← Back to Contents

2.4.1 Displacement and Gravity

In the following outline of the basis of gravity, it is described as a consequence of the curvature of the continuum other than that due to structural curvature. It is convenient, therefore, to assume, initially, that gravity is associated with 'displacement' of the continuum which is otherwise flat.

Fortunately, Newtonian gravity is spherically symmetrical which simplifies the problem of visualization. Our personal experience of gravity is that it is an accelerating force which exists in the space surrounding the Earth and directed towards its center. If we define distance from the Earth's center as radius, \(r\), then the varying strength of gravitational acceleration above the Earth can be represented as a system of displaced spherical shells, the smallest being at the Earth's surface. Above the Earth's surface, gravity is an inverse square phenomenon such that gravitational acceleration in the shell at radius \(r\), is proportional to the inverse square of radius, \(r\). \begin{align} Gravitational\ acceleration\ =\frac{MG}{r^2} \nonumber \end{align} where \(M\) is the central mass, \(r\) is the radius from its center, and \(G\) is Newtons' Universal Gravitational constant.

These imaginary spherical shells are concentric with the Earth's surface which is displaced from the otherwise flat continuum, so that each shell is 'pushed' into a fourth direction by an amount inversely proportional to its radius. As radius increases the displacement becomes less and less until it merges with the general curvature of the continuum. It is as if the mass of the Earth has depressed the local continuum so that it curves inwards. The slope (gradient) of this displacement is related to gravitational acceleration.

In analyzing this situation, is convenient to define a displacement function, \(N(x,y,z)\), in an xyz coordinate system having its origin at the center of the central mass. \(N\) is a scalar variable which represents the displacement of the continuum surrounding the Earth, into a fourth dimension perpendicular (normal) to the radius of each spherical shell. Using this concept, gravitational structure can be represented by only two variables, the radius, \(r\), of each point from the gravitational center and its displacement, \(N\). For simplicity, the displacement function, \(N\), can be written in terms of radius \(r\), (\(N(r)\)), where, \(r^2 =(x^2+y^2+z^2)\).

If each spherical shell is displaced by an amount, \(N\), and assuming that the gravitational acceleration is related to the gradient of \(N\), we have \begin{align} Gradient\ of\ N =\frac{\partial N(r)}{\partial r}\ related\ to\ Gravitational\ acceleration = \frac{MG}{r^2} \nonumber \end{align}

Since both gradient of \(N\) and gravitational acceleration both decrease with increase in radius \(r\), its logical to assume that they are proportional to each other. Hence, the whole gravitational structure surrounding a central mass, \(M\), can be related to displacement function, \(N(r)\). This relationship can be found via the relationship,
\begin{align}
\frac{\partial N(r)}{\partial r} \propto \frac{MG}{r^2} \ \
so\ that\ \ N(r) \propto \int \frac{MG}{r^2} \partial r\ \nonumber \\ integrating,\ \ N(r) \times constant\ = \frac{-MG}{r^2}\ r \nonumber \\ Hence\ \ \frac{N(r)}{r} \times constant = - \frac{MG}{r^2} \nonumber \end{align}

Since both \(M\) and \(G\) are constants, the simplest function for \(N(r)\) is, \(-A/r \), where \(A\) is a constant for a given mass. The presence of a minus sign denotes that displacement, \(N=-A/r\) is against the direction of expansion. The units of \(N(r)\) are length, so that the units of \(A\) will be length squared. Both \(N(r)\) and radius, \(r\), have units of length, so that the constant of proportionality will have units of acceleration. The value of this constant will be derived in the following section.

← Previous Next →