Gravity around a black hole
\(g=-g_{eo}e^{-\cfrac{1}{r_{eo}}x}\)
where,
\(g_{eo}\) = 5.07071e18 ms-2 and \(r_{eo}\) = 0.00886 m
If we approximate this gravity curve at \(x=0\) with a straight line, we have for small \(x\), \(\Delta x\),
\(\Delta g=\cfrac{g_{eo}}{r_{eo}}e^{-\cfrac{1}{r_{eo}}(x=0)}\Delta x=\cfrac {g_{eo}}{r_{eo}}\Delta x\)
From the posts "In, Up, Out and Away..." and "A Right Turn, A Wrong Turn" we know that photons are repelled away from strong gravitational region, as such
\(\Delta g=-\cfrac {g_{eo}}{r_{eo}}\Delta x\)
If we compare this to Hooke's Law,
\(F=ma=-kx\) or \(a=-\cfrac{k}{m} x\)
We find that, for small \(x\), \(\Delta x\)
\(\cfrac{k}{m}=\cfrac{g_{eo}}{r_{eo}}\)
That is to say, photons around a black hole or massive planet with high gravitational field have a resonance frequency of,
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{1}{2\pi}\sqrt{\cfrac{g_{eo}}{r_{eo}}}\)
At some nominal surface gravity \(g\), as \(\Delta x\) increases \(\Delta g\) increases in the opposite direction, some photons fall back with less acceleration, as \(\Delta x\) decreases \(\Delta g\) decreases in the opposite direction and the same photons accelerate forward again. The result is photons cluster into groups or packets.
For the case of a black hole,
\(f\)=1/(2pi)*(5.07071e18/0.00886)^(1/2) = 3.807e9 Hz = 3.807 GHz
For any other planet,
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{g_{p}}{r_{p}}}\)
where \(g_p\) is the planet surface gravity and \(r_p\) the planet radius.
In the case of Earth,
\(f\)=1/(2pi)*(9.80665/6371000)^(1/2) = 0.000197 Hz or 1 pulse every 5064.3 s
This frequency is not to be confused with light frequency but pulses of photons that escapes the pull of a strong gravitational field. It is possible that such pulses be detectable with our naked eye given the right values for high \(g_p\) and small \(r_p\).
Twinkle, Twinkle Little Stars.
