Woo Hoo, LHC is Working

[Moontanman]

Solving the Schwarzschild radius,

[math]r_s=\frac{2GM}{c^2}[/math]

for the mass needed to fit inside the radius of a proton in order to make a black hole,

[math]m=\frac{10^{-15}c^2}{2G}[/math]

is 6.7E11 kilograms

 

That's one heavy proton :)

 

~modest

 

Modest, what does the notation 6.7E11 kilograms mean?

[Buffy]

Modest, what does the notation 6.7E11 kilograms mean?

 

[imath]6.7\times10^{11}\ Kg=670,000,000,000 Kg[/imath]

 

Big!

 

The creator of the universe works in mysterious ways. But he uses a base ten counting system and likes round numbers, :phones:

Buffy

[Moontanman]

[imath]6.7\times10^{11}\ Kg=670,000,000,000 Kg[/imath]

 

Big!

 

The creator of the universe works in mysterious ways. But he uses a base ten counting system and likes round numbers, :phones:

Buffy

 

If my figuring is correct that would be the mass of asteroid several miles across. So a proton sized black hole would have the mass of an asteroid several miles across? This would seem to indicate that making a micro black hole is not within the grasp of our technology. I would think the LHC couldn't possibly accelerate a particle until it has even a small fraction of that mass.

[alexander]

6.7E11

that's like the anti gravitational constant :) 6.67x10^-11

 

proton is not the smallest a black hole can be thought, according to hawking's equations, the smallest mass of a black hole is somewhere in order of a plancks constant, or around [math]2\times10^{-8}kg[/math]

 

lets consider that we have 2 bunches of 1.15x10^11 protons totaling .0000000000000003847029450 kg in mass

 

hawking states that the time (in seconds) it takes a black hole to evaporate can be calculated by using the following formula:

[math]t_{ev}=\frac{5120\pi G^2M^3_0}{\hbar c^4} s[/math]

 

thus

[math]t_{ev}=\frac{5120\times3.14159265\times (6.67300\times10^{-11})^2\times(1.67262158\times10^{-27}\times1.15\times10^{11})^3}{\frac{6.62606896\times10^{-34}}{2\times3.14159265}\times299792458^4}[/math]

 

so i'm running this in bc: (make sure i have enough brackets for this to run... going to set scale to 100 decimal places)

(5120*3.14159265*((6.67300*10^(-11))^2)*((2*(1.6726215*10^(-27))*(1.15*10^(11)))^3))/(((6.62606896*10^(-34))/(2*3.14159265))*(299792458^4))

or in slightly calculated terms (so i would hope)

(16084.954368*.0000000000000000000044528929*.000000000000000000000000000000000000000000000056934633960794713694558625000)/(.0000000000000000000000000000000001054571629456797971563881778243910775637955480956450544280462331741*8077608713062490229263800746151696)

 

 

.0000000000000000000000000000000000000000000000000000000000000047871795922645116102703687541065837913

 

scaled back to an ever-so-saner answer, a black hole made of all of the protons masses combined upon collision in an LHC chamber with the mass of all of them, would dwindle away in [math]4.787\times10^{-62}s[/math]

 

checking back, with the mass formula [math]M_0=2.28271\times10^5*\left(\frac{t_ev}{s}\right)^{1/3} kg[/math]

just google this, you should see the mass we started with

(2.28271*10^5)*.0000000000000000000000000000000000000000000000000000000000000047871795922645116102703687541065837913 ^(1/3)

 

you can calculate black body radiation given off by such a black hole

[math]P=3.56345\times10^{32}*\left(\frac{kg}{M_0}\right)^2 W[/math]

 

check my math, comment, i'll do more reading in the meanwhile

[alexander]

and the radius of such a black hole should be [math]5.7126\times10^{-43}[/math] meters

[alexander]

oh another thought, the black hole would not survive long enough to see the next bunch collide (75ns apart and eventually to be 25ns apart) thus limiting my need to actually calculate what the size of a black hole of all of the nucleons circling would be...

 

anyways, any input is welcome, i think there is a base here to disproving concerns, so yeah, let's continue in that direction :)

[alexander]

oh wait, i made a big doo-doo, and contradicted myself.... a black hole of the size calculated can not exist....

 

let me run some calculations, methinks that the black hole of that size would require more energy quanta then we have available, though i have to check this because we do have a fair amount of energy available to us from collision, thus it may just have enough to hold it together for some time.... off reading on black hole thermodynamics...

[modest]

If my figuring is correct that would be the mass of asteroid several miles across. So a proton sized black hole would have the mass of an asteroid several miles across? This would seem to indicate that making a micro black hole is not within the grasp of our technology. I would think the LHC couldn't possibly accelerate a particle until it has even a small fraction of that mass.

 

Correct

 

a black hole of the size calculated can not exist....

 

I should think a black hole smaller than a plank length is most impossible - otherwise point particles would turn into black holes as a matter of course.

 

I also think it would be more sensible to concentrate on one collision. As a problem of geometry, how would all these protons get together in an area small enough to make a black hole? Or, I guess that's probably your point. Even as a worst case scenario, it will evaporate before it has time to absorb more energy.

 

Nicely done Sir.

 

And... do I see latex quoting? :)

 

~modest

[alexander]

i'm looking for that code... unfortunately there's not much reference, so i will keep on looking -_-

 

it will evaporate before it has time to absorb more energy.

as far as i'm reading, it will not exist because the amount of energy it will contain is less then the amount of energy it will release... (i know its rather confusing, but i am pretty confused here too :hihi: )

 

i have to figure out the worsest of all the worse case scenarios, all of the atoms forming one hole is one, then there is a collision with a magnet to consider... for now though, i will have to figure out the mass gain with 244KJ of heat energy added to the mix (say if all the kinetic energy turned into heat energy... in an instant...) I'd like to see if the mass gain, predicted by Schwarzschild, would be enough to break the minimum mass barrier. This would get us to some extreme probabilities, but the worsest of this particular scenario. Mind you even if we break 2x10^-8 barrier, the lifetime of the hole would still be far bellow an order of a nano second, if my memory serves me right that is...

 

i'll run numbers sometime tomorrow...

[modest]

[math]t_{ev}=\frac{5120\pi G^2M^3_0}{\hbar c^4} s[/math]

 

It's quoting. You should have taken credit :)

 

~modest

[max4236]

If the beam starts punching through into higher dimensions does that evaporation time equation still hold? Wouldn't further physical removal from the Higgs field that permeates our universe cause something like this to gain greater stability than it normally would, or would it make it pop out of existence faster? Wow, 4.787x10^-62 seconds. Can time be divided into intervals that small? You'd almost think the universe would quantum jump past a moment that small and the process could never even happen just like an electron can't fall all the way into the nucleus of an atom.

[modest]

Using the sr formula [math]E_k=mc^2(1/\sqrt(1-(v/c)^2)-1)[/math] at 99.999999% of the speed of light, 1.15x10^11 protons would have approximately 122224.711701J of energy totalling at about 686413980.912816 J for 2 beams

 

I started to check your math and I got stopped somewhat early. Setting aside for a sec the speed of the beam, we should be able to trust the 7TeV figure. This means each collision (proton / proton collision) will be 14 TeV.

 

There are (according to wiki) [math]1.15 \times 10^{11}[/math] protons per bunch. If we assume (and this is a completely ridiculous assumption) that every proton in a bunch simultaneously collided with every proton from an opposing bunch then the total energy is very straightforward:

[math]14 \ TeV = 22.4 \times 10^{-7} \ Joules[/math]

times the number of collisions in a bunch

[math]22.4 \times 10^{-7} J \times 1.15 \times 10^{11} = 2.58 \times 10^5 \ J[/math]

This is more or less what you have above for one bunch multiplied by two. But then I get a bit confused where you go from here. Your next post has:

 

lets consider that we have 2 bunches of 1.15x10^11 protons totaling .0000000000000003847029450 kg in mass

 

Where did you get this? Wouldn't the proper way to get mass be to convert our value of Joules above to kg via e=mc^2. I should note here that I don't mean the relativistic mass to energy equation as both the energy calculated above and the mass we're considering are both in our rest frame.

 

By e=mc^2, 258000 J is [imath]2.87 \times 10^{-12} Kg[/imath].

 

Which disagrees with the value you put in Hawking's evaporation equation by a few orders of magnitude. So, I'm perplexed - through every fault of my own, I have no doubt.

 

~modest

[modest]

[math]t_{ev}=\frac{5120\times3.14159265\times (6.67300\times10^{-11})^2\times(1.67262158\times10^{-27}\times1.15\times10^{11})^3}{\frac{6.62606896\times10^{-34}}{2\times3.14159265}\times299792458^4}[/math]

 

Ok... I see what you did. [imath]1.67 \times 10^{-27}[/imath] is the rest mass of a proton. That didn't even occur to me. I don't think that's right. I think if you were to fire a proton into a black hole at near the speed of light, the black hole would gain more mass than if it swallowed a proton at rest to the hole. Perhaps I'm missing something, but I don't think the rest mass of a proton is the right value for this equation. I think the total mass for this equation would be around [imath]2.87 \times 10^{-12} Kg[/imath] vs the roughly [imath]2 \times 10^{-16} Kg[/imath] you have.

 

You also only put one bunch in this equation... I can't figure that either. Would it not be two colliding bunches? I'm still probably missing something...

 

~modest

[alexander]

you are right, i was writing a reply to that, and i forgot to account for relativistic mass... this will change the order of mass i am talking about... i will rerun the calculations at lunch using relativistic mass (stupid me)

 

[math]m_{rel}=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

[alexander]

oh i was writing a reply, but my cookie expired, and yeah i lost it all :(

[buddyzen]

O_O that will cause some setbacks O_O

[alexander]

ok, so letasa starta with this equation, and lets properly modify it for using with a stationary mass value...:

 

[math]t_{ev}=\frac{5120\pi G^2M^3_0}{\hbar c^4} s[/math]

where

[math]M_0=\frac{M_0}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

is the function to determine the true mass of the protons, moving at 299792455.00207542 m/s, it is therefore how we should determine if we break the 2x10^-8kg needed for the miniature black hole

 

this also weilds us a function

[math]t_{ev}=\frac{5120\pi G^2M^3_0}{\frac{h}{2\pi} c^4\left(1-\frac{v^2}{c^2}\right)^{3/2}} s[/math]

to figure out the life span of a black hole made of some colliding objects, based on their stationary mass :)

 

gonna run some numbers in a few