# 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).