I posted this video on youtube on January 28, 2020 showing how the inverse square law really is an extremely good approximation of a geodesic on a Schwarzschild Metric:
https://www.youtube.com/watch?v=KUKcp_2rd-Y
The graphs I use in the video are: (dy/dx)^2 = (2*3.985*10^14)/y for the inverse square law and (dy/dx)^2= (1-0.0089/y)^2*(299,000,000^2 – 299,000,000^2*(1-0.0089/y)) for the Schwarzschild metric geodesic
For the zoom of the graph, I first use the ranges:
X range is -6K to 6K
Y range is 0 to 1.4 x 10^7
Because the distance, Y is measured from the center of the
earth instead of the earth’s surface, and earth’s radius is pretty big, 6.371
million meters.
I also zoom WAY in close later in the video,
X range is -5 to 8
Y range is 0.0079 to 0.0099
To show the graphs aren't EXACTLY the same.
My general strategy was to take the inverse square law d/dx of dy/dx = k/y^2 where x is time and y is distance from the center of gravity, and turn it into a graph. Then I took the relevant part of the Schwarszchild metric, ds^2=(1-rs/r)c^2*dt^2
– dr^2/(1-rs/r), and tried to figure out how to do a geodesic on it.
Going to ideas from Sam Lilley’s book, it gave me a
kind of formula for what a geodesic on a metric is. If you draw a graph
with a r axis and a t axis, you can think of dr and dt as infinitesimal
distances on this graph. On a graph
where r and t would be flat spacetime, dr and dt would be equal
everywhere. But warped space has varying
dr and dt at different points.
We can think of ds as the hypotenuse of a triangle
completing the distance moved dt in the t direction and then moved dr in the r
direction. This isn’t entirely accurate with
the Schwarzschild metric because I think ds is actually smaller than either dt
or dx here, but it’ll at least give us a rough idea of what’s going on. If we take this picture here, and imagine
that ds is ALWAYS proportional to dt, no matter how long or short dt is, we’ll
get a geodesic – a line that doesn’t bend to the left or right as you move
across it. So we just need ds to be
equal to some constant times dt and plug that into our metric to get a geodesic
on that metric.
So the next part of the strategy was to set
ds equal to a constant B times (1-rs/r)dt.
Then plug it into the metric.
After messy algebra, the equations for the graphs used in the video popped out.
