Can we create a small black hole? [duplicate]

Suppose we create very large spherical body by using gamma rays generator and they will concentrate on a single point at the centre of sphere.We will place this spherical body thousands of kms above the sun so that it can draw energy from sun.Can the centre of the sphere create a Mini Black hole by the energy from the concentration of gamma rays?

[Just applying my 14 yrs old brain..if im wrong then please correct me...]

-

marked as duplicate by Dilaton, Chris White, Qmechanic♦Aug 3 '13 at 21:48

In fact we can make micro-blackholes(as we constantly do at LHC). But the point is, if blackholes are lighter than a limit, they will just evaporate using Hawking radiation. I assure you, the whole process won't take more than a glimpse. – Ali Aug 3 '13 at 13:35
@Ali Do we have evidence for production of micro black holes in the LHC? I'd like a reference. – Will Aug 3 '13 at 14:04
@Will Not really the point of my comment. I had this work and this in mind, when I wrote my comment. However, these are not experimental works; so maybe there is no evidence for creating blackholes at LHC. – Ali Aug 3 '13 at 14:28
@Ali it doesn't really matter what your point is, we should be careful with what we say. Saying "we constantly do at LHC" indicates that it has been experimentally observed in the LHC. It may have been better to indicate that you are talking about a so far untested theory. (I'm appy you have effectively explained that now) – Will Aug 3 '13 at 14:45

Yes, taking any object and decreasing its volume, while keeping the mass constant, will result in creating a black hole - the question is how long is it going to "live", because there is such thing as evaporation due to Hawking's radiation. For example, turning Earth into a black hole requires squeezing it down into a ball with the radius of about $8\times 10^{-4}$ meters.

-
This is half-sided. Look at user1504's answer. The probability is very small . – centralcharge Aug 10 '13 at 12:37

What you propose is theoretically but not practically possible. Or rather, we can not really tell the difference. Shining a bunch of lasers at a common point is a good example of a rather complicated scattering experiment. You're creating huge numbers of photons which all have the same energy and bouncing them off each other.

In quantum physics, we don't actually get to watch the details of the scattering; that would interfere with the experiment. Instead all we can do is count the frequency of fixed inputs creating various outputs, and compare these frequencies to probabilities we compute using the rules of quantum physics. In quantum physics, this probability is the norm of a complex number, called the amplitude. We can compute the amplitude for a particular input -> output by adding up amplitudes for every possible way the inputs might turn into the outputs, summing over all possible histories.

So, if you can imagine black holes being formed when you concentrate enough photons in a small area and then being unformed by quantum leakage like Hawking radiation, then you ought to include an amplitude for this in the sum over histories.

One of the big questions in theoretical physics right now is: "Exactly precisely what number should this 'a black hole appears and then disappears' amplitude be?" We really don't know how to compute it precisely.

But we can estimate its order of magnitude in various ways.

For example: A photon in a blue laser has a wavelength of about $\lambda =450$ nm, equivalently an energy $E = \hbar \nu = \hbar c/\lambda$ of around $2$ eV $\sim 10^{-19}$ Joules. We know gravity is pretty weak in distance scales near $\lambda$ on our planet. Newton's constant $G$ is very small; equivalently the Planck mass $M_P \sim 1/\sqrt{G} \sim 10^{30}$ eV is huge. We can try thinking of photons forming a black hole in terms of photons exchanging gravitons.

The methods of quantum field theory tells us that if two photons collide/approach closely the effective coupling constant for a 2-photons and 1 graviton is going to vary with the energy $E$ roughly as $E/m_P$. So for photons in a blue laser, we're going to see everything we calculate multiplied by powers of $10^{-30}$. This makes the contribution of the black hole to the scattering amplitude very small, which makes it all but impossible to measure with current technology.

-