Where did the expansion for particle accelerators come from?

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In summary, the potential expansion for the potential of axial traveling beam of particles through an electrostatic lens, as seen in the book "Particle Accelerators" by Livingston and Blewett, is derived from the Laplace Equation in cylindrical coordinates. This approximation is useful for approximating paraxial electric fields and can be seen on page 111 of the free downloadable ebook "Principles of Charged Particle Acceleration" by Humphries.
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steve_rb
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In "Particle Accelerators" book of Stanly Livingston I saw attached potential expansion for the potential of axial traveling beam of particles throght a electrostatic lense. I want to know where this expansion came from? Which expansion is being used and how?
 

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In the book "Particle Accelerators" by Livingston and Blewett on page 101, Eq 5-16 is derived from the Laplace Equation in cylindrical coordinates on page 98 Eq 5-6 with ρ=0.

In a charge-free region, the potential V(r,z) with axial symmetry must satisfy Laplace's equation with charge density ρ =0

Laplace's equation in cylindrical coordinates is then

[tex]\frac{1}{r}\cdot\frac{\partial}{\partial r}[r\frac{\partial V(r,z)}{\partial r}]+\frac{\partial^2V(r,z)}{\partial z^2} = 0[/tex]

The following approximation is useful for approximating paraxial electric fields:

[tex]V(r,z)=V(0,z)+A r\frac{\partial V}{\partial z}+ B r^2\frac{\partial^2 V}{\partial z^2} + C r^3\frac{\partial^3 V}{\partial z^3}+ D r^4\frac{\partial^4 V}{\partial z^4}[/tex]

where the partial derivatives are evaluated on the axis. Substituting the second equation into the first yields A = C = 0, and B = -1/4 and D = +1/64.
To see further discussion, look at page 111 Eq 6.2 in the free downloadable ebook by Humphries "Principles of charged Particle Acceleration":

http://www.fieldp.com/cpa.html

Bob S

test Latex equations here http://www.equationsheet.com/textoimage.php
 
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Thanks Bob. Great job.
 

FAQ: Where did the expansion for particle accelerators come from?

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