An infinite plane with a finite thickness creates a uniform gravitational field, and the math shows that $g=2 \pi G \sigma h$ ($G$ is gravitational constant, $\sigma$ is plane density, $h$ is thickness). Since gravity bends the path of light, any light emitted from the surface will bend back toward the plane. Therefore, someone standing on the plane would see the surface in the sky. What I wish to know is the exact shape of the path this light takes (preferably in an equation I can put on my graphing calculator). I know very little physics, and my highest level of math is Calculus 1. The rest of my knowledge is self-taught from the Internet. I've looked at this from a Newtonian perspective and assumed a parabolic path. I've tried looking into Rindler coordinates and compared this to a constant accelerating observer. I've asked physicists what the path is and was told it's not simple but does indeed bend down. What exactly do I see from the surface?
1 Answers
To calculate light trajectories under gravitational fields, you must use Einstein's General Relativity, so the classical solution $g=2πG\sigma h$ is meaningless here. Making a quick search, I have found a paper that solves Einstein's equations for an infinite plane without thickness (I guess this solution is accurate for you, with $h \ll 1$), but to solve them for a finite plane, you might need numeric calculations.
They calculate the metric and the trajectories for a massive particle fixing $x$ and $y$. This is not what you are searching for, but it is an initial step. Calculating the light trajectory is not trivial; you must find the geodesic of the metric. If you want to give it a try, the metric they find is:
\begin{equation} \mathrm{d}s^2 = e^{\displaystyle2g|z|}(-\mathrm{d}t^2+\mathrm{d}z^2) + \mathrm{d}x^2 + \mathrm{d}y^2 \end{equation}
where $g = 2\pi G \sigma$. To find the light trajectories, set $\mathrm{d}s^2 = 0$ and solve the equation (not trivial).
