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Help Understanding Wakefield When You Integrate Through Path

by jasonpatel
Tags: integrate, path, wakefield
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jasonpatel
#1
Jun19-14, 07:40 PM
P: 28
Hi All,

I am trying to understand the some of the properties of wakefields, namely the energy change. So, as a preface I am interested in primarily the wakefield due to electron beams as they progress through a curved section (the eletcrons radiate strongly when they are in circular motion). There are two main/simple regimes:

1. When the wakefield is constant and not dependent on how far Δθ the electrons have traversed, we have a wakefield like so:

[tex]\frac{dE}{sds}(z) [/tex]

Which to my understanding (which I am very certain of) describes the [tex]\frac{dE}{ds}[/tex] (the change in energy per distance traveled along its curved trajectory) for a given z (position along the eletcron beam, where zero is defined as the certer of the eletcron beam which we can consider to be gaussian-ly distributed).

2. When the wakefield is not constant and is dependent on how far Δθ the eletcrons have traversed, we have the wakefield like so:

[tex]\frac{dE}{ds}(z,θ) [/tex]

Where this describes the [tex]\frac{dE}{ds}[/tex] (the change in energy per distance traveled along the curved trajectory) for a given z (position along the eletcron beam) and θ (the amount the eletcrons have traversed). Now, the main difference between 1 and 2 is the fact that for two [tex]\frac{dE}{ds}[/tex] is changing wrt θ (is some function of theta).

If you were to integrate regime 1, wrt ds from 0 to L (the total path length), then you would get the total energy change through the curved region as a function of z: [tex] Etotal(z,θ) [/tex].

Now this is where I start getting confused: If we turn our attention to regime two with θ dependence.

WHat do we have when:

1. We integrate [tex]\frac{dE}{ds}(z,θ) [/tex] wrt to θ from 0 to θ (thet total travesrved angle of teh eletcron beam)? We would have something like [tex]\frac{dE}{ds}(z) [/tex] which is still a function of ds (the path travesersed).

2. We integrate [tex]\frac{dE}{ds}(z,θ) [/tex] wrt to the path ds? We would now have something like [tex]Etotal(z,θ) [/tex] which is still a function of θ.

Any help on conceptually understanding this would be greatly appreciated! I have been at it for days!
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