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Relativity at cern - deflection

  1. Sep 15, 2010 #1
    obviously an electron will change path when passing through a field

    could anyone point me to an equation which suggests how much electrons change path by - i guess it depends on mass, velocity (maybe in the form of momentum for the relativistic case im think of?), field strength and the charge of the particle

    thanks in advance
     
  2. jcsd
  3. Sep 15, 2010 #2
    i just thought
    could you approximate the LHC as circular motion, then use f=mv^2/r to calculate the force needed to give the particle a centrepetal acceleration? the using some sort of particle in field equation, calculate the field strength needed to provide the centrepetal force as momentum increases?

    i just dont know where i can find an equation telling you how much force fields will apply to a particle, ca anyone help there?
     
  4. Sep 15, 2010 #3
    Use the Lorentz-force equation:

    F = q(v X B)

    F is the centripital acceleration to keep the particle in a circular orbit. For v in the LHC, use c = 3 x 108 m/s.

    Here are approx numbers. LHC circumference is ~26 Km. It takes ~ 9 Tesla (average) to keep a 7 TeV proton in orbit.

    Bob S
     
  5. Sep 15, 2010 #4
    i'm not familiar with the cross product yet (im 17, starting further maths right now, and cross product is a few months away)... is v X B just constant times constant?

    when you say "F is the centripital acceleration to keep the particle in a circular orbit.", do you mean i substitute in the expression f=mv^2/r ?
     
  6. Sep 15, 2010 #5
    The math is complicated, see https://www.physicsforums.com/blog.php?b=1887 [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Sep 15, 2010 #6
    eugh.. i think i should give up this project its way out of my league just yet :(
     
    Last edited by a moderator: May 4, 2017
  8. Sep 15, 2010 #7
  9. Sep 15, 2010 #8
    Here is a short-hand formula used by high energy accelerator physicists (βγ Mc2 = relativistic momentum in pc units):

    Beam rigidity Bρ = βγ Mc2/c Tesla-meters, where ρ is bending radius.

    Here ρ = 26,000m/2 pi = 4140 meters. B = [STRIKE]9[/STRIKE] 6.6 Tesla*, so Bρ = [STRIKE]37,300[/STRIKE] ~27,300 Tesla meters,

    So βγ Mc2 = 3 x 108 x 27,300 T-m = ~8.1 TeV. Actual number is 7 TeV per beam.

    * the bend magnet field at full energy is ~8.3 Tesla, and the magnets occupy ~ 80% of the ring, so the average field is ~6.6 Tesla.

    Bob S

    [added] From the LHC Design Report, the bend field is 8.33 Tesla, and the bend radius is 2804 meters, so cBρ = 7.00 TeV

    See www.cern.ch/lhc Chapter 2 in Vol 1.
     
    Last edited: Sep 15, 2010
  10. Sep 16, 2010 #9
    So beam rigidity is just the product of the bending radius and the field strength? Is this property just as it suggests - the rigidity of the beam being how hard the stream of particles is to curve into a circle?

    how innaccurate is this short-hand in comparison with using the lorentz force law or whatever it is you need?

    i can't see a volume 1 or chapter link on that page anywehrre :/

    you guys here are great, youre so helpful.. pretty selfless to spend your time helping strangers you dont know prepare for physics degrees
     
  11. Sep 16, 2010 #10
    Yes and yes.
    The rigidity equation is exact for both relativistic and non-relativistic charged particles. See equation 2.6 in

    See http://books.google.com/books?id=S8...ook_result&ct=result&resnum=3&ved=0CBwQ6AEwAg

    Here p is the relativistic momentum, which for a particle of charge e and total energy E is sqrt[E2 - (m0c2)2]/c = βγ m0c
    www.cern.ch/LHC
    click on Design Report
    click on Beam Parameters

    Bob S
     
  12. Sep 26, 2010 #11
    i just came up witha question: what on earth are pc units?
     
  13. Sep 26, 2010 #12
    For relativistic particles with total energy E , E2 = (pc)2 +(mc2)2

    where mc2 is the rest mass and pc is the momentum times the speed of light. So it is very easy to use the same units for momentum as for the rest mass and the total energy. The actual momentum p might be in GeV/c ("GeV per c") units, but using GeV for pc is easier.

    As stated in a previous post, the particle rigidity is Bρ = (pc)/c = p Tesla-meters, which is in momentum (i.e., GeV/c) units.

    Bob S
     
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