My axiomatization attempt of Newtonian spacetime

Question

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1. What the question is asking (in plain language)

The student wants to re‑formulate Newtonian space‑time without introducing an absolute time function (t:M\to\mathbb R).
Instead he proposes:

a unit vector field (X) that points “into the future’’,
a symmetric bilinear form (g) on the tangent bundle such that

[ g(Y,Y)=g(Y,X)^2\qquad\text{for every vector field }Y . ]

He then shows (or tries to show) that this forces

[ g(Y,Z)=\omega(Y)\,\omega(Z),\qquad\omega:=g(X,\cdot ) . ]

The problem is to explain, step by step, why this construction cannot serve as a proper Newtonian space‑time structure and what the correct axiomatization looks like (the one used by Schuller and the standard Newton‑Cartan formulation).

In short we have to:

Derive rigorously the consequence (g=\omega\otimes\omega) from the postulated property.
Show that such a (g) is degenerate (rank 1) and therefore cannot give a spatial metric.
Explain why the usual Newtonian axioms use a clock 1‑form (dt) and a degenerate spatial metric (h) rather than a unit vector field.
Summarise the correct set of axioms and why the student’s attempt fails.

2. Detailed step‑by‑step solution

2.1 From the “unit‑speed’’ condition to a rank‑1 bilinear form

Assumption (A).
There exists a smooth vector field (X) on (M) with

[ g(X,X)=1, ]

and for every vector field (Y),

[ \boxed{g(Y,Y)=g(Y,X)^2}. \tag{1} ]

Define the 1‑form

[ \omega(Y):=g(Y,X)\qquad\text{(so } \omega = g(X,\cdot)\text{)}. \tag{2} ]

We want to prove

[ g(Y,Z)=\omega(Y)\,\omega(Z) \qquad\forall Y,Z . \tag{3} ]

2.1.1 Polarisation identity for a quadratic form

For any symmetric bilinear form (g) we have the polarisation identity

[ 2g(Y,Z)=g(Y+Z,Y+Z)-g(Y,Y)-g(Z,Z). \tag{4} ]

This identity holds without any further hypothesis.

2.1.2 Use the hypothesis (1) in the polarisation identity

Insert (1) into (4):

[ \begin{aligned} 2g(Y,Z) &=g(Y+Z,Y+Z)-g(Y,Y)-g(Z,Z)\[2mm] &=\bigl[g(Y+Z,X)\bigr]^{2} -\bigl[g(Y,X)\bigr]^{2} -\bigl[g(Z,X)\bigr]^{2}. \end{aligned} ]

Because of the definition (2),

[ g(Y+Z,X)=\omega(Y+Z)=\omega(Y)+\omega(Z) , ]

since (\omega) is linear (it is a 1‑form). Hence

[ \begin{aligned} 2g(Y,Z) &=(\omega(Y)+\omega(Z))^{2}-\omega(Y)^{2}-\omega(Z)^{2}\[2mm] &=2\,\omega(Y)\,\omega(Z). \end{aligned} ]

Dividing by 2 yields exactly (3):

[ \boxed{g(Y,Z)=\omega(Y)\,\omega(Z)} . ]

Thus the only symmetric bilinear form that satisfies (1) is the outer product of a single 1‑form with itself.

2.2 Consequences of (g=\omega\otimes\omega)

Rank and degeneracy.
For any vector (V) that satisfies (\omega(V)=0) we have

[ g(V,\cdot)=\omega(V)\,\omega(\cdot)=0 . ]

Hence the kernel of (g) is precisely

[ \ker g={V\in TM\mid \omega(V)=0}, ]

a subbundle of codimension 1. Consequently (\operatorname{rank} g=1); the form is highly degenerate and cannot be used to measure spatial distances (which require a non‑degenerate metric on the three‑dimensional “space‑like’’ subspaces).
No information about spatial geometry.
In Newtonian physics we need a spatial metric that tells us how far apart two simultaneous events are.
The tensor (\omega\otimes\omega) only measures the component along the distinguished direction (X) (the “time direction’’). It says nothing about the geometry orthogonal to (X).
Incompatibility with a generic force density.
Newton’s second law in the Newton‑Cartan formulation reads

[ \nabla_{\dot\gamma}\dot\gamma = \frac{1}{m}F , \qquad \text{with } dt(F)=0 . ]

The condition (dt(F)=0) means the force is spatial: it lives in (\ker dt).
In the student’s set‑up we have only (\omega), no separate “spatial’’ subbundle, so we cannot express the requirement “force has no time component’’ in a coordinate‑free way.

2.3 Why the standard Newton–Cartan axioms are different

Feature	Student’s attempt	Standard Newton–Cartan (Schuller)
Temporal structure	A unit vector field (X) (norm 1 w.r.t. (g))	A clock 1‑form (\tau) (or (dt)) with (\tau\neq0) everywhere
Metric	(g=\omega\otimes\omega) (rank 1, degenerate)	Two independent objects: • (\tau) (temporal 1‑form) • (h) (spatial metric) of rank 3, defined on (\ker\tau)
Compatibility	Implicit: (\nabla\tau=0) would force (\nabla X=0) (very restrictive)	Explicit: (\nabla\tau=0) and (\nabla h=0) (torsion‑free) give the Galilean connection
Physical meaning	“unit speed into the future’’ is forced, but spatial distances cannot be defined	(\tau) tells how much time passes between events; (h) tells how far apart simultaneous events are.
Force condition	Not expressible	(dt(F)=0) (force lives in (\ker dt))

The key point is that in Newtonian physics time and space are independent: you have a one‑form that selects a preferred foliation (the absolute time slices) and a spatial metric on each leaf of the foliation. Trying to encode both pieces of data in a single vector field and a degenerate bilinear form inevitably loses the spatial metric.

2.4 The correct set of axioms (concise version)

Absolute time: a nowhere‑vanishing 1‑form (\tau) (often written (dt)) with (\tau\wedge d\tau=0) (so the kernel integrates to a foliation by three‑dimensional slices).
Spatial metric: a symmetric, positive‑definite tensor (h) of rank 3 defined on (\ker\tau) (i.e. (h(\tau,\cdot)=0)).
Galilean connection (\nabla): a torsion‑free affine connection satisfying

[ \nabla\tau =0,\qquad \nabla h =0 . ]
Newton’s second law for a test particle of mass (m):

[ \nabla_{\dot\gamma}\dot\gamma = \frac{1}{m}F ,\qquad \tau(F)=0 . ]

These axioms reproduce the familiar Newtonian physics and avoid the pitfalls of the attempted “unit‑vector‑field’’ construction.

3. Final answer (summary)

From the postulated condition (g(Y,Y)=g(Y,X)^2) one must have

[ g = \omega\otimes\omega,\qquad \omega:=g(X,\cdot). ]
This tensor is rank‑1 and degenerate; it can only measure the component of a vector along the distinguished direction (X). It provides no spatial metric on the three‑dimensional slices orthogonal to (X).
Consequently the student’s formulation cannot encode Newtonian physics, because Newtonian spacetime requires both a temporal 1‑form (absolute time) and a non‑degenerate spatial metric on the simultaneity slices.
The standard Newton‑Cartan (or Schuller) axiomatization uses a clock 1‑form (\tau) and a spatial metric (h) together with a torsion‑free connection preserving both. This correctly separates time from space and allows a well‑defined notion of spatial distances and forces orthogonal to the time direction.

Hence, the attempt to replace absolute time by a unit vector field together with a bilinear form satisfying (g(Y,Y)=g(Y,X)^2) leads inevitably to a degenerate metric that cannot serve as the spatial geometry of Newtonian spacetime; the correct axioms must keep a separate time 1‑form and a spatial metric.

4. Common mistakes in this type of problem

Mistake	Why it’s wrong	How to avoid it
Treating (\omega) as a metric	A metric must be non‑degenerate; (\omega\otimes\omega) has a whole 3‑dimensional kernel.	Always check the rank of the bilinear form; a genuine spatial metric must be invertible on the spatial subbundle.
Assuming (g(Y+Z,X)=g(Y,X)+g(Z,X)) without justification	This uses linearity of the 1‑form (\omega), which follows only after you have defined (\omega=g(X,\cdot)).	Define (\omega) first and verify its linearity (it is a covector).
*Confusing a unit vector* field with a unit covector field**	In Newtonian spacetime the “clock’’ is a covector (1‑form) (dt), not a vector; a vector field cannot encode the foliation uniquely.	Remember that a foliation is defined by the kernel of a 1‑form, not by a vector field.
Forgetting the force orthogonality condition	Newton’s second law requires the force to be spatial ((dt(F)=0)). Without a separate time 1‑form this condition cannot be expressed.	Keep a distinct (\tau) (or (dt)) and impose (\tau(F)=0).
Believing that any torsion‑free connection preserving a degenerate metric is “the” Newtonian connection	There are many connections preserving a rank‑1 form; Newtonian physics imposes additional conditions (preserving the spatial metric, curvature constraints).	Use the full set of compatibility conditions (\nabla\tau=0,\ \nabla h=0) and the curvature conditions of Newton–Cartan theory.

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