Question about simple linear accelerator

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  • #1
new_r
33
0
Hello,
Lets say we have linear accelerator with
uniform electric field E
and length L.

Lets an electron enters it with high speed v0.

How to calculate speed v the electron will have in output?

Thank you.
 
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  • #2
There are no real linear accelerators utilising static, uniform field.

But if you want to consider one - just calculate its relativistic energy on input, add E*e*L, and calculate back its speed.
 
  • #3
Thank you for the answer.
But it is still hard to believe that additional energy do not depends on the speed of the particle.
Can you advise a laboratory or a scientist who really work with it to ask?

Thank you.
 
  • #4
new_r said:
But it is still hard to believe that additional energy do not depends on the speed of the particle.
Why is it hard to believe? When you drop something in a uniform g-field from height h, the kinetic energy gained over h is m*g*h and also not dependent on the initial speed.
 
  • #5
As I told you - there are no real linear accelerators utilising static E field - so I can't direct you to any publications nor labs.
If you want to see how real linear accelerators work - the most famous is Stanford Linear ACcelerator http://www.slac.stanford.edu/
 
  • #6
to: A.T.

Yes, but if force F will be less by some factor (1-(v/c)^n)^m
you will not notice this with your experiment.

to: xts

What about real accelerators, do here additional energy depends on input speed?

Thank you.
 
  • #7
A.T. said:
Why is it hard to believe? When you drop something in a uniform g-field from height h, the kinetic energy gained over h is m*g*h and also not dependent on the initial speed.

excuse me? mgh is for potential energy, energy with respect to position H (height) or D (distance). KE is 1/2 mv^2.
 
  • #8
121910marj said:
mgh is for potential energy
which is converted into kinetic energy when you drop something.
 
  • #9
new_r said:
to: A.T.
Yes, but if force F will be less by some factor (1-(v/c)^n)^m
you will not notice this with your experiment.
I don't know what you mean. If both fields are uniform, then the force is constant in both cases.
 
  • #10
new_r, the energy gained from the field is independent of v. The energy required to accelerate a particle, however, grows as v0 increases. That's also where the relativistic correction will enter.

The change in kinetic energy of the particle will be L*E*q regardless of v0. The initial kinetic energy is given by:

[tex]KE=(\gamma_0-1)mc^2[/tex]

With

[tex]\gamma_0=\frac{1}{\sqrt{1-v_0^2/c^2}}[/tex]

Similar equation governs the final kinetic energy at some vf. Knowing the difference between the two, you should be able to find vf.

Edit: This DOES NOT take into consideration the energy lost due to synchrotron radiation.

Edit 2: Re: Your concerns about force on particle in particle's frame. Yes, the force changes, but so does the distance traveled by particle. The two corrections cancel to give you the same change in energy in either frame.
 
  • #11
What about real accelerators, do here additional energy depends on input speed?
Yes, it depends, but not in the way you mean.
Real accelerators are unable (or at least inefficient) to operate with very slow particles. So large accelerators (like SLAC, and especially ring accelerators, like LHC), are feeded by particles initially accelerated by some smaller accelerator.
 
  • #12
K^2 thank you for your post.
Really I was willing step by step to collect info without asking my main question.
But because you mentioned kinetic energy I will do it.

I have hypothesis that kinetic energy can be calculated by different equation than you have wrote.
This is my version:

KE = m*Integral( v/(1-v/c) )dv

If you plot it you will get similar results. The biggest difference is at speed about 0.5-0.75c
The main question is to find real experimental facts against it.

My initial question was only first step to find where I can be wrong.
Do not explore it too deep.
I do not state that additional energy depends on v0.
Just I don't want to miss any possibility, even if it looks not much realistic.
 
  • #13
An addition:
in circular motion I also predict such synchrotron radiation

P=k * Integral ( v/(1-v/c) - v/(1-v^2/c^2)^1/2 )dv

This is very close to Larmor's v^4 when v < c/2
http://en.wikipedia.org/wiki/Larmor_formula#Relativistic_Generalisation

I hope there are enough data to find experimental refutation or confirmation.
Because it slightly differs from the predictions of the relativity.
Any experimental data you know would be much helpful.

Thank you
 
  • #14
Please reread the post labeled: IMPORTANT! Read before posting

This forum is meant as a place to discuss the Theory of Relativity and is for the benefit of those who wish to learn about or expand their understanding of said theory. It is not meant as a soapbox for those who wish to argue Relativity's validity, or advertise their own personal theories.
 

Related to Question about simple linear accelerator

1. What is a simple linear accelerator?

A simple linear accelerator is a type of particle accelerator that uses electromagnetic fields to accelerate charged particles, typically electrons, in a straight line. It is used in various fields of science, including nuclear physics, medical research, and material science.

2. How does a simple linear accelerator work?

A simple linear accelerator works by using a series of electrically charged metal cavities to create alternating electric fields. These fields accelerate the charged particles, which are then focused and guided by magnets into a straight line. The particles gain energy as they travel down the accelerator.

3. What are the main applications of a simple linear accelerator?

A simple linear accelerator has a wide range of applications, including medical treatments such as radiation therapy for cancer, particle physics research, and industrial applications such as materials testing and sterilization. It is also used in the production of medical isotopes for diagnostic imaging.

4. How is a simple linear accelerator different from other types of particle accelerators?

A simple linear accelerator differs from other types of particle accelerators, such as circular accelerators, in its design and function. While circular accelerators use a circular path to accelerate particles, a simple linear accelerator uses a straight path. This makes it more compact and less expensive to build, but also limits the energy that can be achieved.

5. What are the benefits of using a simple linear accelerator?

There are several benefits to using a simple linear accelerator, including its compact design, lower cost, and versatility in applications. It can also produce high-energy particles at a high rate, making it useful for experiments and medical treatments that require a large number of particles. Additionally, it can be easily modified and upgraded for different uses.

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