# Newtonian spacetime - MTW cap. 12

I have some questions regarding Newtonian spacetime; reference is MTW chap. 12.

MTW translate the Lagrange e.o.m. for Newtonian mechanics (with a potential phi derived from a mass density rho via Poisson eq.) into a geodesic equation in 4-dim. spacetime. They explicitly construct the connection Gamma, Riemann curvature R and Ricci tensor Ric.

They present an exercise to prove that the connection cannot be derived from a metric on spacetime. So the first consequence is that this Newtonian spacetime 4-manifold is not a Riemann manifold.

They show that the Ricci tensor contains the mass density only. That means that in vacuum the manifold is Ricci-flat Ric = 0.

1st question: what type of manifold is this? not Riemann due to the missing metric; not affine b/c it's not globally flat (see 3rd question)
2nd question: how can I further study the curvature? I would proceed with the Weyl tensor C, but I was not able to find a definition how to extract C from R w/o using the metric
3rd question: what does it mean that a manifold is flat? vanishing of R is too strong b/c it does contain coordinate effects; vanishing of Ric is too weak b/c it misses the Weyl curvature

Thanks
Tom

martinbn
I have some questions regarding Newtonian spacetime; reference is MTW chap. 12.

MTW translate the Lagrange e.o.m. for Newtonian mechanics (with a potential phi derived from a mass density rho via Poisson eq.) into a geodesic equation in 4-dim. spacetime. They explicitly construct the connection Gamma, Riemann curvature R and Ricci tensor Ric.

They present an exercise to prove that the connection cannot be derived from a metric on spacetime. So the first consequence is that this Newtonian spacetime 4-manifold is not a Riemann manifold.

They show that the Ricci tensor contains the mass density only. That means that in vacuum the manifold is Ricci-flat Ric = 0.

1st question: what type of manifold is this? not Riemann due to the missing metric; not affine b/c it's not globally flat (see 3rd question)
2nd question: how can I further study the curvature? I would proceed with the Weyl tensor C, but I was not able to find a definition how to extract C from R w/o using the metric
3rd question: what does it mean that a manifold is flat? vanishing of R is too strong b/c it does contain coordinate effects; vanishing of Ric is too weak b/c it misses the Weyl curvature

Thanks
Tom

1. Are you asking about the terminology? I don't know but I don't think there is a specific name, it is a differentiable manifold with a linear connection.
2. I don't think the Weyl tensor is defined for non-Riemannian manifolds. You have to stick with the Riemann tensor.
3. It means that the Riemann tensor is zero.

1. Are you asking about the terminology? I don't know but I don't think there is a specific name, it is a differentiable manifold with a linear connection.
2. I don't think the Weyl tensor is defined for non-Riemannian manifolds. You have to stick with the Riemann tensor.
3. It means that the Riemann tensor is zero.
1. Yes, just to find other resources than standard Riemannian geometry
2. I couldn't find any, but that proves nothing :-(
3. I don't think so; it's sufficient but not necessary; there are coordinates with non-zero R on flat manifolds; what about polar coordinates on a plane?

martinbn
I don't understand the last point! If a tensor is zero in one set of coordinates it will be zero in any other. We are talking about the Riemann tensor not the Christoffel symbols.

Matterwave
Gold Member
3. I don't think so; it's sufficient but not necessary; there are coordinates with non-zero R on flat manifolds; what about polar coordinates on a plane?

Since the Riemann tensor is a tensor, if it vanishes in one coordinate system, it must vanish in all coordinate systems. The transformation from one coordinate system to another is:

$$R^{a'}_{~~b'c'd'}=\frac{\partial x^{a'}}{\partial x^e}\frac{\partial x^f}{\partial x^{b'}}\frac{\partial x^g}{\partial x^{c'}}\frac{\partial x^h}{\partial x^{d'}}R^e_{~~fgh}$$

uh, hm, sorry, 3. was nonsense

but 1. and 2. remain valid questions

martinbn
but 1. and 2. remain valid questions

I think the answer to 1. is that there is no special name. It's just a manifold. Four-dimensional with a chosen connection. For 2. I don't know, my guess is the Weyl tensor is not defined unless the manifold is Riemannian.

bcrowell
Staff Emeritus
Gold Member
Newtonian spacetime would qualify as a smooth manifold, which is slightly more specialized than a manifold in general. Locally, its structure is that it's a direct product of two metric spaces, R^1 and R^3, with the latter having a Euclidean metric.

Matterwave
Gold Member
1. If you really want a name for this manifold, I believe it's a type of vector bundle ##E##. The base space is ##B=\mathbb{R}## (time) and at each point is connected a fiber ##F=\mathbb{R}^3## which is a linear vector space, and as bcrowell says, locally it is a product space. However, I believe it is not only locally trivial, but it is globally a product space ##E=B\times F## which would make it a trivial vector bundle (is there a reason you specified locality bcrowell? Am I missing something?). The projection ##\pi:E\rightarrow B## is given by the time coordinate which is constant over every fiber ##\pi(\mathscr{P})=t(\mathscr{P})##. There's no real need to define a structure group since the bundle is trivial.

2. You already know the Riemann curvature tensor. That is all the intrinsic curvature there is to know. All other intrinsic curvatures are derived from it. I am not sure how you could "study further" the curvature.

vanhees71
robphy
Homework Helper
Gold Member
The Riemann curvature tensor is associated with a connection (or a related object, a derivative operator). If it is derivable from/compatible with the metric, it is called a Levi-Civita connection.

Note that there is a metric-structure but is degenerate (so no inverse exists... so you have to take care raising and lowering indices, if you can).

Possibly useful..
http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf Malament's book (see the last chapter)
http://arxiv.org/abs/0907.2645 Ehlers's paper uses the field equations to determine a Newtonian limit of Weyl.

Another question regarding the geodesic equation; usually this is derived from an action S[C] measuring the length of the curve C via the line element ds.

A length is not available w/o a metric. My question is whether there is an action formulation from which the geodesic (as straightest line) can be derived. This action shall be constructed from the connection Gamma.

PAllen
I've seen manifolds with connection but no metric described as affine manifolds. I can't recall, off the top of my head, whether there are other requirements before a manifold is properly labeled as an affine manifold.

An affine manifold has a flat connection, whereas the Newtonian spacetime has a non-flat connection.

haushofer
Another question regarding the geodesic equation; usually this is derived from an action S[C] measuring the length of the curve C via the line element ds.

A length is not available w/o a metric. My question is whether there is an action formulation from which the geodesic (as straightest line) can be derived. This action shall be constructed from the connection Gamma.

There is, Tom. Take a look at this paper, eq. 2.28

http://arxiv.org/abs/1206.5176

with references therein. The explicit variation of the non-rel. particle action is a bit cumbersome and much more involved than the rel. case. Note the appearance of the vector field m_{\mu}, which can both be interpreted as the gauge field belonging to the central extension of the Galilei algebra and the ambiguity in the connection Gamma as defined by the "Newton-Cartan"metric compatibility conditions.

PAllen