Nature of the Inner Gravitational Field

The physical interpretation of the function $\nu(r)$ is that the proper time interval at the radial position $r$, between two events occurring at the same spatial point, is given by $d\tau=\exp[\nu(r)]dt$, where $dt$ is the time interval between the two events as seen at spatial infinity, where spacetime is flat. If we consider a photon traveling in the radial direction, either inwards or outwards, this means that the proper frequency $f(r)$ of the photon changes with position, between a first point $r_{a}$ and a second point $r_{b}$, according to

$$f(r_{a}) = \,{\rm e}^{-\nu(r_{a})}f_{\infty},$$ (72)

$$f(r_{b}) = \,{\rm e}^{-\nu(r_{b})}f_{\infty},$$ (73)

where $f_{\infty}$ is the frequency of the photon at radial infinity. Dividing these two equations and making the two points very close together, so that $r_{a}=r$ and $r_{b}=r_{a}+\delta r$, we have

  $$\frac{f(r+\delta r)}{f(r)} = \,{\rm e}^{-[\nu(r+\delta r)-\nu(r)]}.$$ (74)

For sufficiently small $\delta r$ we may write the variation of the function $\nu(r)$ in terms of its derivative $\nu'(r)$, so that we get

  $$\frac{f(r+\delta r)}{f(r)} \simeq \,{\rm e}^{-\delta r\,\nu'(r)}.$$ (75)

Since the energy $hf(r)$ of a photon, $h$ being the Planck constant, is proportional to its frequency, we have an interpretation of the red and blue shifts of the frequency of the photons as decreases or increases in their energies, respectively. We thus observe that, if a photon is going outward, so that $\delta r>0$, and if the derivative $\nu'(r)$ is positive, then we will have that $f(r+\delta r)<f(r)$, and therefore a red shift in the frequency. If it is going outward but the derivative is negative, then we will have that $f(r+\delta r)>f(r)$ and hence a blue shift. On the other hand, if the photon is going inward, so that $\delta r<0$, and the derivative is positive, then we will have a blue shift, and finally, if it is going inward and the derivative is negative, then we will have a red shift. Let us write down the derivative of $\nu(r)$ in the inner and outer vacuum regions,

$$\nu'(r) = \left\{ \begin{array}{rcl} -\, {\displaystyle\frac{\displaystyle ... ...displaystyle r(r-r_{M})}} & \mbox{for} & r_{2}\leq r<\infty. \end{array}\right.$$ (76)

Let us now consider the consequences of Equation (75) in more detail in each one of these two regions, starting with the outer vacuum region. As one can see above, in the outer vacuum region, since we have that $r>r_{2}>r_{M}>0$, the derivative $\nu'(r)$ is always positive. Therefore, photons traveling outward undergo red shifts, while those traveling inward undergo blue shifts. This can be interpreted in energetic terms as the statement that when traveling inward the photons gain energy from the gravitational field, and when traveling outward they lose energy to it. This is characteristic of a gravitational field that is attractive towards the origin.

However, in the inner vacuum region the situation is reversed. Since we have that $r_{\mu}>0$, the derivative is everywhere negative in that region. This means that photons traveling outward within this region are blue shifted, and therefore gain energy from the gravitational field, while photons traveling inward within this region are red shifted, and therefore lose energy to the gravitational field. This is characteristic of a gravitational field that is repulsive, driving matter and energy away from the origin. This is the exact opposite of what happens in the outer vacuum region. It is important to note that this repulsion is not from the matter in itself, but from the origin, consisting therefore of an outward attraction towards the shell of matter.