Proton vs Graphite
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Re: Proton vs Graphite
Grist for the mill ...
LHC not good for cooking frozen pizzas http://www.telegraph.co.uk/science/larg ... pizza.html
A nice little 3 pager on the design of the LHC beam dump. http://epaper.kek.jp/e92/PDF/EPAC1992_1545.PDF
Regardless of the actual number there is a lot of energy to dissipate in a short amount of time. You might cook a frozen pizza with the LHC if you turned off the water cooling and put the pie on top of the beam dump. A one shot affair I imagine.
LHC not good for cooking frozen pizzas http://www.telegraph.co.uk/science/larg ... pizza.html
A nice little 3 pager on the design of the LHC beam dump. http://epaper.kek.jp/e92/PDF/EPAC1992_1545.PDF
Regardless of the actual number there is a lot of energy to dissipate in a short amount of time. You might cook a frozen pizza with the LHC if you turned off the water cooling and put the pie on top of the beam dump. A one shot affair I imagine.
Re: Proton vs Graphite
Wow....
Intuition is a bad thing in this discussion...
These things MUST be VERY difficult for common people to follow..
Look at the last article...
"Another estimate of beam energy is 10 trillion watts" and a few sentences later "protons will release an amount of energy comparable to that of two colliding mosquitoes."
I can see where any normal person would be very confuzed by all this..
So, the protons do have a unprecedented amount of speed/energy going on here. This is obvious because of the almost insane levels of joules released into the beam dump. The fact you gotta spin the beam with a magnet to make sure it does not stay on one spot too long or it will cause crazy high power bad things to happen to that microscopic spot on the graphite seems to indicate truly immense energy transfer from the protons to the graphite.
I mean seriously. One serious collision does occur. Huge amounts of joules do seem to transfer from the beam to the dump. In fact the energy dissipated seems to match what you would expect a 3.5TeV bunch of 8E9 protons to release... And it seem this energy is released as the protons hit the graphite.
Speed of light or not, conservation of energy does apply here. How that energy is converted from the beam to the graphite is what ? a collision ?
OK, so the graphite does some insane, ummmm,,,,, temporal multidimensional expansion to accommodate these hyper warp drive speeded up protons ? And this does not count as a collision ?
Hmmmm...... Usually the laws of physics tend to be elegant and simple. I see a bit of matter moving real damn fast and hitting a object and dissipating a reasonably comparable amount of energy to what was put into the bit of moving matter.
Its hard to get my head around the idea that a collision did not occur at the energies expected.
OK so I get the idea of a expanding medium sort of catching the protons and slowing them down slowly and dissipating the energy in this slow down ? But that initial contact before the medium is moving with the proton, slowing it, must occur at a high rate.
Look it would seem, unless the absorbing medium can anticipate the oncoming particle and speed up to its velocity, that the accelerated proton must first strike a non moving object and speed it up. If your riding on the particle in its frame of reference you would still see a wall that was not moving.. From the frame of reference of the proton, that wall is not moving. From the frame of reference of the wall, that proton is moving at horrific speed... Close to light speed in fact.. And low and behold a HUGE amount of energy is dissapiated...
Yea.. I still dont get this... This sure looks like a collision to me..
I must just be very stupid for not following this...
Intuition is a bad thing in this discussion...
These things MUST be VERY difficult for common people to follow..
Look at the last article...
"Another estimate of beam energy is 10 trillion watts" and a few sentences later "protons will release an amount of energy comparable to that of two colliding mosquitoes."
I can see where any normal person would be very confuzed by all this..
So, the protons do have a unprecedented amount of speed/energy going on here. This is obvious because of the almost insane levels of joules released into the beam dump. The fact you gotta spin the beam with a magnet to make sure it does not stay on one spot too long or it will cause crazy high power bad things to happen to that microscopic spot on the graphite seems to indicate truly immense energy transfer from the protons to the graphite.
I mean seriously. One serious collision does occur. Huge amounts of joules do seem to transfer from the beam to the dump. In fact the energy dissipated seems to match what you would expect a 3.5TeV bunch of 8E9 protons to release... And it seem this energy is released as the protons hit the graphite.
Speed of light or not, conservation of energy does apply here. How that energy is converted from the beam to the graphite is what ? a collision ?
OK, so the graphite does some insane, ummmm,,,,, temporal multidimensional expansion to accommodate these hyper warp drive speeded up protons ? And this does not count as a collision ?
Hmmmm...... Usually the laws of physics tend to be elegant and simple. I see a bit of matter moving real damn fast and hitting a object and dissipating a reasonably comparable amount of energy to what was put into the bit of moving matter.
Its hard to get my head around the idea that a collision did not occur at the energies expected.
OK so I get the idea of a expanding medium sort of catching the protons and slowing them down slowly and dissipating the energy in this slow down ? But that initial contact before the medium is moving with the proton, slowing it, must occur at a high rate.
Look it would seem, unless the absorbing medium can anticipate the oncoming particle and speed up to its velocity, that the accelerated proton must first strike a non moving object and speed it up. If your riding on the particle in its frame of reference you would still see a wall that was not moving.. From the frame of reference of the proton, that wall is not moving. From the frame of reference of the wall, that proton is moving at horrific speed... Close to light speed in fact.. And low and behold a HUGE amount of energy is dissapiated...
Yea.. I still dont get this... This sure looks like a collision to me..
I must just be very stupid for not following this...
- CharmQuark
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Re: Proton vs Graphite
Well am completely lost
Anyone who doesn't take truth seriously in small matters cannot be trusted with large ones either by Albert Einstein.
Re: Proton vs Graphite
Regarding center of mass energy, imagine a grain of sand suspended in space compared to grain of sand mounted to a steel plate. When you hit both by a bullet, the suspended one will probably survive as it will just bounce or stick to the bullet, while the one mounted to a wall will get shattered.
Carbon nuclei in the dump carbon block are not mounted to a steel plate. They are suspended in space and held in place by electromagnetic forces. These nuclei are positively charged as well as protons are. If a dumped proton arrives, one of the following happens:
1/ the proton passes near some nucleus without touching it. In that case there is no collision but there is repulsive force that transfers a bit of the proton energy on the nucleus - proton gets slowed down a bit, the nucleus gains some speed -> heat.
2/ the proton crashes into a nucleus
2a/ the proton bounces off the nucleus giving it some of its energy -> heat
2b/ the proton is captured in the nucleus. In that case the electromagnetic forces are too weak to keep the nucleus in place and it is kicked off its position and continues flying in original proton direction like a freely lying styrofoam bubble when hit by a bullet. Some (relatively small part) of the energy is consumed on releasing the nucleus off its electromagnetic bed (giving heat impulse to surrounding atoms) and the rest of the energy gets distributed among all new nucleus particles, which means significantly less energy per particle for further collisions.
So the dump is something like a bag of sand into which we are shooting bullets. All bullets get stopped but the sand is mostly untouched by it, it just gets warmer.
Carbon nuclei in the dump carbon block are not mounted to a steel plate. They are suspended in space and held in place by electromagnetic forces. These nuclei are positively charged as well as protons are. If a dumped proton arrives, one of the following happens:
1/ the proton passes near some nucleus without touching it. In that case there is no collision but there is repulsive force that transfers a bit of the proton energy on the nucleus - proton gets slowed down a bit, the nucleus gains some speed -> heat.
2/ the proton crashes into a nucleus
2a/ the proton bounces off the nucleus giving it some of its energy -> heat
2b/ the proton is captured in the nucleus. In that case the electromagnetic forces are too weak to keep the nucleus in place and it is kicked off its position and continues flying in original proton direction like a freely lying styrofoam bubble when hit by a bullet. Some (relatively small part) of the energy is consumed on releasing the nucleus off its electromagnetic bed (giving heat impulse to surrounding atoms) and the rest of the energy gets distributed among all new nucleus particles, which means significantly less energy per particle for further collisions.
So the dump is something like a bag of sand into which we are shooting bullets. All bullets get stopped but the sand is mostly untouched by it, it just gets warmer.
- chriwi
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Re: Proton vs Graphite
Taken the case the proton gets stuck in the carbon nucleus and smashes it out of its lattice alltogheter continue moving still almost speed of light:
Lets see it like that, there is really a nonelastic collission of an object carrying 3.5TeV with a object not moving. Taking only this fact you really have your 3.5TeV collission here, but the important thing for particle physics is not howmuch energy is actuaslly transferred, but how much of the nergy is not again mtionenergy (including relativistic mass) afterwards but can be used to create new particles or rays. In the headoncollissin a significant part of the 3.5TeV goes int o creating particles, because the debree is hardly moving compared to the 2 Protons before. In the dump most of the energy stays motionenergy at least for the first collission and also all later collissions. The whole aqmount of energy is converted by billions of successive collissions in the dumpblock, but in all off theses collissions more energy is onlytransferred to motionenergy of ohter particles and only a very minor part may b e usd to create any lite weight particles.
Lets see it like that, there is really a nonelastic collission of an object carrying 3.5TeV with a object not moving. Taking only this fact you really have your 3.5TeV collission here, but the important thing for particle physics is not howmuch energy is actuaslly transferred, but how much of the nergy is not again mtionenergy (including relativistic mass) afterwards but can be used to create new particles or rays. In the headoncollissin a significant part of the 3.5TeV goes int o creating particles, because the debree is hardly moving compared to the 2 Protons before. In the dump most of the energy stays motionenergy at least for the first collission and also all later collissions. The whole aqmount of energy is converted by billions of successive collissions in the dumpblock, but in all off theses collissions more energy is onlytransferred to motionenergy of ohter particles and only a very minor part may b e usd to create any lite weight particles.
bye
chriwi
chriwi
Re: Proton vs Graphite
I think I see your misunderstanding. Of course if you send a 3.5 TeV proton into a beam dump, that energy will not disappear - it will end up in the beam dump, making the beam dump hotter. It will end up there like the previous 2 posts describe - a cascade of collisions and scatterings, a bit like a cosmic ray shower in the atmosphere (just happening in a smaller space because of the material being a dense solid).
BUT that is not what you are interested in when comparing with the head on collision of two beams. There, what counts is the energy that can be converted into new stuff in the first collision (when there is one, and the incoming proton is not just deflected). This "collision energy" has to be worked out like i described in my earlier post, and it is less than you'd expect for relativistic reasons, and because a lot of the energy of the beam goes into keeping the collision products moving further into the dump by momentum conservation. (Since you asked: Once you understand it, it is actually quite elegant and simple, just conservation of energy and momentum, plus special relativity. But special relativity isn't that easy to understand...)
The other confusion arises from mistaking the energy of a single LHC collision (TeV, tiny in everyday terms (kinetic energy of mosquitoes), while a lot for a single proton) with the energy stored in the beam as a whole (huge, because there are so very many protons in a fully loaded beam). It helps to play around with the conversions into your favourite units yourself - e.g. with this website: http://www.wolframalpha.com/input/?i=7+TeV+in+Joules
I don't understand your comments about the moving reference frame. If you were riding along with a beam proton, you would see the beam dump coming towards you at tremendous speed. The fact that the beam dump is not moving relative to the earth does not change this. It's just a different point of view. It's no different for you sitting in a car heading towards a wall - you see the wall coming at you.
BUT that is not what you are interested in when comparing with the head on collision of two beams. There, what counts is the energy that can be converted into new stuff in the first collision (when there is one, and the incoming proton is not just deflected). This "collision energy" has to be worked out like i described in my earlier post, and it is less than you'd expect for relativistic reasons, and because a lot of the energy of the beam goes into keeping the collision products moving further into the dump by momentum conservation. (Since you asked: Once you understand it, it is actually quite elegant and simple, just conservation of energy and momentum, plus special relativity. But special relativity isn't that easy to understand...)
The other confusion arises from mistaking the energy of a single LHC collision (TeV, tiny in everyday terms (kinetic energy of mosquitoes), while a lot for a single proton) with the energy stored in the beam as a whole (huge, because there are so very many protons in a fully loaded beam). It helps to play around with the conversions into your favourite units yourself - e.g. with this website: http://www.wolframalpha.com/input/?i=7+TeV+in+Joules
I don't understand your comments about the moving reference frame. If you were riding along with a beam proton, you would see the beam dump coming towards you at tremendous speed. The fact that the beam dump is not moving relative to the earth does not change this. It's just a different point of view. It's no different for you sitting in a car heading towards a wall - you see the wall coming at you.
Last edited by photino on Tue Mar 23, 2010 3:36 pm, edited 1 time in total.
- DCWhitworth
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Re: Proton vs Graphite
OK I can't claim to have absorbed all this thread (failure to turn brain on) but one thing struck me this morning.
Firstly to summarise this thread (as I've grasped it anyway) -
a. Two protons travelling at relativistic speeds collide head on = Massive energy collision.
b. One proton travelling at relativistic speed collides with a more or less stationary proton = much, much lower energy collision.
OK, assuming I'm roughly right with this summary how is it that cosmic rays ploughing into the earth's atmosphere can be used to prove the safety of the LHC when surely they resemble collision type b. rather than collision type a. ?
Firstly to summarise this thread (as I've grasped it anyway) -
a. Two protons travelling at relativistic speeds collide head on = Massive energy collision.
b. One proton travelling at relativistic speed collides with a more or less stationary proton = much, much lower energy collision.
OK, assuming I'm roughly right with this summary how is it that cosmic rays ploughing into the earth's atmosphere can be used to prove the safety of the LHC when surely they resemble collision type b. rather than collision type a. ?
DC
The LHC - One ring to rule them all !
The LHC - One ring to rule them all !
Re: Proton vs Graphite
It's all in the LHC safety report http://lsag.web.cern.ch/lsag/LSAG-Report.pdf:
Proton-proton collisions are foreseen at an energy of 7 TeV per
beam. An equivalent energy in the centre of mass would be obtained in the
collision of a cosmic-ray proton with a fixed target such as the Earth or some
other astronomical body if its energy reaches or exceeds 10^8 GeV, i.e., 10^17 eV
cosmic rays with energies of 10^17 eV or
more that hit each square centimeter of the Earth’s surface is measured to be
about 5x10^–14 per second [5]. The area of the Earth’s surface is about 5x10^18
square centimeters, and the age of the Earth is about 4.5 billion years.
Therefore, over 3x10^22 cosmic rays with energies of 10^17 eV or more, equal to
or greater than the LHC energy, have struck the Earth’s surface since its
formation. This means [6] that Nature has already conducted the equivalent
of about a hundred thousand LHC experimental programmes on Earth
already – and the planet still exists.
- DCWhitworth
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Re: Proton vs Graphite
Right so the energies of some cosmic rays are waaaay in excess of the LHC energies (about 100 PeV ?), that would explain it.
DC
The LHC - One ring to rule them all !
The LHC - One ring to rule them all !
Re: Proton vs Graphite
Ok things are starting to get clearer... Thanks...
So a collision occurs. Well billions of them, in cascades, but the first ones are pretty serious. Makes perfect sense.
BUT its the relativistic part that is really my problem obviously. Everything else makes perfect sense.
I just dont understand how something going REALLY REALLY fast has less impact energy then what you would expect. Well wait, in fact it DOES have that energy, its released as joules into the block? So, what energy is released into the block in total ? the 3.5TeV or 290GeV ? And if its 290GeV where did all the other energy go. We know we put it into the proton..
hehehehe... this is great fun...
So a collision occurs. Well billions of them, in cascades, but the first ones are pretty serious. Makes perfect sense.
BUT its the relativistic part that is really my problem obviously. Everything else makes perfect sense.
I just dont understand how something going REALLY REALLY fast has less impact energy then what you would expect. Well wait, in fact it DOES have that energy, its released as joules into the block? So, what energy is released into the block in total ? the 3.5TeV or 290GeV ? And if its 290GeV where did all the other energy go. We know we put it into the proton..
hehehehe... this is great fun...
Re: Proton vs Graphite
Excellent
Yes! No energy is ever lost... it all has to go somewhere...So, what energy is released into the block in total ? the 3.5TeV?
Re: Proton vs Graphite
I guess it's the 'relative' part rather than relativistic. I'll try a few more examples.Xymox wrote:BUT its the relativistic part that is really my problem obviously. Everything else makes perfect sense.
I just dont understand how something going REALLY REALLY fast has less impact energy then what you would expect.
Example #1: Two cars crash to each other, each goes 100 km/h
They stop at the place and get some amount of damage corresponding to their energy.
Example #2: A car goes 100 km/h and crashes into a solid wall
In this case the effect on that single car is almost exactly the same as in example 1 including the amount of damage it takes, although the speed at which the two objects (wall and car) originally approach each other is half compared to the example 1. And there is one more difference - the crash actually makes a tiny little change to the earth rotation speed. In fact, it is example of crashing the car with whole earth.
Example #3: A car going 100 km/h crashes to a stationary car of the same weight.
In this case, if we ignore friction, the two crashed cars will continue moving at 50 km/h after the crash and only half of the first car's original energy is used to damage the car.
Example #4: A car going 100 km/h crashes to a stationary styrofoam plate (imagine it is not attached to anything)
In this case, the plate will get attached to the car's front, it will slow down the car a little bit and damage the car a little bit and sustain some slight damage but the car will continue moving forward at almost the same speed - with the plate - and the plate will just acquire the car's speed.
Well... now we can call the styrofoam plate a carbon atom, the car a beam proton and damage to the car collision energy - and that's pretty much what happens in the dump.
The relativistic part is why the proton can be as heavy as a car and carbon atom as light as styrofoam in this case (before the collision) - and that at the moment when the "styrofoam plate" is finally attached to "car's front", the "styrofoam" will suddenly become much heavier and the "car" much lighter.
Re: Proton vs Graphite
GOOD simple explanation !
However...
Maybe the collision with the graphite causes the graphite atom to move and hit other graphite atoms and so on and so forth until it runs out of energy ? A big shower of collisions turning into heat ?
So the reason the initial first collision is elastic more then a expected 50/50 is ? 3.5TeV to .2TeV ? Mass of the object struck ? I would think a few collisions would hit heavy bits in the graphite atom.
I think im slowly getting this.. slowly...
The proton hits a whole mess of Styrofoam plates till its lost all its energy ? It does not simply transfer all its energy in its first collision ? It has a whole mess of collisions ?
The amount of elastic behavior depends on what your hitting ? Mass ?
And I thought I had some sort of brain up there.. Apparently not
However...
But in this case ALL the energy is absorbed by the block. Nothing moves and mostly the only loss is heat. So unlike the Styrofoam plate the graphite is fixed to the earth and the protons hit a mostly stationary target.Well... now we can call the Styrofoam plate a carbon atom, the car a beam proton and damage to the car collision energy - and that's pretty much what happens in the dump.
Maybe the collision with the graphite causes the graphite atom to move and hit other graphite atoms and so on and so forth until it runs out of energy ? A big shower of collisions turning into heat ?
So the reason the initial first collision is elastic more then a expected 50/50 is ? 3.5TeV to .2TeV ? Mass of the object struck ? I would think a few collisions would hit heavy bits in the graphite atom.
I think im slowly getting this.. slowly...
The proton hits a whole mess of Styrofoam plates till its lost all its energy ? It does not simply transfer all its energy in its first collision ? It has a whole mess of collisions ?
The amount of elastic behavior depends on what your hitting ? Mass ?
And I thought I had some sort of brain up there.. Apparently not
Re: Proton vs Graphite
Basically yes. The dump block still gets pretty heavy kinetic impulse from the beam.Xymox wrote:The proton hits a whole mess of Styrofoam plates till its lost all its energy ?
Depends on what you consider transfer of energy. When the proton is captured in the carbon nucleus, all of its energy is transferred to the carbon (or rather among the newly created nitrogen nucleus) - but most of it is just used to make the nucleus move as fast as the proton was moving. Only small part is transferred to make the nucleus explode and eventually create some new particles. If we use our styrofoam analogy, it will stick to the car (proton) crack on a few places and continue moving with the car.Xymox wrote:It does not simply transfer all its energy in its first collision ?
Yes and absolute majority is just elastic scattering.Xymox wrote:It has a whole mess of collisions ?
That's too advanced question for me. Under certain energy all collisions are elastic because the particle won't overcome electromagnetic repulsion. But even over this energy, the space between individual carbon nuclei is enormous compared to their radii so the beam goes through mostly empty space and is rather affected electromagnetically.Xymox wrote:The amount of elastic behavior depends on what your hitting ? Mass ?
- DCWhitworth
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Re: Proton vs Graphite
Would it make any difference if the car was a Honda instead of a Proton ?
DC
The LHC - One ring to rule them all !
The LHC - One ring to rule them all !