So say I have a black hole and have setup six measurement stations away from the black hole event boundary to measure Hawking Radiation.
I have taken this picture from the internet to represent the setup. The center red dot in the picture is the black hole, and each of the six points (East, West, North, South, Zenith and Nadir) represent the location of the measurement stations.
Say at time t = 0, the black hole absorbs a body (e.g. star) at the zenith point.
Will all the six measurement stations show an equal rise in Hawking Radiation at the same time ?
There is no such thing as “same time” for events separated in space. Events that are simultaneous in one frame of reference will not be simultaneous in another.
Hawking radiation is a different thing than the radiation emitted by matter falling in to the black hole.
If you had a black hole big enough to absorb a star, the amount of Hawking radiation it would emit would be so low that it would be swamped by the cosmic background radiation.
Hawking radiation is when a virtual particle pair pops into existence out of nothing. Usually they’d mutually annihilate each other. But what if one of those particles crossed the black hole event horizon while the other escaped?
But stellar mass black holes aren’t currently evaporating due to Hawking radiation, because they are absorbing more particles from the cosmic background radiation than they emit by Hawking radiation.
Okay fine. Say I have a recorder at all the six measurement stations and they measure they record the time when the black hole absorbs the body at the time rate of Hawkins Radiation change. Years later, I collect the data and compare them. Do they show the same rate of increase of the radiation ?
Agree to everything you say but its not relevant to the OP.
The event horizon of a black hole is representative of the temperature or entropy of the black hole (is that understanding correct ?). So, when a body (star or anything else) enters a black hole, its event horizon expands. My question is - does this expansion happens smoothly and instantly or is it more wobbly like when a small drop of water joins another large drop ?
In case of the water drops joining - the disturbance on the bubble surface travels at the speed of sound governed by the surface tension.
Whats an equivalent mechanism for the black hole event horizon ? To me, this is the equivalent of heat transfer on the horizon.
No they don’t because absorption of matter doesn’t increase Hawking radiation.
The amount of Hawking radiation emitted depends only on the size of the event horizon. And the size of the event horizon depends on the mass of the black hole. And since when matter crosses the event horizon the mass of the black hole increases, the size of the event horizon increases. That means less Hawking radiation, not more.
Are you trying to ask: if we increase the temperature of the black hole by dumping in a hot star, how fast can we measure the change in that temperature increase?
The answer is that a black hole’s temperature is only due to it’s size. And the bigger it is, the lower the temperature. A black hole of stellar mass has a temperature much much lower than the 3K background radiation, which means it is absorbing more energy than it radiates.
But, I am asking, if the reduction in radiation is the same at all sides with no time effects. To put it in more “relatable” terms - does the event horizon behave more like a stretched membrane which stretches/deforms when the mass is absorbed or is the event horizon totally something different.
So to put it in more “understandable” terms. Say I have two 1 ft sphere’s - one made of jello and another of ballistics gel - both suspended by a thread from the ceiling. Next I throw in a 1/2 inch ball bearing in each one of them. The surface of the ballistics gel vibrates and deforms differently than that of the jello. It depends on the elasticity / other physical properties of the jello / gel.
How does the surface (event horizon) of the black hole behave ? Does it instantly expands in all directions ? Are there vibrations ? (I am guessing that these perturbations will show up as measurement changes of the Hawking radiation assuming it could be measured)
Well, the event horizon isn’t a physical thing. As has often been said, if you’re falling into a black hole and cross the event horizon you won’t notice anything different.
The event horizon just means that events inside the horizon will never influence events outside it, because they’d have to travel faster than light to do so.
Agreed - and hence the measurement of Hawking radiation which is a real thing. So, back to the OP, if Hawking radiation is measured in orthogonal directions, will we see different behaviors in different directions ?
Before you can say “same time” you have to say how you went about synchronizing your clocks. It’s less trivial than you may think.
And in any case, what you would detect would be the black hole getting colder, not hotter, as it absorbs mass.
Let’s say that you do have synchronized clocks at equal distances from the black hole, ignoring that both of those presumptions kinda make relatively mad. You would detect a decrease in the temperature of the hawking radiation (ignoring for the moment that these are temperatures that require an unreasonable number of zeros between the . and the first significant digit), in the direction that the mass entered first. But, in going along with the unreasonable things to detect, the temperature would go down by something pretty tiny, and the effect would last for periods of time that are rough to measure. The event horizon will smooth out pretty much at the speed of light.
That’s the hard part to understand about relativity. The speed of causality is always the same, c. If that means your measurements of time, space, and dimension have to stretch like taffy, so be it.
Like so many other aspects of black holes, the event horizon smoothing out quickly is a consequence of the No-Hair Theorem. A black hole at equilibrium can be described using only a handful of parameters: Its mass, its angular momentum, its electric charge, and its magnetic charge. Any way which a black hole is out of equilibrium will approach equilibrium on timescales comparable to its mass (i.e., the time it would take for light to propagate across it). This means that if you have two black holes with the same mass, angular momentum, and charges, they are identical, or at least will very quickly become identical. If a black hole looked different where it had absorbed a star, that would be “hair” that would distinguish it from other black holes.
A cite about what part, that the event horizon will smooth out rapidly? Here’s this.
But no, light doesn’t slow down, or speed up, it can only go at light speed (in a vacuum).
If you were sitting right at the edge of the event horizon, and accelerating at some ungodly amount as you slowly dipped in, you would, in a sense, be racing the photons that were trying to escape, and if you could actually see the photon, and your time was not being dilated by the extreme acceleration that you are under, then sure, the photon would seem to be hovering there right at the event horizon, but you have to break at least two laws of relativity to accomplish that, maybe more, so it’s really not a good visualization.
ETA: Ninja’d by chronos, had a customer for the last 20 minutes.
Thank you Chronos and k9bfriender. I read the Wikipedia link for the no-hair phenomenon.
However, the No-hair theory is for the equilibrium state of the black hole not how fast it gets to equilibrium (or did I read it wrong ?). For example - if I have two containers one with 12 lbs of Carbon + 32 lbs of Oxygen and a second container with 44 pounds of CO2, they will both get to the same equilibrium concentration (at the same temperature and pressure ) per Gibbs free energy minimization. However getting to this equilibrium will take several years.
As per Chronos above, the equilibrium state will propagate at c. It can’t go faster than that because c is the speed of causality, but if it went slower than that the black hole would have hair.
It might turn out that the no-hair theory is wrong for some reason.
