# Question about simple linear accelerator

1. Aug 29, 2011

### new_r

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.

2. Aug 29, 2011

### xts

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. Aug 30, 2011

### new_r

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. Aug 30, 2011

### A.T.

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. Aug 30, 2011

### xts

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. Aug 30, 2011

### new_r

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. Aug 30, 2011

### 121910marj

excuse me? mgh is for potential energy, energy with respect to position H (height) or D (distance). KE is 1/2 mv^2.

8. Aug 30, 2011

### A.T.

which is converted into kinetic energy when you drop something.

9. Aug 31, 2011

### A.T.

I don't know what you mean. If both fields are uniform, then the force is constant in both cases.

10. Aug 31, 2011

### K^2

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:

$$KE=(\gamma_0-1)mc^2$$

With

$$\gamma_0=\frac{1}{\sqrt{1-v_0^2/c^2}}$$

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. Aug 31, 2011

### xts

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. Aug 31, 2011

### new_r

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. Sep 2, 2011

### new_r

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. Sep 2, 2011