Newton-Cartan theory

1. Apr 9, 2010

JustinLevy

Newton-Cartan theory has been brought up many times on this forum as an example of any theory being able to be written in "general covariance". I've even used the example myself after hearing about it here. However, when I actually went to look up the definition:
http://en.wikipedia.org/wiki/Newton–Cartan_theory
it was not what I expected. I was expecting literally Newtonian gravity, just written in tensor formalism somehow. But instead it is a metric theory ... ie. spacetime is curved.

So either I've misused the example, or I am misunderstanding the formulation. Either way I am wrong. I may have just misunderstood what people were casually referring to when using this as an example and took it too literally.

Although I've also seem people here make the argument that operationally defined coordinate systems are geometric objects and thus even all the Newtonian physics is technically, albeit implicitly, defined in geometrically invariant terms.

So what is everyone's take on this?

----------------
Now, that aside, I'd like to actually play with this theory a bit.
In Newtonian gravity, I've seen the escape velocity argument to discuss newtonian black holes. So I'm curious if that remains in Newton-Cartan.

Using the somewhat bizarre (to me at least) notation in wiki
$$R_{\mu \nu} = 4 \pi G \rho {\delta^0}_\mu {\delta^0}_\nu$$
Which I assume means this?
$$R_{\mu \nu} = 4 \pi G \rho \left ( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right)$$
But it seems to be saying this is only correct in inertial coordinate systems. Which due to the curvature can't be global anymore, so I'm not understanding now to actually interpret this as an equation for the metric to solve.

Any help? Are there some papers that work out some simple cases in Newton-Cartan (the equivalent of Schwartzschild and Kerr maybe?)

EDIT:
Still looking, but this is interesting: http://arxiv.org/abs/gr-qc/9604054 Newton-Cartan Cosmology

2. Apr 9, 2010

bcrowell

Staff Emeritus
Re: Newton-Cartan

Newtonian physics can be formulated with or without general covariance, with or without spacetime curvature. GR can be formulated with or without general covariance, with or without spacetime curvature.

The Newton-Cartan theory has spacetime curvature, but it's nonrelativistic in the sense that there is no maximum velocity, simultaneity is absolute, horizons don't exist, etc.

3. Apr 9, 2010

JustinLevy

Re: Newton-Cartan

Using this presentation here
http://williewong.wordpress.com/2009/12/05/newton-cartan-gravity/
Newton-Cartan simplifies it to:
$$R_{00} = 4\pi G \rho$$
and notes
$${\Gamma^0}_{\mu\nu} = {\Gamma^\mu}_{\nu j} = 0$$
where the j index is only over the spatial coordinates in that last equation.

So using the definitions:
$$\Gamma^m_{ij}=\frac12 g^{km} \left( \frac{\partial}{\partial x^i} g_{kj} +\frac{\partial}{\partial x^j} g_{ik} -\frac{\partial}{\partial x^k} g_{ij} \right)$$
$$R_{ij}={R^\ell}_{i\ell j}=g^{\ell m}R_{i\ell jm}=g^{\ell m}R_{\ell imj} =\frac{\partial\Gamma^\ell_{ij}}{\partial x^\ell} - \frac{\partial\Gamma^\ell_{i\ell}}{\partial x^j} + \Gamma^\ell_{ij} \Gamma^m_{\ell m} - \Gamma^m_{i\ell}\Gamma^\ell_{jm}$$
gives:
$$R_{00}=\frac{\partial\Gamma^\ell_{00}}{\partial x^\ell} - \frac{\partial\Gamma^\ell_{0\ell}}{\partial x^0} + \Gamma^\ell_{00} \Gamma^m_{\ell m} - \Gamma^m_{0\ell}\Gamma^\ell_{0m}$$
$$=\frac{\partial\Gamma^\ell_{00}}{\partial x^\ell} - \frac{\partial\Gamma^0_{00}}{\partial x^0} + \Gamma^0_{00} \Gamma^0_{00} - \Gamma^0_{00}\Gamma^0_{00}$$
$$=\frac{\partial}{\partial x^1} \Gamma^1_{00} + \frac{\partial}{\partial x^2} \Gamma^2_{00} + \frac{\partial}{\partial x^3} \Gamma^3_{00}$$

Noting:
$$\Gamma^\ell_{00}= \frac12 g^{k\ell} \left( \frac{\partial}{\partial x^0} g_{k0} +\frac{\partial}{\partial x^0} g_{0k} -\frac{\partial}{\partial x^0} g_{00} \right) = g^{k\ell} \frac{\partial}{\partial x^0} \left( g_{k0} -\frac12 g_{00} \right)$$
It is clear there is quite a bit of freedom to set the spatial-spatial terms of the metric. So restricting to coordinate systems with orthogonal spatial coordinates and constant spatial-spatial terms in the metric, yields:
$$\frac{\partial}{\partial x^1} \Gamma^1_{00}= g^{11} \frac{\partial}{\partial x^1} \frac{\partial}{\partial x^0} \left( g_{10} -\frac12 g_{00} \right) + \frac{\partial}{\partial x^1} \frac{\partial}{\partial x^0} g^{01} \left( g_{00} -\frac12 g_{00} \right) = \frac{\partial}{\partial x^1} \frac{\partial}{\partial x^0} \left[ g^{11} (g_{10} -\frac12 g_{00}) + \frac12 g^{01} g_{00} \right]$$

Hmm... maybe I did something wrong. I was expecting something that looked more like Poisson's equation.

Last edited: Apr 9, 2010
4. Apr 9, 2010

Fredrik

Staff Emeritus
Re: Newton-Cartan

Both $\mu$ and $\nu$ need to be =0 for $\delta^0_\mu\delta^0_\nu$ be ≠0.

5. Apr 9, 2010

JustinLevy

Re: Newton-Cartan

Doh! Yes. Obviously.
Thanks.

6. Apr 11, 2010

JustinLevy

Re: Newton-Cartan

Well, I found another error, which helps me get at least a little further now.

I think up to here was alright:
$$4\pi G \rho = R_{00}=\frac{\partial}{\partial x^1} \Gamma^1_{00} + \frac{\partial}{\partial x^2} \Gamma^2_{00} + \frac{\partial}{\partial x^3} \Gamma^3_{00}$$

But I made a typo when writing the Christoffel symbols, where I should have written:
$$\Gamma^\ell_{00}= \frac12 g^{k\ell} \left( \frac{\partial}{\partial x^0} g_{k0} +\frac{\partial}{\partial x^0} g_{0k} -\frac{\partial}{\partial x^k} g_{00} \right)$$

Looking for a Schwarzschild type solution we have:
$$ds^2=-A(r)\,dt^2 + B(r)\,dr^2+r^2\,d \theta^2+r^2 \sin^2 \theta \,d \phi^2$$

So simplifying further, for [itex]\ell \neq 0[/tex]:
$$\Gamma^\ell_{00}= -\frac12 g^{\ell\ell} \frac{\partial}{\partial x^\ell} g_{00}$$
$$4\pi G \rho = R_{00} = -\frac12 \frac{\partial}{\partial x^1} g^{11} \frac{\partial}{\partial x^1} g_{00} -\frac12 \frac{\partial}{\partial x^2} g^{22} \frac{\partial}{\partial x^2} g_{00} -\frac12 \frac{\partial}{\partial x^3} g^{33} \frac{\partial}{\partial x^3} g_{00} = \frac12 \frac{\partial}{\partial r}B(r) \frac{\partial}{\partial r} A(r)$$

That is:
$$2\pi G \rho = \frac{\partial}{\partial r} B(r) \frac{\partial}{\partial r} A(r)$$

And outside of the mass (ie. the vacuum solution),
$$0 = \frac{\partial}{\partial r} B(r) \frac{\partial}{\partial r} A(r)$$
$$B(r) \frac{\partial}{\partial r} A(r) = K$$
where K is a constant.

Something appears wrong, since A(r)=1 and B(r)=1 is a valid solution ... meaning a spherical mass wouldn't curve spacetime.

EDIT:
I think I see, there are other restrictions, coming from many of the Christoffel symbols being zero.

$$\Gamma^0_{00}=\frac12 g^{00} \left( \frac{\partial}{\partial x^0} g_{00} +\frac{\partial}{\partial x^0} g_{00} -\frac{\partial}{\partial x^0} g_{00} \right)= \frac12 A(r) \frac{\partial}{\partial t} A(r) = 0$$
which unfortunately doesn't tell us anything.

$$\Gamma^0_{10}=\frac12 g^{00} \left( \frac{\partial}{\partial x^1} g_{00} +\frac{\partial}{\partial x^0} g_{10} -\frac{\partial}{\partial x^0} g_{10} \right)= \frac12 A(r) \frac{\partial}{\partial r} A(r) = 0$$

$$\Gamma^0_{11}=\frac12 g^{00} \left( \frac{\partial}{\partial x^1} g_{01} +\frac{\partial}{\partial x^1} g_{10} -\frac{\partial}{\partial x^0} g_{11} \right) = \frac12 A(r) \frac{\partial}{\partial r} B(r) = 0$$

$$\Gamma^1_{10}=\frac12 g^{11} \left( \frac{\partial}{\partial x^1} g_{10} +\frac{\partial}{\partial x^0} g_{11} -\frac{\partial}{\partial x^1} g_{10} \right) = \frac12 B(r) \frac{\partial}{\partial t} B(r) = 0$$
which doesn't tell us anything unfortunately.

$$\Gamma^1_{11}=\frac12 g^{11} \left( \frac{\partial}{\partial x^1} g_{11} +\frac{\partial}{\partial x^1} g_{11} -\frac{\partial}{\partial x^1} g_{11} \right) = \frac12 B(r) \frac{\partial}{\partial r} B(r) = 0$$

So all together we have:
$$B(r) \frac{\partial}{\partial r} A(r) = K$$
$$A(r) \frac{\partial}{\partial r} A(r) = 0$$
$$A(r) \frac{\partial}{\partial r} B(r) = 0$$
$$B(r) \frac{\partial}{\partial r} B(r) = 0$$

These seem to require:
A(r) = Constant
and
B(r) = Constant

Which would mean no curved spacetime outside a spherically symmetric mass!?

Clearly I made some more mistakes here (or maybe those restrictions on the Christoffel symbols from http://williewong.wordpress.com/2009/12/05/newton-cartan-gravity/ aren't quite right)? I'm assuming I just made some stupid mistakes here though.

How can I see that (especially the no maximum velocity)?
As for the horizons, I think they must exist. Since Newton-Cartan theory has the same vacuum equation as Einstein's Field Equations, I would expect the same vacuum equations: minkowski spacetime with no masses, Schwartzschild with a static spherical mass, and Kerr with a rotating spherical mass.

This would also explain why the naive idea of a "black star" where the escape velocity = c on the surface in Newtonian theory, gives the same critical radius as Einstein's black holes.

Last edited: Apr 11, 2010
7. Apr 12, 2010

haushofer

Re: Newton-Cartan

The fact that there is no maximum velocity can for instance be seen from the symmetry group of Newton-Cartan: it's the Galilean group which leaves both the "timelike" and the "spacelike" metric invariant. Here the speed v is just a parameter of the group and there are no restrictions on it.

If you're looking for more literature on Newton-Cartan I could provide you with some; I think I have pretty much most of the articles on this subject written for the last 50 years here :P For instance, one paper I really liked is Dautcourt's "On the Newtonian limit of general relativity", in which the link between Newton Cartan and the Newtonian limit of GR is explained very nicely. Actually, this doesn't give exactly NC but a slightly more general type of theory.

Have you already taken a look at Misner,Thorne, Wheeler chapter 12? They provide a very pedagogical explanation of NC. They start with the geodesic equation and show from there that the only surviving connection coefficient is

$$\Gamma^{i}_{00} = \partial^{i}\phi$$

where phi is the gravitational potential appearing in the Poisson equation. This Poisson equation is then equal to the EOM you wrote down.

If you have any questions, I would be happy to try to answer them!

8. Apr 12, 2010

haushofer

Re: Newton-Cartan

The notation of Wiki is by the way not that bizar; it just says that

$$R_{00} = 4 \pi G \rho, \ \ \ \ \ \ \ \ \ \ R_{ij}=0$$

In 3+1 dimensions this says that the spatial part of your curvature is flat; in 3 dimensions a vanishing Ricci tensor implies a vanishing Riemann tensor.The only dynamics is in the 00 component of the Ricci tensor. The first equation is nothing more than the Poisson equation.

9. Apr 12, 2010

haushofer

Re: Newton-Cartan

How would you write down GR without spacetime curvature? You mean by some sort of embedding?

10. Apr 12, 2010

JustinLevy

Re: Newton-Cartan

That notation looked weird to me because it had "dangling" uncontracted tensor indices that were put in by hand:
$$R_{\mu\nu} = 4 \pi G \rho {\delta^0}_\mu {\delta^0}_\nu$$
Why is the thing on the right a valid geometric object?

When searching, I found this:
http://arxiv.org/abs/gr-qc/9506077
with the same title. Is that any good?

I assume this is the paper you are talking about:
http://th-www.if.uj.edu.pl/~acta/vol21/pdf/v21p0755.pdf [Broken]
I'll ty to look over that later. I borrowed MTW to check out chap 12. I think I'm still missing something, because I still don't see what I'm doing wrong in my calculation above. I'm probably missing something quite basic.

Also, in MTW, based on their requirement for x^0 to be the global time label, while it is conceptually obvious from actual Newtonian mechanics why t' = t + constant is allowed, it is unclear to me why it is allowed in this geometric picture. Can we actually answer: what operations can we do to a global time label and yield another "valid" global time label? I have no clue how to specify what a "valid" global time label is.

I need to play with stuff for quite awhile before I feel comfortable with it. So if you could help me understand how to play with these equations, that would be wonderful. Then I can hopefully learn enough to fully understand the articles in much more depth.

What I'm conceptually running into right now, is that both these theories have the same vacuum equation:
$$R_{\mu\nu} = 0$$
And, as far as I understand, the constants in Schwartzschild and Kerr GR solutions were fixed by matching assymptotics with newtonian theory. So ... why doesn't any theory with the same Vacuum equations for spacetime yield the same dynamics outside a static spherical mass, or rotating spherical mass, etc.

Since I failed miserably above to derive the equations outside a spherical mass (I just got that the metric was a constant), I'm having trouble "brute-forcing" my way through these questions either.

EDIT:
Just to be sure, when they speak of rho, they mean
$$\rho = T^{00}$$
up to factors of c, right?

Last edited by a moderator: May 4, 2017
11. Apr 12, 2010

haushofer

Re: Newton-Cartan

Because it transforms as a second rank tensor, you can check that for yourself. I'll come back to the rest later, sleeping time for now :)

12. Apr 13, 2010

haushofer

Re: Newton-Cartan

For a first exposure to NC I wouldn't say that's helpful. I can't remember I got anything useful from that article in understanding NC, to be honest.

That's the article indeed. I think if you've read MTW a lot of things will become clear :)

Well, because NC replaces the "relativistic spacetime" by "Cartan spacetime". This is basically a foliation of spacetime by a global time coordinate, and in this spacetime it's the Galilei group which acts as symmetrygroup.

A valid global time label is, I guess, the same thing as it is in GR: a coordinate which is integrable.

Because the spacetime structure is completely different. Remember, only one connection component is non-zero for inertial observers! (Going to non-inertial observers introduces Coriolis-like forces, as you can check by yourself by writing down how connection coefficients transform under the Galilei group)

It's the mass density which appears in the Poisson equation, such that the EOM reproduce the Poisson equation. But Dautcourt shows that this EOM can be obtained via the relativistic field equation with a simple perfect fluid by some limiting process; he explains that from equation (64) onwards in his article. So in the non-relativistic limit I guess the mass density will indeed come down to the T^00 which you wrote down. But be carefull; the equations of motion are in the first place NOT written down in terms of an energy momentum tensor! They are motivated by their link to the Poisson equation!

I suggest that you carefully read through MTW and try to digest the Dautcourt article :) If you're looking for more articles on this, try to look for authors like Ehlers, Künzle, Trautman, or Duval. An article which is a bit more philosophical but carefully classifies all the "classical theories of gravity" in a geometric way is "Theories of Newtonian gravity and empirical indistringuishability" of Jonathan Bain.

Good luck with it and let me know if things are not clear!

Last edited by a moderator: May 4, 2017
13. Apr 13, 2010

JustinLevy

Re: Newton-Cartan

Thank you for the guidance. I think I'll try to work through MTW chap 12, then Dautcourt, and then if I actually can digest it all... that Jonathan Bain article you suggest sounds really interesting!

However, there is a very fundamental piece I am missing here, that is preventing me from being able to get much out of MTW.
Since, as in GR, it is curvature in spacetime not just some kind of spatial curvature (like people imagine with the curved rubber sheet analogy in science museums), I don't understand what makes the time coordinate special. MTW seems to just take the time coordinate and say: it is special, now that I've declared it so it obviously can only transform like t' = t + constant.

I don't understand. Let's say you take the Newton-Cartan theory and solve for a vacuum spacetime. What time labels are allowed? It must be more restricted than what you say above, otherwise we could do a Lorentz transformation and get a new set of "time" labels.

Newton-Cartan requires that we foliate spacetime into constant time slices. Fine. But I don't understand how this imposes anything. It seems like it only imposes something if we require these foliations are somehow unique ... otherwise we could choose any foliation we want, as long as we label the foliation with a coordinate, and then demand this coordinate is called time.

Even worse than conceptual issues, I can't even brute force my way through the equations (hoping understanding will come after working some examples). For instance, MTW (as did that web link I tried to work everything out from) says that:
$${\Gamma^0}_{10} = 0$$
which means:
$${\Gamma^0}_{10} = \frac{1}{2}g^{0m} \left(\frac{\partial g_{m1}}{\partial x^0} + \frac{\partial g_{m0}}{\partial x^1} - \frac{\partial g_{10}}{\partial x^m} \right) = 0$$
If we consider a solution for a static spherical stationary mass, there should be no time dependence and the metric will be diagonal so:
$$g^{00} \frac{\partial g_{00}}{\partial r} = 0$$
This gives that g_{00} can only be a constant!

Similarly for g_{11}. And the others are fixed by spherical symmetry.
So something is wrong here!

I must be missing something basic about what these equations mean.

Last edited: Apr 13, 2010
14. Apr 13, 2010

haushofer

Re: Newton-Cartan

Because you were considering solutions of the field equations, maybe the following article is also useful for you:

"Examples of Newtonian limits of relativistic spacetime" by Jürgen Ehlers. But first try to understand the basics! ;)

Because of the "absolute status" of time, you can't do any Lorentz transformations. Maybe the following helps.

We want to describe Newtonian gravity geometrically. We know that the symmetry group of Newtonian gravity is the Galilei group. This Galilei group can be seen as the group which keeps a "timelike metric" and a "spatial metric" both independently invariant. This means that in 4 dimensions our metric splits up into a "one-dimensional metric" and a "three dimensional metric". This means that our 4-dimensional metric becomes degenerate, the crucial fact which makes things very subtle!

That the metric becomes degenerate can be described as follows: the 4-dimensional metric can be written downs as

$$h^{\mu\nu}$$

and that it's degenerate means there is an covariant eigenvector t such that

$$h^{\mu\nu}t_{\nu} = 0$$

It turns out that you can build the timelike metric with this covariant eigenvector t via a direct product.

What's very important for you, and the fundamental reason why your calculations went horribly wrong, is the following aspect of NC:

The metric is NOT the fundamental geometric object anymore!

What I mean by this is the following: In MTW you can see how the connection is derived. Ofcourse, it's tempting to see this as a usual Levi-Civita connection and relate it to the metric.However, in MTW it's given as an exercise to actually show that if you do this, the Riemann tensor doesn't have the right symmetries anymore. That's no surprise now: as I said, the metric becomes degenerate! You cannot write down a non-degenerate, 4-dimensional metric in Newton Cartan. And that's the reason for a lot of confusion!

However, like I said: this is not the end of the world, for we can write down 2 metrics h and t. We still demand that they're covariantly constant with respect to some connection:

$$\nabla_{\rho}h^{\mu\nu} = 0, \ \ \ \ \ \ \ \ \nabla_{\rho}t_{\mu} = 0$$

Some conditions on the metric (like being symmetric in the lower indices in order to avoid non-classical effects like torsion) give then that this one-form t is actually closed:

$$\partial_{[\rho}t_{\mu]} = 0 \ \ \ \rightarrow \ \ \ dt =0$$

This means that we can write this one-form t, by Poincare's lemma, as

$$t_{\rho} = \partial_{\rho}t$$

of some scalar function t, and the one-form t is called "integrable". This scalar function t can now act as the absolute time t which you know of Newtonian mechanics. Note that in this coordinate system we get

$$t_{\rho} = \partial_{\rho}x^0 = \delta_{\rho}^0 \ \ \ \rightarrow \ \ \ t_{\rho} = (1,0,0,0)$$

and this implies that

$$h^{\mu0} = 0$$

so only the spatial part of the metric h survives.

I hope this has become clear now :)

I think you're missing the understanding of the geometric objects in NC and the status of the metric. :)

15. Apr 13, 2010

haushofer

Re: Newton-Cartan

Also, take a look at equation (29) of Dautcourt's article if you're curious how one can write down a connection in terms of these two metric h and t. The derivation of this connection is not given in any text as far as I know, but right now I'm writing an article about NC from a gauge-point of view in which we actually can derive these expressions quite elegantly.

Dautcourt first derives single components of the connection terms by using the metric conditions on h and t I gave you and some more. This is a more general theory than Newton Cartan itself.

By now I think it has become clear that you cannot simply write down the connection in terms of one single nondegenerate metric anymore. If you have more questions, let me know, but first give MTW some time :)

16. Apr 13, 2010

bcrowell

Staff Emeritus
Re: Newton-Cartan

No, there are alternative formulations of GR in which space is Minkowskian, and yet the field equations are exactly the same, and the predictions of experimental results are all exactly the same. MTW call this the "spin-two viewpoint," and discuss it in box 18.1, p. 437.

17. Apr 13, 2010

bcrowell

Staff Emeritus
Re: Newton-Cartan

GR can be expressed in a background-independent way (standard geometrical formulation) or on a fixed Minkowski background (MTW, "spin-2 viewpoint," pp. 436-437).

Newtonian gravity can be expressed in a generally covariant way (Cartan) or in a non-generally covariant way (standard formulation).

Re the philosphical/interpretational side of all this, here are some notes I wrote on a paper by Giulini, which might be helpful. I think the conclusion is that there are some aesthetic reasons why the Newton-Cartan formulation of Newtonian gravity is not preferable to the standard formulation, and why the standard geometrical formulation of GR is preferable to nongeometrical ones -- but -- the criteria are rather subtle and difficult to define rigorously, so the whole thing really is aesthetic, not objective. (I don't really buy Giulini's claim that "the principle of general covariance is devoid of any physical content.")

Giulini, "Some remarks on the notions of general covariance and background independence"
http://arxiv.org/abs/gr-qc/0603087

Kretschmann's 1917 critique: any theory can be put into generally covariant form.
That's what Cartan did with Newtonian gravity. But (1) it may be overly complex, and (2) it may be background-dependent.
Argues that Cartan gravity isn't overly complex (as Einstein apparently thought it would be), so the real issue is background-independence.
Claims that it's widely accepted that (a) "the most outstanding and characteristic feature of General Relativity is its manifest background independence," and (b) "the principle of general covariance is devoid of any physical content."
Heuristic criterion: background dependence shows up when solutions have degrees of freedom that divide up into classes A and B, where A (e.g., Newtonian spacetime's geometry) acts on B, but B doesn't act on A.
Conclusion is that there is still no satisfactory way to define background-independence.

18. Apr 13, 2010

Old Smuggler

Re: Newton-Cartan

The spin-2 viewpoint describes linearized GR, not the full theory. What is described in MTW pp. 436-437
is a method of deriving the EFE by starting with linearized theory and then doing a bootstrapping procedure
by requiring mathematical consistency at ever higher levels of approximation.

19. Apr 13, 2010

bcrowell

Staff Emeritus
Re: Newton-Cartan

I think the idea is that the approximations converge to an exact result which is completely equivalent to GR. Linearized gravity is just the first level of approximation.

20. Jun 9, 2010

Passionflower

Re: Newton-Cartan

I think that the Newton-Cartan formulation is an excellent stepping stone towards understanding GR. Just like GR, NC spacetime is curved, but only the time part, the space part curvature, or better the space and time curvature, as they are equivalent, is unique to GR.

Last edited: Jun 9, 2010