The event horizon is the boundary surrounding a black hole from which not even light can escape. There's a certain negative potential energy level associated with the event horizon. Now, some shorter-wavelength photons may have enough energy to escape the black hole's gravity well starting off at this level, whereas some longer-wavelength ones might not – they would have to get infinitely redshifted and more.

Does this mean that the event horizon is actually not the same for different wavelengths of light?


1 Answer 1


No, it's all the same.

GR does not care about the mass of something, nor does it care about its energy, when it calculates its path. It's always a geodesic and for massless particles it's always a lightlike geodesic. Makes not difference what its freq or energy is.

If it did when light is deflected by gravity, such as the many light deflection experiments done over the years, they would split up into its colors, I.e., it would show dispersion. It does not.

It would be possible if light or gravity is dispersive, i.e., if the speed of light c was different for each freq., or at least in non-flat spacetimes. There is an alternative to GR that says it is dispersive in the large scales of the universe, but alas, it's never been founded in measurements. Do not remember the error bounds, but they've been pretty good. See Experimental bounds on Lorentz-violating dispersion relation See also https://en.m.wikipedia.org/wiki/Variable_speed_of_light for a bunch of variable speed of light theories, none of which have had any support from measurements.


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