\({v}_{r}(x)=c\sqrt{1-e^{-\cfrac{x}{r}} }\)
The perpendicular component is unaffected by gravity. If we let an initial approach of -45 degrees at light speed,
\({v}_{op}\) = \(c.cos{\cfrac{\pi}{4}}\) = 150 scaled units
\({v}_{or}\) = \(c.sin{\cfrac{\pi}{4}}\) = -150 scaled units
Then the path of such a photon is given by,
\(\triangle S = \sqrt{({v}_{op}\triangle t)^2 +[({v}_{r}(x)+{v}_{or})\triangle t]^2 }\)
\(\triangle S = \sqrt{({v}_{op})^2 +[({v}_{r}(x)+{v}_{or})]^2 }.\triangle t\)
Integrating the above curve will involve both \(t\) and \(x\). Instead the integration of \({v}_{r}(x)\) is plotted on a graph, where the x-axis is traces out the path of a constant velocity component ie. a line \(y\) = \(c\) represent a velocity \(c\) in the x-axis direction and traces out a path \(c.x\), \(x\) here represent time. And the integration of a curve \(f(x)\) represent the distance traveled by the vertical velocity component spread along the x-axis, through time. That is to say, the path of the velocity equation scaled in the x-axis which is now the time-axis also.
The actual equation plotted are, 300*(1-e^(-x/6.3))-150-225 in blue, the plot has been shifted down by 225. The new reference axis is plotted yellow; the integral of (300*(1-e^(-x/6.3))-150)/100 in red.
The blue curve shows that the radial velocity component of an approaching photon actually reverses direction. The red curve shows the path of the same photon bouncing out back into space. Where the radial velocity component is zero (yellow line and blue curve intersect) the red photon path reaches its lowest point.
