Newtonian limit of GR?

1. Aug 14, 2015

pseudoriemann

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

In this limit ga b (x) = ηa b (x) + ha b (x) where O(h2→0). Then, according to GR, it's straightforward to demonstrate that the Einstein equation Gt t = k*Tt 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 Gr r ≠ k*Tr r.

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

2. Aug 14, 2015

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.

3. Aug 14, 2015

pseudoriemann

Thank you Markus Hanke,

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. You may write the expressions of these components of the Einstein tensor in terms of Christoffel symbols and check that Gr r Gθ θ, Gφ φ are not neglected in this limit and neither null.

4. Aug 14, 2015

Mentz114

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.

Last edited: Aug 14, 2015
5. Aug 14, 2015

pseudoriemann

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??

6. Aug 14, 2015

Mentz114

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

7. Aug 14, 2015

pseudoriemann

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.

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??

Last edited: Aug 14, 2015
8. Aug 14, 2015

Mentz114

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

Last edited: Aug 14, 2015
9. Aug 14, 2015

pseudoriemann

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?

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.

Last edited: Aug 14, 2015
10. Aug 14, 2015

pseudoriemann

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.

Last edited by a moderator: May 7, 2017
11. Aug 14, 2015

pseudoriemann

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

12. Aug 14, 2015

PAllen

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

13. Aug 14, 2015

Staff: Mentor

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 $v << 1$.

Last edited: Aug 14, 2015
14. Aug 14, 2015

Staff: Mentor

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 $dt$.

15. Aug 15, 2015

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.

16. Aug 15, 2015

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...

17. Aug 15, 2015

pseudoriemann

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.

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.

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.

18. Aug 15, 2015

pseudoriemann

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)).

19. Aug 15, 2015

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 :).

20. Aug 15, 2015

Staff: Mentor

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 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.

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).

21. Aug 15, 2015

Staff: Mentor

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.

22. Aug 15, 2015

pseudoriemann

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

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.

23. Aug 15, 2015

pseudoriemann

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 :)

24. Aug 15, 2015

Staff: Mentor

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.

Is there a reference that discusses how this is done?

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

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.

25. Aug 16, 2015

Markus Hanke

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.