# Thread: Solving the Schwarzschild metric

1. ## Solving the Schwarzschild metric

Here is a Wiki link to some methods for solving the metric. I don't understand half of them and the other half don't make sense to me. For instance, I don't see that it is mathematically rigorous to divide by ds or dτ when either is zero for a photon. Even if we were to consider a particle that travels just under c, we are not actually using the particle as the origin as we would for SR, where the distance of the particle from the particle's own origin would be zero, leaving just ds^2 = c^2 dτ^2 in terms of the proper time of the particle, but rather we are using the distance of the particle from the gravitating body, so even in the particle's own frame, this distance would be non-zero and so should be included in the metric. Anyway, I think I have found a very simple way to solve the metric. If L is the locally measured angular momentum, and if this quantity is conserved, then we have

Since m is also a constant, we can drop that and we have

where the primed values are locally measured and unprimed is measured by a distant observer. Since L is constant, then at the point of closest approach b, where the photon travels perfectly tangent to the gravitating body, we have

and solving for dt,

The metric is

where and

and substituting what we found for dt, we get

Integrating this from r=b to r=∞ gives the amount of bending for the gravitational lensing of a photon. It seems much simpler and somewhat more mathematically rigorous than any of the other methods, don't you think? Now I just need to try to find a way to prove that the angular momentum is conserved as measured locally. Does anybody know a way to show that?

2. Here's something very interesting. If we just start with the already known solution to the metric and reverse engineer it, we'll end up with

and substituting according to the metric,

and re-arranging gives

and transforming c to the tangent speed measured by the distant observer so that all values will be as measured by the distant observer, we have

so according to the metric and its solution, this relation must be conserved regardless of whether we consider it angular momentum or whatever, but that's definitely what I would call it.

That's not the interesting part though. The angular momentum was found at the point where the photon travels tangently at a distance b. For a photon that approaches the body and then escapes, that's fine. But in the case of a photon orbitting the body, there will be two tangent points, at the aphelion and perihelion, with the photon travelling tangent to the body with a locally measured speed of c at both points. If the angular momentum is conserved at each of those points, then we have

and transforming to the locally measured speeds,

But the locally measured tangent speeds at each of these points is just

, giving

And so it would appear that a photon cannot have an elliptical orbit. A circular orbit is still allowed, but only if it is perfectly circular, so it looks like photons do not orbit a body at all, only approach and escape or fall in.

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And so it would appear that a photon cannot have an elliptical orbit. A circular orbit is still allowed, but only if it is perfectly circular, so it looks like photons do not orbit a body at all, only approach and escape or fall in.
Actually, they can only orbit at the radius of the photon sphere, with . Thus a photon orbit must be circular. Nothing can orbit a black hole closer than this radius.

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Originally Posted by grav
But the locally measured tangent speeds at each of these points is just

, giving

Actually, all you showed was that, for a photon, for some constant
at the two extrema. This gives us:

By the Descartes rule of signs, this has one negative root and either 2 or 0 positive roots. Since a represents the radius, we are interested in the two positive roots.

Examining the dynamics, the imaginary solutions likely represent an angular momentum too high for a circular orbit at the photon sphere, a double root represents an angular momentum exactly of a circular orbit at the photon sphere and the two positive roots represents an elliptical orbit with less angular momentum than a circular orbit with one solution inside p_s and the other outside p_s.

This last class, I believe, is not viable. This might be an interesting place to explore the meaning of angular momentum in the Schwarzschild metric.

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Originally Posted by utesfan100
For a=3m we get . The quoted equation then factors to .

6. Originally Posted by utesfan100
Actually, they can only orbit at the radius of the photon sphere, with . Thus a photon orbit must be circular. Nothing can orbit a black hole closer than this radius.
Right, for a perfectly circular orbit, we would have

This would be the only possible stable orbit that may be permitted, but it would have to be absolutely circular. Looking around, I found that there can be no stable orbits (elliptical orbits at least I suppose) for photons, planets, moons, or even satellites orbitting the Earth. That is interesting. I didn't know that. They must all move steadily inward or outward.

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Originally Posted by grav
Now I just need to try to find a way to prove that the angular momentum is conserved as measured locally. Does anybody know a way to show that?
The metric doesn't depend on , therefor it provides a killing vector field, which turns out to represent the conservation of specific angular momentum.

For a highly symmetric spacetime such as schwarzschild it's easiest to consider getting the equations of motion through killing vector fields, which also gets you the conservation laws (ie the metric independence on gets you conservation of specific energy, and conservation of specific angular momentum).

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Originally Posted by grav
Right, for a perfectly circular orbit, we would have

This would be the only possible stable orbit that may be permitted, but it would have to be absolutely circular. Looking around, I found that there can be no stable orbits (elliptical orbits at least I suppose) for photons, planets, moons, or even satellites orbitting the Earth. That is interesting. I didn't know that. They must all move steadily inward or outward.
You (and utesfan) are not finding photon orbits, but the limiting case of stable massive orbits. The problem is that you're implicitly using to parametrize the orbits, which doesn't work for null geodesics (proper time is identically zero over those geodesics). For that you need to introduce some other parameter at the start (it's usually called ETA: i see that in that wiki article you linked to they call it ). You can't just use proper time and then set it to zero (that would be getting limiting cases over timelike geodesics, which is not the same as getting null geodesics).

Basically what you two have been doing is this:
Finding the timelike geodesics (implicitly) on the schwarzschild metric. Then setting the speed of that massive particle to be c, and finding the orbit which conforms to those constraints. The result is the innermost stable orbit (smaller orbits give higher orbital speed, so that's why you're getting the limiting innermost orbit by setting the speed to the maximum, ie c). That's also why you're only getting circular orbits, because you put in the constraint that the speed is c over the entire orbit, which precludes other types of orbits.

So while your calculation is correct for that, it does not do what you think it does. You'll need to parametrize the geodesics differently at the start of your derivation (using - implicitly in your case - already constrains you to timelike geodesics from the start)
Last edited by caveman1917; 2012-Aug-11 at 05:33 AM. Reason: added eta

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Originally Posted by grav
Right, for a perfectly circular orbit, we would have

This would be the only possible stable orbit that may be permitted, but it would have to be absolutely circular. Looking around, I found that there can be no stable orbits (elliptical orbits at least I suppose) for photons, planets, moons, or even satellites orbitting the Earth. That is interesting. I didn't know that. They must all move steadily inward or outward.
Your double use of and got me on the wrong track, it appears also stands for the mass of the black hole, not the mass of the orbiting particle. So your derivation is correct after all (the limit didn't make sense which at first appeared like a problem with the parametrization of the orbit as i was thinking of it as the mass of the particle, not the black hole).

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Originally Posted by grav
This would be the only possible stable orbit that may be permitted, but it would have to be absolutely circular.
Since the orbital speed depends on the radius, constraining an orbit such that the orbital speed is the same at perihelion and aphelion naturally constrains it to be circular (which goes for photons as well as massive particles).

Looking around, I found that there can be no stable orbits (elliptical orbits at least I suppose) for photons, planets, moons, or even satellites orbitting the Earth. That is interesting. I didn't know that. They must all move steadily inward or outward.
Why? Are you sure you're using the term "stable" correctly? Stability refers to whether an extremum in the potential is concave or convex. In general the effective potential for the schwarzschild spacetime will have two extrema, a stable and an unstable one.

11. Originally Posted by caveman1917
Since the orbital speed depends on the radius, constraining an orbit such that the orbital speed is the same at perihelion and aphelion naturally constrains it to be circular (which goes for photons as well as massive particles).
Right, the photon's speed must be measured locally at c at all points in the orbit, so with locally measured conservation of angular momentum must be circular, but I don't see that that directly follows for massive particles since they are not required to travel at the same speed.

Why? Are you sure you're using the term "stable" correctly? Stability refers to whether an extremum in the potential is concave or convex. In general the effective potential for the schwarzschild spacetime will have two extrema, a stable and an unstable one.
Right, okay, but eventually the unstable end should open and the orbit would be lost. Or perhaps I just should say that the photon cannot orbit in a regular forseeable way. Some erratic and not easily predictable nature of its orbit may still cause its orbit to vary in such a way that it always remains closed, hard to say.

The metric doesn't depend on , therefor it provides a killing vector field, which turns out to represent the conservation of specific angular momentum.

For a highly symmetric spacetime such as schwarzschild it's easiest to consider getting the equations of motion through killing vector fields, which also gets you the conservation laws (ie the metric independence on gets you conservation of specific energy, and conservation of specific angular momentum).
Hmm. That makes sense. I'll have to look into killing vector fields. I can see that the energy depends upon the local time t, since the observed frequency of a photon depends only upon the local gravitational time dilation. If a hovering observer emits a photon, then another photon a time t later, the first photon will follow some path, curved or otherwise, to some other point in the gravitational field, and the second will follow exactly the same path and arrive at the same point a time t later, so the frequency that the photons pass any point remains the same to the observer, but a observer at the point where they arrived will measure a different frequency that depends only upon his own rate of time. So we'll have

for a photon travelling from a radial distance p to a radial distance q in the field, although the path can be curved and non-radial. So there is also a dependence upon r. Since and are directly proportional to the locally measured energy of the photon, we have

Since this ratio of energy holds for light, we can try it for massive particles as well and see how that holds up. Before this, I originally tried to use the equivalence principle such that a freefaller will always measure the same frequency of light emitted by a hovering distant observer. This seems to work well at first, according to the resulting equation, with the freefaller even always measuring the same speed v of a line of freefallers as measured anywhere locally in the field by the freefaller when the line of freefallers begins with a greater relative speed of v than the freefaller, very promising, but it didn't reduce to Newtonian, so it was out. Disappointing, but anyway, if we just take this ratio of energies directly for massive particles, we get

K = 0 for a photon. Of course, that must also reduce to Newtonian gravity, so let's find that for a massive body travelling from p to q = p - dp. The scalar speeds for the energy are locally measured while the distances are measured by a distance observer, whereas and .

and dropping infinitesimals at this point, the locally measured acceleration at r = p with instantaneous speed v' is

We can see that for a body falling from rest at r with v' = 0, the acceleration reduces to Newtonian to first order. Okay, but this is an acceleration meant for a scalar speed. For instance, according to the equation, the acceleration of a photon with v' = c is zero, so only its direction changes, but not its scalar speed. So we'll want to divide that up into its radial and tangent components. We can use the equivalence principle for that. If a rod freefalls at r with original radial speed , matching the radial component of a particle that also freefalls with some tangent speed, then in order for the equivalence principle to hold, the radial acceleration of both must match over an infinitesimal time of freefall. With no tangent component, the radial speed of the rod is its scalar speed, so for the rod we have just

The radial component of the particle must match that, so we have

There should be no tangent acceleration, so the apparent tangent component of the acceleration comes from the tidal gradient I think. For instance, for a circular orbit, as a particle passes the x axis at a vertical distance y, its tangent speed begins to drop while its radial speed increases, although the scalar speed stays the same. Technically, then, we would really be extending the accelerations over some distances in the x and y directions instead of radially and tangent, but the apparent acceleration along infinitesimal y should still match the instantaneous radial acceleration if gravity only acts in that direction. So then, from that equation, a photon travelling radially will not accelerate at all, always measured at c locally in the radial direction, while a photon travelling in a circular orbit will have zero radial speed, so reduces to just

giving r = 3m as we found before.

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Originally Posted by utesfan100
Examining the dynamics, the imaginary solutions likely represent an angular momentum too high for a circular orbit at the photon sphere, a double root represents an angular momentum exactly of a circular orbit at the photon sphere and the two positive roots represents an elliptical orbit with less angular momentum than a circular orbit with one solution inside p_s and the other outside p_s.

This last class, I believe, is not viable. This might be an interesting place to explore the meaning of angular momentum in the Schwarzschild metric.
Examining the equation, there is a maximum in at the photon sphere. This rules out the imaginary case. Whether we go outward or inward from here a photon will have a smaller than this value, and thus a smaller value for your definition of angular momentum. Thus for some neighborhood of the photon sphere there will be an inner value and outer value for a given less than the maximum.

When we examine the second derivatives of the orbits at these points, however, we see that they can not be a closed orbit because the curvature of the inner orbit is smaller than a circular orbit, showing that this would be the outermost point in the orbit. Likewise, the curvature of the outer orbit is larger than a circular orbit, showing that this would be the innermost point in the orbit.

Thus, while there exist distinct radii where the orbit of a tangentially moving photon has the same value using your version of angular momentum, it is not possible to connect these points to produce a closed orbit.

The lone exception is a circular orbit, which appears to be a very unstable equilibrium.

13. Originally Posted by caveman1917
Why? Are you sure you're using the term "stable" correctly? Stability refers to whether an extremum in the potential is concave or convex. In general the effective potential for the schwarzschild spacetime will have two extrema, a stable and an unstable one.
Originally Posted by grav
Right, okay, but eventually the unstable end should open and the orbit would be lost. Or perhaps I just should say that the photon cannot orbit in a regular forseeable way. Some erratic and not easily predictable nature of its orbit may still cause its orbit to vary in such a way that it always remains closed, hard to say.
I'm sure you're probably referring to what I commented about planets and satellites as well, but I haven't found for those yet, only photons. When I Googled before, that's when I read that the orbits of planets and satellites cannot be completely stable, which I found interesting so mentioned it. As far as photons go, though, I've been thinking about it some more, and I can see that as a photon falls toward a body, it spirals inward. If it doesn't pass a tangent point, it will fall all the way in. If it passes a tangent point, however, then it will spiral back out and eventually escape. It cannot turn around to spiral back in again because that would require that it passes another tangent point at a greater radius than the first, so angular momentum wouldn't be conserved. So other than a perfectly circular orbit at r = 3m, which would require that it be "born" there, and so precise with no disturbances as to make it virtually impossible, these are really the only two options for photons.

14. Originally Posted by utesfan100
Examining the equation, there is a maximum in at the photon sphere. This rules out the imaginary case. Whether we go outward or inward from here a photon will have a smaller than this value, and thus a smaller value for your definition of angular momentum. Thus for some neighborhood of the photon sphere there will be an inner value and outer value for a given less than the maximum.

When we examine the second derivatives of the orbits at these points, however, we see that they can not be a closed orbit because the curvature of the inner orbit is smaller than a circular orbit, showing that this would be the outermost point in the orbit. Likewise, the curvature of the outer orbit is larger than a circular orbit, showing that this would be the innermost point in the orbit.

Thus, while there exist distinct radii where the orbit of a tangentially moving photon has the same value using your version of angular momentum, it is not possible to connect these points to produce a closed orbit.

The lone exception is a circular orbit, which appears to be a very unstable equilibrium.
Yes, exactly. Looks like we were thinking the same thing.

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Originally Posted by grav
Yes, exactly. Looks like we were thinking the same thing.
Well, then, I suggest you have your mental health examined.

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Originally Posted by grav
Hmm. That makes sense. I'll have to look into killing vector fields.
You may find these lecture notes (pdf) interesting. It derives the geodesics through killing vector fields and related conservation laws and it also goes into the stability issues with orbits.

17. Originally Posted by caveman1917
You may find these lecture notes (pdf) interesting. It derives the geodesics through killing vector fields and related conservation laws and it also goes into the stability issues with orbits.
Thank you caveman. On page 3, the link says that in deriving the metric, it is assumed that "the angular part of the metric will be unchanged from its form in flat space because of the spherical symmetry" and on page 6, "θ and φ have similar interpretations as for flat space". What does that mean?

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Originally Posted by grav
Thank you caveman. On page 3, the link says that in deriving the metric, it is assumed that "the angular part of the metric will be unchanged from its form in flat space because of the spherical symmetry" and on page 6, "θ and φ have similar interpretations as for flat space". What does that mean?
It means that the metric on a 2-sphere of constant radius is minkowski (as you can see by setting ). Alternatively it means that you can view the spatial part of the schwarzschild spacetime as a collection of concentric minkowskian 2-spheres, with the curvature effects showing up between different spheres, not on any single sphere.

19. Originally Posted by caveman1917
It means that the metric on a 2-sphere of constant radius is minkowski (as you can see by setting ). Alternatively it means that you can view the spatial part of the schwarzschild spacetime as a collection of concentric minkowskian 2-spheres, with the curvature effects showing up between different spheres, not on any single sphere.
Oh right, okay. I thought the author was claiming that the distant observer and local observer would measure the same angles, since the metric transforms between them. I can see that observations would be Minkowski between observers on the 2-sphere, sure. I'm not sure what the observations between two observers at different locations at the same r would have to do with deriving the metric though, since it transforms distances and times between an observer at r and a distant observer. I can see that two observers at different locations at r would measure the same distances along the 2-sphere, but how does that translate to assuming that the distant observer will measure the same distance as well? For instance, two observers on the 2-sphere will also measure the same time between events at r, but that doesn't translate to saying that the distant observer will.

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Originally Posted by grav
Oh right, okay. I thought the author was claiming that the distant observer and local observer would measure the same angles, since the metric transforms between them. I can see that observations would be Minkowski between observers on the 2-sphere, sure. I'm not sure what the observations between two observers at different locations at the same r would have to do with deriving the metric though, since it transforms distances and times between an observer at r and a distant observer. I can see that two observers at different locations at r would measure the same distances along the 2-sphere, but how does that translate to assuming that the distant observer will measure the same distance as well? For instance, two observers on the 2-sphere will also measure the same time between events at r, but that doesn't translate to saying that the distant observer will.
I'm not quite following what you're saying. All that is being said by the author is that by spherical symmetry we can introduce the angular coordinates the same way as in minkowski space, and thus only need to worry about the time and radial coordinate. Since the metric is also time-independent we can put all that information together to constrain our solution to be of the form

where take their usual interpretation

It's not saying anything about transforming between measurements between local and far-away observers.

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Originally Posted by caveman1917
I'm not quite following what you're saying. All that is being said by the author is that by spherical symmetry we can introduce the angular coordinates the same way as in minkowski space, and thus only need to worry about the time and radial coordinate. Since the metric is also time-independent we can put all that information together to constrain our solution to be of the form

where take their usual interpretation

It's not saying anything about transforming between measurements between local and far-away observers.
I assume grav is considering accelerated observers that are static relative to the Schwarzschild chart. Isn't this saying that our metric does not change relative to , so the only relevant parameter is r?

Then the metric is invariant among the concentric spheres, but the observers would differ in radial distances and time dilation.

The outer observer would see the inner observer running slower and stretched radially. The inner observer would see the outer observer running faster and compressed radially. I may be wrong about the compression. It should be noted that the speed of light is not constant (or even isometric) in the Schwarzschild chart.

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Originally Posted by utesfan100
I assume grav is considering accelerated observers that are static relative to the Schwarzschild chart. Isn't this saying that our metric does not change relative to , so the only relevant parameter is r?

Then the metric is invariant among the concentric spheres, but the observers would differ in radial distances and time dilation.
Stationary observers would be locally rindler, which only depends on their acceleration, which in turn only depends on their radial coordinate. So in that sense, yes.

23. Originally Posted by caveman1917
It means that the metric on a 2-sphere of constant radius is minkowski (as you can see by setting ). Alternatively it means that you can view the spatial part of the schwarzschild spacetime as a collection of concentric minkowskian 2-spheres, with the curvature effects showing up between different spheres, not on any single sphere.
I would add to above, that it is best explained in Geroch Decomposition (Geroch splitting) witch I have learned recently.
Geroch decomposition uses Killing fields' properties, but is more general.

For gravity under this decomposition, spacetime is made of collection of Minkowski spheres, where every spot is minimally coupled to scalar field (scalar is equal to gravitational time dilation factor).

P.S.
Hi caveman1917 :-)

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Originally Posted by pogono
I would add to above, that it is best explained in Geroch Decomposition (Geroch splitting) witch I have learned recently.
Geroch decomposition uses Killing fields' properties, but is more general.

For gravity under this decomposition, spacetime is made of collection of Minkowski spheres, where every spot is minimally coupled to scalar field (scalar is equal to gravitational time dilation factor).
I'm not familiar with Geroch decomposition, do you have a reference? Is it related to the ADM formalism?

P.S.
Hi caveman1917 :-)
Good evening pogono.

25. Originally Posted by caveman1917
I'm not quite following what you're saying. All that is being said by the author is that by spherical symmetry we can introduce the angular coordinates the same way as in minkowski space, and thus only need to worry about the time and radial coordinate. Since the metric is also time-independent we can put all that information together to constrain our solution to be of the form

yes, where take their usual interpretation
I can see that the functions f and g would be angle independent and time dependent, leaving only functions of r, but I still don't see why there isn't another function h at the end of the metric, which would also be r dependent only. From replies in another forum, I'm getting the impression that h(r) = 1 isn't so much an assumption, but a given. It is derived that way to make the derivation simpler, determining the metric that would keep h(r) = 1, which can then be changed afterward to other coordinate systems where h(r) is not unity, such as with Eddington's isotropic coordinates. A recent post claims that if we start with three functions, they can be manipulated to leave only two that are tied to dt and dr, and they showed it, but with no explanations given as to what they were doing, so I'm still not sure.

If Schwarzschild is just an arbitrary simple coordinate system to derive, though, then I have another dilemma. I always thought of Schwarzschild as the "true" coordiante system, and all others as psuedo coordinate systems that may be easier in some way to work out a solution, but then the solution would have to be changed back to Schwarzschild to give the actual answer. In other words, there should be only one true coordinate system in GR.

In SR, there is no absolute way to synchronize clocks, so we can make any coordinate system we want depending upon how we synchronize clocks in a frame. But in general, we don't change ruler lengths in an inertial frame, nor do we change the tick rates of clocks. We could, but it would not be natural, nor defined as time dilation and length contraction. For instance, if we are transforming between our frame and another in SR, we would take our own clocks and rulers and transport them to the other frame "as is", keeping the tick rate and ruler length they have when they get there.

In GR, however, all hovering observers are static to each other, so there is no synchronization issue between them. If we want to compare ruler lengths, we just send the same ruler directly between different radiuses. We can only gain a different coordinate system by changing the ruler lengths, but that wouldn't be any more natural than it would be in SR. Let's say the distant observer has a lot of very small rulers. He sends them to the radius r, where observers there directly measure the circumference at r by laying out those rulers end to end along the circumference. The distant observer infers the circumference to be 2 pi r, while the local observers measure the same or something else. That would be the true comparison of tangent length. Or to compare two sets of direct physical measurements, the same set of rulers are then sent to the radius s and laid out and we can compare tangent distance measurements between those two. The distant observer just infers the radiuses r and s as in Euclidean flat space, since he is remote from the gravitating body. There should only be one function of r that correctly relates the tangent distance measurements made by the same rulers at r and s. That would tell us what the tangent length contraction of the rulers is at r and s according to the distant observer.

It's not saying anything about transforming between measurements between local and far-away observers.
But that's what the metric is, transforming from a local hovering observer at the same place as the photon to the distant observer. Starting with a local hovering observer's measurements that coincides with the photon, we have

where v_r' is the radial speed that the local observer measures and v_t' is the tangent speed. That becomes

If we assume the quantity dθ' r' remains unchanged, then we have dθ' r' = dθ r that the distant observer measures also. Since dr' = dr / sqrt(1 - 2m/r) and dt' = dt sqrt(1 - 2m/r) when transforming the time and radial distance measurements between the local observer and the distant observer, we get

which is the metric.

26. Oh wait, here's a surprisingly simple mathematical manipulation that would easily get rid of the function tied to the angle.

v_r'^2 + v_t'^2 = c^2

(dr' / dt')^2 + (dθ' r' / dt')^2 = c^2

dr'^2 + dθ'^2 r'^2 = c^2 dt'^2

f(r) c^2 dt^2 - g(r) dr^2 - h(r) dθ^2 r^2 = 0

where all three functions are time independent and angle independent, leaving them only as functions of r. From there we can gain

[f(r) / h(r)] c^2 dt^2 - [g(r) / h(r)] dr^2 - dθ^2 r^2 = 0

A(r) = f(r) / h(r) and B(r) = g(r) / h(r)

A(r) c^2 dt^2 - B(r) dr^2 - dθ^2 r^2 = 0

I don't know how I didn't see that before. But then, even after solving for A(r) and B(r) and gaining the actual metric by plugging in those values, we still have the problem of not knowing what h(r) really is to begin with. The overall metric must still work out using A(r) and B(r), and if we leave the metric as is, we have Schwarzschild, but how do we know if the tangent length contraction is really unity or something else?

27. ## Geroch decomposition references

Originally Posted by caveman1917
I'm not familiar with Geroch decomposition, do you have a reference? Is it related to the ADM formalism?
No, it is not ADM related (btw, I have ADM on my list to learn up to end of 2012 :-)
Here you may find some references to Geroch decomposition, that should make it familiar :-)
I recommend to learn it a bit - many ATM threads (including this thread, mine own last thread and many other historical) are speculations that may be explained much easier while discussed with Geroch framework.

"Global structure of spacetimes, in General Relativity, An Einstein Centenary Survey,", Geroch, R., Horowitz, G. T. Cambridge University Press, Cambridge, 1979 (big part of the publication is about that)

“A space‐time calculus based on pairs of null directions”, R. Geroch, A. Held, and R. Penrose, J. Math. Phys. 1973 (almost all is about that)

"Global Aspects in Gravitation and Cosmology", Joshi, P. S. Oxford University Press, 1993 (partly about that)

And some more popular or more available publications:

"General relativity", Schmidt B.G., Springer, 2000r, page 43 (but only short general explanation)

Wikipedia short definition

Casual hierarhy of spacetimes on arXiv

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Originally Posted by pogono
No, it is not ADM related (btw, I have ADM on my list to learn up to end of 2012 :-)
Here you may find some references to Geroch decomposition, that should make it familiar :-)
I recommend to learn it a bit - many ATM threads (including this thread, mine own last thread and many other historical) are speculations that may be explained much easier while discussed with Geroch framework.

"Global structure of spacetimes, in General Relativity, An Einstein Centenary Survey,", Geroch, R., Horowitz, G. T. Cambridge University Press, Cambridge, 1979 (big part of the publication is about that)

“A space‐time calculus based on pairs of null directions”, R. Geroch, A. Held, and R. Penrose, J. Math. Phys. 1973 (almost all is about that)

"Global Aspects in Gravitation and Cosmology", Joshi, P. S. Oxford University Press, 1993 (partly about that)

And some more popular or more available publications:

"General relativity", Schmidt B.G., Springer, 2000r, page 43 (but only short general explanation)

Wikipedia short definition

Casual hierarhy of spacetimes on arXiv
You mean Geroch's theorem. It proves the existence of a certain "nice" foliation for a certain class of spacetimes. Which is why i asked you about the ADM formalism, it somewhat builds on it in the sense that it presupposes the existence of such a foliation to apply hamiltonian mechanics to the spacetime. At least that's where i've seen it used before. I thought you meant something more general by your reference to "gravity" in general.

In this case i disagree that an application of geroch's theorem would be better. The conditions it requires for getting that "nice" foliation (ie homeomorphic cauchy surfaces coupled to a real interval that can be identified with a temporal function) are quite strong, and they will fail in all except the most symmetric and simple spacetimes. Relying on the existence of specific foliations is not a good idea, though i'll grant you that if such a foliation exists then the analysis becomes quite simple.

29. Order of Kilopi
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3,590
Originally Posted by grav
If Schwarzschild is just an arbitrary simple coordinate system to derive, though, then I have another dilemma. I always thought of Schwarzschild as the "true" coordiante system, and all others as psuedo coordinate systems that may be easier in some way to work out a solution, but then the solution would have to be changed back to Schwarzschild to give the actual answer. In other words, there should be only one true coordinate system in GR.
The only "true" things in GR are its invariants, ie the bit. Coordinate systems are (somewhat) arbitrary. All of them are. However some of them are "natural" for a specific observer, they constitute an orthonormal tetrad at the location of the observer. In the sense that the coordinate would be what would be measured by this observer in the direction, etc. As you can see the schwarzschild chart in the limit of becomes an orthonormal tetrad (it looks just like the minkowski metric). So that's why we say that the schwarzschild chart is that for an observer at infinity, ie it is a "natural" coordinate system for him. However, natural or not, any observer can still use any coordinate system he wants. None are more "true" than any other.

In GR, however, all hovering observers are static to each other, so there is no synchronization issue between them. If we want to compare ruler lengths, we just send the same ruler directly between different radiuses. We can only gain a different coordinate system by changing the ruler lengths, but that wouldn't be any more natural than it would be in SR. Let's say the distant observer has a lot of very small rulers. He sends them to the radius r, where observers there directly measure the circumference at r by laying out those rulers end to end along the circumference. The distant observer infers the circumference to be 2 pi r, while the local observers measure the same or something else. That would be the true comparison of tangent length. Or to compare two sets of direct physical measurements, the same set of rulers are then sent to the radius s and laid out and we can compare tangent distance measurements between those two. The distant observer just infers the radiuses r and s as in Euclidean flat space, since he is remote from the gravitating body. There should only be one function of r that correctly relates the tangent distance measurements made by the same rulers at r and s. That would tell us what the tangent length contraction of the rulers is at r and s according to the distant observer.
But we have three different rulers, one for each direction. So if the distant observer sends his set of three rulers to the near observer, the radial ruler will be contracted but the others won't.

But that's what the metric is, transforming from a local hovering observer at the same place as the photon to the distant observer.
Not exactly, the metric is a certain function on the spacetime (it assigns distances between points). We can certainly use the metric to transform between different observers, but that's not what it is.

30. Originally Posted by caveman1917
Originally Posted by grav
Hmm. That makes sense. I'll have to look into killing vector fields.
You may find these lecture notes (pdf) interesting. It derives the geodesics through killing vector fields and related conservation laws and it also goes into the stability issues with orbits.
I know it is tempting, people even talk about "killing fields", but it would help if we capitalized the K. Just like for Newton's laws. Rindler too. :)

http://en.wikipedia.org/wiki/Wilhelm_Killing

We're already on enough lists in government basements as it is!

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