# Problem points. Two frequency RF system. Consider a storage ring negative η τ and the RF system operating at two frequencies: ( ) - PDF

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Homework 17. Due November 11 Problem points. Two frequeny RF system. Consider a storage ring negative η and the RF system operating at two frequenies: ( sin( )) de ds = ev o C sin h k rf o Find stationary

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Homework 17. Due November 11 Problem points. Two frequeny RF system. Consider a storage ring negative η and the RF system operating at two frequenies: ( sin( )) de ds = ev o C sin h k rf o Find stationary point on the phase diagram, draw harateristi phase-spae trajetories (approximately is fine) and show the diretion of the motion by arrows. Problem. 4x5 points. For a single frequeny RF system with Hamiltonian with α indiating an energy loss/gain, H s = η π + 1 C os( ) + α ; η Define the stationary points (RF phases) in the phase spae and indiate level of α when stationary points are no longer exists.. Draw phase spae trajetories for α = 1 1. Show the diretion of the C motion by arrows. 3. Define the depth of the RF buket, e.g. the differene between the maximum and minimum π staying within a single RF separatrix (e.g. being loalized). Express it through the RF voltage, the slip fator and the value of stationary phase. Note onsider the entral separatrix around = Find period of the osillation as funtion of H s inside the entral separatrix (around = 0 ). Solution: Problem 1. Adding the orresponding term into the longitudinal Hamiltonian gives: H s = η π + ev o C os os Fig. 1 shows that graph of the RF potential with minima at nπ and maxima at ±1.047 rad. The orrespond to the stationaty points d ds = η π = 0 π = 0; ϕ = ; dπ ds = ev o ( sinϕ sinϕ ) = 0 ϕ = nπ;ϕ = ±1.047 rad. C Fig.1. RF potential. It has minima at zero and ±π. It has maxima at ±1.047 rad. π φ Fig.. RF separatries for negative slip fator. Expanding the Hamiltonian around the stationary points we an find if the motion is stable around it or unstable (separatrix point). Be we know that for positive mass (slipfator) minima will be stable, but for negative mass (slip-fator) the maxima will be stable. Hene, Fig. show the RF separatries with stable point at ±1.047 rad and unstable at zero and ±π. Problem. With given Hamiltonian H s = η π C + 1 C os( ) + α ; η 0. we an easily find stationary points: d ds = η π = 0 π = 0; ϕ = ; dπ ds = ev o C sinϕ α = 0 sinϕ o = l f = C α ev o. ϕ o± = nπ +ϕ o ( n +1)π ϕ o ; ϕ o = sin 1 ( l f ); Stationary point do not exist (e.g. it requires impossible sinϕ o 1 ) if the energy loss (gain) per turn exeed the maximum energy hange in the RF avity: ΔE = α C ev o We now assume that the energy loss α is positive (typial for synhrotron radiation losses or aeleration), we an see that the potential in the Hamiltonian has maxima at ϕ o = nπ + sin 1 l f π sin 1 l f and minima at ϕ o = n +1 slip fator (typial for eletron storage rings) ϕ o = nπ + sin 1. It means that for negative ( l f ) will be stable. For positive slip fator, ϕ o = ( n +1)π sin 1 ( l f ) will be stable point. For l f =1/ ϕ o+ = sin = π 6 trajetories will be as shown below., e.g. 30 degrees. For a negative slip fator the phase-spae π φ The depth of the RF buket depends on the differene between the maxima and minima of the potential see fig. below. Potential depth for negative (left) and positive (right) slip fator it is obvious that the just different by sign. Partiles outside of the RF buket are not limited in the motion and slip either toward positive or negative infinity Finding the differene is straightforward: 1 C ( + sinϕ o ϕ) ϕo Δ os ϕ ϕ o+ = 1 C ( osϕ o + sinϕ o ( ϕ o π )); and the maximum π an be alulated by simply equating the depth of the potential well with kineti energy η π C : π a = η 1/ ( osϕ o + sinϕ o ( ϕ o π )) It an be rewritten in many forms. Without energy loss the RF buket depth is maximums and equal to π a = 1 Cη To find the small amplitude osillation frequeny, we need to expand the potential around the stationary point ϕ o± to the seond order. Sine α is a linear funtion, it does not ontribute to the seond order ter. Hene, notiing that os ϕ o + δϕ = osϕ o osδϕ sinϕ o sinδϕ = osϕ o 1 δϕ 1/ + sinϕ oδϕ + O δϕ we an simply onlude that our result will be different from what we derive in lass by oeffiient osϕ o : Ω = 1 C η osϕ o We used here the fat that osϕ o = osϕ o+. For large amplitude osillations, we an use the fat that the Hamiltonian is invariant and η π C + 1 os + α = H o = inv C π = η d ds = ± ds = η s = s o ± η ± H o H o H o d os where 1, are stopping points defined by π = 0 : 1 1 C os 1, 1/ os( ) Cα ; d os Cα ; Cα + α 1, = H o. Sine the period of osillations omprises of travel bak and forth, we have: 1 C os( 1, ) + α 1, = H o ; d P = η. H o ev 1 os k RF o Cα
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