1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic motion in a particle accelerator: Rate of Energy Loss

  1. Apr 5, 2012 #1
    1. The problem statement, all variables and given/known data
    (a) Consider a 10-Mev proton in a cyclotron of radius .5m. Use the formula (F1) to calculate the rate of energy loss in eV/s due to radiation.
    (b) Suppose that we tried to produce electrons with the same kinetic energy in a circular machine of the same radius. In this case the motion would be relativistic and formula (F1) is modified by an extra factor of [itex]\gamma[/itex][itex]^{4}[/itex]. Find the rate of energy loss of the electron and com¬pare with that for a proton.

    2. Relevant equations
    (F1):
    P = [itex]\frac{2kq^{2}a^{2}}{3c^{3}}[/itex]

    (F1): (modified to have the "extra factor")
    P = [itex]\frac{2kq^{2}a^{2}\gamma^{4}}{3c^{3}}[/itex]

    [itex]\gamma[/itex] = [itex]\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]

    a = [itex]\frac{v^{2}}{r}[/itex]

    KE = .5mv[itex]^{2}[/itex]


    3. The attempt at a solution
    I did part (a) as best I could. I set kq[itex]^{2}[/itex] = (2.307•10[itex]^{28}[/itex] J•m), solved for v by using the Kinetic Energy formula and converting the 10-MeV value into Joules, doubling it, divide by the proton's mass, and take the square root (my velocity was 43738998.62 m/s). Plug it all in to the equation, the answer comes out in J/s. Convert to eV/s and I get 5.21[itex]^{-4}[/itex] eV/s. The book's answer is P = 5.23[itex]^{-4}[/itex] eV/s. I disregarded the small error as due to rounding numbers throughout the equation.

    Part (b) is what's grinding my gears. I do the same thing I did to get v as before. 10-MeV into Joules, double, divide by the electron's mass, and square root:

    v = [itex]\sqrt{\frac{2(1.6•10^{-12}J)}{9.11•10^{-31}kg}}[/itex]

    I get 1.874•10[itex]^{9}[/itex] m/s. That isn't possible at all. But I run with it, solve for gamma (I got an imaginary number given that v was bigger than c), then solve the equation, convert the answer from J/s to eV/s and I get 1.2eV/s. The answer should be 2.05•10[itex]^{5}[/itex]eV/s. The equation I'm plugging all my numbers into looks like this:

    P = [itex]\frac{2kq^{2}v^{4}\gamma^{4}}{3c^{3}r^{2}}[/itex]

    I understand it may appear that I only tried once, and in laziness decided to post the question on here, but I assure you all that I have tried the problem many times. I'm sure the velocity is wrong, but I don't know how it is wrong. Any help is much appreciated.
     
  2. jcsd
  3. Apr 6, 2012 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For relativistic motion, the energy is

    [tex]E = \gamma m c^2,[/tex]

    so the kinetic energy is

    [tex]T = \gamma m c^2 - mc^2.[/tex]

    The fact you found [itex]v>c[/itex] was a hint that you were using an invalid nonrelativistic expression. As an aside, the equation above lets you determine [itex]\gamma[/itex] directly.
     
  4. Apr 6, 2012 #3
    Mother of mercy, that's it! Solving for Kinetic Energy to get gamma, then solving for v using the equation I got, then plugging it all into the last equation got me my answer. It also helped me out on the two questions after the one I posted, where my velocities were all just a hair below the speed of light.
    Thank you very much. I recognized the equations you posted and was able to get the answers right away. Perhaps I should pay better attention to my textbook, huh?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relativistic motion in a particle accelerator: Rate of Energy Loss
Loading...