If light cannot escape from a black hole, how do we detect its presence? I mean there is nothing that can be faster than light so if light can not escape from the black hole there should be nothing that can escape from it.
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1$\begingroup$ In some cases, we can take pictures of them - or at least the accretion disk and other gases surrounding them. $\endgroup$– controlgroupCommented Nov 14 at 15:39
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1$\begingroup$ We could not easily detect a black hole, but astronomers can. Just search the web, for example: 'duckduckgo.com/…' $\endgroup$– my2ctsCommented Nov 14 at 17:04
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$\begingroup$ Is this a duplicate of physics.stackexchange.com/questions/408/… $\endgroup$– ProfRobCommented Nov 14 at 22:51
3 Answers
Actually, detecting a black hole has many ways to be done.
If material is falling into a black hole, it travels at such high speeds that it glows very brightly due to temperature conditions and can be detected.
We can also find black holes by watching the movements of visible objects around them. For larger black holes, their pull has a strength enough that nearby stars will orbit around them, so we can look for stars behaving strangely around a patch of “empty” space. From this, we can calculate exactly how heavy that black hole must be. That is how Nobel Prize winner Andrea Ghez and her team detected the supermassive black hole at the center of our own galaxy.
Furthermore, we could detect black holes by detecting the ripples in spacetime created when two of them crash into each other. From that signal, we can tell how massive the black holes were, how far away they were, and how fast they were traveling when they collided.
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3$\begingroup$ How about gravitational lensing of objects behind the black hole? $\endgroup$– BarmarCommented Nov 15 at 15:30
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2$\begingroup$ Yes, gravitational lensing can also show the presence of a black hole even if it is not actively emitting radiation. $\endgroup$ Commented Nov 15 at 15:49
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1$\begingroup$ You can also find them in the form of Quasars. Rather then just watching material fall in this black hole forms an electromagnetic gun by accelerating charged particles as they spiral in. Accelerating charged particles will make them emit electromagnetic radiation perpendicular to the spiral. If that happens to be pointing at you a black hole can actually be the cause of one of the brightest lights in the sky. $\endgroup$ Commented Nov 16 at 7:38
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$\begingroup$ @candied_orange Quasars and other active galactic nuclei produce light primarily via thermal emission from gas orbiting in an accretion disk around the black hole. The gas is heated by friction (gas at smaller radii orbits faster than gas at larger radii); gas at smaller radii is hotter. While there are other emission mechanisms involved, it is mostly not "accelerating charged particles". $\endgroup$ Commented Dec 8 at 9:57
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$\begingroup$ @PeterErwin that argument fails to explain the directionality of the gun. All you're saying is that things glow when they get hot. Sure, but a hot iron doesn't care where you stand when you look at it. Why do quasars? $\endgroup$ Commented Dec 8 at 13:23
In some cases, we can detect the orbital motion of an object near the black hole. For a distant black hole, the distance the object moves back and forth will be too small to see. But we can detect a periodic Doppler shift as it moves toward and away from us.
In other cases, a star is near enough that tidal forces tear it apart and suck in the debris. This forms a ring around the black hole. The debris speeds up to near light speed as it approaches the black hole. Different parts of the debris violently rubbing and bumping into each other heat it to millions of degrees. It glows very brightly.
You are right that a black hole as such is almost impossible to detect. The black hole itself does not radiate anything much except Hawking radiation which is "darker" than the ubiquitous cosmic microwave background. A nearby large back hole, if one could withstand the spacetime distortions, would be a "dark patch" in front of the faintly "glowing" (in the microwave range) background.
Another way large black holes betray their existence is by distorting light, even acting as a cosmic magnification glass when the stars align just right. But this effect is visible for large black holes only and needs scrutiny to detect.
Yet another detectable "direct" effect is the spacetime vibration caused by collisions of black holes. It is a bit like being inside the pool when somebody butt-dives into it on the other end. But we have only been able to detect these vibrations for less than a decade (and the Caltech and MIT guys got a Nobel prize for it, like, immediately).1
Because a black hole as such (and the reason for this repeated caveat is below) is practically invisible there is even speculation whether there is a much larger number of them than generally assumed.
The reason I say that black holes "as such" are invisible is that there is a class of black holes which make their existence unmistakably known: They are quasars, the brightest objects ever, by far. They spit out radiation and matter at relativistic speeds over many lightyears, enough to disturb their host galaxies and put their imprint on entire cosmic neighborhoods. You are right: The light and matter in these jets does not emanate from within the Schwarzschild radius, that would be impossible. It emanates form the extreme conditions around the black holes, which fuel extreme interactions with and between matter in its vicinity. The black hole is a bit like your blender when you forget to put the lid on: It ruins the entire kitchen, the more is in it, the merrier. The blender as such isn't doing anything, really ;-).
But when it has things to shred, it becomes very visible.
1 Well, almost immediately after their successful experiment. But that was the result of decades of preparation.
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$\begingroup$ Can you please explain how a quasar can be a black hole if it is the brightest object ever; i.e., seen? Thank you. $\endgroup$– DDSCommented Nov 17 at 2:57
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1$\begingroup$ @DDS As I said, what can be seen is not the black hole proper but its surroundings (and, sometimes, the jets of plasma these surroundings produce). If this is what you are aiming at: Indeed, nobody has "seen" a black hole yet because they are "black". But whether you see a car that hits you is irrelevant. $\endgroup$ Commented Nov 17 at 5:04