# Thread: Solving the Schwarzschild metric

1. Originally Posted by caveman1917
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.
That would depend upon our coordinate choice. For instance, in Eddington's isotropic coordinates, the tangent length of the ruler is not the same, so is contracted also. All three rulers will be contracted by the same amount, allowing the distant observer to infer the same speed of light at r in any direction, although less than c that the local observer measures.

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.
Oh right, it actually involves three transformations that combine to form an invariant metric.

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.
Okay, thanks. That has given me a new perspective, one that may help to resolve a problem that has been nagging at me ever since I began working with GR.

In every problem I've ever dealt with, a complex mathematical result describes an impossible scenario, something that cannot occur. Yet in the Schwarzschild coordinate system, we have a coordinate radius beyond which all sorts of complex results arise, below the event horizon. This says nothing should exist within this space, since we cannot have clocks that tick at a complex rate and we cannot have complex lengths. This space below 2m, then, really shouldn't exist. Yet using Schwarzschild coordinates, we can still plot coordinate distances between zero and 2 m according to a distant observer, that a freefaller falls into within finite proper time, although we can't describe what happens below 2 m.

As I mentioned, I had always thought of Schwarzschild coordinates as the "real" coordinate system, while all others were psuedo-systems for ease of derivation that must be switched back to Schwarzschild to gain the "actual" results. But if Schwarzschild is just as arbitrary a coordinate system as any other, then we can just as easily make a different coordinate choice. Of course, even in deriving Schwarzschild, we must follow certain rules: the locally measured speed of light is always c, SR is valid locally, the equivalence principle holds, the EFE equations are valid, locally measured angular momentum is constant, the locally measured energy of a particle per local time dilation is constant, etc. I will add one more which will limit my choice of coordinate systems even further. It is simply that we will only measure "real" space with our coordinate system.

My choice of coordinate systems, at least the simplest I've found so far, then, is

r1 = r (1 - 2 m / r), r = r1 (1 + 2 m / r1) which transforms the metric to

ds^2 = c^2 dt^2 / (1 + 2 m / r1) - dr1^2 (1 + 2 m / r1) - dθ^2 r1^2 (1 + 2 m / r1)^2

I have now turned the event horizon at r = 2 m in Schwarzschild into a point singularity at r1 = 0. Nothing below r1 = 0 exists within this coordinate system, so complex space is eliminated. Problem solved. This is the way I believe it should be, and if I am allowed to use any coordinate system I choose, this is definitely the one that makes sense to me. If a freefalling observer were to fall all the way to the point singularity in this coordinate system, in a finite time according to his own watch, then since that is also where the point mass must exist, the freefaller will immediately strike the mass rather than falling further into some complex space. If the mass were not struck somehow, and the freefaller could keep going, he would simply pass through the point in real space and begin decelerating on the other side.

Of course, there are still plenty of other coordinate systems that create a point singularity, r1 = r sqrt(1 - 2 m / r) for instance. The only other rule that must apply is that r1 approximates r as r goes to infinity. If we want to describe -r as the same as r but in the opposite direction, so that r and -r must gives the same time dilations and length contractions, then we could use r1 = sqrt(r^2 - (2 m)^2). Even if others might not agree, I truly believe this condition of only allowing real space should be included in the coordinate choice. Other conditions might narrow it down further as well. Do you or anyone else have any other conditions for the coordinate choice in mind that might help things make more sense to you?

2. I now want to use K to show that ds = c dτ for the proper time of a particle or freefaller, as I wasn't sure about that earlier in the thread. For a freefaller passing a static observer at r with a scalar speed of v' in any direction, the locally measured time dilation of the freefaller's clock is sqrt(1 - (v'/c)^2). Since there are no simultaneity differences between static observers, only a gravitational time dilation between them, the time dilation that the distant observer measures of the freefaller's clock becomes dτ / dt' = sqrt(1 - (v'/c)^2) slower than for the local static observer that coincides in the same place, with the static observer's clock ticking dt' / dt = sqrt(1 - 2 m / r) slower than the distant observer's own clock, so the distant observer measure's a time dilation for the freefaller of (dτ / dt) = (dτ / dt') (dt' / dt) = sqrt(1 - (v'/c)^2) sqrt(1 - 2 m / r). Okay, so we have

and the metric becomes

and substituting for v'^2,

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Originally Posted by grav
I now want to use K to show that ds = c dτ for the proper time of a particle or freefaller, as I wasn't sure about that earlier in the thread.
This is always true, you don't have to show that seperately.

Just note that in its own coordinates, the coordinate time of an object equals its proper time, and it is always at the center of its own coordinates so it doesn't move spatially in them.
Then the metric in its own coordinates will always reduce to and since is an invariant, it is the same as that in any other coordinates.

4. Originally Posted by caveman1917
This is always true, you don't have to show that seperately.

Just note that in its own coordinates, the coordinate time of an object equals its proper time, and it is always at the center of its own coordinates so it doesn't move spatially in them.
Then the metric in its own coordinates will always reduce to and since is an invariant, it is the same as that in any other coordinates.
That would be true in SR since the particle (or observer) is always at the origin of its own coordinate sytem, so the distance between any two events where the particle coincides for both is zero in the particle's own coordinate system, whereby dx = dy = dz = 0, leaving only its proper time for the Minkowski metric, giving ds = c dτ. But in GR, the particle is not the origin of its own coordinate system, the center of the gravitating mass is, so I wasn't sure if the metric should include dr and dθ according to the change in radial distance that the particle (observer) measures for the center of the gravitating mass. Apparently, though, dr and dθ still refer only to the change for the particle itself, which makes sense to me now that I think about it, since we are not actually referring to anything having to do with the entire radial distance r of the particle from the center of the gravitating mass, but just some change in distance dr in that direction that the particle itself travels, same as for SR, which the particle (observer) would measure as zero.

By the way, you wrote ds^2 = c^2 dt^2 = c^2 dτ^2. What is the dt there?

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Originally Posted by grav
But in GR, the particle is not the origin of its own coordinate system
A particle is always the origin of its own coordinate system. That's what it means to attach a coordinate system to an observer after all.

By the way, you wrote ds^2 = c^2 dt^2 = c^2 dτ^2. What is the dt there?
The coordinate time (which for a coordinate system attached to an observer/clock obviously equals its proper time).

6. Originally Posted by caveman1917
A particle is always the origin of its own coordinate system. That's what it means to attach a coordinate system to an observer after all.
Right, in SR. But in GR, the origin of the coordinate system is at the center of the gravitating body, at the point where r = 0. That's why a distant observer can place himself an infinite distance from the center of the gravitating body without having to incorporate that distance into the metric, because we are not considering coordinates in respect to him, but in respect to the center of the body.

The coordinate time (which for a coordinate system attached to an observer/clock obviously equals its proper time).
The coordinate time isn't ds = c dt, though, at least applying dt for the coordinate time of a distant observer as in the metric. That would make the coordinate time and the proper time the same, unless you are considering the coordinate time according to the freefaller which is just his proper time, right, although somewhat confusing to identify it that way if we don't just use dτ for the time that passes upon his clock according to any observers.
Last edited by grav; 2012-Aug-24 at 11:23 PM.

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Originally Posted by grav
Right, in SR. But in GR, the origin of the coordinate system is at the center of the gravitating body, at the point where r = 0. That's why a distant observer can place himself an inifinite distance from the center of the gravitating body without having to incorporate that distance into the metric, because we are not considering coordinates in respect to him, but in respect to the center of the body.
The observer at infinity is not truly an observer either, what we have is the schwarzschild coordinate chart which we identify with a presumed "bookkeeping observer at infinity".

But your statement is simply wrong, there is nothing telling you that in GR you must put the origin of a coordinate system at the center of a gravitating body. Using a particle's "own coordinate system" specifically means attaching a coordinate system to the particle, who is in effect dragging the origin of that coordinate system with him. He would consider himself to be stationary and the black hole speeding up towards him.

The coordinate time isn't ds = c dt, though, at least applying dt for the coordinate time of a distant observer as in the metric. That would make the coordinate time and the proper time the same, unless you are considering the coordinate time according to the freefaller which is just his proper time, right, although somewhat confusing to identify it that way if we don't just use dτ for the time that passes upon his clock according to any observers.
I think you're missing the entire point somehow. We can always attach a coordinate system to a particle such that that particle remains at the origin of that coordinate system and those coordinates represent the measureables of that particle/observer. Specifically the coordinate time in that coordinate system equals the proper time for that particle.

We're not using "any observers", we're using only the particle itself. The t is the coordinate time for his coordinate system, which obviously equals , ie its proper time. Given that he always remains at the origin of his own coordinate system (or at least doesn't move spatially, even if we put the origin somewhere else), the metric for his own worldline in his own coordinates is simply and thus

Now only at this point do we consider other observers, and we note that is an invariant, so the for the particle's coordinate system over his own worldline equals the for any other observer over that particle's worldline, and thus we have that for every observer over that particle's worldline. Which is what you wanted to prove.

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Perhaps there has been some misunderstanding because you're looking specifically at the schwarzschild spacetime and schwarzschild coordinate chart (therefor identifying t with the t of the schwarzschild observer), whereas the argument i'm making is completely general and not contingent on any specific spacetime, be it schwarzschild, kerr or whatever. So the use of "t" is the time coordinate of the coordinate system attached to the particle irrespective of which spacetime it is in.

9. Originally Posted by caveman1917
But your statement is simply wrong, there is nothing telling you that in GR you must put the origin of a coordinate system at the center of a gravitating body.
Not that it must be, just that it is. The origin in Schwarzschild lies at r = 0. To place the origin at some observer, we would have to consider the distance between the observer at r_obs and the particle at r and incorporate that distance between them into the metric somehow. However, one simple way around this that I had not considered is to just place the observer at the same radius as the particle, whereby r_obs = r. So considering both observer and particle lie upon the same 2-sphere, and if the metric upon the 2-sphere reduces to Minkowski, then ds = c dτ for the particle the same as for Minkowski, right. Perhaps that is the much simpler way you were looking at it, just looking at the local Minkowski 2-sphere where ds = c dτ must naturally follow, while I had to work through the maths to find the same thing.

EDIT - Well, actually the particle's path most likely wouldn't remain upon the same 2-sphere, even infinitesimally, only starting or ending there, say, but still SR is valid locally if we just consider that the observer and particle originally coincide in the same place, and then especially if we consider that the particle is the observer, right, so there you go.

EDIT AGAIN - Wait, yup, you're absolutely right. In a way we both are. I've been trying to place the origin at the observer, and realized it works out to the same difference. In both SR and GR, since the metrics only use changes in distance, not distances themselves, we get the same result regardless of where the origin is actually placed, it doesn't matter. In SR, for instance, if we place the origin 10 meters in the + x direction from the observer, whereas it would normally be placed at the observer, the change in distance dx that a particle travels along x still remains the same, and that is what is used in the metric. The same goes for GR. I've just gotten used to the idea of placing r = 0 at the origin, but it doesn't change the metric if I place it elsewhere, such as at an observer, although I'm still considering only static observers since it's easier to visualize while keeping the origin static.

Perhaps there has been some misunderstanding because you're looking specifically at the schwarzschild spacetime and schwarzschild coordinate chart (therefor identifying t with the t of the schwarzschild observer), whereas the argument i'm making is completely general and not contingent on any specific spacetime, be it schwarzschild, kerr or whatever. So the use of "t" is the time coordinate of the coordinate system attached to the particle irrespective of which spacetime it is in.
Right, okay.
Last edited by grav; 2012-Aug-25 at 03:05 AM.

10. Okay, now I want to work out what the proper time of the freefaller will be upon reaching a radius h when falling from rest from a radius b with an original clock reading of zero. So we have

where and for a particle falling radially, giving

where

for a particle falling from rest at b, where also in that case, so

When falling to h = 2 m where b >> 2m, the atan approaches π / 2 and the first term in brackets helps to make up the difference, so we can approximate (the same as using h = 0) with

11. Geesh. All I really needed to show that ds = c dτ, without any reference to energy, is

v'^2 = v_r'^2 + v_r'^2

for the locally measured radial and tangent speed of a freefaller with scalar speed v', which transforms to

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

v'^2 dt'^2 = dr'^2 + dθ'^2 r'^2

c^2 dt'^2 - [v'^2 dt'^2] = c^2 dt'^2 - [dr'^2 + dθ'^2 r'^2]

c^2 (1 - (v'/c)^2) dt'^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2

where dτ = sqrt(1 - (v'/c)^2) dt' as the clock of the freefaller is observed locally, so

c^2 dτ^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2

ds^2 = c^2 dτ^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2 = c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2

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