Relativistic motion in a particle accelerator: Rate of Energy Loss

Click For Summary
The discussion focuses on calculating the rate of energy loss for protons and electrons in a cyclotron. For the proton, the participant successfully calculated the energy loss as approximately 5.21 x 10^-4 eV/s, which is close to the book's answer. However, the participant struggled with the electron's calculation, initially obtaining an impossible velocity greater than the speed of light, indicating a misunderstanding of relativistic effects. After receiving guidance, they realized the need to use the correct relativistic expressions to find gamma and velocity, ultimately leading to the correct energy loss for the electron of 2.05 x 10^5 eV/s. The conversation highlights the importance of understanding relativistic motion in particle physics.
JackFlash
Messages
13
Reaction score
0

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 \gamma^{4}. Find the rate of energy loss of the electron and com¬pare with that for a proton.

Homework Equations


(F1):
P = \frac{2kq^{2}a^{2}}{3c^{3}}

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

\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

a = \frac{v^{2}}{r}

KE = .5mv^{2}


The Attempt at a Solution


I did part (a) as best I could. I set kq^{2} = (2.307•10^{28} 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 into the equation, the answer comes out in J/s. Convert to eV/s and I get 5.21^{-4} eV/s. The book's answer is P = 5.23^{-4} 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 = \sqrt{\frac{2(1.6•10^{-12}J)}{9.11•10^{-31}kg}}

I get 1.874•10^{9} 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^{5}eV/s. The equation I'm plugging all my numbers into looks like this:

P = \frac{2kq^{2}v^{4}\gamma^{4}}{3c^{3}r^{2}}

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.
 
Physics news on Phys.org
For relativistic motion, the energy is

E = \gamma m c^2,

so the kinetic energy is

T = \gamma m c^2 - mc^2.

The fact you found v>c was a hint that you were using an invalid nonrelativistic expression. As an aside, the equation above let's you determine \gamma directly.
 
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?
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
8
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K