The initial value formulation of the $\lambda$-R model

TL;DR

This paper explores the initial value problem of the $\lambda$-R model, a modified gravity theory, deriving conditions for solutions and demonstrating its non-equivalence to general relativity except in specific cases.

Contribution

It generalizes the Lichnerowicz-York equation for the $\lambda$-R model and analyzes solution existence depending on the coupling $\lambda$, revealing non-equivalence with GR.

Findings

Solutions depend on the value of $\lambda$ and the trace of momentum $\pi$.

Existence and uniqueness are established for $\lambda>1/3$ with $\pi eq0$.

The $\lambda$-R model generally cannot match GR's initial data and evolution, except in special cases.

Abstract

We apply the conformal method to solve the initial value formulation of general relativity to the $\lambda$-R model, a minimal, anisotropic modification of general relativity with a preferred foliation and two local degrees of freedom. We obtain a generalised Lichnerowicz-York equation for the conformal factor of the metric and derive its properties. We show that the behaviour of the equation depends on the value of the coupling $\lambda$. In the absence of a cosmological constant, we recover the existence and uniqueness properties of the original equation when $\lambda>1/3$ and the trace of the momentum of the metric, $\pi$, is non-vanishing. For $\pi=0$, we recover the original Lichnerowicz equation regardless of the value of $\lambda$ and must therefore restrict the metric to the positive Yamabe class. The same restriction holds for $\lambda<1/3$, a case in which we show that the spatial Ricci scalar must also be large enough to guarantee the existence of at least one solution. Taking the equations of motion into account, this allows us to prove that there is in general no way of matching both constraint solving data and time evolution of phase space variables between the $\lambda$-R model and general relativity, thereby proving the the non-equivalence between the theories outside of the previously known cases $\lambda=1$ and $\pi=0$.