Centre-of-mass Energy of Colliding protons

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The discussion centers on calculating the center-of-mass energy (Ecm) of colliding protons at the LHC, which can accelerate proton beams to 8 TeV. For two colliding protons, the total energy is assumed to be 16 TeV, but clarification is needed regarding the inclusion of rest mass energy, which is negligible. In collisions with a stationary target, the energy required is greater because it must account for both the center-of-mass energy and the motion of the center of mass. Understanding the difference between the center-of-mass frame and the lab frame is crucial for accurate calculations. The conversation highlights the complexities involved in energy conservation during particle collisions.
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Homework Statement


LHC can accelerate proton beams to energies of 8Tev.
1. What is the centre-of-mass energy of two colliding protons?
2. What would be the beam energy needed if one wanted to reach the same centre-of-mass energy when colliding beam protons with protons in a target at rest?

Homework Equations



E2 = (pc)2+(E0)2

KE = E - E0

The Attempt at a Solution



Common sense tells me that two particles of equal mass and energy colliding head on results in all the energy of the particle transferring to "collision energy". So I assume an Etot = (8+8)TeV = 16TeV sounds reasonable unless I need to include rest mass energy, which i would have though is negligible. Does this Etot relate simply to Ecm?

I am having trouble understanding the loss of collision energy in the rest target case, surely all energy supplied to the particle for momentum must be conserved/transferred to the rest proton and there is no center-of-mass energy? I guess my understanding of what Ecm (center-of-mass energy) may be throwing me off slightly?
 
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LHC can accelerate proton beams to energies of 8Tev.
This is wrong, the achieved value is 4 TeV and the design value is just 7 TeV. Anyway, does not matter here.

joakley said:
Common sense tells me that two particles of equal mass and energy colliding head on results in all the energy of the particle transferring to "collision energy". So I assume an Etot = (8+8)TeV = 16TeV sounds reasonable unless I need to include rest mass energy, which i would have though is negligible. Does this Etot relate simply to Ecm?
Right.
The rest energy is already included in the 8 TeV (negligible of course...).

I am having trouble understanding the loss of collision energy in the rest target case, surely all energy supplied to the particle for momentum must be conserved/transferred to the rest proton and there is no center-of-mass energy? I guess my understanding of what Ecm (center-of-mass energy) may be throwing me off slightly?
The center-of-mass system is moving here, there is always a reference frame where both protons have an opposite velocity of equal magnitude. In the lab, you need much more energy as you have to provide the CMS energy PLUS the energy for the motion of the center of mass.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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