Question about the LHC

  • #1

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I know they are going to be doing experiments where they send two 7 TeV beams in opposite directions and observe collisions but do they send two beams, one 7 TeV and one 3.5 TeV, in the same direction and observe collisions?

How would these results differ? We should see relativistic effects from this different frame of reference right?
 

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  • #2
Redbelly98
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I know they are going to be doing experiments where they send two 7 TeV beams in opposite directions and observe collisions but do they send two beams, one 7 TeV and one 3.5 TeV, in the same direction and observe collisions?

How would these results differ? We should see relativistic effects from this different frame of reference right?
In the 2nd case, there would be less energy in the center-of-mass reference frame, hence less energetic collisions. You're correct that the center-of-mass proton speeds would still be relativistic.
 
  • #3
In the 2nd case, there would be less energy in the center-of-mass reference frame, hence less energetic collisions. You're correct that the center-of-mass proton speeds would still be relativistic.
I know it'll be less energetic from it's frame but what would the difference between the same energy experiment, one with motionless center of mass frame and one with relativistic center of mass frame, be?
 
  • #4
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I know it'll be less energetic from it's frame but what would the difference between the same energy experiment, one with motionless center of mass frame and one with relativistic center of mass frame, be?
but what is making up the hard scattering event is the quarks and gluons (partons) not the protons themselves, hence there will be almost none centre of mass in rest i the lab frame....
 
  • #5
Vanadium 50
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but what would the difference between the same energy experiment
I know it'll be less energetic from it's frame
I think you just answered your own question.
 
  • #6
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I like the idea. It is unlikely to find new physics but it might be worth doing. But I think from a practical point of view it is not doable. In a head on collision of equal energies the products can spew out over the whole 4-pie steradians but in an unbalanced collision all the products will be spewed out in a narrow cone in the direction of the high energy particle travel. I think it would be hard to get enough resolution to see the event (it being too tightly packed).
 
  • #7
Borek
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I wonder if it will be possible to precisely control collision place, I have a gut feeling that in the case of beams travelling in the same direction it will be much more difficult.

However, I learnt to not believe my intuition when it comes to anything that is not perfectly classical, so I can be wrong.
 
  • #8
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I wonder if it will be possible to precisely control collision place, I have a gut feeling that in the case of beams travelling in the same direction it will be much more difficult.

However, I learnt to not believe my intuition when it comes to anything that is not perfectly classical, so I can be wrong.
The beams are traveling in opposite directions.

The main detectors are the LHC, CMS and ATLAS, who will be doing most of the data collecting over the next week or so (1 million times more data than they have now) are very precise and can detect very concentrated collisions. None of the data will be usable even at the LHC because of all the data coming in, and the need for it to be cleaned of bad data sets.

In other news as of this posting the LHC is at 3.5TeV It is go time!!!!
 
  • #9
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I could be doing the math wrong but I get that in the lab frame the two bunches will close on each other at a velocity of 9m/sec.

That is a 3.5Tev bunch hits a 7.0Tev bunch both going in the same direction.
 
  • #10
Borek
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The beams are traveling in opposite directions.
Have you actually read first post in the thread?
 
  • #11
Redbelly98
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I like the idea. It is unlikely to find new physics but it might be worth doing.
Well, if it did find new physics that would violate Einstein's special relativity theory, which states that the laws of physics are the same in all inertial reference frames.

I could be doing the math wrong but I get that in the lab frame the two bunches will close on each other at a velocity of 9m/sec.
I get 0.6c. Did you use the relativistic velocity addition/subtraction formulas?

It would require much less energy to accelerate two beams to +-0.6c (proton energy of 0.23 GeV per beam) than it does to accelerate proton beams to 7000 GeV and 3500 GeV.

So doing what is suggested makes very little sense; there is no advantage (according to the theory of relativity), and there is the disadvantage of it requires several thousand GeV of energy when the same center-of-mass energy could be replicated with counterpropagating beams of only a fraction of a GeV.

Edit: the beams would not be +-0.6c in the cm frame. See my Post #15 for clarification.
 
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  • #12
Well, if it did find new physics that would violate Einstein's special relativity theory, which states that the laws of physics are the same in all inertial reference frames.


I get 0.6c. Did you use the relativistic velocity addition/subtraction formulas?

It would require much less energy to accelerate two beams to +-0.6c (proton energy of 0.23 GeV per beam) than it does to accelerate proton beams to 7000 GeV and 3500 GeV.

So doing what is suggested makes very little sense; there is no advantage (according to the theory of relativity), and there is the disadvantage of it requires several thousand GeV of energy when the same center-of-mass energy could be replicated with counterpropagating beams of only a fraction of a GeV.
Oh yea, duh. Thanks, that's basically the answer I was looking for.
 
  • #13
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but do they send two beams, one 7 TeV and one 3.5 TeV, in the same direction and observe collisions?
The IP asks two beams in the same direction
 
  • #14
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I get 0.6c. Did you use the relativistic velocity addition/subtraction formulas?
I should have stated what I calculated. I calculated the speed of the 7 Tev bunch in the lab frame of reference and I calculated the speed of the 3.5Tev bunch in the lab frame of reference and I found the difference in speeds in the lab frame of reference to be 9m/sec. I did not calculate the closing speed as seen from the frame of reference of the 3.5Tev bunch nor of the 7Tev bunch.
 
  • #15
Redbelly98
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I see now, you did say earlier you were calculating the relative velocity in the lab frame. However, to see how much energy is available to the collision (in terms of forming new particles, for example), we need to know the beam energies in the center-of-mass frame.

Come to think of it, I botched things somewhat in Post #11. I got the velocity of one beam in the reference frame of the other beam ... but that is different than the velocity of each beam in the center-of-mass frame. It wouldn't simply be ±1/2 of the 0.6c value, it would be somewhat more. Still, the lab-energy-cost of doing an experiment at ±that velocity would be 4 or so orders of magnitude less than the 7+3.5 TeV we have been discussing.
 
  • #16
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Are you talking about the differences in energy between the lab and the accelerator itself because of time-dilation? If so, all of the computers in the control rooms were unable to collect data during the ramping of the beams to 3.5TeV because of minute clock changes to keep the time the same in both the accelerator and the control rooms.

If not, sorry for the interruption!
 
  • #17
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Red, I now agree this experiment is silly. But it is interesting to see the effect of relativity at these high energies. 3.5Tev on 7Tev (same direction) is NOT the difference 3.5Tev in fact it is close to zero (four orders of mag. lower!).
 

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