We have only about 50 days to stop LHC. (12.09.09)
From my page http://darkenergy.narod.ru/magtren.html

Magnetic trap of Devil. Calculation.
The first approach. Nucleon Embedding.
Introductive amendment.

The creation of a magnetic hole is similar to composing of nuclei from nucleons.

Let the mass of nucleon consists from two parts: the field mass of nucleon and the condensed mass of nucleon. Look at the fig. a: the field mass is depicted by the brown dots; the condensed mass - by blue circle. The field mass corresponds to the field of p-mesons, or to gluon field in more fresh interpretation. The condensed mass corresponds to the mass of naked nucleon or to the total mass of quarks in the nucleon.
The sum of masses of two free nucleons (proton and neutron) is bigger than the mass of deuteron, made from proton and neutron. The difference between the initial and final states corresponds to the binding energy of the created nuclei, deuteron. If we neglect the kinetic energies of particles in the initial and final states, then we can say, that the binding energy of nuclei is equal to the energy of emitted photons. In order to restore the equality of masses between initial and final states, we can put the interacting particles into a box with mirror walls. Thus we can add the mass of emitted radiation into the side of equation, corresponding to the final state.
At the process of further connecting of nucleons, we'll see, that the specific condensed mass of the resulting nuclei will be diminishing, and the field mass will be growing. The box will also be filled by more photons. Look at the fig. b and c, where blue circles becomes smaller, but the density of brown dots becomes bigger.
If there were only the nuclear forces inside the nuclei, then the moment will come, when the condensed mass would achieve the zero value. That is showed at the fig. d, where the radiuses of blue circles are already zero. But in the nuclei there are also the repulsive electric forces. These forces give no possibility to create the infinitely big nuclei. These forces are responsible for the iron peak in the periodical system of elements and for the existence of the final stabile element in periodical system. If there were no electric repulsive forces, then all nucleons would drop into the growing "nuclear hole", which would have no the limit of growth.
The growth of magnetic hole is similar to the growth of "nuclear hole". But here the proton, capturing by the hole, emits a positron; the neutron, dropping into the hole, emits an antineutrino.
Proton and neutron can be imagined as current pretzels with three poles:

u, u, d in proton

and d, d, u in neutron .

If the nucleon enters into a magnetic field, its current pretzel is undergo to the straightening forces. The more field - the more straightening forces - the more field mass - the less condensed mass. At last, when the field riches the critical value (10^16 Tl), the nucleon undergo decay - its current pretzel with three mixed poles transforms into a vacuum current vortex, which is shown on the following figure in the form of a ring with two magnetic poles N and S.
The "memory about the proton's knot" is brought out by positron.


The "memory about the neutron's knot" is brought out by antineutrino.

In the above process the couple of photons are also emitted.
It is possible to compare the growth of the "nuclear hole" with the growth of the gravitational black hole.


The blue circles are masses, dropping to the event horizon, or moving around it on circumferences. If they are dropping, then their radiuses becomes less, and the density of brown dots will grow in the center of our figure. In this case the density of brown dots corresponds to the "power and depth" of a gravity funnel of eternally forming black hole. It is known that there are no completely formed black holes in the Universe. From the point of view of external observer, the object drops into the black hole infinitely long. So, it is possible to say that the mass of eternally forming black hole consists of two parts: the constantly diminishing mass of dropping objects (condensed mass, blue circles) and the mass of a black hole's funnel, or mass of BH's gravity field (field mass, brown dots). It is clear that at the time of a gravity collapse there will be the emission of photons, but I can not say now the fraction of it.
In the case of magnetic hole there are two eternally growing funnels, corresponding to N and S poles. The bottom calculation shows that at the time of magnetic collapse, the one thirds of rest energy of the captured matter goes into the growth of magnetic hole, and two thirds of captured energy are spent on radiation and on kinetic energy of emitted shell.
Calculation.

To define the minimum mass of a magnetic hole precisely is not possible, because we do not know the magnetic moment of a particle, which will remain after the capture of the nucleon by magnetic hole.

Below I will talk about "imbedding".
Imbedding of nucleon into another nucleon, should not be taken literally. This is an approximate calculation. Magnetic hole is not composed of nucleons. It captures a neutron, and emits an antineutrino. It captures a proton, and emits a positron. In both these cases it can spend its energy on the kinetic energy of these ejected leptons and on radiation of photons. Magnetic hole consists of bosons, quanta of magnetic field. The magnetic moment of this boson is of course different from the magnetic moment of nucleon. Therefore, our numerical result is an approximation.
Nucleon has a magnetic field. Consequently we can divide the energy of nucleon per two parts: the magnetic field energy and the energy connected with the restlet of nucleon, - condensed energy. The condensed energy is connected with its condensed-mass [math]$m_R$[/math]. The total condensed-mass of several connected particles depends from their quantity. Compare: the rest-mass of nucleus is also depend from the quantity of its nucleons, and this dependence in not linear.
At some critical number of captured nucleons, the hole's condensed-mass can be turned to be zero. In this case all its energy is concentrated in its magnetic field. That will be the minimal possible magnetic hole. It can be created at once, in the collision of two particle, but here we create it stage by stage, successively imbedding nucleons into it.
Let's neglect the rest mass of emitted positrons and neutrinos.
Let's imagine the nucleon as a ring contour with an electrical current, creating the magnetic dipole moment, equal to experimental value of a nucleon's dipole magnetic moment, [math]$p$[/math].
The energy of a ring contour is [math]$E = pB$[/math].
It is concentrated in a magnetic field
[math]$B = \frac{\mu_0I}{2R}$[/math].
Magnetic dipole moment is
[math]$p = IS$[/math].
After a simple transformation we can write the energy of magnetic field
[math]$E = \frac{\mu_0pRI^2}{2}$[/math].
We supposed that the electric current was created by the movement of the elementary charge along the circular orbit of radius [math]$R$[/math] with the speed of light. (The radius [math]$R$[/math] is by [math]$p/p_n$[/math] times greater then the Compton radius of nucleon. Here [math]$p_n$[/math] is the nuclear magneton.)
Let's take a single nucleon. Its rest mass consists of two parts, [math]$m_{mag.field}$[/math] - the mass of magnetic field and of the condensed mass of nucleon [math]$m_R$[/math]:
[math]$m = m_{mag.field} + m_R$[/math].
The energy of a nucleon is also consists of two parts:
[math]$mc^2 = \frac{\mu_0pRI^2}{2} + m_Rc^2$[/math].

Let's "imbed" two nucleons, one in one.
Neglect polarization and related changes in the dipole moment.
The energy of dinucleon's magnetic field will increase by 4 times because of a the square over [math]$I^2$[/math]. In the right side of the equation, we must add the energy of emitted photon, which is equal to [math]$pB=\frac{\mu_0pRI^2}{2}$[/math]. It's origin: acceleration of one nucleon's magnet by another nucleon's magnet; growth of kinetic energy and the loss of condensed mass; transition of kinetic energy into the energy of the emitted radiation in the moment of nucleons collision.
[math]$2mc^2 = 2^2 \frac{\mu_0pRI^2}{2} + m_Rc^2 + \frac{\mu_0pRI^2}{2}$[/math].

Let's "imbed" the third nucleon. The energy of the thee-nucleon’s magnetic field will increase by 9 times. In the right side of the equation, we must add the energy of emitted photon, which is equal to [math]$p\cdot 2B$[/math]. The energy of this photon is two times bigger than the energy of the first photon.
[math]$3mc^2 = 3^2 \frac{\mu_0pRI^2}{2} + m_Rc^2 + \frac{\mu_0pRI^2}{2} + 2 \frac{\mu_0pRI^2}{2}$[/math].
Let's "imbed" the [math]$N$[/math]-th nucleon. The energy of magnetic field of [math]$N$[/math]-nucleon will increase by [math]$N^2$[/math] times. In the right side of the equation, we must add the energy of emitted photon, which is equal to [math]$p (N-1) B$[/math]. The resulting condensed mass, [math]$m_R$[/math], becomes equal to zero.
[math]$Nmc^2 = N^2 \frac{\mu_0pRI^2}{2} + \frac{\mu_0pRI^2}{2} + 2 \frac{\mu_0pRI^2}{2} + ... + (N-1) \frac{\mu_0pRI^2}{2}$[/math].
[math]$mc^2 / \frac{\mu_0pRI^2}{2} = (N^2 + 1 + 2 + 3 + ... + (N-1)) / N$[/math].

Further absorption of nucleons can not lead to an increase in the magnetic field, because the total condensed mass will be negative. Consequently, we had received the value of critical magnetic field. The further capture of nucleons by magnetic hole leads to the growth of the radius of a magnetic hole.

The right side in the last formula for large [math]$N$[/math] is equal to [math]$\frac{3}{2}N$[/math]. Therefore, we can write
[math]$N = \frac{2}{3}mc^2 / pB$[/math];
[math]$B_{max} = NB$[/math].
2/3 of nucleon’s energy goes to the growth of the hole.
1/3 goes to radiation.
The last corresponds to the binding energy of the magnetic hole.
If we use the received formula for protons, we'll receive the total energy [math]$E$[/math], needed to create the minimal possible magnetic trap, and [math]$E$[/math] will be equivalent to the rest energy of [math]$N$[/math] protons. In this case [math]$N=510$[/math] and [math]$E=510mc^2$[/math], or about [math]$255mc^2$[/math] per colliding proton.
If in the received formula for [math]$p$[/math] we will use the magnetic moment of neutron then [math]$N=350$[/math].
The formula for [math]$N$[/math] can be transformed to [math]$N=\frac{4}{3} \frac{p}{p_n \alpha}$[/math], where [math]$\alpha$[/math] is the fine structure constant, [math]$1/\alpha = 137.036$[/math]. If we suppose that the magnetic moment of resulting boson is equal to the nuclear magnetic moment, then the total energy, needed to create the minimal possible magnetic hole will be [math]$N = \frac{4}{3\alpha} = 183$[/math]a.u.m.
The results shows that the minimal possible magnetic hole can be made at collisions with the energy about one-thirds of TeV per particle.
This energy is three times smaller than can be achieved at Tevatron!
Why it was not been received yet?
Here are some possible explanations.
1. We do not know the magnetic moment of the resulting boson.
2. It is possible that magnetic hole could be born in a collision of two quarks.
3. At the Tevatron they collide the protons with antiprotons and this lead to the annihilation of the baryon number.
4. We do not know the magnetic polarization dependence.
5. It is not excluded that microscopic magnetic traps were already created, and they are growing now somewhere inside the Earth.
The second approach. Creation of Magnetic Holes.

Let's find the value [math]$pB$[/math], where [math]$p$[/math] is dipole magnetic moment of the first proton, and [math]$B$[/math] is the magnetic induction, created inside the first proton by the second proton, flying beside the first proton. Let's solve the problem in the coordinate system, connected with the first proton. Magnetic induction is the variable value in time and in space.

[math]$B = \frac{\mu_0qv\sin{(v,r)}}{4 \pi r^2}$[/math]

Let's the second proton fly along the straight line, lying in the plane of the contour [math]$S$[/math] of the first proton. The minimal possible distance [math]$r_{min}$[/math] between the line and the border of the contour is defined by the equality of kinetic energy to the potential energy of two elementary charges [math]$E_{kin} = \frac{q^2}{4\pi\epsilon_0r_{min}}$[/math].
The formula [math]$E = pB$[/math] must by changed by the following one
[math]$dE = p (BdS) / S$[/math].

Let's create a computer program and the computer will compute and will sum up all [math]$dE$[/math], giving us in the result the total energy [math]$pB$[/math] of magnetic field and its ratio to the rest energy.
At high energies the kinetic energy is the product of the particles momentum into the velocity of light, [math]$E_{kin}=Pc$[/math].
[math]$E^2 = m^2c^4 + P^2c^2$[/math];
[math]$E^2 = m^2c^4 + E_{kin}^2$[/math];
[math]$E_{kin} = \sqrt{E^2 - m^2c^4}$[/math].
The bottom figure corresponds to quite small energy of flying by proton, [math]$E = 1.3mc^2$[/math].

The next figure: [math]$v=0.968c$[/math], [math]$E=4mc^2$[/math].

The next figure: [math]$v=0.999999998c$[/math], [math]$E=160000mc^2$[/math]. In the coordinate system of collider: [math]$E'= 126mc^2$[/math], or slightly more than 0,1 TeV.


It was found (look tables at my page http://darkenergy.narod.ru/magtren.html ) that magnetic hole can be creared at [math]$E_{total} = 510mc^2$[/math], which was received in the first approach for the total minimal energy, needed to create the magnetic trap. Dividing this energy by 2, we'll receive the energy per one colliding proton in the collider, [math]$E' = 255.2mc^2$[/math]. In the system of resting proton the flying-by proton must have the energy 65107mc^2.

Conclusion. The global catastrophe can be switched by magnetic trap, which can be made at proton collisions with 0.25TeV energy per proton.

Ivan, I'm not a scientist, so I don't understand much of all those calculations.
Our only hope lies not with the general subject journalists, who have no understanding of this scientific subject either, but with science journalists.
They're the only ones who have enough scientific knowledge to understand and evaluate what you're saying.
And accordingly ask CERN the pertinent questions and see if CERN's answers are assuring enough. Here is one contact:

World Federation of Science Journalists
28, Caron Street
Gatineau (QC) Canada J8Y 1Y7

Telephone: +1 819 770-0776
Fax: +1 819 595-2458
Email: info@wfsj.org
Website: http://www.wfsj.org


I too will forward your latest posting to as much scientific publications and media as possible.

Since i am still very afraid of the LHC soon switch on, could some physicist (ore people that have even degrees in physics) look carefully and honestly at Ivan Gorelik's calculations and assumptions ?

Do they raise serious issues (so the LHC must be halted) or are they crack ?


Thank by advance for your help !

Where have all the 'briljant minds' on this board gone

Thank you, Free Thinker.



Dear Shin, I found the similar quetion on the forum http://www.physforum.com/index.php?showtopic=26582

There were no arguments, but were the discussion how a bad man I am.

Here is my answer posted there today.

Hi all!

Dear physicists, details of my person do not deny the conclusions, made by me on the basis of logic and computation. Have you a single argument, proving that my conclusions are wrong?

What will happen if pB will be equal to mc^2, where p – magnetic moment of a proton; B – magnetic field, created by another flying by proton (or the field of already made magnetic hole) mc^2 – the rest energy of proton?

At what energy of collisions the equality pB=mc^2 is valid? My answer: 0.255 TeV.

Note 1:
Simplest definition of magnetic hole can be made on the following comparison:
Newtonian black hole: GMm/r = mc^2/2;
Magnetic hole: pB=mc^2/2.

Note 2:
The time needed the microscopic black hole capture the whole Earth is mach greater than million years. Consequently, they are safe for now living people on the Earth.
The time needed the magnetic hole capture the whole Earth is about 1000 seconds. Consequently, magnetic holes can explode the Earth and Solar system.

Note 3:
We know two types of explosions chemical and nuclear.
Astronomers see the third type of explosions – collaptical.
At nuclear explosion only few thousand’s parts of explosive rest energy is realized in the form of radiation and kinetic energy of fragments.
At some stellar collaptical explosions about a half of rest energy of the star is realized in the form of radiation and kinetic energy of the rejected stellar shell.
What type of collaptical explosions do astronomers observe: gravitational into black hole or magnetic into magnetic hole?
Is it possible to switch the magnetic collapse at collider?
My answers: The observed collapses are magnetic ones. Magnetic collapse can be switched on by powerful colliders.
Your answers: …

Physicists, what will happen at pB=mc^2, or at 0.255 TeV–collisions of protons at LHC.

This energy is smaller than can be achieved at Tevatron!

- Why it was not been received yet?
Here are some possible explanations.
1. We do not know the magnetic moment of the resulting boson.
2. It is possible that magnetic hole could be born in a collision of two quarks, for example by d and d.
3. At the Tevatron they collide protons with antiprotons and this lead to the annihilation.
4. We do not know the magnetic polarization dependence.
5. It is not excluded that microscopic magnetic traps were already created, and they are growing now somewhere inside the Earth.

LHC will be switched on in the mead-November; the first collisions are waited in a few weeks later. Do we survive or do we start into outer space?

Computations: http://darkenergy.narod.ru/magtren.html
Reproduction of biospheres and civilizations: http://darkenergy.narod.ru/civilen.html
Arguments proving that..: http://darkenergy.narod.ru/argen.html

Letters to presidents and authorities. http://darkenergy.narod.ru/presid.html This page is in Russian. You can translate it with the help of Google. I wrote these letters because i fear the creation of magnetic holes.

The starter of this topic is “Very Affraid By Ivan Gorelik's Theory On Lhc”.
I also fear of my theory, I fear it drastically. I see that it is almost impossible to deny the launch of the LHC, or to prevent our common global suicide. According to my assumptions the probability of global catastrophe, switched by LHC, is about of dozens of percents.

This is Prometheus's answer to Ivan's post on http://www.physforum.com:

FYI, I'm a theoretical physics postgraduate student.

Essentially what this guy is doing is pulling equations out of his anal orifice and claiming they are relevant. For example his formula pB=mc^2. Apart from the fact that mc^2 is only relevant when particles are at rest, the magnetic field would have to be absolutely huge. In fact, nearly 4 x 10^7 T for the LHS to equal the RHS for the proton. For comparison, the largest magnet at the LHC produces just over 8 T.

He is also wrong in his assertions. He says "Physicists, what will happen at pB=mc^2, or at 0.255 TeV–collisions of protons at LHC. This energy is smaller than can be achieved at Tevatron!" which is simply wrong whichever way you count the energy. The Tevatron can achieve 0.98 TeV in a single beam and 1.96 TeV collision energy.

He also seems to like to make up words to try and give what he's writing a technical sounding jargon. A collaptical[sic] explosion is a phrase I've never heard anyone use, and I've done a fair bit of work on astronomy and cosmology.

Basically Ivan Gorelik is a failure and crackpot at physics. You wouldn't be scared if I said that monsters living on Uranus were coming to destroy the planet would you? Basically, that is what Gorelik is doing - making stuff up.

No they don't. You honestly do not need to worry about the LHC. There are far more pressing problems for the world that you could be worried about like global warming and arms proliferation.

My pleasure.
PS, I got my info on the Tevatron and LHC from wikipedia. Both pages are worth a read if you're interested.
[/quote]

This is moderator's insertion in front of my text.
http://www.physforum.com/index.php?showtopic=26582&st=0
Do you see physics in his text?
Are LHC safe?

From: http://www.physforum.com/index.php?showtopic=26582&st=0

People, I am in shock from your silence!

Humans, like a herd of obedient sheep, are lead to the abyss. I, as a mortally frightened lamb, run from side to side of the herd, but the herd do not hear me. They laugh at me and moves slowly, plucking the juicy grass.

People, please, wake up! Open your eyes!

November the 30-th, according to the preliminary plan for the Large Hadron Collider, the collision of protons with energies of 0.45 TeV will be implemented, 14 December - collision with energies of 3.5 TeV. If the magnetic collapse starts, it will be impossible to stop it. It will end with ejection of the Earth’s shell into space.

I give the probability of about 10% that magnetic collapse will start at 0.45 TeV; 50% at 3.5 TeV; 70% - at 7 TeV.
30% I leave to my mistake.

Why is such confidence, - you would ask.

Compare Newtonian black hole and magnetic hole:
GMm / r = mc^ 2 / 2;
pB = mc ^ 2.

It was obtained two independent proofs that the equality pB = mc^2 is reached at the energies, achievable in modern colliders. http://darkenergy.narod.ru/magtren.html
There are many arguments in favor of the fact, that magnetic holes exist, and astronomers see them but call them on the error as black holes.
http://darkenergy.narod.ru/argen.html
It was found several great errors and criminal lies in the assessment documents about the safety of LHC.

Humans, are you really obedient sheep, ready to be killed?

so what do you want to do about it?