Proton vs Graphite

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Xymox
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Proton vs Graphite

Post by Xymox » Sun Mar 21, 2010 9:56 pm

I have a question which needs answering...

I have been told that a 3.5TeV beam dump does not result in 3.5TeV collisions.. By 2 different very smart people...

I do not understand why..

If a 3.5TeV proton happens to hit a nucleon of a graphite atom in the beam dump wouldn't a 3.5TeV collision occur ?

It would seem to make sense that this may well have occurred ?

No ?

I really don't follow why hitting a nucleon at rest with a 3.5TeV proton results in a lesser energy collision then a proton / proton center-of-mass collision at 2.36TeV ?

The debate is if man made 3.5TeV collisions have already occurred in the beam dumps but obviously not measured or verified.

I am awaiting a answer from very smart people now but thought I would throw this question out here. This will also serve to show why I am wrong in my thinking. Others have wondered this as well... So this will answer these questions.

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Re: Proton vs Graphite

Post by oxodoes » Sun Mar 21, 2010 10:30 pm

See here: http://www-bd.fnal.gov/public/relativity.html and keep in mind that the center of mass energy is the energy that can go into the creation of new particles.

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Re: Proton vs Graphite

Post by Kasuha » Sun Mar 21, 2010 11:46 pm

Awww... yeah makes sense. I guess at 3.5 TeV protons are way heavier than carbon nuclei too so whole dump is more like braking foam than like a wall for them.

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Re: Proton vs Graphite

Post by Xymox » Mon Mar 22, 2010 12:43 am

ok guys...

Can I get a simple translation ?

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Re: Proton vs Graphite

Post by tswsl1989 » Mon Mar 22, 2010 1:52 am

Treating the situation classically (i.e. ignoring relativity for now)

Take a 3.5TeV moving particle (A) and stationary particle (B) (One of the nuclei in the absorber)

The momentum of A is non zero and the momentum of B is zero, so the total momentum is non-zero, and has the same direction as A.

When A and B collide, not all of the energy will be dissapated (either as new particles or whatever), as the total momentum after the collision will be the same as the total momentum before.


Particle physics collisions are generally described from the "Centre of Mass frame", the reference frame where the total momentum is zero. The energy measured in this frame will differ from the energy measured in the lab frame. The whole situation would also be dealt with using relativistic formulae, due to the velocities and momenta involved.

This is hardly scientifically rigourous, but I hope it answers your question Xymox :)

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Re: Proton vs Graphite

Post by JNW » Mon Mar 22, 2010 5:02 am

When colliding a high energy relativistic proton at a stationary proton, the center-of-mass energy goes approximately as the square root of the energy of the moving proton. You need a hundred times as much energy to get ten times as much energy in the center-of-mass frame.

A 3.5TeV proton hitting a stationary proton will have about the same center-of-mass energy as two 40GeV protons colliding. The SPS at CERN had about ten times that much energy in the 1970s.

Of course, this was the first time that a 3.5TeV beam of protons has been collided with a fixed target. It's just that the physics involved was studied over 30 years ago.

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Re: Proton vs Graphite

Post by Xymox » Mon Mar 22, 2010 6:44 am

Ok wow... This is a ruff one..

Something that sounds so simple in concept, seems to be highly complex.

A very smart person wrote me back and said
The effectiveness of the collision is much higher when colliding head on. Imagine two cars hitting head on vs a car hitting a brick wall. Both make a big mess of the car but much more energy is involved in the head on case.

Don’t forget that the Carbon atom has an effective energy of 12GeV vs the proton with an energy of 3500GeV. The effective centre of mass energy is therefore around 290GeV.
So lets stick with the car into another car or a wall...

If the second car is stationary its the same impact energy as hitting the wall ? Assuming the 2 are aligned. If the other car is going 12MPH and the accelerated car is going 3500MPH then its a 3512MPH head on collision not 290MPH ?

Im still so confuzed... I know everybody is right, I just don't conceptually follow this.

Its this whole center-of-mass thing im not understanding ?

This is a really amazingly weird thing. You would think if you smack something into another thing the resulting energy of the collision would be simple addition. But its not in this case. Why ?

Sorry for my stupidity here.

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Re: Proton vs Graphite

Post by Kasuha » Mon Mar 22, 2010 9:18 am

Xymox wrote:If the second car is stationary its the same impact energy as hitting the wall ? Assuming the 2 are aligned. If the other car is going 12MPH and the accelerated car is going 3500MPH then its a 3512MPH head on collision not 290MPH ?
Um... no. It's like a 3500 kg car crashing into a stationary 12 kg styrofoam car model.

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Re: Proton vs Graphite

Post by chriwi » Mon Mar 22, 2010 9:18 am

I also have problems understanding this, but simply try to put ist this wa,evenso I am not sure if I am right:
The relativistic an thereby effective mass of the 3.5TeV proton is much higher than the mass of the carbonatom. So even assuming a totally unelastic collission (all parts stick togehter and take more or less the same way after collission) which leaves the most energy for creating new particles compared to a fullyelastic collissin (leaving nothing) or partly elastic (leving less) the resulting lumb made of the proton and the carbonatom (or maybe only its nucleus) will still move with with a velocety that allows that most of the 3.5TeV of energy are still stored in the resulting lumb of the mass of about 13 nuclear mass units.

On the other hand we have to take another look at the dump itselfe, it is not only the carbonblock but before this also another strong magnet forcing the protons on a spiral track which makes them lose some of their energy by radiating. So the question is also how much of the enregy is converted to radiation in this way and how much is left for the impact with the stationary carbonatom.
bye

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Re: Proton vs Graphite

Post by Kasuha » Mon Mar 22, 2010 9:33 am

chriwi wrote:On the other hand we have to take another look at the dump itselfe, it is not only the carbonblock but before this also another strong magnet forcing the protons on a spiral track which makes them lose some of their energy by radiating. So the question is also how much of the enregy is converted to radiation in this way and how much is left for the impact with the stationary carbonatom.
That magnet does not decelerate the beam, it only bends it pretty much like all bending magnets all around the circle. There of course is some synchrotron radiation but I believe it's negligible.

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Re: Proton vs Graphite

Post by DCWhitworth » Mon Mar 22, 2010 12:24 pm

Kasuha wrote:
chriwi wrote:On the other hand we have to take another look at the dump itselfe, it is not only the carbonblock but before this also another strong magnet forcing the protons on a spiral track which makes them lose some of their energy by radiating. So the question is also how much of the enregy is converted to radiation in this way and how much is left for the impact with the stationary carbonatom.
That magnet does not decelerate the beam, it only bends it pretty much like all bending magnets all around the circle. There of course is some synchrotron radiation but I believe it's negligible.
IIRC (rare these days !) isn't there some gas in the dump tunnel before the beams hit the actual dump ?
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Re: Proton vs Graphite

Post by DCWhitworth » Mon Mar 22, 2010 12:28 pm

DCWhitworth wrote:
Kasuha wrote:
chriwi wrote:On the other hand we have to take another look at the dump itselfe, it is not only the carbonblock but before this also another strong magnet forcing the protons on a spiral track which makes them lose some of their energy by radiating. So the question is also how much of the enregy is converted to radiation in this way and how much is left for the impact with the stationary carbonatom.
That magnet does not decelerate the beam, it only bends it pretty much like all bending magnets all around the circle. There of course is some synchrotron radiation but I believe it's negligible.
IIRC (rare these days !) isn't there some gas in the dump tunnel before the beams hit the actual dump ?
Yup, just checked http://lhc-machine-outreach.web.cern.ch ... m-dump.htm

It ploughs through nitrogen at above atmospheric pressure, that'll probably disperse the energy a lot I would think.
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Re: Proton vs Graphite

Post by Kasuha » Mon Mar 22, 2010 12:56 pm

Yes there is a titanium plate and nitrogen chamber but main decelerating medium is still the carbon block and I think majority of protons will make it there (almost) undisturbed - otherwise the carbon would be unnecessary.
The purpose of the nitrogen is to prevent fire in case of fault of the shaping magnet causing the whole beam to hit single spot on the block (causing melting/rupture of the block due to extreme heat). Yes, it has some effect on the beam. I believe its effect is comparable to a few milimeters of carbon though.

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Re: Proton vs Graphite

Post by photino » Mon Mar 22, 2010 8:18 pm

The details of the beam dump are not really important for Xymox' question, because for the comparison with head-on collisions only the first collision of each proton matters. The reason his intuition is failing him is because the beam is moving so close to the speed of light, so relativistic effects come into play.

To compare the collision energy with that of a head-on collision, you have to find the equivalent head on collision (so you are comparing like with like). This is done by changing to a moving frame of reference - changing to the point of view of an observer moving towards the beam dump. The frame of reference to pick (as pointed out by others) is the center of mass frame, where total momentum is zero (like in a head-on collision).

Because the beam is moving so close to the speed of light, so will the center of mass frame. In switching to this frame a lot of the energy of the collision will "disappear" compared to your nonrelativistic intuition. Trying to find a way to picture this... is tricky. In a very handwaving way it is the reverse of the effect that you have to expend more and more energy to accelerate a massive object the closer it gets to the speed of light (so much so it is impossible to actually reach the speed of light). Another way to see that "simple addition and subtraction" cannot hold when you get close to the speed of light is to recall that if you see a photon zipping past at speed c, and change to a moving reference frame following it with speed v, from the point of that reference frame the photon will _still_ be moving with speed c (not c-v), because the speed of light is constant. Anyway, the result is what JNW describes.

Another way of describing what is going on is that (as tswsl points out) from the point of view of an observer standing next to the dump, a lot of the beam energy must always go into the kinetic energy of the collision products due to momentum conservation (the products will in general keep on going deeper into the dump), and hence is "already spoken for". While for a head-on collision the total momentum of the products is zero, so in principle all the collision energy is available to be turned into new particles.
Last edited by photino on Mon Mar 22, 2010 9:04 pm, edited 1 time in total.

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Re: Proton vs Graphite

Post by jmayes » Mon Mar 22, 2010 8:50 pm

So if I can put a spin on this (hopefully correct), the graphite is not a hard wall to the protons but more like a net that expands as the protons shoot into it. The resulting kinetic energy would be much lower then a true non moving wall and way less then proton-proton head on as they are the same mass exactly. Just to clarify, I know the graphic does not actually expand but rather the carbon atoms in it have that effect because their center of gravity is way lower then the protons- right?

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