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Special relativity and particle accelerator

  1. Mar 22, 2017 #1
    1. The problem statement, all variables and given/known data
    (a)
    A spaceship at rest in a certain reference frame S is given a speed increment of 0.50c. Relative to its new frame, it is then given a further 0.50c increment.This process is continued until its speed with respect to its original frame S exceeds 0.999c. How many increments does this process require?
    (b)
    The job of the Stanford Linear Accelerator Center (SLAC) in the United States is to accelerate electrons to very high energies, to crash these electrons into things and then to see what happens. The acceleration is accomplished with a 3.2 km long linear accelerator (linac) which is capable of accelerating electrons such that each electron will have energy of 50 GeV at the exit point of the accelerator.i. Assuming that the electrons accelerate constantly down the linac, what is the speed of the electron after going 1 m down the accelerator?
    ii. When the electrons exit the linear accelerator, magnets are used to curve the electron beams. Construction plans indicate that the radius of the curvature is 280 m. By considering the momentum of these now relativistic electrons, what is the minimum magnetic field required to curve the electron beams?

    2. Relevant equations
    v = (v' + u)/(1 + uv/c2) where v is velocity measured by stationary observer v' velocity measured by moving observer and u is the relative velocity between s and s' in this case 0.5c

    Etotal = ϒ mo c2\
    vf = √(2 a Δx) where a is acceleration and Δx is distance covered when starting from 0 velocity

    3. The attempt at a solution
    ok for the first part my attempt was just to write down the velocities after three increments and soon realized it was too tedious is there some kind of geometric progression that the velocities after increments follow.
    not putting my working here because it just pure math mess.

    for part b) i used the fact that the final energy is related to its final velocity by
    vfinal =√(1 - mo2c4/Etotal2) c
    is there anything wrong with the equation on top because when i plug in i get 0.followed by 10 9s.
    assuming that was correct although i highly doubted it i continued and calculated the acceleration in the accelerator by the above equation in relevant equations.
    then i calculated velocity at 1m to get 0.017c which is completely off can someone help me what is wrong with my working if you dont understand my working please ask me i will gladly make it clearer to you thanks for your help!
     
  2. jcsd
  3. Mar 22, 2017 #2

    BvU

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    Forgot a quote (also in your calculations ?)
    v = (v' + u)/(1 + uv'/c2).
    Do it all in terms of c so you avoid a 'tedious pure math mess' :
    After 1 kick, it is 0.5
    After two kicks it is (0.5 + 0.5) / (1+0.52 ) = 1/1.25
    Give this 1.25 the name ##z##
    three (0.5 + 1/z) /z
    four
    ...
    See a patterrn emerge ?
     
  4. Mar 22, 2017 #3
    i get what you say but i dont how to get to your expression for third boost

    this what i got
    v = (v' + u)/(1 + uv'/c^2).
    v' = 0.5 and u based on what you said is 1/z
    hence
    (0.5 + 1/z)(1 + 0.5/z) = (0.5z + 1)(z + 0.5) = (0.5 + (1-0.5^2)/z)this is the best i got dont quite get how to manupulate like yours
     
  5. Mar 22, 2017 #4

    BvU

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    Rightly so ! I got it wrong o:) !
     
  6. Mar 22, 2017 #5
    is there still an pattern
     
  7. Mar 22, 2017 #6

    BvU

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    With excel it's easy, but I wonder if that's the exercise composer's intention....
    An= (0.5 + An-1) / (1+ 0.5 An-1)
     
  8. Mar 22, 2017 #7
    wait i will try it out for a little longer because this is junior physics olympiad paper i am pretty sure the solution would not be that hard be must just be missing something i will try out till tomorrow meanwhile you can help me in part b because i think i may have conceptual error for tha
     
  9. Mar 22, 2017 #8

    BvU

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    On a calculator it isn't all that terrible either; otherwise I don't know:
    Code (Text):

    0.5
    0.8
    1.3/1.4 = 0.928571
    ...
    For part b): with ##\displaystyle {\ \beta = {v\over c} \ }## and ##\ \displaystyle { \gamma = {1\over \sqrt {1-\beta^2} } = {50 \times 10^{9} \over 0.511 \times 10^6} = 0.98 \times 10^5 } ##
    I get ##\ \beta^2 = 1 - 10^{-10} \Rightarrow \beta = 1 - 10^{10}/2 \ ## just like you

    For the second part you can't use a non-relativistic formula. Read the assignment carefully (doesn't help, I know):
    I think it is supposed to mean that the kinetic energy increases linearly. That makes this part rather easy.
     
  10. Mar 23, 2017 #9
    wait how does it mean kinetic energy increases linearly as ke = moc2(ϒ-1)
    how is that increasing linearly
     
  11. Mar 23, 2017 #10

    BvU

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    Has to do with the way such a thing works: a series of pipe sections at alternating voltages. The electrons are sucked into a section at one end and pushed out at the other, because the polarity has reversed in the time they went through. Every time they cross a gap they get a kick of so and so many eV. They reach a speed close to c very quickly because they are so light (you'll be surprised!).

    I grant you the exercise formulation is not very helpful on this -- but then again, it is an olympiad (so it needs hurdles) and it is one of the last exercises...
     
  12. Mar 23, 2017 #11
    no what i was asking is that how are saying that the kinetic energy increases linearly but based on the equation posted on top by me it does not appear so to me
     
  13. Mar 23, 2017 #12

    BvU

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    'based on the equation posted on top by me' is longer than ##E_k = (\gamma-1)m_0 c^2## which is the one you are referring to, I assume ?

    With ##m_0c^2 = 511## keV, ##\gamma## is quickly > 1 and you can ignore the -1.

    I couldn't find the specific details (initial ##E_k##, ##\gamma({\rm distance})\ ## ) in a didactically effective link, but there's a lot of history to read. They mention on page 90: initial ##E_k \approx 80## keV (##v/c## already 0.5!), and after some bunching, 250 keV. Then normality: standard sections of 3 m, 30 MeV/section. so two mile 30 GeV (this is 1966 -- now they are at 50 GeV I understand from the exercise). And you aren't supposed to worry about the gory details at the beginning, but expected to assume the linear accelerator (pun intended) accelerates ##E_k## linearly (pun intended).

    I repeat that I find the exercise formulation unhelpful: if you think too much you run into trouble.
    Does my answer address your concern, or am I misinterpreting what you mean ?
     
  14. Mar 23, 2017 #13
    no sir i really appreciate your effort in finding these articles but what i want is for you explain using your own words without much equations as i want to understand your thought process behind how you formulated the thought not the engineering of SLAC without much equation after i have gotten the logic i promise i will try my best to answer the question
    what do you mean γ quickly more than 1 and hence i can ignore the -1 because i am not really comfortable with approximations. and still how does that imply ke is linearly related to distance because i dont see the link between constant acceleration and linear increase in ke sorry for not understanding what you mean and thanks for your help.

    on a side note i finally found the correct gamma symbol yay!!!
     
  15. Mar 23, 2017 #14

    BvU

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    I can't ! Other than saying that 0.511 MeV can be ignored after a few kicks of the order of 100 kV when the speed is close to light speed.
    And assuming you can pump in so and so many V/m isn't all that far-fetched. What would be the alternative ?
     
  16. Mar 23, 2017 #15
    OK i guess i will try now slowly seeing what you mean by the approximation give me some time to work out the problem again using your approximation
     
  17. Mar 23, 2017 #16
    sir wow i did what you said out popped the answer for both parts of b thanks a lot!!!
    after a bit of googling i see what you mean by linear increase in k.e now
    is it because after some time once the electron is almost close to c the graph of relativistic kinetic energy is almost a steep straight line hence the linear approximations would work am i correct please respond.
    and the only thing left is part a which i think i will just use calculator because i dont see any recurrence relation so all in thanks for your reply please tell me if my understanding of relativistic k.e approximation is correct or not thankss!
     
  18. Mar 23, 2017 #17

    BvU

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    How do you know o_O ?
    Correct
    I think it is
    I did some too and found a similar exercise (8.2b) that says
    and finds a ##\gamma = 92000## for 47 GeV.
     
  19. Mar 24, 2017 #18
    How do you know o_O ?

    oh i have the answer sheet just no working

    i think i should try out stanford problems because they seem to be close my olympiad style questions thanks for the link
     
  20. Mar 26, 2017 #19
    ooh sir i found out after much googling the final velocity after n increments can be calculated easily with the help of hyperbolic functions. i am familiar them but dont see their link with special relativity. the formula was,
    reltive velcity after n increments = tanh( n * arctanh(0.5) )
    where the 0.5 is the boost in moving frame
    plugging in i got 7 which was what i needed
     
  21. Mar 26, 2017 #20

    BvU

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    Bravo ! Well done and my compliments for your tenacity !
     
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