Formulating High School Gravity with General Relativity

1. The Inverse Problem

We seek a Lorentzian manifold $(M, g_{μ ν})$ of signature $(+, -)$ that exactly reproduces the trajectories of "high school physics"—each object traces a parable—as geodesic curves.
The curves described by test particles are parametrized by:

t (λ) = λ

x (λ) = x_{0} + v_{0} λ - \frac{1}{2} g λ^{2}

Differentiating with respect to the affine parameter $λ$ , we obtain the velocity vector fields $u^{μ} = (1, v_{0} - g λ)$ . And the accelerations are given by $\ddot{t} = 0$ and $\ddot{x} = - g$ :

On the other hand, the acceleration vector is required to satisfy the geodesic equation $\nabla_{u} u = 0$ . In the coordinate basis, this implies:

{\ddot{x}}^{μ} + Γ_{α β}^{μ} {\dot{x}}^{α} {\dot{x}}^{β} = 0

i.e.,

\ddot{t} = - Γ_{t t}^{t} \dot{t} \dot{t} - Γ_{t x}^{t} \dot{t} \dot{x} - Γ_{x x}^{t} \dot{x} \dot{x}

\dot{x} = - Γ_{t t}^{x} \dot{t} \dot{t} - Γ_{t x}^{x} \dot{t} \dot{x} - Γ_{x x}^{x} \dot{x} \dot{x}

Substituting the kinematic constraints $\ddot{t} = 0$ and $\ddot{x} = - g$ :

For the $t$ -component: $Γ_{α β}^{t} = 0$ .
For the $x$ -component: $- g = - Γ_{t t}^{x} (1)^{2} ⟹ Γ_{t t}^{x} = g$ .
Thus, the phenomenological requirement of exact parabolic motion imposes a constant, non-vanishing Christoffel symbol $Γ_{t t}^{x} = g$ , with all other connection coefficients vanishing in this chart.

2. The Riemann tensor

With the connection coefficients fixed ( $Γ_{t t}^{x} = g$ , others zero), we compute the Riemann curvature tensor:

R_{σ μ ν}^{ρ} = \partial_{μ} Γ_{ν σ}^{ρ} - \partial_{ν} Γ_{μ σ}^{ρ} + Γ_{μ λ}^{ρ} Γ_{ν σ}^{λ} - Γ_{ν λ}^{ρ} Γ_{μ σ}^{λ}

Since $Γ_{t t}^{x}$ is constant and the others are 0 we get:

R_{σ μ ν}^{ρ} = 0

Conclusion: The spacetime is locally isometric to Minkowski space $M^{1, 1}$ . The "gravity" observed is purely an inertial force arising from the choice of coordinates, not from intrinsic curvature.

3. The Metric Construction

Since the manifold is flat, the metric $g_{μ ν}$ must be the pullback of the Minkowski metric $η_{A B}$ via a diffeomorphism $Φ : R^{2} \to M^{1, 1}$ .
Let $(T, X)$ be inertial Cartesian coordinates with metric $η = d T^{2} - d X^{2}$ . We identify the transformation that maps inertial lines to parabolic curves in the $(t, x)$ chart:

T = t

X = x + \frac{1}{2} g t^{2}

The differentials transform as:

d T = d t

d X = d x + g t d t

Substituting these into the line element $d s^{2} = d T^{2} - d X^{2}$ :

d s^{2} = d t^{2} - (d x + g t d t)^{2}

d s^{2} = (1 - g^{2} t^{2}) d t^{2} - 2 g t d t d x - d x^{2}

The resulting metric tensor in the $(t, x)$ chart is:

g_{μ ν} = (\begin{matrix} (1 - g^{2} t^{2}) & - g t \\ - g t & - 1 \end{matrix})

4. Coordinate Validity vs. Physical Interpretation

We verify the validity of this chart by computing the determinant:

det (g) = - (1 - g^{2} t^{2}) - (g t)^{2} = - 1 + g^{2} t^{2} - g^{2} t^{2} = - 1

Since $det (g) = - 1$ everywhere, the coordinate chart $(t, x)$ is a valid global chart on Minkowski space. The mapping is a diffeomorphism; it is smooth and invertible everywhere. Mathematically, the coordinate system is flawless.

However, the physical interpretation of the coordinates is not clear:
Let $ξ = \partial_{t}$ be the coordinate basis vector representing the time evolution of the grid. Its norm is:

g (\partial_{t}, \partial_{t}) = g_{t t} = (1 - g^{2} t^{2})

Region I ( $| t | < 1 / g$ ): $ξ$ is timelike. The coordinates represent a frame where the "lab walls" (constant $x$ ) are moving subliminally.
Region II ( $| t | > 1 / g$ ): $ξ$ becomes spacelike. The coordinate grid lines of constant $x$ are now moving superluminally relative to the inertial frame $(T, X)$ .

Therefore, while this chart is valid everywhere (covering the entirety of $M^{1, 1}$ ), it is highly artificial. It cannot corresponds to a team of observers taking measures of space and time.

Related: a constantly accelerating coordinate system in a flat spacetime is better described by Rindler coordinates.