Black holes aren't totally black, and other insights from Stephen Hawking's groundbreaking work

March 16, 2018 by Christoph Adami, The Conversation
Credit: NASA Goddard, CC BY

Mathematical physicist and cosmologist Stephen Hawking was best known for his work exploring the relationship between black holes and quantum physics. A black hole is the remnant of a dying supermassive star that's fallen into itself; these remnants contract to such a small size that gravity is so strong even light cannot escape from them. Black holes loom large in the popular imagination – schoolchildren ponder why the whole universe doesn't collapse into one. But Hawking's careful theoretical work filled in some of the holes in physicists' knowledge about black holes.

Why do black holes exist?

The short answer is: Because gravity exists, and the speed of light is not infinite.

Imagine you stand on Earth's surface, and fire a bullet into the air at an angle. Your standard bullet will come back down, someplace farther away. Suppose you have a very powerful rifle. Then you may be able to shoot the bullet at such a speed that, rather than coming down far away, it will instead "miss" the Earth. Continually falling, and continually missing the surface, the bullet will actually be in an orbit around Earth. If your rifle is even stronger, the bullet may be so fast that it leaves Earth's gravity altogether. This is essentially what happens when we send rockets to Mars, for example.

Now imagine that gravity is much, much stronger. No rifle could accelerate bullets enough to leave that planet, so instead you decide to shoot light. While photons (the particles of light) do not have mass, they are still influenced by gravity, bending their path just as a bullet's trajectory is bent by gravity. Even the heaviest of planets won't have gravity strong enough to bend the 's path enough to prevent it from escaping.

But are not like planets or stars, they are the remnants of stars, packed into the smallest of spheres, say, just a few kilometers in radius. Imagine you could stand on the surface of a black hole, armed with your ray gun. You shoot upwards at an angle and notice that the light ray instead curves, comes down and misses the surface! Now the ray is in an "orbit" around the black hole, at a distance roughly what cosmologists call the Schwarzschild radius, the "point of no return."

Thus, as not even light can escape from where you stand, the object you inhabit (if you could) would look completely black to someone looking at it from far away: a black hole.

But Hawking discovered that black holes aren't completely black?

The short answer is: Yes.

No light can be seen coming from a black hole outside the Schwarzschild radius. Credit: SubstituteR, CC BY-SA

My previous description of black holes used the language of classical physics – basically, Newton's theory applied to light. But the laws of physics are actually more complicated because the universe is more complicated.

In classical physics, the word "vacuum" means the total and complete absence of any form of matter or radiation. But in quantum physics, the vacuum is much more interesting, in particular when it is near a black hole. Rather than being empty, the vacuum is teeming with particle-antiparticle pairs that are created fleetingly by the vacuum's energy, but must annihilate each other shortly thereafter and return their energy to the vacuum.

You will find all kinds of particle-antiparticle pairs produced, but the heavier ones occur much more rarely. It's easiest to produce photon pairs because they have no mass. The photons must always be produced in pairs so they're moving away from each other and don't violate the law of momentum conservation.

Now imagine that a pair is created just at that distance from the center of the black hole where the "last light ray" is circulating: the Schwarzschild radius. This distance could be far from the surface or close, depending on how much mass the black hole has. And imagine that the is created so that one of the two is pointing inward – toward you, at the center of the black hole, holding your ray gun. The other photon is pointing outward. (By the way, you'd likely be crushed by gravity if you tried this maneuver, but let's assume you're superhuman.)

Now there's a problem: The one photon that moved inside the black hole cannot come back out, because it's already moving at the speed of light. The photon pair cannot annihilate each other again and pay back their energy to the vacuum that surrounds the black hole. But somebody must pay the piper and this will have to be the black hole itself. After it has welcomed the photon into its land of no return, the black hole must return some of its mass back to the universe: the exact same amount of mass as the energy the pair of photons "borrowed," according to Einstein's famous equality E=mc².

This is essentially what Hawking showed mathematically. The photon that is leaving the black hole horizon will make it look as if the black hole had a faint glow: the Hawking radiation named after him. At the same time he reasoned that if this happens a lot, for a long time, the black hole might lose so much mass that it could disappear altogether (or more precisely, become visible again).

Do black holes make information disappear forever?

Short answer: No, that would be against the law.

Many physicists began worrying about this question shortly after Hawking's discovery of the glow. The concern is this: The fundamental laws of physics guarantee that every process that happens "forward in time," can also happen "backwards in time."

A pair of photons that annihilate each other is labeled A. In a second pair of photons, labeled B, one enters the black hole while the other heads outward, setting up an energy debt that is paid by the black hole. Credit: Christoph Adami, CC BY-ND

This seems counter to our intuition, where a melon that splattered on the floor would never magically reassemble itself. But what happens to big objects like melons is really dictated by the laws of statistics. For the melon to reassemble itself, many gazillions of atomic particles would have to do the same thing backwards, and the likelihood of that is essentially zero. But for a single particle this is no problem at all. So for atomic things, everything you observe forwards could just as likely occur backwards.

Now imagine that you shoot one of two photons into the black hole. They only differ by a marker that we can measure, but that does not affect the energy of the photon (this is called a "polarization"). Let's call these "left photons" or "right photons." After the left or right photon crosses the horizon, the black hole changes (it now has more energy), but it changes in the same way whether the left or right photon was absorbed.

Two different histories now have become one future, and such a future cannot be reversed: How would the laws of physics know which of the two pasts to choose? Left or right? That is the violation of time-reversal invariance. The law requires that every past must have exactly one future, and every future exactly one past.

Some physicists thought that maybe the Hawking radiation carries an imprint of left/right so as to give an outside observer a hint at what the past was, but no. The Hawking radiation comes from that flickering vacuum surrounding the black hole, and has nothing to do with what you throw in. All seems lost, but not so fast.

In 1917, Albert Einstein showed that matter (even the vacuum next to matter) actually does react to incoming stuff, in a very peculiar way. The vacuum next to that matter is "tickled" to produce a particle-antiparticle pair that looks like an exact copy of what just came in. In a very real sense, the incoming particle stimulates the matter to create a pair of copies of itself – actually a copy and an anti-copy. Remember, random pairs of particle and antiparticle are created in the all the time, but the tickled-pairs are not random at all: They look just like the tickler.

This copy process is known as the "stimulated emission" effect and is at the origin of all lasers. The Hawking glow of black holes, on the other hand, is just what Einstein called the "spontaneous emission" effect, taking place near a black hole.

Now imagine that the tickling creates this copy, so that the left photon tickles a left photon pair, and a right photon gives a right photon pair. Since one partner of the tickled pairs must stay outside the black hole (again from momentum conservation), that particle creates the "memory" that is needed so that information is preserved: One past has only one future, time can be reversed, and the laws of physics are safe.

In a cosmic accident, Hawking died on the birthday of Einstein, whose theory of light, it just so happens, saves Hawking's theory of black holes.

Explore further: Black holes dissolving like aspirin: How Hawking changed physics

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gculpex
1 / 5 (2) Mar 17, 2018
Thank you.
julianpenrod
5 / 5 (1) Mar 17, 2018
A demonstration of an engineered fame.
Among other things, black holes, as defined, are black. Stephen Hawking did not depict them to emit. He said photons near, but not on or in the event horizon would spontaneously divide into particle-antiparticle pairs. He said only those with charges opposite the black hole would enter it, couple would opposite charges in the black hold, causing it to lose mass. Among other things, this is an illusion with some facets of mass loss and radiation as viewed from this universe, based on actions in this universe! Close enough to the event horizon, the gravity can be strong enough to pull both charged particles in. It is said that, from the point of view of this universe, particles take infinitely long to pass the event horizon so no effect would be seen. And who know how particles behave in the universe inside the black hole?
granville583762
1 / 5 (1) Mar 17, 2018
Gravities Light Radius R=2GM/C*

A black hole has an escape velocity of speed of light and a particle travelling at the speed of light travels an infinitely long distance before Gravity brings it to a halt. The implication, a particle inside the light radius is able to travel out the light radius and travel an infinitely long distance before gravity brings it to a halt. A black hole has the escape velocity of the speed of and no greater and gravity travels at speed of light and consequently can only compress matter to the escape velocity of light and only has a gravitational escape velocity of the speed of light which is gravities light radius
granville583762
1 / 5 (1) Mar 18, 2018
Gravity is zero at its centre of mass

A black hole is defined by R=2GM/C* where gravities velocity G is accurately confirmed by LIGO in GW170817! Gravity only compress matter to speed of light and no greater where its Light Radius R is proportional to its mass R=2GM/C* where gravity falls to zero at its centre of mass (exactly as a transport system through the centre of the earth to Australia where gravity is zero at the centre of earth)

As a singularity is an infinitesimally small radius, and gravity is zero at the centre of mass because gravity only compress's matter to the speed of light, it cannot squeeze matter smaller than R=2GM/C*.

This is borne out in the Universe we inhabit has an escape velocity of the speed of light where it has a 15billion Light Year Radius! A black hole with a 15billion light year radius singularity!
mackita
1 / 5 (1) Mar 18, 2018
Black Holes Could Actually Be Bizarre Quantum Stars. As such they of course violate general relativity, on which black hole theory is based...
mackita
not rated yet Mar 18, 2018
Stephen Hawking did not depict them to emit. He said photons near, but not on or in the event horizon would spontaneously divide into particle-antiparticle pairs. He said only those with charges opposite the black hole would enter it, couple would opposite charges in the black hold, causing it to lose mass.
That's correct, in Hawking's way the black holes could never radiate any energy by their very own. At the end of life he finally switched his opinion on behalf of self-radiating black holes, though.
mackita
not rated yet Mar 18, 2018
See also Vacuum polarization in Schwarzschild spacetime and Stellar equilibrium in semiclassical gravity for context.
Conservative physicists already protest indeed: "It's both mathematical nonsense & in conflict with the equivalence principle."
Of course, that every model which violates black hole model MUST also violate general relativity, i.e. the postulates on which this theory is based. There's no other way around. After all, quantum mechanics violates equivalence principle from its very ground, as it predicts nonzero acceleration and momentum for every massive body constrained in motion from uncertainty principle. Some dark matter theories like MOND already account to it.
mackita
1 / 5 (1) Mar 18, 2018
In addition, quantum mechanics predicts only repulsive forces, whereas in general relativity the gravity is always repulsive force. So that at the moment, when you incorporate some quantum mechanical effect into general relativity based theory, then the result MUST violate general relativity postulates or you got something wrong in your derivation.

Ironically the author of comment spent whole her in/productive life by development of quantum gravity "phenomenology" - so that one would expect, that she should know already something about it - yet she still gets surprised with it again and again. This is realistic sample of thinking of mainstream physics community today.
mackita
1 / 5 (1) Mar 18, 2018
The truth being said, the string theorists are whining once some theory predicts violation of Lorentz symmetry instead (despite every quantum gravity theory MUST violate it too from the same reason). In this way the proponents of mainstream theories dismiss and downrate their own ideas mutually, despite that both they're inventing the basically same untestable stuffs. Occasionally these frog&mice battles would be even entertaining to watch, if only we - tax payers - wouldn't sponsor whole this fun from our own pockets.

granville583762
not rated yet Mar 19, 2018
Blackholes have observational data – GW170817 and spin-axis outflows and accretion disk Quasars and……..

Why can you not use Karl Schwarzschild light radius and gravity to explain Blackholes and their spin axis outflow properties to explain Blackholes, Gravity attracts every other particle as Isaac Newton laws of gravity explains why gravity is zero at the centre of mass of black holes!

mackita:- The truth being said, the string theorists are whining once some theory predicts violation of Lorentz symmetry instead (despite every quantum gravity theory MUST violate it too from the same reason). In this way the proponents of mainstream theories http://gurumagazi...-theory. Occasionally these frog&mice battles would be even entertaining to watch, if only we - tax payers - wouldn't sponsor whole this fun from our own pockets.


granville583762
not rated yet Mar 19, 2018
mackita:- A blackhole is so called according to Karl Schwarzschild light radius escape velocity R=2GM/C*

In other words it is a Star with an escape velocity of the speed of light and no greater which is nature absolute limit of the universe the speed of light C which is why Karl Schwarzschild used it in formula and George's Lemaitre used it in his cosmic egg which is a Blackhole!
granville583762
not rated yet Mar 19, 2018
mackita:- Even at natures limit at the speed of light gravity behaves exactly the same in a Blackhole as it does when the proverbial Apple fell on Isaac Newton's head!

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