### Newton-Cartan Gravity

A few weeks ago I discussed Einstein-Cartan geometry, with a focus on relaxing the “torsion-free” condition on a Levi-Civita connection. In this post I will talk about the Newton-Cartan theory of gravity, which is in some sense the Newtonian limit of general relativity, and which relaxes the metric-compatibility of a Levi-Civita connection (while keeping the torsion-free condition).

Newtonian gravity and the naive formulation
(Note: the material in this section is re-hashed from Section 12.1 of Misner, Thorne, and Wheeler’s black-covered bible.)

Consider first Newtonian theory of gravity. The space-time is $\mathbb{R}^1\times\mathbb{R}^3$ with Galilean symmetry, and gravitational interaction is represented by the gravitation potential $\Phi(t,x)$. In Newtonian theory, the gravitational field is given by minus the gradient of the potential $\vec{F}_G = - \vec{\nabla} \Phi$ (I will put the arrows over symbols to denote the fact that they are three-dimensional vectors, and the derivative symbols should be interpreted in the sense of three-dimensional vector calculus). The force on a particle is given by the product of the gravitational field and the gravitational charge of the particle $m_G\vec{F}_G$. By Newton’s second law, the force is also equal to the product of the inertial mass and the acceleration of the particle $m_I \vec{a}$. Now, by the principle of equivalence (or the observation that the gravitational charge is equal to the inertial mass), we have that the gravitational field is equal to the acceleration of the particle.

Now consider a particle traveling in the gravitational field in free fall. Write its trajectory in $\mathbb{R}^3$ as $\vec{\xi}(t) = (\xi_1(t), \xi_2(t), \xi_3(t))$. Lifting to the space-time the world line of the particle is given by $(t,\vec{\xi}(t))$. (For people familiar with General Relativity already: in GR the world-line is usually given as a geodesic with unit speed. Under the 3+1 split in Newtonian theory, “proper time” is not defined, so the natural parametrization is by the global/invariant time.) The velocity vector in the space-time is $(1,\dot{\vec{\xi}}(t))$ and the acceleration vector is $(0,\ddot{\vec{\xi}}(t))$. The Newtonian equation of motion then is described by

Equation 1
$\displaystyle \frac{d^2}{dt^2} (t, \vec{\xi}(t)) + (0, \vec{\nabla}\Phi(t, \vec{\xi}(t))) = 0$

Now, observe that if we do an affine change of variables $t \to t(s)$ (affine means here $d^2t/ds^2 = 0$), and notice that the chain rule gives $d/ds = dt/ds . d/dt$, (and by abuse of notation we write $\vec{\xi}(s) = \vec{\xi}(t(s))$)

Equation 1′
$\displaystyle \frac{d^2}{ds^2}\left(t(s), \vec{\xi}(s)\right) + \left(0, \vec{\nabla}\Phi(t(s),\vec{\xi}(s))\right) \left(\frac{dt}{ds}\right)^2 = 0$

Suppose we want to now “geometrize” Equation 1′. What’s the right equation to compare the above to? $s$ here in the equation is an arbitrary affine (re-)parametrization (relative to global time $t$) of the world-line of a particle. The physical situation–a particle in free-fall–under the “geometrization” procedure of the principle of equivalence, should be identified with a geodesic in space-time. Now let $\nabla$ (N.b. without the arrow on top) denote an affine connection on $\mathbb{R}^1\times\mathbb{R}^3$, and $\Gamma^a_{bc}$ its corresponding Christoffel symbols in the natural coordinates, the geodesic equation in affine parametrization should read

Equation 2
$\displaystyle \frac{d^2 \chi^a}{d\tau^2} + \Gamma^a_{bc}\frac{d \chi^b}{d\tau}\frac{d \chi^c}{d\tau} = 0$

By visual inspection, to identify Equations 1′ and 2, we need ($0\leq a,b\leq 3, 1\leq i,j\leq 3$)

Equation 3
$\Gamma^0_{ab} = \Gamma^a_{bi} = 0$ and $\Gamma^i_{00} = \frac{\partial}{\partial x^i} \Phi$

If we impose the further constraint

Equation 3′
$\frac{\partial}{\partial x^j}\Gamma^i_{00} = \frac{\partial}{\partial x^i}\Gamma^j_{00}$

then Equations 3 and 3′ together gives and identifies Newtonian gravitational potentials with a class of non-trivial affine connections.

Before we proceed, first let us calculate the curvature of this affine connection. Recall the formula for the curvature operator of a connection:

$R(v,w)u = \nabla_v\nabla_w u - \nabla_w\nabla_v u - \nabla_{[v,w]}u$

Inserting the natural coordinate vector fields (whose commutators vanish), we have

$R_{abc}{}^d = \partial_a\Gamma^d_{bc} - \partial_b\Gamma^d_{ac} + \Gamma^f_{bc}\Gamma_{af}^d - \Gamma^f_{ac}\Gamma_{bf}^d$

where repeated indices are assumed to be summed. By Equation 3 and 3′, the quadratic terms all vanish, and the only non-zero components are

Equation 4
$R_{i00}{}^j = \partial^2_{ij}\Phi = - R_{0i0}{}^j$

which implies that the only non-zero contracted curvature is

Equation 4′
$R_{00} = \triangle \phi =: 4\pi \rho$

where $\rho$ is the mass-density. I paraphrase from MTW: equations 3, 3′, 4, 4′, and the equation for geodesic motion are “the full content of Newtonian gravity, rewritten in geometric language”.

Galilean change of coordinates
Newtonian gravity is invariant under the Galilean transformations. These transformations consist of changes of coordinates $(t,\vec{x})\to(t',\vec{x}')$ generated by

• Temporal translations: $t' = t + t_0$; temporal reflections $t' = - t$
• Spatial translations: $\vec{x}' = \vec{x} + \vec{x}_0$; spatial reflections $\vec{x}' = - \vec{x}$
• Spatial rotations: $\vec{x}' = A\vec{x}$ where $A$ is a $3\times 3$ matrix satisfying the condition $AA^T = I$ (in other words $A$ is an orthogonal matrix).
• Shear transformation: $\vec{x}' = \vec{x} + \vec{v}_0 t$

The middle two of the above are the generators of the affine-group of symmetries on Euclidean 3-space. The first is the generator of the symmetry group of Euclidean 1-space. The one that characterizes the “Galilean aspect” is the last.

Let us stop here and think a bit about what it means to have these coordinate changes. Consider a bunch of observers just floating around in space. Each of them carries a stopwatch (since we are in Newtonian picture with global time, all the stopwatches run at the same rate), and three rulers held mutually perpendicular (we’ll also assume that all the rulers are marked the same way). Now put a bunch of particles also in space, and have the observers look at the particles. Now, the observers cannot see the particles from afar. All they can sense is when a particle or another observer gets very close to them (say they are conscious of everything within a foot of them). Every time a particle comes close to an observer, he notes where (relative to his rulers) and when (relative to his stopwatch) the particle enters and leaves his sensory field. And he jots down in his notebook something like

At time 12543.3, a particle appears at coordinates (0,10,0); at time 12548.3, it leaves from coordinates (-10,0,0). Hence it is traveling with velocity vector (-2,-2,0).

Now imagine observers A and B drifts close to each other, and a particle C shoots through their region. Afterward, they compare their entries. Okay, the times are different, but that can be attributed to their starting their stopwatch at different times. The duration of the particle’s presence is different, but that’s probably because the particle stayed closer to observer A, and hence is in his sensory field a smidge longer. The coordinate measurements are different, and that is again due in part to the A and B becoming aware of C at different times, and also due in part to the fact that A and B are not holding their rulers in the same directions. But finally, the speeds they computed are different, even after factoring in how the rulers are held–ah, they are moving relative to each other.

Now looking at the picture in space-time, when the observers A, B, and the particle C gets close, we can take it to mean that they are physically at the same event $p$ in our space-time. The velocity vector of C is then a vector in the tangent space $T_p(\mathbb{R}^1\times\mathbb{R}^4)$. The fact that A and B get different measurements corresponds to them reading off the coordinates of the velocity vector of C in two different bases of the tangent space: each observer has three vectors that corresponds to unit spatial directions, and one vector that corresponds to unit temporal direction.

The Galilean symmetry puts constraints on how these bases can differ. That there is a global time implies that the span of the spatial directions are always the same, and that the difference between the the temporal directions must be in the span of the spatial directions (this is the relative velocity of the two observers). That each observer agrees on what constitutes as a unit length implies that the spatial directions are just rotations of each other (let’s ignore reflections for now; they correspond to discrete, not continuous, symmetries and in the language of Gauge theory should be discarded as they are not connected to the component containing identity).

Another way of arriving at this conclusion would be to look at the infinitesimal transformations in the tangent and co-tangent spaces induced by the Galilean coordinate changes above at a fixed-point of the coordinate change.

We will call a set of bases in the tangent space that corresponds to an observer an “admissible frame”. Each admissible frame has 1 time-like direction and 3 space-like directions. Given two admissible frames $\{ e_t, e_1, e_2, e_3\}$ and $\{e'_t, e'_1,e'_2,e'_3 \}$, the rules of Galilean symmetry implies that there exists a 3-by-3 orthogonal matrix $A_{ij}$ and a 3-vector $v_i$ such that

Equation 5
$e'_t = e_t + \sum_{i = 1}^3 v_ie_i~,\qquad e'_i = \sum_{j = 1}^3 A_{ij}e_j$

A simple calculation shows that if the change from frames $e_* \to e'_*$ uses transformation $(A_{ij}, v_i)$, and if a change from $e'_* \to e''_*$ uses $(B_{ij}, w_i)$, then a change from $e_* \to e''_*$ goes through the transformation $((BA)_{ij}, (v + wA)_i)$ where products are matrix products: $(BA)_{ij} = \sum_k B_{ik}A_{kj}$ and $(wA)_i = \sum_k w_kA_{ki}$.

In other words, the set of all “admissible frames” is a $\mathfrak{G}$torsor for the group $\mathfrak{G} = \mathbb{R}^3\rtimes_\phi SO(3)$, the semi-direct product given by $\phi_A(v) = vA$.

Structure invariants
The question I want to ask now is this: consider the set of all possible quartets of non-zero tangent vectors. This is a 16 dimensional space. Given an arbitrary quartet, consider the $\mathfrak{G}$-torsor generated by it: this is a 6 dimensional smooth submanifold. We should be able to find a 10-dimensional set of invariants whose joint level surface is precisely our torsor. (A good exercise is to try to work this out for Riemannian geometry, i.e. for the manifold of Euclidean frames in a fixed vector space.)

The requirement that in Galilean symmetry a global time is preserved implies that one can find a co-vector $\epsilon_t$ such that for every admissible frame, $\epsilon_t(e_t) = 1$. A bit of short computation shows that, given an initial basis $\{e_t,e_1,e_2,e_3\}$, the $\epsilon_t$ for the $\mathfrak{G}$-torsor generated is uniquely given by the co-vector satisfying two properties: (a) $\epsilon_t(e_t) = 1$ and (b) $\epsilon_t(e_i) = 0$ if $i = 1,2,3$. Since $\epsilon_t$ is a co-vector, this uses up 4-dimensions of freedom. The other invariant is simply the one corresponding to the $SO(3)$ portion of the structure group. Take the symmetric tensor given by $\sigma = e_1\otimes e_1 + e_2\otimes e_2 + e_3\otimes e_3$. Consider the action of an element of $\mathfrak{G}$ on this tensor:

$\displaystyle \sigma' = \sum_{i = 1}^3 e'_i\otimes e'_i = \sum_{i,j,k = 1}^3 A_{ij}e_j\otimes A_{ik}e_k = \sum_{i,j,k = 1}^3 A^T_{ki}A_{ij}e_j\otimes e_k = \sum \delta_{kj} e_j\otimes e_k = \sigma$

where we used that $A_{ij}$ is an orthogonal matrix. For an arbitrary quartet, the corresponding $\sigma$ is by construction a symmetric tensor with 6 degrees of freedom. This completes all the degrees of freedom.

Indeed, one can check that

Proposition 6
A set of admissible frames is uniquely specified by a co-vector $\epsilon_t$ and a symmetric two-tensor $\sigma$ such that

• $\sigma$, as a quadratic form on the space of co-vectors, is positive semi-definite.
• $\epsilon_t$ spans the kernel of $\sigma$.

I outline the proof here: since $\sigma$ is positive semi-definite and symmetric, we can construct a basis of eigen-co-vectors $\{ \epsilon_t, \epsilon_1, \epsilon_2,\epsilon_3\}$ where $\epsilon_t$ is as given and $\sigma(\epsilon_i,\epsilon_j) = \delta_{ij}$ if $i,j = 1,2,3$. Let $\{e_t,e_1,e_2,e_3\}$ be the dual basis to $\epsilon_*$ in the tangent space. It is simple to check that the $\mathfrak{G}$-torsor generated by this basis is an admissible frame. Next it suffices to check that, with $\epsilon_t$ fixed, any other basis of eigen-co-vectors $\epsilon'_*$ can be related to the original $\epsilon_*$ by the dual-action of $\mathfrak{G}$, which implies that the associate vector basis $e'_*$ is related to $e_*$ by the action of $\mathfrak{G}$.

Galilean geometry
The problem with the naive formulation given in the first section is that it depends on the fixed 1+3 splitting of the space-time. In particular, the gravitational potential $\Phi$ is obtained by solving the Poisson equation on the three dimensional slice, and is not, a priori speaking, geometric. Furthermore, the statement in Equations 3 and 3′ relate to a coordinate definition of the connection coefficients, which may or may not change nicely under changes of coordinates. Ideally, we want a fully covariant way of writing the assumptions of the theory.

What we will do, now, is model our construction on the Cartan way (method of moving frames) of formulating Riemannian geometry. (See the Wikipedia article on Frame Bundles for more information.) Now consider an arbitrary 4-dimensional smooth manifold $M$. Associated to the tangent bundle $TM$ is the frame bundle $F_{GL}(M)$ of bases of $TM$. The subscript $GL$ denotes the fact that the most general structure group allows arbitrary changes of bases, which is given by the Lie group $GL(4)$. Any affine connection $\nabla$ can then be described as a $GL(4)$ connection on the principle $GL(4)$-bundle $F_{GL}(M)$. In the case of Riemannian geometry, however, we can restrict to the ortho-normal frame bundle $F_O(M)$ with the group structure given by the orthogonal group $O(4)$ associated to the Riemannian metric. Now, the Riemannian metric is precisely the algebraic invariant for the structure group. A compatible connection must preserve the algebraic invariant across fibers, and so we have that the compatible connection is the Levi-Civita connection under which the Riemannian metric is parallel.

Similarly, to use the local algebraic descriptions developed in the above two sections to extend to an arbitrary smooth manifold, the geometry for a Newtonian gravitational theory should be described by having a connection which preserves the algebraic invariants given in section 3.

Definition 7
A Galilean manifold is the ordered quartet $(M, \epsilon_t, \sigma, \nabla)$ where $M$ is a smooth $(n+1)$-dimensional manifold, $\epsilon_t\in\Gamma T^*M$ is a one-form, $\sigma\in\Gamma T^{2,0}M$ is a symmetric two-tensor, and $\nabla$ is a torsion-free affine connection such that

• $\sigma$ is positive semi-definite as a quadratic form on $T^*M$
• $\epsilon_t$ spans the kernel of $\sigma$
• The connection is compatible with the structure invariants, i.e. $\nabla \epsilon_t = 0$ and $\nabla \sigma = 0$.

(N.b. If you compare the definition given in this paper of Brauer, Rendall, and Reula, which formalism the authors cite to Ehlers, what they define as the spatial metric $h$ is my $\sigma$, and the time metric $g$ is equivalent to $\epsilon_t\otimes\epsilon_t$. At this level the two definitions are equivalent.)

So far we have only described the general geometry of space-time. (Compare to general relativity, what we have done so far is define the analogue of Lorentzian manifolds. We have not specified an equation of motion that couples in matter, e.g. an analogue of Einstein’s equation.) Let us consider some direct geometric consequences that we have gotten so far. First, the fact that $\epsilon_t$ is parallel, and that $\nabla$ is torsion free, implies that $d\epsilon_t = 0$. But rather then assuming that we have a suitable topology which allows lifting to a global time function and constructing a time-foliation thereof, we’ll instead make the observation that $\epsilon_t\wedge d\epsilon_t = 0$ by definition. Therefore applying Froebenius’ theorem, the three-dimensional distribution defined as the kernel of $\epsilon_t$ is integrable. Therefore locally there exists a hypersurface $\Sigma \subset M$ such that $\sigma \in \Gamma T^{2,0}\Sigma$ and defines a Riemannian metric on $\Sigma$. In other words, the underlying geometry already contained an intrinsic 1+3 splitting of the manifold into temporal and spatial directions. Furthermore, we can compute the “extrinsic curvature” of $\Sigma$ in $M$ (this is not exactly the second fundamental form as usually seen in Riemannian geometry per se, but we have some similar notions here).

Recall that the second fundamental form in Riemannian geometry is a measure of how much the ambient parallel transport tend to twist a tangent vector out of its hypersurface. Given that $\Sigma$ is defined by orthogonality to $\epsilon_t$, we can do something similar. Let $V,W\in T_p\Sigma$, consider the object $\epsilon_t(\nabla_VW)$. If it is non-zero this means that the parallel transport given by the connection $\nabla$, when acting on $W$ in the $V$ direction, will tend to twist $W$ out of the hypersurface $\Sigma$. And as it happens, for $W$ a vector field tangent to $\Sigma$ and $V$ an arbitrary vector, we have

Equation 8
$\epsilon_t(\nabla_VW) = V( \epsilon_t(W) ) - (\nabla_V\epsilon_t)(W) = V(0) - 0 = 0$

This in particular implies that $\Sigma$ is totally geodesic in $M$, and that $\nabla$ restricts to a connection on $T\Sigma$ (without projection!). (This captures the statement in Equation 3 that $\Gamma^0_{aj} = 0$.) Now, restricting $\nabla, \sigma$ to $\Sigma$, we see that we now have an intrinsic connection on a manifold that is torsion free and preserves the Riemann metric. Therefore the restriction of $\nabla$ to $\Sigma$ coincides with the induced Levi-Civita connection. We summarize our computations in the following

Lemma 9
If $(M, \epsilon_t, \sigma, \nabla)$ is a Galilean manifold, then $M$ is foliated by hypersurfaces $\Sigma_\tau$ satisfying

• $\epsilon_t(T\Sigma_\tau) = 0$ for any $\tau$,
• $\Sigma_\tau$ is totally geodesic with respect to the connection $\nabla$, and
• $(\Sigma_\tau, \sigma)$ is a Riemannian manifold, whose Levi-Civita connection coincides with the connection induced from $\nabla$.

Curvature and Newton’s equations
The remainder of the Equations 3, 4, and 4′ (3′ is subsumed in the others), as we shall see, will be captured in Newton’s equations.

First we recall that the Riemann curvature tensor $R_{abc}{}^d$ is still well-defined once we are given an affine connection. Furthermore, tensor contraction affords us also the Ricci curvature tensor. By totally geodesy, $\epsilon_t(R(V,W)U) = 0$ if $V,W,U\in T\Sigma$, and the restriction of the curvature tensor to $\Sigma$ agrees with the intrinsic curvature given by the Riemannian metric $\sigma$. (Essentially a version of the Gauss and Codazzi equations.)

Also, it is important to note that as $\nabla$ is assumed to be torsion free, the first Bianchi identity still holds via the Jacobi identity. That the second Bianchi identity still holds is a general property of affine connections.

Definition 10
A Newton-Cartan theory consists of a Galilean manifold $(M, \epsilon_t, \sigma, \nabla)$ and a symmetric (2,0)-tensor $T[\Psi]$ corresponding to the stress-energy for some matter fields with the requirement that

• The matter equation is satisfied $\nabla_aT^{ab} = 0$.
• The Newtonian gravity coupling (with cosmological constant) $Ric = [4\pi T(\epsilon_t,\epsilon_t)- \Lambda ] \epsilon_t\otimes \epsilon_t$
• The symmetry property $R(\sigma(\cdot),V)W$ is a symmetric (2,0)-tensor for any vectors $V,W$.

The first condition gives the minimally coupled evolution equation for the matter fields. The second condition is a generalization of Equation 4′, where the “temporal” component of Ricci is taken to be equal to the matter energy density minus cosmological constant. The third condition is a generalization of Equation 4.

Let us consider the consequences of the definition: the second condition implies that $Ric(V,W) = 0$ for any $V,W$ tangent to $\Sigma$. This directly implies that $(\Sigma,\sigma)$ is Ricci-flat. The third condition, by total geodesy, implies the same condition holds for the curvature operator of $\Sigma$. Using the induced metric on $\Sigma$ to raise and lower indices, the condition becomes $R^{(\sigma)}_{ijkl} = R^{(\sigma)}_{ilkj}$. Applying the first Bianchi identity we see that

Lemma 11
In a Newton-Cartan theory, the spatial hypersurfaces $\Sigma_\tau$ with the Riemannian structure induced by $\sigma$ are flat.

For now, I’ll end here.