Special relativity and particle accelerator

[timetraveller123]

Homework Statement

(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?

Homework 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

E_total = ϒ m_o c²\
v_f = √(2 a Δx) where a is acceleration and Δx is distance covered when starting from 0 velocity

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
v_final =√(1 - m_o²c⁴/E_total²) 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 don't understand my working please ask me i will gladly make it clearer to you thanks for your help!

 

[BvU]

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

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 ?

 

[timetraveller123]

i get what you say but i don't 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 don't quite get how to manupulate like yours

 

[BvU]

how to get to your expression for third boost

Rightly so ! I got it wrong !

 

[timetraveller123]

is there still an pattern

 

[BvU]

With excel it's easy, but I wonder if that's the exercise composer's intention...
A_n= (0.5 + A_n-1) / (1+ 0.5 A_n-1)

 

[timetraveller123]

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

 

[BvU]

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

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):

Assuming that the electrons accelerate constantly down the linac,

I think it is supposed to mean that the kinetic energy increases linearly. That makes this part rather easy.

 

[timetraveller123]

wait how does it mean kinetic energy increases linearly as ke = m_oc²(ϒ-1)
how is that increasing linearly

 

[BvU]

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...

 

[timetraveller123]

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

 

[BvU]

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

'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.

it does not appear so to me

Does my answer address your concern, or am I misinterpreting what you mean ?

 

[timetraveller123]

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 don't 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!

 

[BvU]

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 ?

 

[timetraveller123]

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

 

[timetraveller123]

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 don't 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!

 

[BvU]

out popped the answer for both parts of b

How do you know ?

after some time once the electron is almost close to c the graph of relativistic kinetic energy is almost a steep straight line

Correct

tell me if my understanding of relativistic k.e approximation is correct

I think it is

after a bit of googling

I did some too and found ahttp://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect8Prob_2014SOLNS.pdf (8.2b) that says

Electrons increase their kinetic energy by approximately equal amounts for every meter traveled along
the accelerator pipe as observed in the laboratory frame

and finds a $\gamma = 92000$ for 47 GeV.

 

[timetraveller123]

How do you know ?

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

 

[timetraveller123]

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 don't 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

 

[BvU]

Bravo ! Well done and my compliments for your tenacity !

 

[timetraveller123]

thanks but most of the work was yours then this is solved thanks for all

 

[mfb]

ooh sir i found out after much googling the final velocity after n increments can be calculated easily with the help of hyperbolic functions.

Instead of speeds, you can calculate the rapidity. While that doesn't tell you directly how fast something is going, rapidity differences are the same for all observers. Each 0.5 c boost of the spacecraft increases the rapidity of it (as seen by Earth) by the same amount. All you need is the rapidity after one step and the final rapidity.

The rapidity is often used in particle physics because differences in rapidities of different particles are more interesting than speeds.

 

[timetraveller123]

sir by no means have i learn special relativity formally meaning using matrices and all those so these terms about rapidity and hyperbolic rotations i have only touched and go on them i have not deeply understood what they are so i am really to sure what your post means but i am quite familiar with the math just that i don't see the link with the maths and special relativity

 

[timetraveller123]

wait once more after some googling i finally managed to learn a little about the link
so relativistic addition of velcities can be thought of as first hyperbolic rotation to speed v then hyperbolic rotation to speed u thus giving a total rapidity ie the angle as the relative velocity/c am i right

 

[PeroK]

wait once more after some googling i finally managed to learn a little about the link
so relativistic addition of velcities can be thought of as first hyperbolic rotation to speed v then hyperbolic rotation to speed u thus giving a total rapidity ie the angle as the relative velocity/c am i right

It's simpler if you express your velocity as a proportion of the speed of light. In fact, this can be done choosing units where $c = 1$. In any case, for addition of $v$ and $u$ you have:

$u' = \frac{v + u}{1+uv}$

And, for some $\theta_1, \theta_2, \theta_3$ we have

$v = \tanh \theta_1$, $u = \tanh \theta_2$ and $u' = \tanh \theta_3$.

This gives:

$\tanh \theta_3 = \frac{\tanh \theta_1 + \tanh \theta_2}{1+ \tanh \theta_1 \tanh \theta_2} = \tanh(\theta_1 + \theta_2)$

Hence:

$\theta_3 = \theta_1 + \theta_2$

 

[mfb]

The same thing but without having to guess the right function: We want a function that increases in equidistant steps when the spacecraft makes its acceleration steps. Let's call this function f. Its argument v is always between -1 and 1 exclusive, but f(v) is unbound because the spacecraft can make as many steps as it wants.

Instead of large individual steps, let's make a very small step $\epsilon$.
$$v' = \frac{v+\epsilon}{1+v \epsilon} \approx v + (1-v^2) \epsilon$$
We want f(v')-f(v)=h for some constant h (it still depends on $\epsilon$) independent of the velocity. If we plug in v' from above, we get $f(v+(1-v^2) \epsilon)-f(v)=h$, and in the limit of epsilon going to zero we get a derivative:
$$f'(v) = \frac{b}{1-v^2}$$
Here b is another arbitrary constant which does not depend on $\epsilon$ any more. It just scales the overall function, so let's set it to 1.
The solution to this differential equation is $f(v)=atanh^{-1} (v)$.

 

[timetraveller123]

@PeroK that is what i exactly read

and
@mfb i like your anayltical method that is what i like to see

thanks to both of you