# Gravity instantaneous?

ChrisVer
Gold Member
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
Gold Member
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
Gold Member
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 Schwarzchild'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
Mentor
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
ChrisVer
Gold Member
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
Mentor
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
ChrisVer
Gold Member
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).

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Matterwave
Gold Member
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
Mentor
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
Gold Member
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...

PeterDonis
Mentor
I tried thinking of some good combination for m(r,t)

For modeling an object that radiates away mass, any such function is going to have to describe several distinct regions, where the functional dependence is different: a starting region where ##m## is time-independent, but is a function of ##r## (describing the initial mass distribution of the body before it starts radiating); then a description of the radiation process, which will require ##m## to be a function of both ##r## and ##t##, describing both the reduction of mass of the body (which will involve the radius of its surface decreasing as well) and the mass carried away by the radiation (which will be, in the idealized case, a spherically symmetric expanding shell whose inner and outer radius are changing with time); and finally a description of the vacuum regions, the one left inside the inner surface of the radiation shell after all the mass is radiated away, where the geometry is flat, and the one outside the outer surface of the radiation shell, where the mass is still the original mass ##m## and the geometry is the standard Schwarzschild geometry.

I don't think there a single closed-form function that describes all of the above; the best you can do is to fit together several different functions with appropriate junction conditions at the boundaries of the different regions.

##r_s = 2 G m(r,t)## doesn't stay constant...

Of course not. How can it if ##m## is changing?

pervect
Staff Emeritus
I think the Vaidya metric, http://en.wikipedia.org/w/index.php?title=Vaidya_metric&oldid=628366498, might be of some interest.

Wiki said:
n general relativity, the Vaidya metric describes the non-empty external spacetime of a spherically symmetric and nonrotating star which is either emitting or absorbing null dusts.

"null dust" can be interpreted as incoherent electromagnetic radiation. (See for instance http://en.wikipedia.org/w/index.php?title=Bonnor_beam&oldid=479934048).

So if you want the metric of a radiating star, the Vaidya metric is the place to start, rather than trying to modify the Schwarzschild metric.

ChrisVer
PeterDonis
Mentor
the Vaidya metric is the place to start, rather than trying to modify the Schwarzschild metric

The Vaidya metric basically is "the Schwarzschild metric, modified to allow ##m## to be a function of ##r## and ##t##". (More precisely, it's that under the assumption that ##m## is changing because of radiation.) The Vaidya metric is usually written in null coordinates, where you can collapse the coordinate dependence of ##m## down to one coordinate (either ##u## or ##v## depending on whether you are looking at the outgoing or ingoing Vaidya metric--outgoing would be the relevant one for this case). But if you write it in ordinary ##t, r## coordinates instead of ##u, v##, then ##m## will be a function of both ##t## and ##r##.

ChrisVer
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.

So there was no "big bang" after all? That's a relief, as that whole "fiat lux" creationist idea never made any sense at all.

PeterDonis
Mentor
So there was no "big bang" after all?

If you mean the actual scientific theory of the "Big Bang", that theory has never said matter and energy suddenly appeared out of nowhere. That's a pop science misunderstanding. The term "Big Bang" is properly used to refer to the fact that the early universe was in a very hot, dense state and was expanding rapidly; that's all. We've had a number of threads on this recently in the Cosmology forum.

pervect
Staff Emeritus
The Vaidya metric basically is "the Schwarzschild metric, modified to allow ##m## to be a function of ##r## and ##t##". (More precisely, it's that under the assumption that ##m## is changing because of radiation.) The Vaidya metric is usually written in null coordinates, where you can collapse the coordinate dependence of ##m## down to one coordinate (either ##u## or ##v## depending on whether you are looking at the outgoing or ingoing Vaidya metric--outgoing would be the relevant one for this case). But if you write it in ordinary ##t, r## coordinates instead of ##u, v##, then ##m## will be a function of both ##t## and ##r##.

The Schwarzschild metric is a vacuum solution - the Vaidya solution isn't. Certainly, they are very similar, as usually the gravitational effects of emitted radiation are tiny - the contributions of the stress-energy tensor of the radiation to the gravitational field are very small. In the limit where the emitted radiation has negligible effects, the Vaidya metric and the Schwarzschild metric should be equivalent. If the emitted radiation has experimentally measurable effects, though, you'd want to use the Vaidya metric.

ChrisVer
Gold Member
So there was no "big bang" after all? That's a relief, as that whole "fiat lux" creationist idea never made any sense at all.

As already said, the Big Bang Theory doesn't look the topic of matter creation out of nowhere.
Also, even if that was once the case (that is a hypothesis), when this happened the Einstein Field Equations (EFE) wouldn't look the same... because you would need something else (=a quantum theory of gravity), since EFEs are classical equations.

ChrisVer
Gold Member
The Schwarzschild metric is a vacuum solution - the Vaidya solution isn't. Certainly, they are very similar, as usually the gravitational effects of emitted radiation are tiny - the contributions of the stress-energy tensor of the radiation to the gravitational field are very small. In the limit where the emitted radiation has negligible effects, the Vaidya metric and the Schwarzschild metric should be equivalent. If the emitted radiation has experimentally measurable effects, though, you'd want to use the Vaidya metric.

I think that's the same here, no? we said that inserting in the Schwarzchild's metric $m=m(t)$ leads to inconsistency/singularity for the EM tensor and this led us to try setting $m=m(r,t)$ to get rid of it. In the first case (also I believe in the second, as a generalization of the first), we obtained a non-vacuum solution (non-vanishing Einstein Tensor/non-vanishing energy-momentum tensor). So indeed we "left" Schwarczhild's metric.
I think by the time we let the S.metric to be time-dependent, we left its special case.

PeterDonis
Mentor
The Schwarzschild metric is a vacuum solution - the Vaidya solution isn't.

Yes, I was only referring to the forms of the two metrics being the same; that is, the Vaidya metric, formally, is the same as the Schwarzschild metric except that ##m## is a function of one of the null coordinates (in the null coordinate chart) or of ##t## and ##r## (in the standard Schwarzschild-type chart).

And on massless objects. Light is affected by gravity.

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.
That is a very helpful reply. Thank you. It is pleasant to read something I can understand.