Google Search

My Debates

Challenge Period

Debating Period

Voting Period

Post Voting Period

Recently Updated

Debate Leaderboard

Voting Leaderboard

Judges Leaderboard

Arts

Cars

Economics

Education

Entertainment

Fashion

Funny

Games

Health

Miscellaneous

Movies

Music

News

People

Philosophy

Places-Travel

Politics

Religion

Science

Society

Sports

Technology

TV

Opinions Leaderboard

Debate.org

Arts

Economics

Education

Entertainment

Funny

Games

Health

History

Miscellaneous

News

Personal

Philosophy

Politics

Religion

Science

Society

Sports

Technology

Forums Leaderboard

## Hawking radiation evaporates black holes.

Wile its not possible do strate up vaporize a black hole eventualy it will disapear. According to the quantum mechanical uncertainty principle, rotating black holes should create and emit particles. Hawking radiation reduces the mass and the energy of the black hole and is therefore also known as black hole evaporation. Because of this, black holes that lose more mass than they gain through other means are expected to shrink and ultimately vanish.

Posted by: haloman1999## Shoot the laser!

I'm not entirely sure, but I'll go with 'yes'. Given that you have an almost endless amount of energy... My argument "in short" is

(1) nothing ever reaches the event horizon as seen from the outside

(2) if the energy you use to vaporize it is large enough, you can neglect the mass of the black hole itself and assume its escape velocity is proportional to the square root of the energy you put in

(3) assuming a Maxwell-Boltzmann distribution (not sure if justified, but I won't investigate the case for degenerate matter for the sake of it), the most probable velocity (mpv) grows proportional to the square root of the energy you put as well.

Thus, it all comes down to whether the mpv of the particles falling onto the black hole grows faster with the energy you put in than the escape velocity. If it does, the mpv will after some amount of energy input be larger than the escape velocity and then it is sure, that we could evaporate a black hole.

Since the quadratic escape velocity grows with 2GE/(rc^2) (r is some kilometers) and the quadratic mpv grows with 2RE/M, where I go with M as the neutron mass, the mpv grows faster. Thus, shooting a black hole with a ridiculously strong laser might evaporate it. Maybe.Posted by: Debaterpillar

## Very sceptical about it.

Black holes are, in some cases, invisible. But if monitored through super computers, it is possible to see the light bend around it, before reaching the Event Horizon. In essence, it is just like trying to destroy a patch of space, by, say, using an atomic bomb. But even if you physically threw anti-matter into the black hole, you would only enlarge it, because it would swallow and absorb anything that approaches it.

Posted by: hinderbolt## No don't try

I say this because the black hole can destroy /suck in light. A laser is just a very hot strong light beam that would not affect the black hole. The more energy you give the black hole the more it will grow.This is why i say no to this question.

## Black holes are so dense that nothing in the universe is powerful enough to destroy them.

The matter contained in black holes is so dense that their gravity overcomes the strong nuclear force so that protons and electrons do not have the space to orbit the nucleus of atoms and all the material within the black hole is compacted into one homogeneous mass.

The gravitational pull of black holes is so strong that not even light can escape (although protons have no mass and are not, therefore, directly affected by gravity, the gravity created by a black hole bends the fabric of space time to such a degree that they cannot escape).

Therefore, any object or force (other than another black hole) that comes into contact with a black hole will be destroyed rather than the black hole itself.

However, quantum mechanics does allow for a very tiny amount of energy to radiate from a black hole. This is called Hawking radiation which has a blackbody (Planck) spectrum with a temperature T given by

kT = hbar g / (2 pi c) = hbar c / (4 pi rs)

where k is Boltzmann's constant, hbar = h / (2 pi) is Planck's constant divided by 2 pi, and g = G M / rs2 is the surface gravity at the horizon, the Schwarzschild radius rs, of the black hole of mass M. Numerically, the Hawking temperature is T = 4 × 10-20 g Kelvin if the gravitational acceleration g is measured in Earth gravities (gees).

The Hawking luminosity L of the black hole is given by the usual Stefan-Boltzmann blackbody formula

L = A sigma T4

where A = 4 pi rs2 is the surface area of the black hole, and sigma = pi2 k4 / (60 c2 hbar3) is the Stefan-Boltzmann constant. If the Hawking temperature exceeds the rest mass energy of a particle type, then the black hole radiates particles and antiparticles of that type, in addition to photons, and the Hawking luminosity of the black hole rises to

L = A (neff / 2) sigma T4

where neff is the effective number of relativistic particle types, including the two helicity types (polarizations) of the photon.

So, we can see that because the amount of energy / mass radiated from the black hole is so negligibly small we can ignore it.Posted by: brian_eggleston

©2017 Debate.org. All rights reserved.