A theory of gravitation

Abstract

The paper presents a theory of gravitation as a continuum dynamical theory, i.e. the equations of motion are first order in the time derivatives. The theory satisfies the laws of conservation of total energy, momentum and mass in the standard sense, i.e. as applications of Stokes’ theorem. A model in this theory is defined by the specification of an energy density function that accounts for the total mechanical and gravitational energy of the system and which in turn defines an action function. The derivation of the equations of motion is based on Hamilton’s principle of least action with the “same” action function as used for the derivation of the Einstein equation of the theory of general relativity, however, not with respect to variations of the metric but with respect to variations induced by local displacements of space, i.e. variations of the conjugate variable of the momentum and variations of the momentum density. Entropy density is introduced as a variable. This makes it possible to determine the thermodynamic equilibrium conditions for a model. Solutions of the equilibrium conditions are the Schwarzschild and Minkowski metrics on space-time, and the metrics defining the spherical and hyper spherical spaces.

Appendix: Geometrical preliminaries

Consider the following setting: the space-time is $R\times X$ where X is be a three-dimensional differential manifold. Coordinates $\left( y^{i}\right)$ and $\left( x^{i}\right)$, with $i=1,2,3$, denote points on X via coordinate charts. In continuum dynamics we have two modes of description; either we describe the flow of the fluid by following a fluid element in space or we observe the variation of the fluid observables at fixed points of the space X. They are referred to as the Lagrange and Euler descriptions respectively. The Lagrange description is thus given by specifying a flow

$$\begin{aligned} \varPsi :R \times X \rightarrow X;\left( t, y^i \right) \mapsto \left( t,\psi ^i _{t}\left( y^i\right) \right) = \left( t,x^i \right) \end{aligned}$$

on X that takes the initial configuration of points in the fluid to the actual configuration while the Euler description is associated with a flow $\varphi _{t}$ that refers back to the initial configuration, i.e. $\psi _{t}\circ \varphi _{t}=id_{X}$.

The space-time is endoved with a quasi-riemannian metric $g_{t\mu \nu }\left( x^i\right)$.

1.1 Variations by spatial displacements

Definition 1

A field of spatial displacement $\delta \psi$ in the Lagrange representation defines a field of displacements

$$\begin{aligned} \delta q^{i}=\delta \psi ^{i}\circ \varphi =-\left( \varphi ^{i}_{,j}\right) ^{-1}\delta \varphi ^{j} \end{aligned}$$

in the Euler representation by the condition $\delta \left( \psi \circ \varphi \right) =0$.

Axiom 2

In the Lagrange representation the metric which is given by

$$\begin{aligned} {\tilde{g}}_{ \mu \nu }\left( y^i\right) =\varPsi ^{\kappa }_{,\mu }\left( t,y^i\right) \varPsi ^{\lambda }_{,\nu }\left( t,y^i\right) g_{t\kappa \lambda }\circ \psi _{t}\left( y^i\right) \end{aligned}$$

is assumed to be invariant under variations, i.e.

$$\begin{aligned} \delta ^{q}{\tilde{g}}_{\mu \nu }\left( y^i\right)= & {} 0 \end{aligned}$$

Remark 4

Technically, a variation is a differentiation on an infinite dimensional manifold. In the following I will use the notation $\delta$ for the total differentiation and $\delta ^{q}$ for the partial differentiation corresponding to variations due to spatial displacements.

Proposition 4

In the Euler representation the variation $\delta ^{q}$ of the metric $g_{\mu \nu }$ is

$$\begin{aligned} \delta ^{q}g_{t00}= & {} -\delta q^{k}g_{t00,k}\\ \delta ^{q}g_{ti0}= & {} -\delta q^{k}_{,i}g_{tk0}-\delta q^{k}g_{ti0,k} \\ \delta ^{q}g_{t0i}= & {} -\delta q^{k}_{,i}g_{t0k}-\delta q^{k}g_{t0i,k} \\ \delta ^{q}g_{tij}= & {} -\delta q^{k}_{,i}g_{tkj}-\delta q^{k}_{j}g_{ti,k}-\delta q^{k}g_{tij,k} \end{aligned}$$

Proof

Since the metric is invariant with respect to variations in the Lagrange representation we get

$$\begin{aligned} \delta ^{q}{\tilde{g}}_{\mu \nu }\left( y\right)= & {} \delta ^{q}\left( \varPsi ^{\kappa }_{,\mu }\left( y\right) \varPsi ^{\lambda }_{,\nu }\left( y\right) g_{\kappa \lambda }\circ \psi \left( y\right) \right) \\= & {} \delta \varPsi ^{\kappa }_{,\mu }\left( y\right) \varPsi ^{\lambda }_{,\nu }\left( y\right) g_{\kappa \lambda }\circ \psi \left( y\right) \\&+\varPsi ^{\kappa }_{,\mu }\left( y\right) \delta \varPsi ^{\lambda }_{,\nu }\left( y\right) g_{\kappa \lambda }\circ \psi \left( y\right) \\&+\varPsi ^{\kappa }_{,\mu }\left( y\right) \varPsi ^{\lambda }_{,\nu }\left( y\right) \delta ^{q}g_{\kappa \lambda }\circ \varPsi \left( y\right) \\&+\varPsi ^{\kappa }_{,\mu }\left( y\right) \varPsi ^{\lambda }_{,\nu }\left( y\right) \delta ^{q}\varPsi ^{\eta }g_{\kappa \lambda ,\eta }\circ \psi \left( y\right) \\= & {} 0 \end{aligned}$$

Thus, in the Euler representation we get

$$\begin{aligned} \delta ^{q}g_{\mu \nu }=-\left( \left( \delta ^{q}\varPsi ^{\kappa }\circ \varphi \right) _{,\mu }g_{\kappa \nu }-\left( \delta \varPsi ^{\kappa }\circ \varphi \right) _{,\nu }g_{\kappa \mu }-\delta ^{q}\varPsi ^{\kappa }\circ \varphi g_{\mu \nu ,\kappa }\right) \end{aligned}$$

Since displacements in time is not included, i.e. $\delta ^{q}\varPsi ^0=0$ and $\varPsi ^0_{,0}=1$ we obtain the above result. $\square$

Corollary 2

Let $g=-\det \left( g_{\mu \nu }\right)$ then $\delta ^{q}\sqrt{g}=-\left( \sqrt{g}q^{i}\right) _{,i}$.

Definition 2

The volume of a domain D is

$$\begin{aligned} V\left( D\right) =\int _{D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k}. \end{aligned}$$

Proposition 5

The variation $\delta ^{q}$ of the volume $V\left( D\right)$ is

$$\begin{aligned} \delta ^{q}V\left( D\right) =-\int _{D}\delta q^{l}\sqrt{g}_{,l}\frac{1}{3!} \varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

Proof

By computation,

$$\begin{aligned} \delta ^{q}V\left( D\right)= & {} \delta ^{q}\int _{D}\sqrt{g}\frac{1}{3!} \varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\= & {} \int _{D}\delta ^{q}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&+\int _{\delta ^{q}D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\= & {} -\int _{D}\delta q^{l}\sqrt{g}_{,l}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

since

$$\begin{aligned}&\int _{\delta ^{q}D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&\quad =\int _{D+\delta ^{q}D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&\qquad -\int _{D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&\quad =\int _{D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}d\left( x^{i}+\delta q^{i}\right) \wedge d\left( x^{j}+\delta q^{j}\right) \wedge d\left( x^{k}+\delta q^{k}\right) \\&\qquad -\int _{D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&\quad =\int _{D}\sqrt{g}\det \left( \delta _{m}^{l}+\delta q^{l}_{,m}\right) \frac{1}{3!} \varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&\qquad -\int _{D}\sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&\quad =\int _{D}\sqrt{g}\delta q^{l}_{,l}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

$\square$

Definition 3

The mass M and entropy S are defined in terms of the mass and entropy densities $\rho$ and s by

$$\begin{aligned} M= & {} \int _{D}\rho \left( x\right) \sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\ S= & {} \int _{D}s\left( x\right) \sqrt{g}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

Axiom 3

The variations of the 3-forms with respect to spatial displacements are

$$\begin{aligned} \rho \left( x\right) \frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\ s\left( x\right) \frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

are zero.

Proposition 6

The axiom implies that the variation of the mass and entropy densities due to spatial displacements are respectively

$$\begin{aligned} \delta ^{q}\rho= & {} -\left( \rho \delta q^{i}\right) _{,i} \\ \delta ^{q}s= & {} -\left( s\delta q^{i}\right) _{,i} \end{aligned}$$

moreover, the variations of the mass M and entropy S due to spatial displacements are

$$\begin{aligned} \delta ^{q}M= & {} \delta ^{q}\int _{D} \left( \rho \left( x^{i}\right) \sqrt{g\left( x^{i}\right) } \delta q^{l}\left( x^{i}\right) \right) _{,l} \frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\ \delta ^{q}S= & {} \delta ^{q}\int _{D} \left( s \left( x^{i}\right) \sqrt{g\left( x^{i}\right) } \delta q^{l}\left( x^{i}\right) \right) _{,l}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

Proof

By computation

$$\begin{aligned}&\delta ^{q}\left( \rho \left( x^{i}\right) \frac{1}{3!}\varepsilon _{ijk}dx^{i} \wedge dx^{j}\wedge dx^{k}\right) \\&\quad = \delta ^{q}\left( \rho \left( x^{i}+\delta q^{i}\left( x^{i}\right) \right) \right) \frac{1}{3!}\varepsilon _{ijk}d\left( x^{i}+ \delta q^{i}\left( x^{i}\right) \right) \\&\qquad \wedge d\left( x^{j}+ \delta q^{j}\left( x^{i}\right) \right) \wedge d\left( x^{k}+ \delta q^{i}\left( x^{k}\right) \right) \\&\qquad -\rho \left( x^{i}\right) \frac{1}{3!} \varepsilon _{ijk}dx^{i} \wedge dx^{j} \wedge dx^{k} =0 \end{aligned}$$

noticing that

$$\begin{aligned} \delta ^{q}\left( \rho \left( x^{i}+\delta q^{i}\left( x^{i}\right) \right) \right) = \delta ^{q}\rho \left( x^{i} \right) + \rho \left( x^{i} \right) _{,l}\delta q^{l}\left( \left( x^{i}\right) \right) \end{aligned}$$

and

$$\begin{aligned}&\frac{1}{3!}\varepsilon _{ijk}d\left( x^{i}+ \delta q^{i}\left( x^{i}\right) \right) \wedge d\left( x^{j}+ \delta q^{j}\left( x^{i}\right) \right) \wedge d\left( x^{k}+ \delta q^{i}\left( x^{k}\right) \right) \\&\quad -\rho \left( x^{i}\right) \frac{1}{3!} \varepsilon _{ijk}dx^{i} \wedge dx^{j} \wedge dx^{k}=\rho \left( x^{i}\right) \delta q^{l}_{,l}\left( x^{i}\right) \frac{1}{3!} \varepsilon _{ijk}dx^{i} \wedge dx^{j} \wedge dx^{k} \end{aligned}$$

moreover,

$$\begin{aligned}&\delta ^{q}\int _{D}\rho \left( x^{i}\right) \sqrt{g\left( x^{i}\right) }\frac{1 }{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\&\quad = \int _{D}\left( \delta ^{q}\rho \left( x^{i}\right) \sqrt{g\left( x^{i}\right) }-\rho \left( x^{i}\right) \sqrt{g\left( x^{i}\right) }_{.l}\delta q^{l}\left( x^{i}\right) \right) \frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

$\square$

Axiom 4

Let $\pi _{i}$ denote the momentum density. The variation of the form

$$\begin{aligned} \pi _{i}\left( x\right) \delta q^{i}\left( x\right) \frac{1}{ 3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

with respect to spatial displacements is zero.

Proposition 7

The variation of the momentum density with respect to spatial displacements is

$$\begin{aligned} \delta ^{q}\pi _{i}=-\delta q_{,i}^{j}\pi _{j}-\left( \delta q^{j}\pi _{i}\right) _{,j} \end{aligned}$$

(5)

Proof

By computation. $\square$

Definition 4

The velocity at time t of a point which was at $y\in D$ at time $t=0$ is given by

$$\begin{aligned} {\tilde{v}}_{t}^{i}=\partial _{t}\psi _{t}^{i} \end{aligned}$$

in the Lagrange representation.

We notice that since $\varphi _{t}=\psi _{t}^{-1}$

$$\begin{aligned} \partial _{t}\varphi _{t}^{i}=-\varphi _{t,j}^{i}v _{t}^{j}\Rightarrow v_{t}^{i}=-\left( \varphi _{t}^{i,j}\right) ^{-1}\partial _{t}\varphi _{t}^{j} \end{aligned}$$

where $v_{t}={\tilde{v}}_{t}\circ \varphi _{t}$ is the velocity in the Euler representation.

Proposition 8

The variation of the velocity $v_{t}$ by spatial displacements is

$$\begin{aligned} \delta ^{q}v_{t}^{i}=\partial _{t}\delta q_{t}^{i}+ {v}_{t}^{j}\delta q^{i}_{,j}-v_{t,j}^{i} \delta q_{t}^{j} \end{aligned}$$

(6)

Proof

By computation,

$$\begin{aligned} \delta ^{q}v_{t}^{i}-\partial _{t}q^{i}=-\delta ^{q}\left( \left( \varphi _{t,j}^{i}\right) ^{-1}\partial _{t}\varphi _{t}^{j}\right) +\partial _{t}\left( \left( \varphi ^{i},_{j}\right) ^{-1} \delta ^{q}\varphi _{t}^{j}\right) =v _{t}^{j}\delta q_{t,j}^{i}-v_{t,j}^{i} \delta q_{t}^{j} \end{aligned}$$

$\square$

1.2 Transport by flows

Definition 5

Transport by a flow $\psi _{t}$ of a function or form on X is the pullback by the inverse $\varphi _{t}$ of $\psi _{t}$. Thus,

$$\begin{aligned} \varphi _{t*}\left( f \right) =f\circ \varphi _{t} = {\tilde{f}}_{t} \end{aligned}$$

for a function. Let ${\tilde{h}}_{t}$ be a time dependent function. Then

$$\begin{aligned} \partial _{t}\left( {\tilde{h}}_{t}\circ \psi _{t}\right) =0 \end{aligned}$$

is a necessary and sufficient condition for the function h to have been transported by the flow $\psi _{t}$.

Axiom 5

The evolution of the densities of mass and entropy and of the metric is by transport of the flow.

Proposition 9

The conservative evolution of the mass and entropy densities are

$$\begin{aligned} \partial _{t}\rho _{t}= & {} -\left( \rho _{t}v_{t}^{i}\right) _{,i} \\ \partial _{t}s_{t}= & {} -\left( s_{t}v_{t}^{i}\right) _{,i} \end{aligned}$$

Proof

According to the Axiom 5

$$\begin{aligned}&\partial _{t}\left( \rho _{t}\circ \psi _{t}\det \psi _{,l}^{k}\right) \frac{1}{3!}\varepsilon _{mno}dx^{m}\wedge dx^{n}\wedge dx^{o}\\&\quad = \left( \left( \partial _{t}\rho _{t}\right) \circ \psi _{t}+\rho _{t,j}\circ \psi _{t}{\tilde{v}} _{t}^{j}+\rho _{t}\circ \psi _{t}\psi _{,j}^{i}{\tilde{v}}_{t,i}^{j}\right) \det \psi _{,l}^{k}\frac{1}{3!}\varepsilon _{mno}dx^{m}\\&\qquad \wedge dx^{n} \wedge dx^{o}=0 \end{aligned}$$

from which the result follows $\square$

Corollary 3

Thus,

$$\begin{aligned} \frac{d}{dt} M= & {} -\int _{D}\left( \rho \left( x\right) \sqrt{g}v^{i}\right) _{,i} \frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \\ \frac{d}{dt} S= & {} -\int _{D}\left( s\left( x\right) \sqrt{g}v^{i}\right) _{,i}\frac{1 }{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

Proof

By computation. $\square$

Proposition 10

The transport of the metric $g_{\mu \nu }$ and the measure $\sqrt{g}$ under $\psi _{t}$ satisfies

$$\begin{aligned} \partial _{t}g_{t00}= & {} -v_{t}^{k}g_{t00,k}\\ \partial _{t}g_{ti0}= & {} -v_{t}^{k,i}g_{tko}-v_{t}^{k}g_{ti0,k} \\ \partial _{t}g_{t0i}= & {} -v_{t}^{k,i}g_{tko}-v_{t}^{k}g_{t0i,k} \\ \partial _{t}g_{tij}= & {} -v_{t}^{k,i}g_{tkj}-v_{t}^{k,j}g_{tki}-v^{k}g_{tij,k}\\ \partial _{t}\sqrt{g_{t}}= & {} -\left( \sqrt{g_{t}}v_{t}^{i}\right) _{,i} \end{aligned}$$

Proof

The proof is similar to the proof of Proposition 4. $\square$

Proposition 11

The variation of volume in time is

$$\begin{aligned} \frac{d}{dt}\int _{D_{t}}\sqrt{g}_{t}\frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k}=-\int _{D_{t}}v_{t}^{l}\sqrt{g}_{t},_{l} \frac{1}{3!}\varepsilon _{ijk}dx^{i}\wedge dx^{j}\wedge dx^{k} \end{aligned}$$

Proof

Similar to the proof of Proposition 5. $\square$