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Balloon-Borne Cosmic Ray Collision Detectors vs Earthbound Colliders

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Islam Hassan
#1
Dec16-12, 06:32 AM
P: 149
If cosmic rays include particles with energies way way beyond what is achievable with classic colliders like the Tevatron and LHC, why not build light collision detectors and have them carried aloft. AFAIK altitudes of 20km-50km are possible with present high-altitude balloon technology. At such altitudes, the incidence of very very high energy collisions should be greatly increased.

Considering the US$ 9 billion total budget of the LHC and US$ 2 billion sunk cost of the SSC when cancelled, US$ 11 billion could have bought you boatloads of such low-cost balloon-borne collision detectors...


IH
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K^2
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Dec16-12, 07:08 AM
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Setting aside all the problems of carrying an enormous weight of hardware up into upper atmosphere, including huge magnets, stacks of detectors, electronics, liquid helium, condensers for the later, and power plant to power it all. The biggest problem is going to be number of events. Modern accelerators give you either a continuous beam or frequent bursts. The number of events is huge. But by the time you consider all the losses, take into account only relevant events, and clean up the data, you sometimes barely have enough events to draw a straight line through and call it a result. This is the main motivation right now into building bigger and better accelerators. Not for the completely new experiments in higher energies, so much as significantly more data on existing experiments that have raised some questions.

You might get some really high energy events up there. Some of them might be the kind that would be useful to study. There might even be a way to get equipment up there. But by the time it gathers enough data to make some sort of a discovery, we could have built the accelerator, got the data, made the discoveries, and moved on to bigger and better things. It just wouldn't be worth it in the end.
mfb
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Dec16-12, 07:39 AM
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It is not enough to have particles with higher energy. The main feature of a collider is the second high-energetic particle as target: This increases the center of mass energy significantly.
To get the same collision as current LHC collisions (4 TeV per proton), you need cosmic rays of ~20000 TeV or ~2*10^16 eV. Those occur about once per year and m^2. If you lift the ATLAS detector (~7000 tons*) in the upper atmosphere, you get a few collisions per year in the inner part**. This is not enough to observe anything interesting. In addition, you would be completely spammed by background: Low-energy collisions somewhere in the detector.
To make things even worse, those collisions are extremely boosted in our reference frame - you would need a detector with a length of much more than 100m, and the spatial resolution would be quite bad.
Oh, and power/cooling/data transmission would be a significant issue as well.

*the other big detectors have even more mass, except LHCb with 5600 tons.
**this has to be compared with 700 million collisions per second in the current setup.

Vanadium 50
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Dec16-12, 08:51 AM
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Balloon-Borne Cosmic Ray Collision Detectors vs Earthbound Colliders

Fundamentally you have two problems.

1) How do you know in advance where in the atmosphere there will be a high energy cosmic ray event.
2) How do you move your detector there in time?
Islam Hassan
#5
Dec16-12, 11:15 AM
P: 149
Quote Quote by K^2 View Post
Modern accelerators give you either a continuous beam or frequent bursts. The number of events is huge.

But by the time it gathers enough data to make some sort of a discovery, we could have built the accelerator, got the data, made the discoveries, and moved on to bigger and better things. It just wouldn't be worth it in the end.
Quote Quote by Vanadium 50 View Post
Fundamentally you have two problems.

1) How do you know in advance where in the atmosphere there will be a high energy cosmic ray event.
2) How do you move your detector there in time?

The simple -if not simplistic- answer to the both your replies is US$ 11 billion. If a light enough contraption of an airborne detector can somehow be engineered, you could make a hell of a lot of them for that magnitude of funding. Just pop 'em up there and wait. You're bound to get some extremely interesting results from some of them within a year or so...

I think the real problem is indicated by mfb:

Quote Quote by mfb View Post
To make things even worse, those collisions are extremely boosted in our reference frame - you would need a detector with a length of much more than 100m, and the spatial resolution would be quite bad.
That one seems to be the irrevocable, monolithic deal-buster, 100m length is out of the question for weight reasons.


IH
Islam Hassan
#6
Dec16-12, 01:07 PM
P: 149
Quote Quote by mfb View Post
To get the same collision as current LHC collisions (4 TeV per proton), you need cosmic rays of ~20000 TeV or ~2*10^16 eV. Those occur about once per year and m^2.
The quoted occurrence rate is at sea level or the altitudes under consideration of 20km-50km?


IH
mfb
#7
Dec16-12, 05:07 PM
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Outer space, and estimated based on this graph.

The simple -if not simplistic- answer to the both your replies is US$ 11 billion. If a light enough contraption of an airborne detector can somehow be engineered, you could make a hell of a lot of them for that magnitude of funding. Just pop 'em up there and wait. You're bound to get some extremely interesting results from some of them within a year or so...
The major LHC detectors cost about US$ 1 billion each. A floating detector would be more expensive to build, data analysis would cost the same.
The answer is the difference in event rates by a factor of ~1016. To get as many events as a single LHC detector and assuming 100% efficiency, you would need to cover a surface of ~1016m^2. This is 20 times the surface of earth.

And in addition, you would not even see those few high-energetic collisions in all the background.
K^2
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Dec16-12, 07:18 PM
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Quote Quote by Islam Hassan View Post
If a light enough contraption of an airborne detector can somehow be engineered [...]
It can't. This is a shot of CEBAF Hall C at J-Lab. That's just the detecting equipment. None of it is related to actual particle acceleration. So all of this hardware, you'd still need up in the air. And this doesn't include power generators and condenser stations that are located elsewhere.
Vanadium 50
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Dec16-12, 09:56 PM
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LHC detectors weigh thousands of tons. Forget balloons - it would take dozens of aircraft to even lift the detectors. And you still haven't explained how you know where and when an energetic cosmic ray event is coming, and how to fly your detector to it in time to record it.
DrDu
#10
Dec17-12, 03:52 AM
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Quote Quote by mfb View Post
To get the same collision as current LHC collisions (4 TeV per proton), you need cosmic rays of ~20000 TeV or ~2*10^16 eV.
Could you elaborate on how to derive the factor 5000 difference?
DrDu
#11
Dec17-12, 03:54 AM
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I remember that there are experiments with extended particle shower detectors in desert areas which allow to detect the tracks of the secondary particles generated in cosmic ray impacts.
K^2
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Dec17-12, 06:15 AM
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Quote Quote by DrDu View Post
Could you elaborate on how to derive the factor 5000 difference?
If we look at p-p collision, center of mass collision at 4TeV means ##\small \gamma \approx 4\times 10^3## of each particle. Or v ≈ 0.99999997c. If target particle is at rest, the relative velocity for the same collision is 2v/(1+vē/cē). If you substitute this velocity into Lorentz boost formula, you get the new ##\small \gamma \approx 3.2\times 10^7##. So for a proton-proton collision with target at rest, you need a 32,000TeV beam to replace a 4TeV beam in head-on collision. If your target is massive, this number will come down a bit. 20,000TeV sounds about right, but feel free to verify it.
I remember that there are experiments with extended particle shower detectors in desert areas which allow to detect the tracks of the secondary particles generated in cosmic ray impacts.
That's basically how we know that such events happen, yes. But by the time that shower propagates to the detector, about the only information we get is the energy of initial particle.
DrDu
#13
Dec17-12, 07:38 AM
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Gee, funny!
So you are telling me that relativistic collisions are highly inefficient when one of the particles is at rest as we have to take the center of energy and not that of mass.
mfb
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Dec17-12, 08:38 AM
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Quote Quote by DrDu View Post
Gee, funny!
So you are telling me that relativistic collisions are highly inefficient when one of the particles is at rest as we have to take the center of energy and not that of mass.
Right. All high-energy particle accelerators are colliders exactly for that reason.

20000 TeV was the result of a rough approximation, but I forgot a factor of 2 (which gets squared afterwards):

##\sqrt{s}=|p_1+p_2|## with the 4-momenta pi should be the same.
- For symmetric head-on collisions, this is just twice the sum of the particles: ##\sqrt{s}=E_1+E_2=2E_1##. It grows linear with the particle energy.
- For collisions with a fixed target 2, the formula can be simplified to ##\sqrt{s}=\sqrt{m_1 + m_2 + 2E_1 m_2}##. Neglecting the first two terms, ##\sqrt{s}=\sqrt{2E_1 m_2}##. It grows with the square root of the energy.

This gives 68200 TeV as required energy to simulate a collision at the LHC.

Quote Quote by K^2
So for a proton-proton collision with target at rest, you need a 32,000TeV beam to replace a 4TeV beam in head-on collision. If your target is massive, this number will come down a bit.
Interesting collisions in the LHC are always parton<->parton, so even if the target is a nitrogen atom or something else heavy, it does not change the relevant center of mass system and its energy.
Islam Hassan
#15
Dec18-12, 09:25 AM
P: 149
Quote Quote by mfb View Post
Right. All high-energy particle accelerators are colliders exactly for that reason.

20000 TeV was the result of a rough approximation, but I forgot a factor of 2 (which gets squared afterwards):

##\sqrt{s}=|p_1+p_2|## with the 4-momenta pi should be the same.
- For symmetric head-on collisions, this is just twice the sum of the particles: ##\sqrt{s}=E_1+E_2=2E_1##. It grows linear with the particle energy.
- For collisions with a fixed target 2, the formula can be simplified to ##\sqrt{s}=\sqrt{m_1 + m_2 + 2E_1 m_2}##. Neglecting the first two terms, ##\sqrt{s}=\sqrt{2E_1 m_2}##. It grows with the square root of the energy.

This gives 68200 TeV as required energy to simulate a collision at the LHC.

Thanx for the info; so not practical at all really. I was also thinking the apparatus could be miniaturised to below 1-2 tons, which isn't the case either.


IH


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