Newtonian Limit of GR: Exploring Einstein Equations

[pseudoriemann]

Hi guys, I don´t understand too much the Newtonian limit of General Relativity. My question is:

In this limit g_{a b} (x) = η_{a b} (x) + h_{a b} (x) where O(h²→0). Then, according to GR, it's straightforward to demonstrate that the Einstein equation G_{t t} = k*T_{t t} in that limit leads to Poisson equation (here we have considered Λ = 0 and the energy-stress tensor of a perfect fluid with null pressure), but what about the other components of these equations??

I've read in many books that they are neglected but I don't agree with this. For example, according to this limit in 4 dimensions we have R = -k*T ≠ 0 and there is not any reason to neglect it, so G_{r r} ≠ k*T_{r r}.

What's happening here?? Thanks in advanced and forgive my english.

 

[Markus Hanke]

In the Newtonian limit ( which means weak gravity, and low velocity ), the $$T^{00}$$ component is larger than the other components of the energy-momentum tensor by several orders of magnitude, which is why those are generally disregarded. Only if you go to relativistic velocities and/or strong sources do these play a substantial role.

 

[pseudoriemann]

Thank you Markus Hanke,

in this limit T_{r r} is neglected but what I'm saying is that according to my opinión G_{r r} is not neglected and the same for G_{θ θ} and G_{φ φ}, so the Einstein equations are not totally valid in this limit. You may write the expressions of these components of the Einstein tensor in terms of Christoffel symbols and check that G_{r r} G_{θ θ}, G_{φ φ} are not neglected in this limit and neither null.

 

[Mentz114]

Thank you Markus Hanke,

in this limit T_{r r} is neglected but what I'm saying is that according to my opinión G_{r r} is not neglected and the same for G_{θ θ} and G_{φ φ}, so the Einstein equations are not totally valid in this limit. You may write the expressions of these components of the Einstein tensor in terms of Christoffel symbols and check that G_{r r} G_{θ θ}, G_{φ φ} are not neglected in this limit and neither null.

Maybe you misunderstand 'neglected'. In taking a weak field limit one chooses to ignore terms which are $O(1/c)$ smaller than others. It is an approximation and
a perfectly valid one.

 

[pseudoriemann]

(My text disappeared)

Maybe you misunderstand 'neglected'. When taking the weak field limit we may choose to ignore terms which are $O(1/c)$ smaller than others. It is a valid approximation.

ok let's do it (I apologize in advance about possible mistakes):

For example, if we considered a static spherically symmetric spacetime in a coordinate system where $ds^2 = g_{t t} (r)c^2dt^2-\frac{dr^2}{g_{tt}(r)}-r^2d\Omega^{2}$ then we have $R_{\theta \theta}=\partial_{r}\Gamma^{r}\,_{\theta \theta}-\partial_{\theta}\Gamma^{\phi}\,_{\theta \phi}-\Gamma^{\phi}\,_{\theta \phi}\Gamma^{\phi}\,_{\theta \phi}$ I think that's the general expression in this case and there isn't any term of order $1/c$ which implies that such a component $R_{\theta \theta}$ may be neglected and smaller than others (in fact, $\Gamma^{r}\,_{t t}$ is the only component of such a spacetime that depends on $c$). Moreover, the Einstein tensor contains a $\frac{R}{2}g_{\theta \theta}=\frac{R}{2}\eta_{\theta \theta}$ term that I think we shouldn't neglect because it's of the same order than the others...

Finally, if we look at the component $R_{t t}$, in the weak field limit we have $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$ where $O(h^{2}) \rightarrow 0$, thus I'm doing $R_{t t} \approx \partial_{r}\Gamma^{r}\,_{t t}+(\Gamma^{\theta}_{r \theta}+\Gamma^{\phi}_{r \phi})\Gamma^{r}_{t t}$, but the standard weak field limit of GR goes further and neglects the $(\Gamma^{\theta}_{r \theta}+\Gamma^{\phi}_{r \phi})\Gamma^{r}_{t t}$ too and suddenly an energy-stress tensor of a perfect fluid with pressure null appears in the action $S$ from nothing. Why??

 

[Mentz114]

in fact, Γrtt\Gamma^{r}\,_{t t} is the only component of such a spacetime that depends on cc

Is that a factor of $c$ ? If so then that implies that ${\Gamma^r}_{tt}$ is larger than the others by a factor of $\approx c$ ?

 

[pseudoriemann]

Is that a factor of $c$ ? If so then that implies that ${\Gamma^r}_{tt}$ is larger than the others by a factor of $\approx c$ ?

But $c$ is only a constant, you can take $c^{2}G_{rr}$ and then this factor and $G_{tt}$ are of the same order in $c$, why do we neglect the Einstein equation $c^2G_{rr}=c^2kT_{rr}$? In fact, in Planck units we can take $c=1$ and $O(v^2) \rightarrow 0$ so there wouldn't be any difference between the above Einstein components.

And what about this??:

Finally, if we look at the component $R_{t t}$, in the weak field limit we have $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$ where $O(h^{2}) \rightarrow 0$, thus I'm doing $R_{t t} \approx \partial_{r}\Gamma^{r}\,_{t t}+(\Gamma^{\theta}\,_{r \theta}+\Gamma^{\phi}\,_{r \phi})\Gamma^{r}\,_{t t}$, but the standard weak field limit of GR goes further and neglects the $(\Gamma^{\theta}\,_{r \theta}+\Gamma^{\phi}\,_{r \phi})\Gamma^{r}\,_{t t}$ too and suddenly an energy-stress tensor of a perfect fluid with pressure null appears in the action $S$ from nothing. Why??

Both terms $\partial_{r}\Gamma^{r}\,_{t t}$ and $(\Gamma^{\theta}\,_{r \theta}+\Gamma^{\phi}\,_{r \phi})\Gamma^{r}\,_{t t}$ are of the same order in $c$ but the second one is neglected. Why??

 

[Mentz114]

But $c$ is only a constant, you can take $c^{2}G_{rr}$ and then this factor and $G_{tt}$ are of the same order in $c$, why do we neglect the Einstein equation $c^2G_{rr}=c^2kT_{rr}$? In fact, in Planck units we can take $c=1$ and $O(v^2) \rightarrow 0$ so there wouldn't be any difference between the above Einstein components.

And what about this??:
Both terms $\partial_{r}\Gamma^{r}\,_{t t}$ and $(\Gamma^{\theta}\,_{r \theta}+\Gamma^{\phi}\,_{r \phi})\Gamma^{r}\,_{t t}$ are of the same order in $c$ but the second one is neglected. Why??

Nothing as far as I can see.

Where did you get this derivation ? Does it not explain why the assumptions are not justified.

I know that $c$ is a constant. If you are making an approximation you cannot set $c=1$. In the metric for example $g_{tt}$ influences matter movement more than the spatial terms by a factor of $\approx 3 \times 10^8$.

Have a look here this explains it simply
http://www.mth.uct.ac.za/omei/gr/chap7/node3.html
http://www.mth.uct.ac.za/omei/gr/chap7/node3.html

 

[pseudoriemann]

Nothing as far as I can see.

Where did you get this derivation ? Does it not explain why the assumptions are not justified.

I know that $c$ is a constant. If you are making an approximation you cannot set $c=1$

The above derivation is mine and I think that the weak field limit is equivalent to consider low velocities and curvature, setting $c = constant$, in fact the value of $c = 3 \times 10^8 m s^{-1}$ doesn't appear in GR, we know this value because of experiments with light (whose mass is zero) so there is nothing in GR that tells you setting this value instead of $c=1ms^{-1}$, the only thing you need to know is that $v \leq c$ and, taking into account the Newtonian limit, $c^2-v^2 \approx c^2$.

Why can't I set $c=1$ in Planck units when I'm making the approximation? The approximation is only $O(h^2), O (v^{2}) \rightarrow 0$, what's wrong with this?

In the metric for example $g_{tt}$ influences matter movement more than the spatial terms by a factor of $\approx 3 \times 10^8$

I think that's the same as saying $c^2-v^2 \approx c^2$ and I've always agreed with this and you can set $c=constant$ but always taking into account this approximation.

 

[pseudoriemann]

Have a look here this explains it simply
http://www.mth.uct.ac.za/omei/gr/chap7/node3.html

Thank you for the link. I think the expression (23) is wrong, because for a spherically symmetric spacetime with the typical coordinates {t, r, theta, phi} we have $\partial_{\lambda}\eta_{\mu \nu} (x) \neq 0$ and the link says that the line element of such a limit is (33) but for the above case it's $ds^2 = g_{t t} (r)c^2dt^2-\frac{dr^2}{g_{tt}(r)}-r^2d\Omega^{2}$ so I think that's wrong.

 

[pseudoriemann]

This weekend I can't post anything so I will see you on Monday, thank you so much !

 

[PAllen]

Thank you for the link. I think the expression (23) is wrong, because for a spherically symmetric spacetime with the typical coordinates {t, r, theta, phi} we have $\partial_{\lambda}\eta_{\mu \nu} (x) \neq 0$ and the link says that the line element of such a limit is (33) but for the above case it's $ds^2 = g_{t t} (r)c^2dt^2-\frac{dr^2}{g_{tt}(r)}-r^2d\Omega^{2}$ so I think that's wrong.

${\eta_{\mu \nu} }$ is by definition, the Minkowski metric not the Schwarzschild metric, so you are doing this wrong. It's partials are all zero.

 

[PeterDonis]

if we considered a static spherically symmetric spacetime in a coordinate system where $ds^2 = g_{t t} (r)c^2dt^2-\frac{dr^2}{g_{tt}(r)}-r^2d\Omega^{2}$

The general metric of a static, spherically symmetric spacetime does not have this form; $g_{rr}$ is not the reciprocal of $g_{tt}$, it's a separate function of $r$. Only in the vacuum case, i.e., Schwarzschild spacetime, does the metric take the form you give.

If you want to consider Schwarzschild spacetime in the Newtonian approximation, it's actually easier to use isotropic coordinates, in which the metric looks like this (note that I'm using units in which $c = 1$, to make it clear that the numerical value of $c$ doesn't affect the reasoning we're about to do):

$$
ds^2 = \frac{\left( 1 - \frac{M}{2r} \right)^2}{\left( 1 + \frac{M}{2r} \right)^2} dt^2 - \left( 1 + \frac{M}{2r} \right)^4 \left( dr^2 + r^2 d \Omega^2 \right)
$$

The Newtonian approximation then implies $M << r$; if we expand the above order by order in $M / r$ and neglect terms of second order or higher, we get

$$
ds^2 = \left( 1 - \frac{2M}{r} \right) dt^2 - \left( 1 + \frac{2M}{r} \right) \left( dr^2 + r^2 d \Omega^2 \right)
$$

From the above we can read off $h_{tt} = - 2M / r$ and $h_{rr} = h_{\theta \theta} = h_{\phi \phi} = 2M / r$. So just from what we've done so far, we don't have any reason to neglect the spatial part of $h$ compared to the time part.

However, the Newtonian approximation also includes the assumption that $v << 1$ (in units where $c = 1$). This means that we are only considering line elements where $dr$, $r d\theta$, and $r \sin \theta d \phi$ are much smaller than $dt$. So the terms which have $1$ multiplying the spatial coordinate differentials is already much smaller than the term which has $1$ multiplying $dt$; and that means that, in this approximation, the terms which have $1$ multiplying the spatial coordinate differentials are of the same order as the term which has $h_{tt}$ multiplying $dt$. The terms which have $h_{rr}$, etc. multiplying the spatial coordinate differentials are much smaller still, which is why they can be neglected.

To see this another way, note that, if we are considering a weak field but we drop the assumption that $v << 1$, we can no longer neglect the spatial terms in $h$. For example, when computing the bending of light by the sun, the spatial terms in $h$ are necessary to get the right answer; if we drop them, we get an answer that's only half as large. But the bending of light is not a "Newtonian" problem, precisely because we do not have $v << 1$.

 

[PeterDonis]

in this limit Tr r is neglected but what I'm saying is that according to my opinión Gr r is not neglected and the same for Gθ θ and Gφ φ, so the Einstein equations are not totally valid in this limit.

The fact that you find yourself thinking "the Einstein equations are not totally valid in this limit" should be a big red flag to you that you are doing something wrong.

Consider a more general static, spherically symmetric spacetime, for which the line element reads, in isotropic coordinates and making a similar approximation to what I did in my previous post:

$$
ds^2 = \left( 1 - 2 \Phi \right) dt^2 - \left( 1 + 2 \Psi \right) \left( dr^2 + r^2 d\Omega^2 \right)
$$

where I have relabeled the time component of $h$ as $\Phi$ and the spatial components (they must all be the same because of spherical symmetry) as $\Psi$. (The vacuum case I analyzed in my last post can be recovered by setting $\Phi = \Psi = M / r$.) Then we can simply apply the same kind of reasoning as I applied in my last post for the vacuum case; even though $\Phi$ and $\Psi$ are not equal, they are both much smaller than $1$, so when we add the Newtonian requirement of $v << 1$, we still get that the spatial terms in $\Psi$ are much smaller than the time term in $\Phi$, because the spatial coordinate differentials are much smaller than $dt$.

 

[haushofer]

As a bit of an offtopic question: does anybody have a reference where the transformation of the newt.potential is given under the residual coordinate transfo's after the Newtonian limit is made? I know how the divergence of it is determined, but somehow I can't seem to get the answer from the transformstion of the metric itself straightaway. Probably overlooking something silly, but no text seems to discuss this.

 

[pseudoriemann]

Thank you for the posts, I'll try to answer quickly because I don't really have too much time before Monday to post, but I'm going to try...

 

[pseudoriemann]

The general metric of a static, spherically symmetric spacetime does not have this form; $g_{rr}$ is not the reciprocal of $g_{tt}$, it's a separate function of $r$. Only in the vacuum case, i.e., Schwarzschild spacetime, does the metric take the form you give.

I know, I'm just considering the form I gave because in the general case if we only have a spacetime with a spherical mass then Birkhoff's theorem determines this form and that's what I'm considering.

If you want to consider Schwarzschild spacetime in the Newtonian approximation, it's actually easier to use isotropic coordinates, in which the metric looks like this (note that I'm using units in which $c = 1$, to make it clear that the numerical value of $c$ doesn't affect the reasoning we're about to do):

$$

ds^2 = \frac{\left( 1 - \frac{M}{2r} \right)^2}{\left( 1 + \frac{M}{2r} \right)^2} dt^2 - \left( 1 + \frac{M}{2r} \right)^4 \left( dr^2 + r^2 d \Omega^2 \right)

$$

The Newtonian approximation then implies $M << r$; if we expand the above order by order in $M / r$ and neglect terms of second order or higher, we get

$$

ds^2 = \left( 1 - \frac{2M}{r} \right) dt^2 - \left( 1 + \frac{2M}{r} \right) \left( dr^2 + r^2 d \Omega^2 \right)

$$

From the above we can read off $h_{tt} = - 2M / r$ and $h_{rr} = h_{\theta \theta} = h_{\phi \phi} = 2M / r$. So just from what we've done so far, we don't have any reason to neglect the spatial part of $h$ compared to the time part.

However, the Newtonian approximation also includes the assumption that $v << 1$ (in units where $c = 1$). This means that we are only considering line elements where $dr$, $r d\theta$, and $r \sin \theta d \phi$ are much smaller than $dt$. So the terms which have $1$ multiplying the spatial coordinate differentials is already much smaller than the term which has $1$ multiplying $dt$; and that means that, in this approximation, the terms which have $1$ multiplying the spatial coordinate differentials are of the same order as the term which has $h_{tt}$ multiplying $dt$. The terms which have $h_{rr}$, etc. multiplying the spatial coordinate differentials are much smaller still, which is why they can be neglected.

I don't know what you mean. If we neglect the terms which have $h_{rr}$ multiplying the spatial coordinate differentials then the metric takes the following form:

$ds^{2} \approx \left( 1 - \frac{2M}{r} \right) dt^2 - \left( dr^2 + r^2 d \Omega^2 \right)$

But you can't reproduce the weak field from this approximation. For example, in this approximation, $G_{tt} \approx 0 \, \forall \, M$ and the rest of diagonal components are not zero; moreover, $R_{tt} \approx \frac{c^2}{2}(\frac{d^{2}h_{tt}}{dr^{2}}+\frac{2}{r}\frac{dh_{tt}}{dr})$so you can't even obtain a Poisson equation doing this way.

To see this another way, note that, if we are considering a weak field but we drop the assumption that $v << 1$, we can no longer neglect the spatial terms in $h$. For example, when computing the bending of light by the sun, the spatial terms in $h$ are necessary to get the right answer;if we drop them, we get an answer that's only half as large. But the bending of light is not a "Newtonian" problem, precisely because we do not have $v << 1$.

I think that I don't have any problem with that, what I'm questioning here is the way to obtain a Poisson equation through Einstein equations in the Newtonian limit.

 

[pseudoriemann]

${\eta_{\mu \nu} }$ is by definition, the Minkowski metric not the Schwarzschild metric, so you are doing this wrong. It's partials are all zero.

I think $\eta$ is a backgroun metric, i. e. a Minkowski spacetime metric or de Sitter (Anti de Sitter) metric and, for example, if we consider a Minkowski spacetime background, then $\partial\eta \neq 0$ in spherical coordinates (you can check it, for example, in the following book: T. Ortín - Gravity and Strings (Section 3.4.1, which it's an introduction to Newtonian limit in GR and $\partial\eta \neq 0$ in this limit)).

 

[pseudoriemann]

Finally, I would like to say that I'm working in an alternative theory which, among other things, it maybe improve this limit (in general, I'm a supporter of GR but I don't agree in 100% with this particular point of the theory and I think this limit could be improved by a better theory than GR).

Thank you so much for everything and please forgive my english once again :).

 

[PeterDonis]

pseudoriemann, please note that I did surgery on your post #17 using magic moderator powers. Otherwise it would have been impossible to reply to.

I'm just considering the form I gave because in the general case if we only have a spacetime with a spherical mass then Birkhoff's theorem determines this form

I'm not sure I understand. Are you only considering vacuum solutions? If so, the Einstein tensor vanishes identically, so trying to extract information from it is not going to work very well.

OTOH, if you are considering the metric within a spherically symmetric gravitating body, i.e., not vacuum, then Birkhoff's theorem does not apply and you need to use the more general form of the metric that I gave.

what I'm questioning here is the way to obtain a Poisson equation through Einstein equations in the Newtonian limit.

See chapter 4 of Carroll's lecture notes on GR:

http://arxiv.org/pdf/gr-qc/9712019v1.pdf

Particularly the section starting with equation (4.35).

 

[PeterDonis]

I would like to say that I'm working in an alternative theory

PF is not for discussion of personal alternative theories; it is for discussion of mainstream physics. The proper venue for getting an alternative theory considered is to get it written up in a scientific paper and have it reviewed by experts in the field through the normal publication process. Please limit discussion here to standard GR.

 

[pseudoriemann]

pseudoriemann, please note that I did surgery on your post #17 using magic moderator powers. Otherwise it would have been impossible to reply to.

Yes, thank you Peter Donis, I'll try to post in this way if you prefer it.

I'm not sure I understand. Are you only considering vacuum solutions? If so, the Einstein tensor vanishes identically, so trying to extract information from it is not going to work very well

I'm considering only vacuum solutions, yes. Vacuum solutions only mean that the stress-energy tensor is zero but there may be a source of gravity in such a spacetime (for example, a M mass). According to my oppinion, the theory should lead from this case to Newton's Gravity in this limit for the same case, but I think it doesn't happen so.

For example, the Boulware-Deser solution leads naturally to GR case from its general expression and you only have to apply the corresponding limit, but this other limit that we are discussing now is, in my oppinion, a different case when we try to do this from the general case with a M mass and we try to obtain the Poisson equation...

I think that the theory should guarantee these steps: 1) let's start with the general solution of a static spherically symmetric spacetime with a mass M (in the typical coordinates: Schwarzschild metric) 2) let's apply the limit and let's obtain Poisson equation.

But, according to my oppinion, this doesn't happen and that's why I'm questioning this particular point of GR.

 

[pseudoriemann]

PF is not for discussion of personal alternative theories; it is for discussion of mainstream physics. The proper venue for getting an alternative theory considered is to get it written up in a scientific paper and have it reviewed by experts in the field through the normal publication process. Please limit discussion here to standard GR.

Thank you for the information. I didn't say 'personal alternative theory', but 'alternative theory'. For example I think that GR and Brans-Dicke theory are different theories of gravitation and there are lots of discussions in PF about Brans-Dicke theory and so on.

Anyway, I didn't want to discuss about this theory and I only wanted to say one of the reasons why I don't totally agree with GR, so according to your post I will limit my discussion here to standard GR and I apologize if someone could understand another different thing.

Good night :)

 

[PeterDonis]

Vacuum solutions only mean that the stress-energy tensor is zero but there may be a source of gravity in such a spacetime (for example, a M mass).

Yes, but if you're looking at the Einstein Field Equation, it's local; it expresses a relationship between curvature (the Einstein tensor) and matter-energy (the stress-energy tensor) at a particular event in spacetime. There is nothing in there about sources of gravity somewhere else; to look at those with the EFE, you would look at the EFE at the event where the source is, i.e., where the stress-energy tensor is nonzero.

If you already have a global solution to the EFE, i.e., an expression for the metric that's valid everywhere in some region of spacetime, then you can look at how sources in one place produce fields in another place. But you can't do that just by looking at the EFE.

the Boulware-Deser solution

Reference, please?

this other limit that we are discussing now is, in my oppinion, a different case when we try to do this from the general case with a M mass and we try to obtain the Poisson equation...

Is there a reference that discusses how this is done?

I think that the theory should guarantee these steps: 1) let's start with the general solution of a static spherically symmetric spacetime with a mass M (in the typical coordinates: Schwarzschild metric) 2) let's apply the limit and let's obtain Poisson equation.

But, according to my oppinion, this doesn't happen and that's why I'm questioning this particular point of GR.

Did you look at the Carroll reference I gave? It does exactly what you're saying, using standard GR.

I didn't say 'personal alternative theory', but 'alternative theory'.

If it's one that's recognized, like Brans-Dicke, you should say which one (and give references, as I asked above). Discussion of those here is fine. You should be aware, though, that all such alternative theories either (a) are experimentally indistinguishable from GR at our current level of experimental technology, or (b) have already been falsified by experiment. (Brans-Dicke, to the best of my knowledge, is still technically in the former category, but only if you adopt a value for its adjustable parameter that makes it basically indistinguishable from GR anyway.)

If it isn't an alternative theory that's recognized, it's a "personal theory" as far as PF posting rules are concerned.

 

[Markus Hanke]

But, according to my oppinion, this doesn't happen and that's why I'm questioning this particular point of GR.

This doesn't seem to make any sense to me - in order to obtain the Schwarzschild solution in the first place, you need to impose the boundary condition that the solution reduces to Newtonian gravity at infinity. After all, that is where the "M" term comes from. Hence, questioning whether Schwarzschild reduces back to Newtonian in the appropriate limit appears somewhat bizarre, for lack of a better term.

 

[pseudoriemann]

Yes, but if you're looking at the Einstein Field Equation, it's local; it expresses a relationship between curvature (the Einstein tensor) and matter-energy (the stress-energy tensor) at a particular event in spacetime. There is nothing in there about sources of gravity somewhere else; to look at those with the EFE, you would look at the EFE at the event where the source is, i.e., where the stress-energy tensor is nonzero.

But if we consider the Schwarzschild solution, we just set the M parameter looking at $r \rightarrow \infty$, using the weak field limit (so the EFE in such a limit implies the Poisson equation, thus we really have a source of gravity there); however, in the Schwarzschild solution, we really have an action $S$ with $T_{\mu \nu}=0$, so these kind of things are what I'm questioning here.

Reference, please?

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.55.2656

Is there a reference that discusses how this is done?

That was "in my opinion" and I don't know if there are references about this point.

Did you look at the Carroll reference I gave? It does exactly what you're saying, using standard GR

Yes, thank you for the link. I knew this standard analysis and I said that I didn't agree totally with it, for all the things I posted in this thread (for example, among others, what I said about the approximation which is below of (4.49) expression).

If it's one that's recognized, like Brans-Dicke, you should say which one (and give references, as I asked above). Discussion of those here is fine. You should be aware, though, that all such alternative theories either (a) are experimentally indistinguishable from GR at our current level of experimental technology, or (b) have already been falsified by experiment. (Brans-Dicke, to the best of my knowledge, is still technically in the former category, but only if you adopt a value for its adjustable parameter that makes it basically indistinguishable from GR anyway.)

If it isn't an alternative theory that's recognized, it's a "personal theory" as far as PF posting rules are concerned

Thank you for the information. I only wanted to note one of the reasons why I don't agree with this particular point of GR but I didn't want to discuss about such a theory neither explain its postulates, even being a theory with lots of scientist references (at first I personally got this theory and then I found it published by other guys on scientist journals), so it was only a simple note.

 

[pseudoriemann]

This doesn't seem to make any sense to me - in order to obtain the Schwarzschild solution in the first place, you need to impose the boundary condition that the solution reduces to Newtonian gravity at infinity. After all, that is where the "M" term comes from. Hence, questioning whether Schwarzschild reduces back to Newtonian in the appropriate limit appears somewhat bizarre, for lack of a better term.

If you impose this limit through the geodesic equation way it's ok, but what I'm questioning here is the Poisson equation in such a limit and the information neglected of the rest of EFE.

 

[PeterDonis]

I knew this standard analysis and I said that I didn't agree totally with it

So what's wrong with it? You have said you don't agree with "the standard analysis", but you haven't actually said anything specific about where you think it goes wrong. Instead, you've tried to do the analysis in your own way and say you can't get an answer. The obvious response to that is "why don't you try doing it the standard way?"

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.55.2656

This paper is behind a paywall so I can't read it, but from the abstract I don't see how it in any way invalidates the standard method of deriving a Poisson equation from the Schwarzschild solution in GR. Saying that GR as it stands might be incomplete and need extension is not at all the same as saying that GR as it stands can't be used to derive a Poisson equation in the static, spherically symmetric case.

if we consider the Schwarzschild solution, we just set the M parameter looking at $\rightarrow \infty$, using the weak field limit

I'm not sure what you mean here, but I assume you are talking about something like deriving the $M$ parameter by observing Keplerian orbits of test objects at large values of $r$ and measuring their orbital parameters?

(so the EFE in such a limit implies the Poisson equation, thus we really have a source of gravity there)

Showing that there is a Poisson equation based on observations at large $r$ is not the same as showing that "we really have a source of gravity there" at small $r$. The only way to know what exactly is at small $r$ is to make observations at small $r$. You might find that there's a black hole there, not an ordinary gravitating body like a planet or a star; a black hole of mass $M$ produces exactly the same effects at large $r$, and therefore the same Poisson equation, as a planet or star of mass $M$.

 

[PeterDonis]

what I'm questioning here is the Poisson equation in such a limit and the information neglected of the rest of EFE.

If we already know that the spacetime is static and spherically symmetric, then the only information in the EFE is two differential equations for the two unknown functions in the metric. If in addition we know the region we are interested in is vacuum, we're down to one unknown function and one differential equation to determine it. Expressing that differential equation as a Poisson equation in the one unknown function does not neglect any information in the EFE; that's all the information there is in the case you are talking about.

 

[pseudoriemann]

So what's wrong with it? You have said you don't agree with "the standard analysis", but you haven't actually said anything specific about where you think it goes wrong.

I've really posted in this thread about things and details which, according to my opinion, are wrong and I've really explained it.

Instead, you've tried to do the analysis in your own way and say you can't get an answer. The obvious response to that is "why don't you try doing it the standard way?"

I didn't ever say "I can't get the answer", please check this thread if you want because that's wrong.

This paper is behind a paywall so I can't read it. This paper is behind a paywall so I can't read it, but from the abstract I don't see how it in any way invalidates the standard method of deriving a Poisson equation from the Schwarzschild solution in GR.

Ok, this a too famous paper in physics and I recommend you to try reading it in detail by some different way if you want and then perhaps you could understand what I said about this solution by yourself.

Saying that GR as it stands might be incomplete and need extension is not at all the same as saying that GR as it stands can't be used to derive a Poisson equation in the static, spherically symmetric case.

I know it and that's the reason why I posted here "among other things", instead of "at all the same" as you are posting here. Please check this thread if you want to understand what I was saying about this here.

I'm not sure what you mean here, but I assume you are talking about something like deriving the M M parameter by observing Keplerian orbits of test objects at large values of r r and measuring their orbital parameters?

This is a too famous result of the standard analysis of GR in the weak field limit (one of the things I said I was questioning here), you can check it on simple (7.27) expression of "Carroll's lecture notes on GR":

$g_{0 0}(r \rightarrow \infty) = - (1+ \frac{\mu}{r})$

$g_{r r}(r \rightarrow \infty) = 1+ \frac{\mu}{r}$

Showing that there is a Poisson equation based on observations at large r r is not the same as showing that "we really have a source of gravity there" at small r.

This is wrong because if such a source of gravity doesn't exist there, then we would just have $R_{\mu \nu \lambda \rho}=0$ and there wouldn't be gravity in such a case. The right answer is simply that in such a spacetime we have a M source of gravity whose information is inside the metric (thus, in absence of torsion and under the condition $\nabla_{\lambda}g_{\mu \nu}=0$, this information is inside curvature tensor and affine connection too and so on) and the $T_{\mu \nu}=0$ (but this $T_{\mu \nu}$ refers to other things which are different from the M mass).

Well, I'm thinking this won't lead anywhere useful for the discussion and I think you didn't understand what I said, so I prefer to leave my contribution to this thread here.

Anyway, thank you for the replies and best regards to everyone!

 

[Dale]

Expressing that differential equation as a Poisson equation in the one unknown function does not neglect any information in the EFE; that's all the information there is in the case you are talking about.

That seems like a pretty clean approach. If the OP is not interested, I am. Do you have a reference?

I see the reason for doing it the Carroll way, since that is not limited to spherical symmetry, but pedagogically the approach you outline seems clean.

 

[PeterDonis]

If the OP is not interested, I am. Do you have a reference?

The fact that a general static spherically symmetric spacetime has only two unknown functions of $r$ in the metric is demonstrated in MTW, and also (IIRC) Wald. It's probably covered in other texts as well.

The fact that, if you add vacuum to the specifications, you're down to one function of $r$, is just Birkhoff's theorem. (The only wrinkle here is that you don't actually need the "static" assumption if you have spherical symmetry and vacuum; Birkhoff's theorem derives the "static" part--more precisely, the fact that there is a fourth KVF--for you.)

Those items, plus Carroll's general discussion of how to derive the Poisson equation, which is a differential equation for one unknown function ($h_{00}$), are sufficient to demonstrate what I said. If there is only one unknown function in the metric, and you have a differential equation that gives a solution for it, that equation must contain all the information that the EFE contains, since the metric is the solution of the EFE.

 

[PeterDonis]

This is a too famous result of the standard analysis of GR in the weak field limit

Ah, ok, this clarifies what you were referring to. It's more or less equivalent to what I was thinking (measuring orbital parameters of Keplerian orbits at large $r$ is equivalent to measuring the quantity $\mu$ in the metric coefficients you give).

This is wrong because if such a source of gravity doesn't exist there, then we would just have $R_{\mu \nu \lambda \rho}=0$ and there wouldn't be gravity in such a case.

You are incorrect. If you were correct, the Schwarzschild solution, and indeed any vacuum solution that is not Minkowski spacetime, would not exist. Since they do exist, your statement is disproved.

The right answer is simply that in such a spacetime we have a M source of gravity whose information is inside the metric

Now you're moving the goalposts; you're defining "a source of gravity" to mean that the metric is not Minkowski, i.e., that there is curvature present. But, as I showed above, the metric can be curved in a vacuum solution, and "vacuum" means $T_{\mu \nu} = 0$ everywhere, i.e., no "source of gravity" present.

I think you didn't understand what I said

I understand enough of what you are saying to see that you have made a number of incorrect statements about standard GR. I would advise taking some time to learn why those statements are incorrect, so that you have a proper understanding of standard GR, before trying to find some alternative theory that you think "works better".

 

[Markus Hanke]

I think it is easier to think about the "M" as a parameter for a family of metrics. Physically speaking, M should be a global property of the entire space-time, not "just" an isolated object in it. That wouldn't even make sense, since there are plenty of vacuum solutions which are everywhere source-free. I think the OP was questioning how a space-time that is globally source-free can still be non-trivial; my amateur answer to that would be that the energy-momentum tensor via the Einstein equations does not in fact determine all aspects of local geometry, but only a certain combination of components of the Riemann tensor, being the Einstein tensor ( which, I believe, is the contracted double dual of Riemann ). All other information needed to uniquely fix the local geometry comes from boundary conditions imposed when solving the system of differential equations - and that's where the M comes from, not the ( vanishing ) energy-momentum tensor.

 

[PeterDonis]

I think it is easier to think about the "M" as a parameter for a family of metrics. Physically speaking, M should be a global property of the entire space-time, not "just" an isolated object in it.

This is correct; $M$ appears in the metric everywhere in the spacetime, so it is a global property, not a local one; and there is an infinite family of metrics parameterized by their value of $M$ (flat Minkowski spacetime is the limiting case of $M = 0$).

I think the OP was questioning how a space-time that is globally source-free can still be non-trivial

This can happen because the Einstein Field equation is nonlinear.

the energy-momentum tensor via the Einstein equations does not in fact determine all aspects of local geometry, but only a certain combination of components of the Riemann tensor, being the Einstein tensor ( which, I believe, is the contracted double dual of Riemann ).

This is correct if we put in the word "locally"; the EFE equates the Einstein tensor and the stress-energy tensor at the same event. The rest of the Riemann tensor (the Weyl tensor) is determined by how spacetime curvature "propagates" (this isn't really the right word, but it's the best I can do) from one event to another.

All other information needed to uniquely fix the local geometry comes from boundary conditions imposed when solving the system of differential equations - and that's where the M comes from

The problem with looking at it this way, if we are looking at the idealized case where the spacetime is vacuum everywhere, is: at which boundary do we impose the condition that gives the value of $M$? As I noted above, since flat Minkowski spacetime is a member of this family of spacetimes (the one with $M = 0$), whatever boundary condition you are referring to can't distinguish between flat Minkowski spacetime and curved Schwarzschild spacetime.

There is a way of defining the "mass" of an asymptotically flat spacetime by looking at the limiting behavior of the metric coefficients as $r \rightarrow \infty$, called the ADM mass. (Actually, there are two ways of doing this, depending on whether you choose spacelike infinity, as the ADM mass does, or future null infinity, as the Bondi mass does.) Wikipedia has a very brief discussion:

https://en.wikipedia.org/wiki/Mass_...ndi_masses_in_asymptotically_flat_space-times

But this isn't really a "boundary condition".