Gravitational Effects & Instantaneous Spacetime Distortion

[TimeRip496]

Mass distort spacetime resulting in gravitational effects on other non massless objects. When I placed a mass in a particular region, will the distortion of spacetime be instantaneous? Or will it be based on the distance away from that mass and the further away from that mass, the longer it takes for spacetime there to be distorted?

 

[PeterDonis]

Mass distort spacetime resulting in gravitational effects on other non massless objects.

And on massless objects. Light is affected by gravity.

When I placed a mass in a particular region, will the distortion of spacetime be instantaneous?

This question can't be answered as you pose it, because you can't just "place mass in a particular region" in GR. That would violate energy conservation. (More precisely, it would violate the law that the covariant divergence of the stress-energy tensor is zero; that is basically the GR version of local energy conservation.) Whatever mass is in a particular region at a particular time got there somehow, and it had effects on the curvature of spacetime while it was getting there; there is no way to have mass and its associated effects on spacetime curvature suddenly appear out of nowhere. Any such model would not be a valid solution of the Einstein Field Equation.

A better way to pose the question is this: suppose I want to know the curvature of spacetime in a particular region. What portion of spacetime must I "sample" (i.e., look at all the matter and energy in it) in order to have enough information to predict the curvature of spacetime in the particular region I'm interested in? The answer is, I only have to know what is in the past light cone of the region I'm interested in. (Actually, since GR is deterministic, it suffices to know what is in the intersection of one particular spacelike hypersurface with the past light cone.) Nothing outside the past light cone of a particular region of spacetime can affect the spacetime curvature in that region. This is the GR version of "causality", and is often stated informally as "gravity propagates at the speed of light". In this sense, gravity is certainly not instantaneous.

 

[Matterwave]

A better way to pose the question is this: suppose I want to know the curvature of spacetime in a particular region. What portion of spacetime must I "sample" (i.e., look at all the matter and energy in it) in order to have enough information to predict the curvature of spacetime in the particular region I'm interested in? The answer is, I only have to know what is in the past light cone of the region I'm interested in. (Actually, since GR is deterministic, it suffices to know what is in the intersection of one particular spacelike hypersurface with the past light cone.) Nothing outside the past light cone of a particular region of spacetime can affect the spacetime curvature in that region. This is the GR version of "causality", and is often stated informally as "gravity propagates at the speed of light". In this sense, gravity is certainly not instantaneous.

Although do note that in the GR the light "cone" is often no longer a cone :D

Perhaps a more precise way of stating this is then to say that an event P can only be affected by events which can be connected to P via past directed (from P) causal (meaning time-like or null) curves.

 

[PeterDonis]

do note that in the GR the light "cone" is often no longer a cone

Yes, the term "past light cone" in GR just means "causal past", which is defined as you say.

 

[TimeRip496]

So can you use general relativity to explain a thought experiment whereby when I remove sun from solar system, Earth will instantaneously leave its orbit(from a book that I read)?

 

[martinbn]

You can ask the question in Newtonian gravity as follows: you are on the Moon and I on Earth. You have very precise instruments, I have a heavy ball in my feet. At a specific time I lift the ball and you measure until you detect the effect. Then we will know whether it is instantaneous or not. In general relativity the above scenario is meaningless, but if you use an initial value formulation you can phrase everything above in a meaningful way and at least set up the question properly. Of course such an experiment is out of the question, but the theory says that the effect you'll see is not instantaneous i.e. gravity has finite speed of propagation.

 

[NTW]

The following is just an speculation of mine, but I think that the action of gravity over a given mass exist continuously where and when 'its web has already been woven'; that 'gravitational web' being the interaction between that mass at that place and time and the rest of the masses of the universe, either distant or close. If, however, a piece of matter appeared suddenly, 'out of the blue', then a gravitational wave will propagate from that mass, at that time and place of its apparition, towards the rest of the universe, slightly 'reweaving the gravitational web', and endowing that newly born mass with a growing amount of inertia, that will reach an asymptotical maximum as that wave propagates through the farthest reaches of the universe...

A similar thing would happen if the distribution of mass were altered at some place in the universe. The change in the distribution of mass would introduce corrections in the established web, that corrections being transmitted by the emission of gravitational waves to the rest of the universe, reweaving the web...

 

[Nugatory]

So can you use general relativity to explain a thought experiment whereby when I remove sun from solar system, Earth will instantaneously leave its orbit(from a book that I read)?

No. For the reasons given in the first paragraph of PeterDonis's #2 above, you cannot just add or remove masses. Either you have misunderstood what that book was saying, or it is wrong.

 

[PeterDonis]

The following is just an speculation of mine

Please review the PF guidelines on non-mainstream theories, here:

https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

What you are saying here is pretty close to a personal theory, which is off limits here on PF. You do say one thing which can be (charitably) reinterpreted to be more or less correct; see below.

If, however, a piece of matter appeared suddenly, 'out of the blue'

This cannot happen, for the reasons I gave in post #2.

The change in the distribution of mass would introduce corrections in the established web, that corrections being transmitted by the emission of gravitational waves to the rest of the universe

This does happen; if we substitute "changes in spacetime curvature" for "corrections in the established web", this is exactly what gravitational waves are: propagating changes in spacetime curvature due to changes in the distribution of masses. This has been experimentally verified by measuring the changes in binary pulsar systems over several decades.

 

[ChrisVer]

No. For the reasons given in the first paragraph of PeterDonis's #2 above, you cannot just add or remove masses. Either you have misunderstood what that book was saying, or it is wrong.

Can't you look at it as a thought experiment? (not caring about other aspects the removal of mass would have, apart from what happens with the spacetime geometry)

 

[PeterDonis]

Can't you look at it as a thought experiment?

Not in the context of GR, because, as I said, there is no valid solution of the Einstein Field Equation that has mass just appearing out of nowhere. Without a valid solution of the EFE, you have no framework in which to construct even a thought experiment.

 

[ChrisVer]

Not in the context of GR, because, as I said, there is no valid solution of the Einstein Field Equation that has mass just appearing out of nowhere. Without a valid solution of the EFE, you have no framework in which to construct even a thought experiment.

Wouldn't that be a discontinuous jump from Schwarzschild's metric to Minkowski metric? [just asking]

 

[Nugatory]

Can't you look at it as a thought experiment? (not caring about other aspects the removal of mass would have, apart from what happens with the spacetime geometry)

You can in classical mechanics. The distinction between time and space allows you to hold your nose while you suspend conservation of energy and say that the mass is present at time $t$ but not $t+\Delta{t}$. You end up with an infinite speed of gravity, as wherever you are in the universe the quantity $Gm_1m_2/r^2$ must fall to zero during that arbitrarily small time interval.

However, in GR we're describing spacetime, not space. There's no curved spacetime solution to the EFE in which the sun is there one moment and not there the next (what's at the "end" of the worldlines of each particle making up the sun, and what does the metric look like in the region of those "ends"?). Instead, as PeterDonis said, the solution has to include the mass moving into position from somewhere else, or moving out of position to somewhere else. Martinbn has described above how you can formulate the question in those terms - do so and you'll get gravitational waves propagating at $c$.

 

[Nugatory]

Wouldn't that be a discontinuous jump from Schwarzschild's metric to Minkowski metric? [just asking]

That's not an EFE solution. You have to be able to join coordinate patches smoothly.

 

[ChrisVer]

That's not an EFE solution. You have to be able to join coordinate patches smoothly.

This will be my last attempt I am not trying to prove that what you say is wrong, but why it is correct.
Take for example the Schwarzschild's metric written as:
$ds^2_{Schw} = \Big( 1 - \frac{r_s}{r} \Big) dt^2 - \Big( 1 - \frac{r_s}{r} \Big)^{-1} dr^2 - r^2 d\Omega^2$
instead of disappearing the mass suddenly, you can take the case that the mass of the generating source starts dropping (Of course that would mean that this Schwarzschild's metric would have coefficients that are time-dependent =not static-but let's say that this happens very slowly so that it doesn't affect badly the "static" part). By doing so you can see that $r_s \rightarrow 0$ and indeed $ds_{Schw}^2 \rightarrow ds_{Mink}^2$.
I think the problem of the OP, is that the metric is written as such and describes the whole spacetime without saying how fast some change in some point will affect the metric in some other point very far away. eg. in my example we see that the whole spacetime changes instantly as you change the mass parameter (without even sending the mass to zero).

$ds^2_{Schw1} = \Big( 1 - \frac{r_{s1}}{r} \Big) dt^2 - \Big( 1 - \frac{r_{s1}}{r} \Big)^{-1} dr^2 - r^2 d\Omega^2$
describes all the spacetime...
$ds^2_{Schw2} = \Big( 1 - \frac{r_{s2}}{r} \Big) dt^2 - \Big( 1 - \frac{r_{s2}}{r} \Big)^{-1} dr^2 - r^2 d\Omega^2$
describes again all the new spacetime.

 

[ChrisVer]

Ah I think I got it...

 

[PeterDonis]

Wouldn't that be a discontinuous jump from Schwarzschild's metric to Minkowski metric? [just asking]

As Nugatory said, that's not a valid solution. You can't have discontinuities like that. (Nugatory used the phrase "coordinate patches", but the requirement does not depend on your coordinate choice.)

Also, the "jump" from Schwarzschild to Minkowski geometry would not just discontinuously change the geometry; it would also have to change the global topology of the spacetime. Minkowski spacetime has topology $R^4$. Schwarzschild spacetime has topology $R^2 \times S^2$.

instead of disappearing the mass suddenly, you can take the case that the mass of the generating source starts dropping

If you're talking about the full Schwarzschild geometry of a black hole (i.e., vacuum everywhere), this can't happen in classical GR, because there is no "generating source". There is no object inside the black hole that is "generating" its gravity. The spacetime geometry is a self-sustaining curved geometry with no stress-energy anywhere.

However, we can consider a geometry consisting of a spherically symmetric massive object like the Sun surrounded by vacuum; the vacuum geometry will be the metric you wrote down, and the non-vacuum geometry inside the object will be the appropriate interior solution for a spherically symmetric mass (its exact form will depend on how the density varies with radius, and I won't try to write it down as it isn't necessary for this discussion). Then we can ask, what happens if the mass of the object starts decreasing?

What happens is, once again, that the stress-energy has to go somewhere; it can't just disappear. For example, the central object could start emitting radiation that escapes to infinity. We'll assume that the radiation is spherically symmetric, so that we can keep things simple. Eventually, all of the mass could be radiated away, so the central object is gone and there is just vacuum there. At that point, the spacetime will consist of a central vacuum region--the geometry of which will, indeed, be Minkowski--surrounded by a spherically symmetric "shell" of radiation moving outward at the speed of light. (Note that the radius of the vacuum region will be increasing as the radiation moves outward.)

What is the "mass" of this spacetime? Well, it depends on where you are. If you are inside the central vacuum region, then you will see zero "mass"--spacetime in your vicinity will be flat. The spherically symmetric radiation that surrounds the central region will not produce any spacetime curvature inside it. (This is the GR version of the Newtonian "shell theorem".)

However, suppose you are very, very far away from all this--so far that none of the radiation emitted by the central object has reached you yet. Then to you, the mass present is exactly what it was when the central object was there, before it emitted any radiation; the fact that that object has radiated away all of its "mass" makes no difference to you, because the radiation is still closer to the center than you are. You will not see any change until the first bit of radiation reaches you--which, of course, it will do at the speed of light.

in my example we see that the whole spacetime changes instantly as you change the mass parameter.

No, it doesn't. See above. Note that the full metric of the spacetime is not what you assumed; it is not just the Schwarzschild geometry with a time-dependent mass parameter. It's more complicated than that.

 

[Nugatory]

You can't have discontinuities like that. (Nugatory used the phrase "coordinate patches", but the requirement does not depend on your coordinate choice.)

Yes - thanks. If there is no possible set of coordinates that work smoothly, we know that we have a problem but that's better viewed as a symptom of the underlying discontinuity which is there no matter how we play with the coordinates.

 

[Matterwave]

This will be my last attempt I am not trying to prove that what you say is wrong, but why it is correct.
Take for example the Schwarzschild's metric written as:
$ds^2_{Schw} = \Big( 1 - \frac{r_s}{r} \Big) dt^2 - \Big( 1 - \frac{r_s}{r} \Big)^{-1} dr^2 - r^2 d\Omega^2$
instead of disappearing the mass suddenly, you can take the case that the mass of the generating source starts dropping (Of course that would mean that this Schwarzschild's metric would have coefficients that are time-dependent =not static-but let's say that this happens very slowly so that it doesn't affect badly the "static" part). By doing so you can see that $r_s \rightarrow 0$ and indeed $ds_{Schw}^2 \rightarrow ds_{Mink}^2$.
I think the problem of the OP, is that the metric is written as such and describes the whole spacetime without saying how fast some change in some point will affect the metric in some other point very far away. eg. in my example we see that the whole spacetime changes instantly as you change the mass parameter (without even sending the mass to zero).

$ds^2_{Schw1} = \Big( 1 - \frac{r_{s1}}{r} \Big) dt^2 - \Big( 1 - \frac{r_{s1}}{r} \Big)^{-1} dr^2 - r^2 d\Omega^2$
describes all the spacetime...
$ds^2_{Schw2} = \Big( 1 - \frac{r_{s2}}{r} \Big) dt^2 - \Big( 1 - \frac{r_{s2}}{r} \Big)^{-1} dr^2 - r^2 d\Omega^2$
describes again all the new spacetime.

Try this solution, now $r_s=r_s(t)$ and plug this metric to find the Einstein tensor. See what you get. I am actually curious what the result is myself. Since the Bianchi identities are a geometric identity, it would seem that they can not be violated by any valid metric to any manifold, not just the solutions to the EFE's. However, the EFE's will use the Bianchi identities to force conservation of the stress-energy tensor. So my guess is that the Bianchi identities will still hold, but there is no way to make the Einstein tensor equal to any valid stress-energy tensor (which would imply that the Einstein tensor is non-zero in this case...). The other possibility that I can think of is that there is a defect with this metric itself, and that it can't be a valid metric to a Lorentzian manifold.

 

[PeterDonis]

I am actually curious what the result is myself.

I ran it through Maxima and obtained the following Einstein tensor:

$$
G^t{}_r = \frac{2 \frac{dm}{dt}}{\left( r - 2m \right)^2}
$$

$$
G^r{}_t = \frac{2 \frac{dm}{dt}}{r^2}
$$

$$
G^{\theta}{}_{\theta} = G^{\phi}{}_{\phi} = \frac{r^2 \frac{d^2 m}{dt^2} + r \left[ \left( \frac{dm}{dt} \right)^2 - 2 m \frac{d^2 m}{dt^2} \right]}{\left( r - 2m \right)^3}
$$

Since the Bianchi identities are a geometric identity, it would seem that they can not be violated by any valid metric to any manifold, not just the solutions to the EFE's.

Actually, since computing the Einstein tensor is just a matter of taking derivatives and multiplying, any metric with continuous first and second derivatives will give a valid "solution" to the EFE, mathematically speaking. You just take the Einstein tensor, multiply it by $8 \pi$, and call the result the stress-energy tensor. Whether the solution makes sense physically is another matter.

the EFE's will use the Bianchi identities to force conservation of the stress-energy tensor.

Yes. But again, that's just a mathematical fact. Whether the stress-energy tensor makes sense physically is another matter.

my guess is that the Bianchi identities will still hold, but there is no way to make the Einstein tensor equal to any valid stress-energy tensor (which would imply that the Einstein tensor is non-zero in this case...).

No, it implies that the stress-energy tensor is nonzero, i.e., that the spacetime is not vacuum. You can see from the results I gave above that indeed this is the case: the "Schwarzschild metric" with $m$ as a function of $t$ gives a non-zero stress-energy tensor, therefore the spacetime is not vacuum. The Bianchi identities still hold (as they must, since, as you say, they are geometric identities that don't depend on the solution being physically reasonable).

Mathematically, this non-vacuum stress-energy tensor is perfectly "valid"; after all, I just wrote it down, and there aren't any expressions in it that are problematic, mathematically speaking. However, there is an obvious issue physically: $G^r{}_t$ is singular at $r = 0$ unless we set $dm/ dt = 0$ at $r = 0$. However, we specified this metric with the assumption that $m$ was a function only of $t$, not $r$. That means that, if $dm / dt = 0$ at $r = 0$, we must have $dm / dt = 0$ at all values of $r$. And, of course, if this is true, then the Einstein tensor vanishes and we just have the ordinary Schwarzschild metric.

In other words, what we have done here is not derive a metric that is physically different from the Schwarzschild metric. What we have done is to prove that, if the metric takes the form of the Schwarzschild metric, then $m$ cannot be a function of $t$, physically speaking. This is why I said, at the end of post #17, that the metric for the case of a mass that is "radiating away" is not the Schwarzschild metric with a time-dependent $m$; it can't be, as we have just shown.

The other possibility that I can think of is that there is a defect with this metric itself, and that it can't be a valid metric to a Lorentzian manifold.

No, the metric is perfectly valid; it just doesn't mean what ChrisVer thinks it means. See above.

 

[Matterwave]

And there we go. Thanks for clarifying! :D

I think another way to look at this would be from Birkhoff's theorem which says that Schwarzschild metric is the only (most general) vacuum, spherically symmetric solution to the EFE's. As such, modifying the Schwarzschild metric to have a time dependent m either destroys the vacuum part or the spherically symmetric part of the solution. It seems the spherical symmetry is manifest in the form of the metric itself, so it must not be a vacuum solution. :)

 

[PeterDonis]

another way to look at this would be from Birkhoff's theorem which says that Schwarzschild metric is the only (most general) vacuum, spherically symmetric solution to the EFE's.

Yes. That's the underlying reason why what we tried didn't work.

modifying the Schwarzschild metric to have a time dependent m either destroys the vacuum part or the spherically symmetric part of the solution.

No, what we found is more than that. We found that a metric with the same form as the Schwarzschild metric cannot have m depend on t alone. We didn't find that such a solution is non-vacuum; we found that it's not possible (because the only way to remove the singularity at $r = 0$ is to make $m$ constant). The fact that, mathematically, the Einstein tensor we computed is non-zero turned out to be a red herring.

The more interesting question is, what happens if we allow $m$ to be a function of both $t$ and $r$ (instead of just $t$ alone)?

 

[ChrisVer]

The more interesting question is, what happens if we allow mm to be a function of both tt and rr (instead of just tt alone)?

And that would be my next question...
I think it can work, but it will not be a vacuum solution 100%.

 

[PeterDonis]

I think it can work, but it will not be a vacuum solution 100%.

You are correct; it can work, and it will not be a solution with vacuum everywhere. But you can, by picking an appropriate function for $m(t, r)$, produce a solution that looks like an object that radiates away all its mass and leaves a central vacuum region with flat spacetime geometry (and with $m = 0$), surrounded by a spherically symmetric region of radiation expanding outward (this is where the behavior of $m(t, r)$ is interesting), surrounded by a vacuum region with Schwarzschild geometry (and with $m$ constant in this region).

 

[Matterwave]

No, what we found is more than that. We found that a metric with the same form as the Schwarzschild metric cannot have m depend on t alone. We didn't find that such a solution is non-vacuum; we found that it's not possible (because the only way to remove the singularity at $r = 0$ is to make $m$ constant). The fact that, mathematically, the Einstein tensor we computed is non-zero turned out to be a red herring.

The more interesting question is, what happens if we allow $m$ to be a function of both $t$ and $r$ (instead of just $t$ alone)?

I don't understand this reasoning, the singularity at r=0 in the Scwharzschild solution itself is non-removable. It is a "real" singularity. So even if we made m constant, as is the case in the Schwarzschild metric, the singularity at r=0 is still there, we haven't removed it.

 

[ChrisVer]

I don't understand this reasoning, the singularity at r=0 in the Scwharzschild solution itself is non-removable. It is a "real" singularity. So even if we made m constant, as is the case in the Schwarzschild metric, the singularity at r=0 is still there, we haven't removed it.

I think that (the r=0 singularity) is not for the energy momentum tensor... If m is constant the Einstein tensor vanishes, and so does the EM tensor.

 

[Matterwave]

I think that is not for the energy momentum tensor...

What do you mean? The essential singularity at r=0 in the Schwarzschild metric is not even part of the manifold. No tensors exist there.

 

[ChrisVer]

What do you mean? The essential singularity at r=0 in the Schwarzschild metric is not even part of the manifold. No tensors exist there.

What I meant is that the singularity at r=0 in this case does not appear for the metric itself, but for the Einstein Tensor...
For the Schwarzschild's metric (where m=const) the Einstein tensor vanishes (take the case of the components PeterDonis gave, and see that). In this case it does not, which means that there should be an Energy momentum tensor on the right hand side for the EFE. That tensor will have to get this singularity at r=0 for a time-dependent alone m...
maybe I'm wrong.

 

[PeterDonis]

the singularity at r=0 in the Scwharzschild solution itself is non-removable

Yes. But that is a singularity in the components of the Riemann curvature tensor. It doesn't show up in the Einstein Field Equation--it can't, because for the standard Schwarzschild solution with constant $m$, the RHS of the EFE vanishes--it's a vacuum solution with zero stress-energy tensor. That's true even at $r = 0$; mathematically, the EFE components still all vanish at $r = 0$ if they are evaluated with $m$ constant (i.e., not a function of any of the coordinates).

The singularity I referred to is a singularity in the components of the EFE; it's saying that the mathematical solution with $m$ as a function of $t$ only is physically unreasonable because the EFE components become singular at $r = 0$, which means the SET components, things like the energy density, would be singular at $r = 0$. In the standard Schwarzschild solution with $m$ constant, the energy density (and all other SET components) are zero at $r = 0$. If you want to be precise and account for the fact that $r = 0$ is not actually part of the manifold, you can say that the limit of all the EFE components for constant $m$ is zero as $r \rightarrow 0$, while the limit of various Riemann tensor components (that don't appear in the EFE) is $\infty$ as $r \rightarrow 0$. If $m$ is a function of $t$, however, then the limit of some of the EFE components becomes $\infty$ as $r \rightarrow 0$, which is a separate issue from the Riemann tensor component issue.

 

[ChrisVer]

Of course I could find some other point in this... For example, why should we take that the Schwarchzild's metric as given (with m=m(t) ) should be valid in all regions? I think that the Schwarczhild's metric changes for $r<r_s$.
I mean it can as well be m=m(r,t) for r<r_s and m=m(t) outside.

 

[PeterDonis]

it can as well be m=m(r,t) for r<r_s and m=m(t) outside.

If this is possible, it will be a special case of the general solution where $m = m(r, t)$--there will just be regions where $\partial m / \partial r = 0$. So solving the general case of $m(r, t)$ will tell you whether this is possible.

 

[ChrisVer]

If this is possible, it will be a special case of the general solution where $m = m(r, t)$--there will just be regions where $\partial m / \partial r = 0$. So solving the general case of $m(r, t)$ will tell you whether this is possible.

Yes... sometimes you think of things after you've written them... I eventually did exactly that by the time I wrote m=m(r,t).

 

[Matterwave]

Yes. But that is a singularity in the components of the Riemann curvature tensor. It doesn't show up in the Einstein Field Equation--it can't, because for the standard Schwarzschild solution with constant $m$, the RHS of the EFE vanishes--it's a vacuum solution with zero stress-energy tensor. That's true even at $r = 0$; mathematically, the EFE components still all vanish at $r = 0$ if they are evaluated with $m$ constant (i.e., not a function of any of the coordinates).

The singularity I referred to is a singularity in the components of the EFE; it's saying that the mathematical solution with $m$ as a function of $t$ only is physically unreasonable because the EFE components become singular at $r = 0$, which means the SET components, things like the energy density, would be singular at $r = 0$. In the standard Schwarzschild solution with $m$ constant, the energy density (and all other SET components) are zero at $r = 0$. If you want to be precise and account for the fact that $r = 0$ is not actually part of the manifold, you can say that the limit of all the EFE components for constant $m$ is zero as $r \rightarrow 0$, while the limit of various Riemann tensor components (that don't appear in the EFE) is $\infty$ as $r \rightarrow 0$. If $m$ is a function of $t$, however, then the limit of some of the EFE components becomes $\infty$ as $r \rightarrow 0$, which is a separate issue from the Riemann tensor component issue.

So you are saying the problem really is that the stress-energy tensor must become arbitrarily large at points of the manifold arbitrarily close to r=0? I guess I'll buy this argument. :D

 

[PeterDonis]

So you are saying the problem really is that the stress-energy tensor must become arbitrarily large at points of the manifold arbitrarily close to r=0?

Yes. And that this problem is a separate problem from the problem of the curvature tensor becoming arbitrarily large at points arbitrarily close to $r = 0$ (which is the "essential singularity" in the standard Schwarzschild geometry, with $m$ constant).

 

[ChrisVer]

I tried thinking of some good combination for m(r,t), but it's not that easy... because I think it gets a self-feeding function, like:
$m(r,t) = m(t) [1 - e^{-r/r_s}]$
But $r_s = 2 G m(r,t)$ doesn't stay constant...