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Particle Accelerators.

  1. Sep 17, 2007 #1
    First off I am an amateur scientist and my questions are for the most part academic and any experiments that I do are always done prudently.

    My questions are concerning particle accelerators.

    1. If I accelerate protons within a field provided simply by two plates by which the protons when accelerated never touch the cathode; by what mechanism is the power transferred from the driving circuit to the protons? I keep thinking about it but it looks to me like it is a static field and no current can flow. The protons as they pass the cathode do not pick up electrons and so the field remains static and no power is consumed. This is of course not the case which is why I have the question.

    2. If I had a ring that charged particles will circulate in and one portion of the ring has a pair of plates to provide an accelerating field with, will simply providing this field in time add power to the beam? Or am I missing something. what I am asking exactly is as the protons pass into the area of the field and the field is activated to 10000v will it increase the energy of the beam by 10000ev or is there a problem with the particles already having a "net" charge of 10000v and not be accelerated by the field?

    The answers you provide will greatly clarify things for me in my head and I do appreciate any attention to these questions that this community may provide. Thank you.
  2. jcsd
  3. Sep 18, 2007 #2
    I just realized that the answer to the second question is rather obvious; I don’t know what I was thinking. of course it would add energy to the particles! That is how the cyclotron works it gradually gives the particles more and more power using alternating charged plates. I feel rather stupid right now but anyway If I new everything I would be teaching at MIT or something like that.

    I do have a new question though:
    The unit of measure for the KE of a particle is eV. This is, as I understand it, because the speed of relativistic particles is difficult to show increased energy by, due to diminishing returns. The closer you get to the speed of light. Anyway is there a chart or rules of thumb that will correlate the speed of a Proton when the eV is known? The math would be nice too.

    Anyway thanks for any replies.
  4. Sep 19, 2007 #3
    Let T be the kinetic energy and m the mass of the proton (or any other particle), then
    [tex] v = \frac{pc}{E}\cdot c = \frac{\sqrt{T^2 + 2 mc^2 T}}{mc^2 + T} \cdot c [/tex].
  5. Sep 19, 2007 #4


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  6. Sep 19, 2007 #5
    Thank you for the response, the math is helpful. But I am having trouble using it. (SQRT(T^2 + (2 * m * c^2 * T))/(m * c^2)+ T) * c this is my Excel formula I am using but it does not test right. It does not seem to account for relativistic speeds, since it does not I am wondering what all of the references to mc^2 are for, and why a simple force calc wouldn’t work instead. But I could be misinterpreting the formula so I posted it. I must say I don’t understand it enough to fallow what it is trying to do.

    Proton mass 1.67E-27 Kg
    c 299,792,458 m/s
    1.60217646E-19 Joules per eV.

    1 eV = 13,841 m/s
    1 KeV = 437,695 m/s
    1 MeV = 13,844,810 m/s
    389 MeV = 299,852,854 m/s
    1 GeV = 541,910,366 m/s

    How correct are the numbers on the lower end?

    I might try an integral using a regular Acceleration formula and Coulomb force while adjusting the mass as the speed reaches the speed of light.

    I keep thinking about the energy consumption problem of a PA. if you put a charged particle in a field provided by two plates I still cannot see how a static field can loose its charge. I picture a proton injected between two plates say with a 1v charge the proton is attracted to the Negative plate, the charge is removed from the plate allowing the proton to drift through a hole in the plate, the proton now has K.E. or 1 eV of energy but no VI/t was transferred from the plate to the particle. I know I am missing something but what?

    I checked out the fnal.gov site, some interesting equipment, it had a basic description on PA in the site.

    Thanks for the help; this is a great place to get specific answers without a professor down the hall available to ask.
  7. Sep 19, 2007 #6
    If you copy-pasted your formula, then you've got an error in it: You divide by mc² and then add T instead of dividing by mc²+T. The appearing m's are the mass - it shouldn't be too surprising that mass comes in when you express velocity as a function of kinetic energy ([tex] v = \sqrt{ 2 T / m } [/tex] in classical mechanics). The c's are conversion factors to get mass and energy expressed in the same units (like in E=mc²). I'm not sure how simple a "force calc" is, force is not really a widely-used concept in particle physics. I gave you velocity as a function of kinetic energy because that's what the question "is there a chart or rules of thumb that will correlate the speed of a Proton when the eV is known?" seemed to ask for.

    And in case it wasn't clear by now: The formula is (supposed to be) the correct relativistic one. Resulting velocities >=c therefore are obviously wrong.
    Last edited: Sep 19, 2007
  8. Sep 19, 2007 #7
    1ev = 13,841
    1KeV = 437,694
    1MeV = 13,830,070
    1GeV = 262,326,093
    1TeV = 299,792,326

    This looks right. Thank you. I will have to study this formula until I understand it. I missed that Typo (I didn’t cut and past, I wrote it out in Excel). The force calc is probably the hard way considering I would be using integrals but I don’t see why it can’t be used, charge produces force, protons have mass, mass increases as it approaches the speed of light. I like this formula, it is much simpler. But I suspect that it is indirect and simplified much like Area = r^2 * pi = (d * pi * r) /2 which reflects the logic of how that formula really works. So I have some thinking time ahead of me.

    Anyway thank you again.
  9. Sep 19, 2007 #8
    There is not so much to understand there. The velocity of a relativistic particle is v = pc²/E, where momentum p and energy E are related via (mc²)² = E² - p²c². E,p and m are the properties usually used for describing relativisic particles. You asked for kinetic energy which is defined as total energy minus the energy at rest, i.e. T = E - mc². The formula I gave follows directly from those mentioned in this post. Perhaps try to obtain the eq yourself (it follows pretty straightforwardly by expressing p and E in terms of T and m) to verify if my equation is correct.
    Last edited: Sep 19, 2007
  10. Sep 20, 2007 #9
    It took a series of "duh" moments but I think I have finally figured it out. see I was stuck on how simple to formula was, I know that you simply did substitutions for the p from K.E. but I was having trouble wrapping my head around how the calc. found the limit to be the speed of light without using integrals, I make things to complicated and I can also be dense some times (I wonder what the Z is for my head anyway). Thanks for the input Timo you have saved me a few hours of loosing hair follicles.

    Now I am going to assume for sake of me knowing practically nothing about accelerators that the number of protons (Np) that an accelerator can hurl is as this

    Np = J/T = J/(eV *1.6E-19)
    I = (J/t)/eV

    Where I is the current in amps, J is the energy available to accelerate the particles, eV is of course the energy of the beam in electron Volts. And t is the duration of the pulse.

    I don’t understand the nature of the acceleration field enough to make assumptions more then this, and I tried not to get carried away with my methods, but would you consider this close?

    150MeV,1 Joule,4.17E10 Protons, 3ns pulse, 2.2A beam current

    150MeV,140 Joule,5.83E12 Protons, 3ns pulse, 311.1A beam current
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