Relativistic motion in a particle accelerator: Rate of Energy Loss

  • #1
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Homework Statement


(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.

Homework 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]


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.
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
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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.
 
  • #3
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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?
 

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