes get closer the actual curvature is more and more poorly approximated by a linear superposition of the individual solutions for each hole.

Thanks.

 

[tflahive]

It's empty space, not very different from the empty space close to the horizon of a single black hole.
What stress are you talking about? What kind of stress do you think empty space can have?

I mean that the area, and what ever is in that area, between two gravity wells must be pulled in two directions.

 

[Ibix]

I mean that the area, and what ever is in that area, between two gravity wells must be pulled in two directions.

Well - the net "pull" (not a great term in the case of general relativity, which doesn't model gravity as a force) will always be in one direction or another. It is the result of the combination of the gravitational effects of both holes, though, yes. But that isn't anything particularly special - you are being "pulled on" by the Earth, Sun and Moon right now.

 

[kimbyd]

The "acceleration due to gravity" due to the two holes might partially cancel, but that is not spacetime curvature. Spacetime curvature will, if anything, be larger in between two black holes than it would be due to either hole individually. But really spacetime curvature is a tensor and doesn't have a single "size".

You can use the Ricci scalar for a coordinate-free description of the total amount of space-time curvature. Obviously this doesn't tell you all of the details of the curvature, but I think it's a reasonable way to make this situation more concrete.

But I guess it makes sense that the space there would be highly curved, for the reason that a small amount of movement towards either black hole will have a large impact on the future trajectory of a particle.

 

[PeterDonis]

You can use the Ricci scalar for a coordinate-free description of the total amount of space-time curvature.

Not in a vacuum spacetime, which is what we're talking about (black holes are vacuum solutions); the Ricci tensor is identically zero in vacuum.

The Kretzschmann scalar would be the nonzero one to use in this case.

 

[Orodruin]

black holes are vacuum solutions

Just to point out that this is not necessarily true, as it has been mentioned a couple of times in this thread. The Reissner-Nordström metric describes a charged black hole and such a black hole comes with a non-zero electromagnetic field and therefore a non-zero energy-momentum tensor.

 

[PeterDonis]

The Reissner-Nordström metric describes a charged black hole and such a black hole comes with a non-zero electromagnetic field and therefore a non-zero energy-momentum tensor.

Yes, that's a valid point; but the stress-energy tensor for this case is traceless, so the Ricci scalar is still zero.

 

[timmdeeg]

I mean that the area, and what ever is in that area, between two gravity wells must be pulled in two directions.

I would agree if you talk about an extended area and if this extended area is placed such that a point within this area is halfway between the two holes , see post #29. Another question is if it makes sense to say that a point (in contrast to an area) is being pulled in different directions.

 

[PeterDonis]

Another question is if it makes sense to say that a point (in contrast to an area) is being pulled in different directions.

First you need to specify what you mean by "pull". As @Ibix has pointed out, gravity is not a force in GR so there is no "pull" in the Newtonian sense.

 

[timmdeeg]

First you need to specify what you mean by "pull".

"pull" in the sense of what happens to a "ball of test particles" as mentioned in post #29, so no forces.

 

[PeterDonis]

"pull" in the sense of what happens to a "ball of test particles" as mentioned in post #29, so no forces.

Ok, then "pull" is a bad term to use since, as you say, there are no forces and all of the test particles are in free fall. The correct term for what happens to the test particles is "tidal gravity". And any given ball of test particles will have one thing happen to it, so it makes no sense to say it has two "pulls" on it.

 

[kimbyd]

I would agree if you talk about an extended area and if this extended area is placed such that a point within this area is halfway between the two holes , see post #29. Another question is if it makes sense to say that a point (in contrast to an area) is being pulled in different directions.

Either way, after looking into the matter, I'm pretty sure the Ricci scalar wouldn't provide much of any information about the curvature caused by the black holes themselves, even in the non-idealized scenarios. There may be some higher-order effects, but it seems like most* of the information about the situation will be suppressed. Most of the source for the Ricci scalar, if any, would stem from matter in the area around the black holes.

* The only reason why the contribution to the Ricci scalar from the black holes might be non-zero would be due to the fact that GR isn't linear, and thus solutions to the field equations don't simply add together.

 

[PeterDonis]

The only reason why the contribution to the Ricci scalar from the black holes might be non-zero would be due to the fact that GR isn't linear, and thus solutions to the field equations don't simply add together.

For an uncharged black hole, which is a vacuum solution, the Ricci scalar is always identically zero.

For a charged black hole, with no other stress-energy present, the SET of the EM field due to the hole's charge is traceless, so again the Ricci scalar is zero.

These statements are true even for solutions with multiple black holes present, because they don't depend on any symmetry of the solution or on a linear approximation when superposing solutions. They depend only on the definition of "vacuum" and the known properties of the SET of any EM field.

 

[kimbyd]

For an uncharged black hole, which is a vacuum solution, the Ricci scalar is always identically zero.

For a charged black hole, with no other stress-energy present, the SET of the EM field due to the hole's charge is traceless, so again the Ricci scalar is zero.

These statements are true even for solutions with multiple black holes present, because they don't depend on any symmetry of the solution or on a linear approximation when superposing solutions. They depend only on the definition of "vacuum" and the known properties of the SET of any EM field.

That's why I said that the primary source for the Ricci scalar would be matter around the black holes, and any contribution from the black holes (if there is any) would stem from non-linear effects.

 

[PeterDonis]

any contribution from the black holes (if there is any) would stem from non-linear effects

And my point is that there is not any contribution from the black holes due to non-linear effects. There can't be, because the two statements I made (vacuum = zero SET = zero Ricci scalar, EM = traceless SET = zero Ricci scalar) are true for any black hole solution whatever, non-linear effects and all.

 

[Orodruin]

And my point is that there is not any contribution from the black holes due to non-linear effects. There can't be, because the two statements I made (vacuum = zero SET = zero Ricci scalar, EM = traceless SET = zero Ricci scalar) are true for any black hole solution whatever, non-linear effects and all.

Put slightly differently:
Take the trace of the EFEs. You now have a linear relationship between the Ricci scalar and the trace of the stress-energy tensor. This is independent of non-linear effects from the black holes. If the SET (or less restricted, its trace) is zero locally, then so is the Ricci scalar.

 

[timmdeeg]

The correct term for what happens to the test particles is "tidal gravity". And any given ball of test particles will have one thing happen to it, so it makes no sense to say it has two "pulls" on it.

Agreed, I was actually referring to the term "pull" with regard to the comment "I mean that the area, and what ever is in that area, between two gravity wells must be pulled in two directions", #32.

Do you agree that depending on whether or not this area is occupied by a body (being hold together by e.g. intermolecular forces) one should talk about tidal acceleration or tidal forces respectively. The term "tidal gravity" doesn't seem to distinguish between tidal acceleration and tidal forces. But please correct if wrong.

 

[Orodruin]

Do you agree that depending on whether or not this area is occupied by a body (being hold together by e.g. intermolecular forces) one should talk about tidal acceleration or tidal forces respectively. The term "tidal gravity" doesn't seem to distinguish between tidal acceleration and tidal forces. But please correct if wrong.

I do not agree. Since there is no gravitational force there can be no tidal force, which in Newtonian gravity are due to differences in gravitational acceleration. Instead, tidal gravity (as Peter mentions) or ”geodesic deviation” is the more appropriate description. For an object that is held together, different parts of the object will deviate from the geodesics due to internal forces. Those forces are exactly that, internal, and not tidal forces. Just as what is preventing you from falling through the surface of the Earth is an electrostatic force between you and the surface, not a gravitational force.

 

[timmdeeg]

I do not agree. Since there is no gravitational force there can be no tidal force, which in Newtonian gravity are due to differences in gravitational acceleration. Instead, tidal gravity (as Peter mentions) or ”geodesic deviation” is the more appropriate description. For an object that is held together, different parts of the object will deviate from the geodesics due to internal forces. Those forces are exactly that, internal, and not tidal forces. Just as what is preventing you from falling through the surface of the Earth is an electrostatic force between you and the surface, not a gravitational force.

Thanks, I hope I understand you correctly.
So ”tidal gravity” is tantamount to ”geodesic deviation”. Both terms refer to test particles in free fall and thus to geodesics.

I thought the term "tidal acceleration" is restricted to particles on geodesics, e.g. the ball of test particles mentioned above and the term "tidal forces" to particles not on geodesics, e.g. those hold together by cohesive forces.

But what exactly distinguishes tidal gravity from tidal acceleration?

 

[PeterDonis]

what exactly distinguishes tidal gravity from tidal acceleration?

Choice of terminology. What you are calling "tidal acceleration" is the same thing as what @Orodruin is calling "tidal gravity" or "geodesic deviation". His terms are the more usual ones in the literature.

the term "tidal forces" to particles not on geodesics

No; as @Orodruin said, there are no such things as "tidal forces". The forces experienced by particles not traveling on geodesics when tidal gravity is present do not come from tidal gravity; they come from non-gravitational interactions between the particles in a bound object, which prevent the particles (or at least those not at the center of mass of the object) from traveling on geodesics.

 

[Ibix]

"Tidal forces" in general relativity are like the centrifugal force (etc) that emerges in a rotating frame. If you want to pretend that your frame is a simple inertial frame, you need to invoke "inertial" or "fictitious" forces to explain why free-falling particles aren't moving in straight lines.

I think this would be an unusual approach in GR since global inertial frames aren't available anyway. Like centrifugal forces, the obvious application is if you are in a box that is only just big enough to show notable non-inertiality (I'm thinking of a spaceship near a neutron star - with thanks to Larry Niven). But a global view would be that particles are either free-falling with non-zero geodesic deviation, or are being pulled off geodesics by electromagnetic (or other) forces.

 

[timmdeeg]

No; as @Orodruin said, there are no such things as "tidal forces". The forces experienced by particles not traveling on geodesics when tidal gravity is present do not come from tidal gravity; they come from non-gravitational interactions between the particles in a bound object, which prevent the particles (or at least those not at the center of mass of the object) from traveling on geodesics.

Indeed unfortunately I didn't realize the meaning of @Orodruins explanation, my fault.
Does one have to distinguish "cause" from "come from"? Is tidal gravity the cause that such particles aren't traveling on geodesics (because in flat spacetime they do unless there is proper acceleration) and do the the forces they experience "come from" non-gravitational interactions?

EDIT
MTW write regarding the head to foot stretching in a gravitational field"page 860:

During the early stage, one can analyze the tidal forces by means of the equation of geodesic deviation ...

 

[PeterDonis]

Is tidal gravity the cause that such particles aren't traveling on geodesics

No.

because in flat spacetime they do unless there is proper acceleration

The same is true in curved spacetime. "Traveling on geodesics" is equivalent to "zero proper acceleration".

do the the forces they experience "come from" non-gravitational interactions?

Yes. This is true in both flat and curved spacetime: "experiencing forces" is equivalent to "experiencing non-gravitational interactions" is equivalent to "has nonzero proper acceleration" is equivalent to "not traveling on geodesics".

 

[Wes Tausend]

Summary:: The area between two black holes before they collide must be under a great deal of stress. That stress change should be measurable by the effect of radiation or light coming through that area.

Recently there have been a lot of studies of black holes colliding and the gravitational waves that they produce. My question is: What is the effect on the space between the two black holes before they collide. The stress must be extraordinary. That stress should be measurable by radiation, or light coming through that area.

In a simple sense, it does seem as though there should be a double lensing effect occurring outside the Swartzchild radii between the two "black stars". In this case, it seems that any stars, or ambient light from behind the two gravity sources, should be compressed into a straight, vertical-to-common axis, bright line that exists up to the event that the two radii meet, precluding any more light from passing between the two black holes.

Wes

 

[PeterDonis]

In a simple sense, it does seem as though there should be a double lensing effect occurring outside the Swartzchild radii between the two "black stars". In this case, it seems that any stars, or ambient light from behind the two gravity sources, should be compressed into a straight, vertical-to-common axis, bright line that exists up to the event that the two radii meet, precluding any more light from passing between the two black holes.

Why do you think these things are true?

 

[Wes Tausend]

Why do you think these things are true?

Because I believe planetary gravitational sources bend light, warping the view in the immediate area. I was supposing the light from stars behind would have the effect of being normally displaced in a ring effect. The reason for choosing the straight line, would only be straight if the two BH's were of equal mass, otherwise curved. In this equal-mass case I expect that the Lagrange points would precisely cancel so that a straight vertical equilibrium 'force' line would exist to 'channel' the flow of light in such a pattern.

 

[PeterDonis]

Because I believe planetary gravitational sources bend light

While this is true, it is a very general statement; but the claims you made were much more specific.

I was supposing the light from stars behind would have the effect of being normally displaced in a ring effect.

First, the ring effect requires a very precise alignment of the star and the black hole; and second, it occurs for a single black hole, not a pair of them. Optical effects from a pair of black holes will be more complicated.

The reason for choosing the straight line, would only be straight if the two BH's were of equal mass, otherwise curved

Since nobody has an exact solution for this case, and since you have shown no results of numerical simulations, this is just personal speculation.

In this equal-mass case I expect that the Lagrange points would precisely cancel so that a straight vertical equilibrium 'force' line would exist to 'channel' the flow of light in such a pattern.

Light is not bent by a Newtonian "force". This viewpoint is already problematic for the simple case of a single non-rotating black hole. It is even more problematic for the more complicated case under discussion.

 

[Ibix]

Since nobody has an exact solution for this case, and since you have shown no results of numerical simulations, this is just personal speculation.

Just thinking about that - this would require some serious computing. You need to do everything the LIGO people do to find the spacetime, and then after that you need to propagate rays through it, which isn't computationally cheap either.

Maybe it's not so bad when the holes are far enough apart that things aren't changing significantly on the scale of light crossing the system? You might be able to pretend the spacetime is static and save yourself a lot of time.

 

[timmdeeg]

The reason for choosing the straight line, would only be straight if the two BH's were of equal mass, otherwise curved.

Saying straight line, have you this scenario in your mind?

Star and two equally sized black holes form an equilateral triangle and on the opposite side the same with an observer instead of the star. Both triangles are in the same plane.

 

[PeterDonis]

this would require some serious computing

Indeed it would. I don't know of any attempt to do it, since, as you say, it would require all the computations that are done for LIGO, plus additional ray propagation computations on top of that.