Rigid motions and generalized Newtonian gravitation

Abstract

In an effort to contribute to a better understanding of General Relativity, here we lay the foundations of generalized Newtonian gravity, which unifies inertial forces and gravitational fields. We also formulate a kind of equivalence principle for this generalized Newtonian theory. Finally, we prove that the theory we propose here can be obtained as the non-relativistic limit of General Relativity.

Appendix

Here, we give a detailed derivation of the transformation of position, velocity and acceleration connecting two Newtonian reference frames. By Chasles’s theorem [7], the most general relative motion connecting any pair of Newtonian reference frames, $\mathcal{K }$ and $\mathcal{K }^\prime$, is the combination of an arbitrary translational motion and an arbitrary rotational motion.

Let $O$ be the origin of $\mathcal{K }$, whose axes are aligned according to the orthonormal base $\{\vec \varepsilon _i \}_{ i=1\ldots 3}$, and let $Q$ be the origin of $\mathcal{K }^\prime$, whose axes are aligned with the orthonormal base $\{\vec {e}_i(t) \}_{ i=1\ldots 3}$. Assume that $\vec {OQ}= X^i(t) \vec \varepsilon _i$.

As both bases are orthonormal, $\vec \varepsilon _i\,\vec \varepsilon _j = \vec {e}_i(t)\,\vec {e}_j(t) = \delta _{ij}, \quad \forall t$, whence it follows that:

$$\begin{aligned} \vec {e}_j(t) = R^i_j(t)\,\vec \varepsilon _i \end{aligned}$$

where $R^i_j(t)$ is an orthogonal matrix.

A point $P$ can be referred to both frames, as $\vec {x}$ in $\mathcal{K }$ and as $\vec {y}$ in $\mathcal{K }^\prime$, and we have that $\vec {x} = \vec {X} + \vec {y}$ which, writing the coordinates explicitly, reads:

$$\begin{aligned} x^i\vec \varepsilon _i =X^i\vec \varepsilon _i +y^j\vec {e}_j \end{aligned}$$

or

$$\begin{aligned} x^i = X^i(t) + R^i_j(t)\,y^j \end{aligned}$$

(41)

The transformation law for the velocity is:

$$\begin{aligned} \vec {v} := \frac{dx^i}{dt}\vec {\varepsilon }_i = \frac{dX^i}{dt}\vec \varepsilon _i + \frac{dy^j}{dt}\,\vec {e}_j + y^j\frac{d\vec {e}_j}{dt} = \vec {V} + \vec {w} + y^j\frac{d\vec {e}_j}{dt} \end{aligned}$$

where

$$\begin{aligned} \vec {w} := \frac{dy^j}{dt}\,\vec {e}_j \,, \qquad \vec {V} := \frac{dX^j}{dt}\,\vec \varepsilon _j \end{aligned}$$

and using that:

$$\begin{aligned} \frac{d\vec {e}_j}{dt} =\dot{R}^k_j\,\vec \varepsilon _k = \vec \Omega \times \vec {e}_j \end{aligned}$$

with

$$\begin{aligned} \vec {\Omega } := \frac{1}{2}\sum _j R_j^k\,\dot{R}_j^l\, \vec {\varepsilon }_k\times \vec {\varepsilon }_l = \Omega ^h \vec {\varepsilon }_h, \end{aligned}$$

that is:

$$\begin{aligned} \Omega _h \equiv \frac{1}{2} \sum _jR_j^k\,\dot{R}_j^l\eta _{klh}\,, \quad \Omega ^{kl} =\sum _j R_j^k\,\dot{R}_j^l\,, \end{aligned}$$

we arrive at:

$$\begin{aligned} \vec {v} = \vec {w} + \vec {V} + \vec \Omega \times \vec {y} \end{aligned}$$

(42)

or, alternatively,

$$\begin{aligned} \dot{x}^k = \dot{X}^k + R^k_l \dot{y}^l + \dot{R}^k_l y^l \quad \end{aligned}$$

Notice that:

$$\begin{aligned} \vec \Omega \times \vec {y} = \frac{1}{2} \sum _j R_j^k\,\dot{R}_j^l\, (\vec \varepsilon _k\times \vec \varepsilon _l)\times (y^i R_i^h \vec \varepsilon _h) = \dot{R}_j^k y^j \vec \varepsilon _k \end{aligned}$$

If we now take the derivative of (42), we obtain that the acceleration is:

$$\begin{aligned} \vec {a}&:= \frac{d^2x^i}{dt^2}\vec \varepsilon _i = \frac{d^2y^i}{dt^2}\vec {e}_i + \frac{dy^i}{dt}\frac{d\vec {e}_i}{dt} + \frac{d^2X^i}{dt^2}\vec \varepsilon _i \\&+\frac{d\vec \Omega }{dt}\times \vec {y} +\vec \Omega \times \left( \frac{dy^i}{dt}\vec {e}_i + y^i\frac{d\vec {e}_i}{dt}\right) \end{aligned}$$

and the transformation law for acceleration reads:

$$\begin{aligned} \vec {a} = \vec {b} + \vec {A} + \,\frac{d\vec \Omega }{dt}\times \vec {y} + \vec \Omega \times (\vec \Omega \times \vec {y}) + 2\vec \Omega \times \vec {w} \end{aligned}$$

(43)

or

$$\begin{aligned} {\ddot{x}}^k = R^k_l {{\ddot{y}}^l} + {{\ddot{X}}^k} +{{\ddot{R}}^k_l} y^l+ 2{\dot{R}}^k_l{\dot{y}}^l \end{aligned}$$

where

$$\begin{aligned} \vec {b} := \frac{d^2y^j}{dt^2}\,\vec {e}_j \,, \qquad \vec {A} {:=} \frac{d^2X^j}{dt^2}\,\vec \varepsilon _j \end{aligned}$$

Let us remark that:

$$\begin{aligned} \frac{d\vec \Omega }{dt}\times \vec {y}+\vec \Omega \times (\vec \Omega \times \vec {y}) = {{\ddot{R}}^k_l} y^l\vec {\varepsilon }_k \quad \mathrm{and} \quad \vec \Omega \times \vec {w} = \dot{R}_j^k \dot{y}^j \vec {\varepsilon }_k \end{aligned}$$