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2.4.5 Gradient and Curvature in Spherical Coordinates

It has already been shown that a gravitational system is essentially a region of the continuum which has been displaced in a fourth direction by a function, \(N(r)=-A/r\). This displacement subtracts from the value of the 'quasi-flat' vector radius, \(R_o\), such that vector radius, effectively becomes, \((R_o - A/r) = (R_o + N(r))\). Since \(R_o\) is constant, that is vector radius of the quasi-flat continuum, the gradient and curvature depend only on \(N(r)\). Hence, \begin{align} gradient\ = \frac{\partial N(r) }{\partial r} = \frac{\partial}{\partial r} \frac{(-A)}{r} =\frac {A}{r^2} \nonumber \end{align}

\begin{align} curvature\ = \nabla^2 N(r) = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \frac{\partial N(r)}{\partial r}) = \ \frac{\partial^2 N(r)}{\partial r^2} + \frac{2}{r} \frac{\partial N(r)}{\partial r} \nonumber
\end{align}

Since \(N(r) = -A/r\), we have that,
\begin{align}
the\ first\ term\ (radial\ curvature) =&\ \frac{\partial^2}{\partial r^2} (\frac{-A}{r}) = \frac{-2A}{r^3} \nonumber \\
and\ the\ second\ term\ (spherical\ curvature)\ =&\ \frac{2}{r} \frac{\partial}{\partial r} (\frac{-A}{r}) = \frac{2A}{r^3} \nonumber \\
Hence\ \frac{\partial^2 N(r)}{\partial r^2} + \frac{2}{r} \frac{\partial N(r)}{\partial r} =&\ 0 \nonumber \end{align}

This is a remarkable result. Even though gravitation structure is obviously curved in the usual sense of the word, its mathematical definition as Euclidean curvature shows curvature to be zero for all spherically symmetrical functions of the kind, \(A/r\), for all values of \(A\). This occurs because the negative curvature along the radius is canceled by the positive curvature of the spherically symmetrical gravitational structure.

On the other hand, the gradient of gravitational structure along the radius is \(A/r^2\), while the gradient perpendicular to radius at each point is zero. Hence, gravity acts centripetally, along the radius, with the inverse square law, \(A/r^2\). As already noted, gravity only acts in the region above the central core of matter, and the parameter, \(A\), is proportional to its mass. This means that mass of any size will displace the continuum in such a way that gradient becomes an inverse square function while Euclidean curvature is not affected.

Given Einstein's intuition that gravity is associated with curvature of space, it is perhaps ironic that the spherical displacement function, \(-A/r\), which has a gradient proportional to gravity, has zero Euclidean curvature. In fact, spherically symmetrical displacement with zero curvature is uniquely associated with gravity, and since curvature is also strongly associated with energy density, this function can be thought of as one that minimizes energy.

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