We can describe light not only as electromagnetic waves, but
also as particles (called photons). What happens if a particle flies
by a very massive star? - You're right, it is pulled in by the star's gravitation.
Does this happen to light too? Is light affected by gravitation?
The scientists already thought about this problem way back in the 18th century. We can already find one of those thoughts in "The Philosophical Transactions of The Royal Society of London" (Vol. XV) in a work called "Distance and Magnitude of the Stars" by John Mitchell (written 1784):
"Let us now suppose the particles of light to be attracted in the same manner as all other bodies with which we are acquainted; that is, by forces bearing the same proportion to their vis interiae; of which there can be no reasonable doubt, gravitation being, as far as we know, or have any reason to believe, a universal law of nature." (Page 466)
What do you think happens if a shining body has an enormous mass? You might have the same thoughts as Peter Simon La Place*:
"A shining star with the same density as the earth, which diameter would be 250 times larger than the diameter of the sun, would let none of its light come to us due to its attraction. It is therefore possible that the biggest shining bodies of the universe are not visible for this reason."
This was probably the first description of a black hole.
A black hole is a body which has such a great mass that not even light can escape from it due to it's gravitation. This strong gravitation creates another problem because it also acts on the black hole itself. The gravitational force is strong enough that even the subatomic particles cannot withstand the pressure. The black hole collapses into one singular point. This point contains all the mass of the black hole. It is called the singularity.
The radius within which the Gravitation is so strong that nothing can escape the black hole is called the Schwarzschild radius. In the theory of relativity, time stops at this distance. Because nothing can reach us from beyond the Schwarzschild radius, it's also called the event horizon.
Normally a black hole is created when a huge star has burned all of its fuel and the heat cannot withstand the pressure of gravitation any longer. Tiny black holes might have been created during the big bang.
Translated from the German
version into English. The German version was translated from the French
original by Johann Karl Friedrich Hanff in 1797. The Quote is from:
P. S. La Place. Darstellung des Weltsystems, translated by J. K. F. Hanff (1797), Volume 4, page 333.
"The Hole Man" is a short criminal story written by Larry Niven, which won a Hugo Award. "The Hole Man" is a short criminal story written by Larry Niven, which won a Hugo Award. It deals with a group of scientists, who discover an abandoned alien base on Mars. On this base there's a communicative device, which sents out gravitational waves using a mini black hole. Lear tries to figure out how the communicator device works. He and Captain Childrey can't stand each other. Captain Childrey doesn't believe in Lear's hypothesis that there's a mini black hole inside the alien's communicator. Finally Lear shows this fact by deactivating the electromagnetic fields, and the mini black hole drops right through Captain Childrey's body. Childrey died an hour later.
We'll use the black hole inside the communicator for example calculation throughout our web pages. We find it suitable for us to use it as a framework for our calculations illustrating various effects of a black hole.
Note: Niven wrote this story (copyright 1973) before Hawking radiation was discovered in March 1974. So he couldn't know that not only a black hole's gravitational force can be deadly, but also its radiation. "The Hole Man" is a very well researched story as our calculations will show.
[Reference: Niven, Larry. "The Hole Man." Anglog Science Fiction & Fact Magazine. (c)1973.]
The size of a black hole is the size of the Schwarzschild radius. It is also called the event horizon. Once something is within this radius, it'll gone; it will inevitably fall into the singularity at the ceter of the black hole. The derivation of the Schwarzschild radius is quite simple.
The Schwarzschild radius is where even the kinetic energy of a photon is not sufficient to escape from the black hole. The kinetic energy is given by the formula:
and the potential energy a particle has at a distance R of a black hole is given by Newton's formula:
From a classical point of view we can assume that a photon is radiated from the singularity at the center of a black hole at the speed of light and while it reaches a greater and greater distance, it's kinetic energy is slowly turned into potential energy. Nothing can get past the radius where the potential energy of a particle or photon is as great as its kinetic energy at the time it was emitted from the singularity.
It's also important to mention that the speed of light in vacuum is constant. You probably know this already. This means that a photon can't be slowed down even if it loses its energy. Is this a contradiction? No! A photon isn't slowed down, but its color shifts to the less energetic red and therefore also loses kinetic energy. Thus we can equate the two energy formulas from above:
|m: mass of the photons radiated from the singularity|
|c: speed of light;|
|G: gravitational constant;|
|M: mass of the black hole||R: Schwarzschild radius|
That was easy. Now we know how big the Schwarzschild radius is. Even though this is only a classical derivation it's result is the exact formula.
The mass of the Earth is about .
A black hole with the mass of the Earth would be less than two centimeters (just 0.7 inches) across!!
The black hole in "The Hole Man"had a mass of .
That's already the mass of a small mountain.
Well, how big (or small) would this black hole be? Another calculation:
This is much smaller than an atom!
You'll need a little bit of quantum physics for this. We'll
In quantum physics the empty space isn't empty at all. In it there are always particles flashing into existence and disappear again. They always come in pairs; one particle and one anti-particle, like an electron and a positron, or a photon (a particle of light) and another photon with opposite spin and impulse. These particles are called "virtual particles". They only exist for a very short time:
; where f is their frequency.
This follows out of the Heisenberg principle of uncertainty:
is the energy of the pair of virtual particles, and
h is the Planck constant.
(a natural constant, ).
The heisenberg Principle of Uncertainty:
The energy of one photon:
The energy of a pair of photons:
Example calculation for a photon of orange light :
(80 attoseconds, 1as is a millionth of a millionth of a millionth of a second.)
The vacuum is defined to have zero energy. Therefore yes, the virtual particles violate the law of energy preservation. But the particles disappear again directly after (e.g. eighty attoseconds for a photon of orange light) and give the energy back to the vacuum.
If one particle of a pair of virtual particles falls into a black hole and the other one doesn't they can't react back to energy and the escaping one becomes a real particle leaving behind a lack of energy in the vacuum. Somehow this "hole" in the vacuum energy has to be filled again - even in quantum physics the law of energy preservation can't be disobeyed for a longer time. So this "hole" draws energy from that black hole. But what kind of energy does a black hole without any spin or charge have? It's mass! Thus a black hole loses some of its mass after Einstein's famous formula . Maybe you have noticed that that's no proof that the energy has to originate from the black hole. In fact we can't really prove it without a lot of quantum physics (It depends on the tunnel-effect). You just have to swallow it as being true. Another, yet similar, way to explain the black hole's loss of mass is that the particle that gets sucked into the black hole gains a negative mass and thus the mass of the black hole decreases...
You've got it! If one virtual particle falls into the black hole and the other one escapes, it escapes as Hawking radiation from the black hole. This radiation shows the same allocation as black body radiation. This assumes that black holes do have a temperature too. In fact this was proven by Bekenstein in 1972, before Hawking discovered that black holes also emit temperature radiation.
As you see the radiation only seems to come from the inside of the black hole. In reality it originates just outside the black hole.
Somehow the energy for the Hawking radiation has to get out of the black hole, doesn't it? The only thing that gets beyond the Schwarzschild radius of a black hole from the inside is gravitation. Therefore Hawking radiation has to depend on the gravitation.
When something is constantly accelerated, for example by gravity, it gets kinetic energy. This energy is proportional to the mass and the distance on which it is accelerated. You might ask how a virtual photon can be accelerated, if they always move at the speed of light. Okay, they are not, but their frequency will change giving them a higher energy.
|m: mass||a: gravitational acceleration||d: distance the mass m falls|
1st: Photons always move at the speed of light, so the distance a virtual photon travels in his lifetime is given by:
We already calculated the lifetime of a virtual Photon:
Putting that formula in here we get:
Let's put this into our formula (1):
2nd: We can calculate the gravitational acceleration using Newton's gravitation formula:
|F: gravitational force|
|r: distance of the two bodies|
|G: gravitational constant|
One of the virtual photons has to fall into the black hole in order to produce Hawking radiation. Thus they must originate close to the Schwarzschild radius. As an approximation we can take the Schwarzschild radius as the distance between the pair of photons and the black hole. Remember the formula for the Schwarzschild radius?
Let's put it in here:
Now we put this into formula (2):
3rd: Now let's try to substitute m:
We know Einstein's famous formula:
The energy of a photon is (as we already used):
The energy of a pair of virtual photons therefore is:
We put this into the formula for the mass:
Putting it into our main formula (3) we get:
This should be the energy of a photon of the Hawking radiation.
A pair of virtual photons gets more energy from a small black hole than from a large one! Very surprising isn't it? But this explains why nobody has ever seen a shining black hole.
We know the Energy of a photon of this Radiation:
This doesn't tell us a lot, does it? The frequency would be more interesting. No problem:
Putting in the energy:
The frequency of Hawking radiation emintted by a black hole the mass of Earth this is:
Such a black hole would send out microwaves as used for radar!
This black hole weights .
Help! This is hard gamma-radiation! We should check if it's a dangerous amount later on.
The frequency of visible light lies between and .
A black hole of will emit a dark red and a black hole of will emit a deep violet light.
As already said, the Hawking radiation will show the same allocation as black body radiation. The average energy of a photon of this radiation is given by the following formula:
|k: Boltzmann constant;|
|T: Temperature of the black body.|
This Energy should be equal to the energy a black hole puts into a pair of virtual photons. So let's bring the formulas together.
Now we can substitute all of the constants and we got:
Looks great! If we put in a mass in kilograms we'll get a temperature in Kelvin. Our formula looks promising, even though we used mostly classical physics.
Well, it's pretty close. In the real formula the divisor 2.821
is replaced by a pi:
We only calculated the energy for virtual photons that are aligned radially to the black hole and that are originated at the event horizon. But all directions have to be considered and all pairs of virtual photons of which one reaches the Schwarzschild radius within it's lifetime can submit to the radiation.
This black hole would be colder than the cosmic background radiation.
I don't know anything to compare such a high temperature to!
The luminosity of a black body is given by the Stefan-Boltzmann formula:
|: Stefan-Boltzmann constant;|
A: Surface area of the black body
T: Temperature in Kelvin
1st: The relevant surface area for the Hawking radiation is the surface area of a sphere with the radius of the Schwarzschild radius, because that's where the radiation originates:
We can put this into the formula for the luminosity:
2nd: We have already calculated the temperature of a black hole:
Putting this into the formula for the luminosity we get:
The luminosity of a black hole with the mass of the Earth equals:
The luminosity of the black hole in "The Hole Man":
35.7 gigawatts of hard gamma-radiation would certainly be absolutely deadly!
How heavy will be a black hole with the luminosity of the sun?
A black hole of only 961kg of mass would have the same luminosity as the sun. And it would only be across! (diameter of the Schwarzschild radius). This black hole wouldn't be visible though, because it emits deadly gamma-radiation.
Just now we have seen that a black hole "shines" with Hawking radiation. The escaping member of a virtual particle pair carried away energy from the black hole, and the black hole loses mass as a result. Eventually the black hole loses all its energy, or equivalently mass, and evaporates. Let's derive a formula to calculate the lifetime of a black hole.
The power of Hawking radiation is just the same as the luminosity of a black hole:
On the other hand, the amount of energy carried away by Hawking radiation is the same as the energy loss by the black hole. Therefore, the power P of Hawking radiation is the rate of loss of total energy E of a black hole, that is:
P = -dE/dt
From Einstein's mass-energy equivalence relation,
(Speed of light is a constant!)
Equating P in equation (1) and (2) gives:
Em... The left-hand side of the equation looks pretty clumsy eh? Let's get rid of the chains of constants by a big constant K:
We now have a nicely written simple equation:
Finally, we are going to make use of a powerful math weapon... Yes, it's integration! As a black hole slowly evaporates, its mass drops from (its initial mass) to zero. The time required for evaporation starts from zero to (total time for evaporation). Integrating both-hand sides of equation (3),
The lifetime of a black hole is:
Cheers! We have derived the formula for calculating the lifetime of a black hole! The formula tells us that the lifetime of a black hole is proportional to the cube of its mass. That means a massive black hole takes proportionally much longer time to evaporate, and the process of evaporation accelerates as the black hole slowly loses its mass. This is known as the "runaway" effect. Moreover, take a look at the temperature formula of a black hole:
h: the Planck's constant;
c: the speed of light;
k: the Boltzmann constant;
G: the gravitational constant;
M: mass of a black hole
We see that as the black hole loses its mass, its temperature increases. When a black hole get very very small, its temperature may become so high that it may burn up and cause an explosion!
Let's do some more calculation.
Wooow! The lifetime of such a black hole is even longer than that of our universe!
Well... the lifetime of this black hole is comparatively "shorter". If this black hole was formed not long after the beginning of our universe, we might be able to see its evaporation.
From: "Black holes aren't black - After Hawking they shine!" /C007571 Presented by Angie, Matthias and Thorsten Team C007571, ThinkQuest Internet Challenge 2000 (http://www.thinkquest.org). Last modified: 2000-08-13.