LHC forces on a proton

  • #1
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I posted an LHC question a while back but did not get a reply. I'll ask a somewhat different, simplified and more specific question. I hope I am asking the question correctly.

How much inward radial force is needed to keep a proton going in a 27 km circular path at 7 Tev or a Lorentz factor of 7500? What is the relativistic formula for it? Is it just mv^2/r with m being relativistic mass?

Supposedly the proton will slow down since it radiates as it is accelerating circularly, so a tangential force behind it is also needed to keep it at 7 Tev. What is this force and the formula for it?
 

Answers and Replies

  • #2
Dale
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Hi teve,

There was a recent thread by upurg which addressed the radial component of the question:
https://www.physicsforums.com/showthread.php?t=466305

Wrt the tangential force required, I don't know the answer there. I guess you could look into bremsstrahlung radiation and find out what direction it points and how much energy it has, then you would have to correct for that with incoming radiation of equal 4-momentum.
 
  • #3
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There was a recent thread by upurg which addressed the radial component of the question:
https://www.physicsforums.com/showthread.php?t=466305

The last post of that thread suggests the radial force is just mv^2/r, but the velocity is multiplied by the Lorentz factor. Is this correct?

Then with the LHC Lorentz factor 7500 and radius 4300m, the radial force on a proton is
((1.67e-27)(7500c)^2)/4300 = 2e-6 Newtons. For 2808 bunches with 1.15e11 protons each (from wiki) and a 27km circumference that's about 25 kN/m of inward radial force. That's 50 kN for 2 beams.

I roughly estimate the weight of the dipole magnets at 4000 kg/meter. From further rough calculations I find that the dipole magnets have to be bolted down or they could tip over (if they were free standing). Is that right?
 

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