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geordief

The spacetime curvature of a point between two massive bodies

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Let's say we have two massive bodies such as two identical black holes  ,each with a radius of 1 light  second  and separated by a distance of 100 light seconds

These bodies are   not spinning and approach each other directly  with a  speed of c/1000

Is it possible with that information to say how the curvature of spacetime at a point midway between the  two bodies would be characterized or quantified? 

 

Would the frame of reference used to calculate this spacetime curvature be that of the point itself,so that it would lie at the origin of the calculation and the two black holes and their motions would be referred to that point?

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For one, the black holes would be accelerating towards each other, not moving at that constant speed. That speed of course is relative to your frame of choice, which in this case is your point in the middle, a sort of saddle-point of unstable equilibrium.  Time would dilate as the two black holes approached, so they'd appear to approach faster than the speed measured by a distant observer.

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5 minutes ago, Halc said:

For one, the black holes would be accelerating towards each other, not moving at that constant speed. That speed of course is relative to your frame of choice, which in this case is your point in the middle, a sort of saddle-point of unstable equilibrium.  Time would dilate as the two black holes approached, so they'd appear to approach faster than the speed measured by a distant observer.

Yes ,they would be accelerating but would there be a point in space and time where the two bodies would be judged by an observer there to be then moving towards each other at that speed?**

And would the observer at that point be at the origin of the frame of reference used to characterise/quantify the spacetime curvature at that point?

If GR is a local theory does it mean that any calculation of  spacetime curvature uses the location of that curvature as the origin of the frame of reference for the purposes of any calculation?

**the observer in this instance would presumably see both bodies moving towards him or her at c/500 at that point in his or her time.

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Posted (edited)
1 hour ago, geordief said:

Yes ,they would be accelerating but would there be a point in space and time where the two bodies would be judged by an observer there to be then moving towards each other at that speed?**

Yes, as they accelerate, there is a time in any frame where the speed crosses that particular rate of the two objects approaching at c/1000.   You can't get from slow to fast without crossing 'medium'. Maybe I'm misunderstanding the question.

Quote

And would the observer at that point be at the origin of the frame of reference used to characterise/quantify the spacetime curvature at that point?

I don't think the spacetime curvature is frame dependent, similar to the way events are absolute, not frame dependent. As I said, the curvature at the point between the masses is a saddle shape: positive curvature in one direction, negative in another. This is true in any frame.

Quote

**the observer in this instance would presumably see both bodies moving towards him or her at c/500 at that point in his or her time.

Something like c/2000 actually, and that depends on which frame measured the c/1000 approach of the two BH's to each other.

Edited by Halc

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Posted (edited)

I have more questions about my understanding of this than I have answers... hopefully someone can correct me.

3 hours ago, geordief said:

two massive bodies [...] separated by a distance of 100 light seconds [...] approach each other directly  with a  speed of c/1000

You're specifying a moment in time here. The curvature of a static spacetime should be frame-independent, but if it's changing over time in different locations, the timing of those changes will be frame-dependent (right?), and you won't have a single description of the spacetime at a given moment (eg. the moment of some event), because there are no frame-independent individual moments that span the space.

The frame is still needed to define the moment. You're specifying the moment, and you have the choice of specifying whose frame of reference you're using. Since you didn't specify any other frame, the midpoint observer's frame (ie. the frame where the two black holes are symmetric) is the only sensible frame of reference that can be assumed by your description. (However, your description is missing some info, like whether you mean the black holes are traveling relative to each other at c/1000, or that's their closing speed according to the midpoint observer, and neither seems like your obvious intention.) 

Changes in spacetime curvature are propagated as gravitational waves, and they propagate when there is a change in acceleration of a mass (right?). When you have two masses approaching each other head-on in freefall, it is symmetric and no gravitational waves are emitted (right?). I'm guessing that means that there's no frame-dependence here, and all observers (all frames) would agree on the mathematical description of the spacetime at the moment that the two black holes are 100 light seconds apart according to the midpoint observer.

Also, I suspect that if what you were describing wasn't symmetric, it would matter that the black holes are approaching each other (rather than eg. moving apart after a flyby). In your example, without any gravitational waves being propagated, I suspect that you'd have the same curvature with the masses at the given separation, regardless of their speed.

(Right?)

Edited by md65536

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12 hours ago, geordief said:

Is it possible with that information to say how the curvature of spacetime at a point midway between the  two bodies would be characterized or quantified?

Spacetime in a small local region about that point would be locally flat. However, the global spacetime is of course not flat, so if you were to place a clock at that Lagrange point, and compare it to a reference clock at infinity, you would find that it is gravitationally time-dilated, even though it remains at rest and experiences no net acceleration in space - until of course it crosses an event horizon, after which it cannot remain stationary any longer.

Note that this whole scenario is a relativistic 2-body problem, so it's not straightforward to treat.

12 hours ago, geordief said:

Would the frame of reference used to calculate this spacetime curvature be that of the point itself,so that it would lie at the origin of the calculation and the two black holes and their motions would be referred to that point?

Spacetime curvature is a tensorial quantity (the Riemann curvature tensor) and is thus covariant, i.e. not frame dependent. In GR it is calculated directly from the connection coefficients and its derivatives, which in turn are functions of the metric and its derivatives.

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Posted (edited)
4 hours ago, Markus Hanke said:

Spacetime curvature is a tensorial quantity (the Riemann curvature tensor) and is thus covariant, i.e. not frame dependent. In GR it is calculated directly from the connection coefficients and its derivatives, which in turn are functions of the metric and its derivatives.

Edit: I'm trying to wrap my head around this. Does this mean that the curvature has a specific value for every event in 4D spacetime (ie. a field), and that at each event, it's the same for all observers? Conversely, something like the magnetic field also has a value for every event in 4D spacetime, but the value at a given event can differ depending on inertial frame.

Then 'globally' would refer to all of 4D spacetime, not 'spatially everywhere at a particular moment' like I've been thinking of it? If so then then my original reply below might not make sense.

----

Is the curvature locally frame independent, or globally? It is time-dependent, right? OP's experiment involves changing curvature?

If you set up two of OP's experiment, say some light years apart, then the relative timing of the experiments depends on inertial frame. Can the global curvature (or the tensor field?) be described without that mattering? Does the concept of a "global curvature at a given time (a Cauchy surface?)" even make sense? Or does the tensor field necessarily extend through time or something? Now I'm confused.

My guess would be that it's all local, and trying to apply it globally to causally disconnected 4D regions would require some arbitrary choice of ... how you want to connect the regions into a global thing.

Basically, OP's experiment describes an event at the midpoint observer's location, and all observers everywhere in the universe agree on the spacetime curvature near that event?

Edited by md65536

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42 minutes ago, md65536 said:

Edit: I'm trying to wrap my head around this. Does this mean that the curvature has a specific value for every event in 4D spacetime (ie. a field), and that at each event, it's the same for all observers? Conversely, something like the magnetic field also has a value for every event in 4D spacetime, but the value at a given event can differ depending on inertial frame.

Then 'globally' would refer to all of 4D spacetime, not 'spatially everywhere at a particular moment' like I've been thinking of it? If so then then my original reply below might not make sense.

----

Is the curvature locally frame independent, or globally? It is time-dependent, right? OP's experiment involves changing curvature?

If you set up two of OP's experiment, say some light years apart, then the relative timing of the experiments depends on inertial frame. Can the global curvature (or the tensor field?) be described without that mattering? Does the concept of a "global curvature at a given time (a Cauchy surface?)" even make sense? Or does the tensor field necessarily extend through time or something? Now I'm confused.

My guess would be that it's all local, and trying to apply it globally to causally disconnected 4D regions would require some arbitrary choice of ... how you want to connect the regions into a global thing.

Basically, OP's experiment describes an event at the midpoint observer's location, and all observers everywhere in the universe agree on the spacetime curvature near that event?

 

It means that it does not have one value but many depending upon which direction you look in.

4 hours ago, Markus Hanke said:

Spacetime curvature is a tensorial quantity

 

14 hours ago, Halc said:

the curvature at the point between the masses is a saddle shape: positive curvature in one direction, negative in another. This is true in any frame

 

 

Note that permittivity mu is also a tensor which only has one value in an isotropic situation. Otherwise you have the same situation in an anisotropic situation. The B and H vectors then point in slightly different directions.

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Posted (edited)
2 hours ago, studiot said:

It means that it does not have one value but many depending upon which direction you look in.

The (Riemann) curvature has one tensor value for each location+time, and that one tensor value is made up of "other stuff" that gives you the different scalar curvature values in different directions?

The "other stuff"'s beyond me and probably not important for the conversation but wikipedia says it's "the Christoffel symbols and their first partial derivatives"... with values corresponding to the 4 dimensions? And there's a scalar curvature in the direction of time as well as any other direction?

Edited by md65536

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Maybe what I am asking is whether  ,when calculating the spacetime curvature at any particular point in spacetime,that point itself is in any sense privileged .

Let's take an event where the Moon impacts the     Earth

That event is visible from more or less anywhere  in the Solar System and each of those vantage points will (If I have understood correctly)  indeed calculate an identical spacetime curvature in the spacetime region around the impact event.

 

So all observers agree on the spacetime curvature  but is the assessment of that curvature  done by an observer at the site of the event  any different (ie   in any sense prefetential) in modality from that carried out by any of the myriad of other potential observers?

It seems to me that the "in situ" observer  has the simplest task 

Is it not easier as a matter of practicality to calculate the spacetime curvature in a region adjacent to where you are sitting  than in a region that is at a remove (possible extremely so) from where you are?

 

And if that observer has the simplest task does that mean he or she has ,in some sense a privileged frame of reference (even though all frames of references do agree ) ?

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12 hours ago, md65536 said:

I'm trying to wrap my head around this. Does this mean that the curvature has a specific value for every event in 4D spacetime (ie. a field), and that at each event, it's the same for all observers?

Spacetime curvature is not a single value, it's a rank-4 tensor; in the most general case, and taking into account the various index symmetries of that tensor, it has 20 functionally independent components. What this means is that every event, all observers agree on the overall tensor (meaning they agree on how its various components are related to one another), while at the same time arriving at different numerical values for each individual component.

12 hours ago, md65536 said:

Then 'globally' would refer to all of 4D spacetime, not 'spatially everywhere at a particular moment' like I've been thinking of it?

Yes, that is what global means.

12 hours ago, md65536 said:

Is the curvature locally frame independent, or globally?

The Riemann tensor is generally covariant, and it's a tensor field, so it doesn't depend on any frames neither locally nor globally.

12 hours ago, md65536 said:

It is time-dependent, right? OP's experiment involves changing curvature?

Yes, it's components can be time dependent, but don't have to be. In this case some of them are.

12 hours ago, md65536 said:

If you set up two of OP's experiment, say some light years apart, then the relative timing of the experiments depends on inertial frame. Can the global curvature (or the tensor field?) be described without that mattering?

Yes. You can in some sense think of it as a global function on spacetime.

12 hours ago, md65536 said:

Does the concept of a "global curvature at a given time (a Cauchy surface?)" even make sense?

Well, there is the difficulty of what 'at a given time' means in a global sense. What you can do though is define what is called a foliation - you can consider spacetime as an ordered set of 3-surfaces, with time being the ordering parameter. Each of these surfaces then can have a purely spatial curvature associated with it. This is called the ADM formalism of GR.
Note that there are infinitely many ways to perform such a foliation, so this does not violate any principles of relativity since there is no preferred time ordering scheme.

12 hours ago, md65536 said:

Or does the tensor field necessarily extend through time or something? Now I'm confused.

If you use the standard 4D formalism, then you use the Riemann tensor to define spacetime curvature, which is generally covariant across all of spacetime.
If you use the above mentioned ADM formalism, then the spatial curvature of each hyperslice is described using a different quantity, which depends on the time ordering parameter.

12 hours ago, md65536 said:

My guess would be that it's all local, and trying to apply it globally to causally disconnected 4D regions would require some arbitrary choice of ... how you want to connect the regions into a global thing.

GR only places a constraint on local geometry, but it has nothing to say about global topology of your spacetime. So for example, the standard Schwarzschild metric equally well describes one multiply connected global spacetime, or two separate singly connected regions of spacetime. GR itself does not distinguish these, so the choice of topology needs to come from elsewhere.

12 hours ago, md65536 said:

Basically, OP's experiment describes an event at the midpoint observer's location, and all observers everywhere in the universe agree on the spacetime curvature near that event?

Yes. But they may not agree on times and distances, which are of course observer dependent.

9 hours ago, md65536 said:

The (Riemann) curvature has one tensor value for each location+time, and that one tensor value is made up of "other stuff" that gives you the different scalar curvature values in different directions?

Yes, this is pretty much precisely the idea :)

9 hours ago, md65536 said:

The "other stuff"'s beyond me and probably not important for the conversation but wikipedia says it's "the Christoffel symbols and their first partial derivatives"... with values corresponding to the 4 dimensions?

It's the various components of the tensor itself (up to 20 of them), and their relationships. You calculate the value of those components from the Christoffel symbols and their derivatives.

9 hours ago, md65536 said:

And there's a scalar curvature in the direction of time as well as any other direction?

Yes. Curvature in the time direction is nothing other than gravitational time dilation, and spatial curvature would manifest as gravitational tidal forces.

4 hours ago, geordief said:

So all observers agree on the spacetime curvature  but is the assessment of that curvature  done by an observer at the site of the event  any different (ie   in any sense prefetential) in modality from that carried out by any of the myriad of other potential observers?

No, there are no privileged vantage points. Any observer anywhere will calculate the same spacetime curvature, which is a tensor field extended throughout all of spacetime. Remember they use the same equations and the same gravitational sources, hence they must arrive at the same answer. 

4 hours ago, geordief said:

Is it not easier as a matter of practicality to calculate the spacetime curvature in a region adjacent to where you are sitting  than in a region that is at a remove (possible extremely so) from where you are?

Sometimes the choice of coordinate system can make a huge difference to how difficult the actual calculation is, so yes, in practical terms there may be differences. But the physics are always the same.

4 hours ago, geordief said:

And if that observer has the simplest task does that mean he or she has ,in some sense a privileged frame of reference (even though all frames of references do agree ) ?

No, it's not physically privileged, it's just that they were lucky enough to use a coordinate frame that makes the calculation easier. The physics are always the same for all observers.

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Posted (edited)
2 hours ago, Markus Hanke said:

 

GR only places a constraint on local geometry, but it has nothing to say about global topology of your spacetime. 

This statement isn't entirely accurate. The FLRW metric is a GR solution of the weak field limit of GR.

[math]g_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}[/math]

 The Schwartzchild metric is a common application of the strong field. The other common class of solutions being the vacuum solution.

( Though I do agree with your overall statements in your last post.)

Edit should also note the energy momentum term being 

[math]T^{\mu\nu}=0[/math]

Edited by Mordred

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22 hours ago, Mordred said:

This statement isn't entirely accurate. The FLRW metric is a GR solution of the weak field limit of GR.

gμν=ημν+hμν

 The Schwartzchild metric is a common application of the strong field. The other common class of solutions being the vacuum solution.

( Though I do agree with your overall statements in your last post.)

Edit should also note the energy momentum term being 

Tμν=0

This is true, but I don’t quite understand the connection to global topology. By its very nature, the GR field equations are constraints on local geometry only, so they cannot fix the global topology of spacetime. In order to arrive at the FLRW metric (a specific kind of spacetime), you need to also impose boundary conditions - and I think it is those boundary conditions, rather than the model (GR) itself which constrain global topology in this case. Or maybe I am misunderstanding your comment...?

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Posted (edited)

Well the situation gets complicated by the commoving coordinate system however the Ricci tensor is still applicable. The steps to describe the FLRW metric for a homogeneous and isotopic universe with and without curvature terms is fairly lengthy. 

 Here is a rather detailed article covering the main aspects. It also shows how the key FLRW formulas are arrived at via the GR formalism.

https://www.google.com/url?sa=t&source=web&rct=j&url=http://icc.ub.edu/~liciaverde/Cosmology.pdf&ved=2ahUKEwj3k6qiv4DqAhW4IjQIHaI8CcEQFjACegQIARAB&usg=AOvVaw3mK36Mj5P8FmJLI4RPNXOX

Mathius Blau also has a decent section on the FLRW metric in his lecture notes on GR. ( much later chapters)

http://www.blau.itp.unibe.ch/newlecturesGR.pdf

My main point is simply that all the key FLRW metric equations can be applied with the GR formalism. I'm point of detail in a sense the FLRW metric is a special solution of GR (Ie vacuum solution of a perfect fluid)

Edited by Mordred

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