Synchrotron radiation

• When a particle with velocity v is deflected it emits radiation : the synchrotron radiation.

• Relativistic particles emits in a characteristic cone 1/g

• The emitted power is strongly dependent from the energy (^4) and inversely proportional to the curvature radius (^2)

• In the accelerators particle emit in the dipoles, but also in the insertion devices. Due to the very interesting characteristics of the emitted radiation special periodic magnetic systems (wigglers, ondulators) are integrated in the storage rings (in dedicated insertion lines) to produce high brillance photon flux.

1/g

f

acceleration

Electron frame

acceleration

q

Lab frame

velocity b

Transformation between frames:-tan q = g-1 sin f (1+b cos f )-1

If f = 900 then q = g-1

SR Properties

• We have : X = gq, wr = bc/r (cyclotron frequency), C = e0 c

𝜔𝑐(𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦) =3

2𝛾3𝜔𝜌 =

3𝛾3𝑐

2𝜌• Frequency spectrum:

𝑑𝑃(𝜔)

𝑑𝜔𝑑Ω=𝛽2𝑞2𝜔2

𝐶 16𝜋3− 𝑒||𝐺(𝜔)|| + 𝑒⊥𝐺(𝜔)⊥

2

With 𝐺(𝜔)|| =2 1+𝑋2

𝜔𝜌𝛾2 3

𝐾 2 3 ξ , 𝐺(𝜔)⊥ =2𝑋 1+𝑋2

𝜔𝜌𝛾2 3

𝐾 1 3 ξ

Respectively the components for the polarization parallel and perpendicular to the orbit plane.

So : 𝑑𝑃(𝜔)

𝑑𝜔𝑑Ω=

3 𝛾2𝑞2

𝐶 16𝜋3𝜔

𝜔𝑐

2

1 + 𝑋2 2 𝐾 2 3

2ξ +

𝑋2

1+𝑋2𝐾 1 3

2ξ

Due to the properties of the asymptotic behavior of the Bessel function the SR is negligible forξ >1. This means that

𝜔 ≤2𝜔𝑐

1 + 𝑋232

, 𝜃 ≤1

𝛾

𝜔𝑐

𝜔

13

So the radiation has a continuous spectrum up to the critical frequency, than it decrease. The radiation for high frequency is confined in a 1/g angle. For big angles -> low frequencies.

parallel perpendicular

• INTEGRATING in dw => Angular distribution

𝑑𝑃

𝑑Ω=

7 𝑞2

96 𝜋 𝑐 𝜖0

𝛾2𝜔𝑐

1 + 𝑋252

1 +5𝑋2

7 1 + 𝑋2

• And INTEGRATING on the full angular range => Energy flux

𝐼 𝜔 =2𝑞2𝛾

9𝜖0𝑐𝑆

𝜔

𝜔𝑐, 𝑤𝑖𝑡ℎ 𝑆(𝑥) =

9 3

8𝜋𝑥

𝑥

∞

𝐾 5 3 𝑗 𝑑𝑗 0

∞

𝑆 𝑥 𝑑𝑥 = 1

• The energy flux gives the instantaneous radiated power:

𝑃𝛾 =1

2𝜋𝜌 0∋∞

𝐼 𝜔 𝑑𝜔 =4𝑞2𝛾𝜔𝑐

36 𝜋 𝜌 𝜖0𝑜𝑟 𝑖𝑛 𝑎 𝑚𝑜𝑟𝑒 𝑐𝑜𝑛𝑣𝑒𝑛𝑖𝑒𝑛𝑡 𝑓𝑜𝑟𝑚:

𝑃𝛾 =𝑐 𝐶𝛾2𝜋

𝐸4

𝜌2~𝐸2𝐵2 𝑤𝑖𝑡ℎ 𝐶𝛾 = 8.85 10−5

𝑚

𝐺𝑒𝑉 3

The critical photon energy is ħwc=0.665E2[GeV]B[T]

The total energy radiated in one revolution is ∶ 𝑈0 =𝐸4𝐶𝛾

2𝜋 𝑑𝑠

𝜌2

and for an isomagnetic ring -> 𝑈0 =𝐸4𝐶𝛾

𝜌, average power -> 𝑃𝛾 =

𝑈0

𝑇0=

𝑐 𝐶𝛾

2𝜋

𝐸4

𝑅𝜌(T0=

𝛽𝑐

2𝜋𝑅= revolution period)

SR Properties

SR Properties

Quantistic point of view

• Emitted radiation quanta with energy u=ħw

• n(u)du=I(w)dw number of photons per unit time and emitted in the interval dw @ w

𝑛 𝑢 =9 3

8𝜋𝑢𝑐2𝑃𝛾

𝑢 𝑢𝑐

∞

𝐾 5 3 𝑗 𝑑𝑗

The total number per second is:

= 0

∞𝑛 𝑢 𝑑𝑢 =

15 3𝑃𝛾

8 𝑢𝑐=

𝛼𝑐𝛾

2 3𝜌

with a= fine structure constant. The average number per revolution is:

N 2 p r/c= 5𝜋

3𝛼𝛾

The moments of the energy distribution are :

𝑢 =1

𝑁 0

∞𝑢 𝑛 𝑢 𝑑𝑢 =

8

15 3𝑢𝑐 , 𝑢2 =

1

𝑁 0

∞𝑢2 𝑛 𝑢 𝑑𝑢 =

11

27𝑢𝑐

2, 𝑢2 =3𝐶𝑢𝐶𝛾

4𝜋

ℏ𝑐2

𝑚𝑐2 3

𝐸7

𝜌3

We can express the distribution of the quantum fluctuation as : 𝑁 𝑢2 = 𝐶𝑜𝑠𝑡𝐸7

𝜌3

Seventh power!!!!

Synchrotron Cooling (heating?)

• Without energy losses the 6D emittance is an invariant of motion

• But particle loose energy by synchrotron radiation. This energy is restored in the RF cavities by a longitudinal field

• In longitudinal motion higher energy particles lose more energy than the lower energy ones. They recover energy with the same field -> The motion is damped -> cooling

• In transverse space the synchrotron radiation is emitted in a 1/g

cone. Loss of transverse momentum. The energy is recovered longitudinally -> Angle damping, position unchanged -> cooling

Longitudinal damping• In one revolution a particle (not the synchronous one) loses U(E) by SR and accelerate under the influence of

the e.m field in the RF cavity E=qV(t) with t=arrival time ∆𝜃

𝜔0. So the energy variation for the non synchronous

particle is:

1)𝑑(∆𝐸)

𝑑𝑡=𝑞𝑉 𝜏 − 𝑈(𝐸)

𝑇0

• Now let’s expand the SR power around the synchronous energy at the first order :

𝑈 𝐸 = 𝑈0 +𝑊∆𝐸, 𝑊 =𝑑𝑈

𝑑𝐸|𝐸0

• Let’s consider that : 𝛼𝑐 =𝐸∆𝐶

𝐶∆𝐸this define the path difference of the non synchronous particle in respect

to the synchronous one. The difference in the arrival time coordinates will be:

∆𝜏 = 𝛼𝑐𝐶∆𝐸

𝑐𝐸= 𝛼𝑐𝑇0

∆𝐸

𝐸. So the time derivative of t is:

2)𝑑𝜏

𝑑𝑡= 𝛼𝑐

∆𝐸

𝐸

• Combining 1 and 2 we get a classical second order damped equation :

𝑑2𝜏

𝑑𝑡2+ 2𝛼𝐸

𝑑𝜏

𝑑𝑡+ 𝜔𝑠

2𝜏 = 0

with 𝛼𝐸 =𝑊

2𝑇0(damping term) and : 𝜔𝑠 = −

𝛼𝑐𝑞

𝑇0𝐸

𝑑𝑉

𝑑𝜏synchrotron frequency

• Remind (longitudinal motion)

𝑑2𝜏

𝑑𝑡2+ 2𝛼𝐸

𝑑𝜏

𝑑𝑡+ 𝜔𝑠

2𝜏 = 0

𝑤𝑖𝑡ℎ 𝛼𝐸 =𝑊

2𝑇0,𝑊 =

𝑑𝑈

𝑑𝐸|𝐸=𝐸0 , 𝑈 = synchrotron radiated energy,

𝜔𝑠 = −𝛼𝑐𝑞 𝑉

𝐸𝑇0, 𝛼𝑐=momentum compaction

Damped solution: 𝜏 𝑡 = 𝐴𝑒−𝛼𝐸𝑡c𝑜𝑠(𝜔𝑠𝑡 − 𝜑0)

Longitudinal damping

Transverse Damping, Vertical motion

y y

In the vertical plane the particle undergoesBetatron oscillations. So the vertical

displacement will be 𝑦 = 𝐴 𝛽𝑐𝑜𝑠𝜃, 𝑦′ = −𝐴

𝛽𝑠𝑖𝑛𝜃

where the amplitude A is given by the Courant Snyder invariant 𝐴2 = 𝛾𝑦2 + 2𝛼𝑦𝑦′ + β𝑦′2

For every turn the lost energy (SR with 1/g angle) is restored by the RF cavity (zero angle).This change the longitudinal momentum. For the zero synchrotron amplitude particle :∆𝑝

𝑝=

∆𝑦′

𝑦′=

𝑈

𝐸with U -> energy lost by SR and E -> nominal energy.

Using the Amplitude definition it is possible to demonstrate that after many kicks, in average,

(averaged on all the betatron phases) the amplitude to the first order will vary as : ∆𝐴

𝐴= −

𝑈

2𝐸

So we will have that 𝑑𝐴

𝑑𝑡= −

𝑈

2𝐸𝑇0𝐴 with a damped solution 𝐴𝑒

−𝑡

𝛼𝑦 with

𝛼𝑦 =𝑈

2𝐸𝑇0-> damping decrement.

Transverse Damping. Horizontal motion

• Here the horizontal displacement is given by

𝑥 = 𝑥𝛽 + 𝑥𝐷, 𝑥𝐷= D(s)∆𝐸

𝐸

𝑥′ = 𝑥′𝛽 + 𝑥′𝐷, 𝑥′𝐷= D′ s∆𝐸

𝐸

• Emitting an energy u by SR the betatron displacement will vary by:

δ𝑥𝛽 = − D s𝑢

𝐸, 𝛿𝑥′𝛽 = − D′ s

𝑢

𝐸

• Always calculating the average of the Amplitude variation we get (after calculations):

∆𝐴

𝐴=𝑈02𝐸

𝐷

𝜌2K(s) +

1

𝜌2𝑑𝑠

𝑑𝑠

𝜌2

−1

= 𝒟𝑈02𝐸

• 1) The second term is positive…so the SR emission amplify the betatron motion….

• 2) Including the RF damping term (see vertical damping)

∆𝐴

𝐴= −(1 − 𝒟)

𝑈02𝐸

• so the Horizontal damping coefficient will be: •

𝛼𝑥 = −(1 − 𝒟)𝑈02𝑇0𝐸

Transverse Damping. Horizontal motion

Summarizing• Three dimensions : solution for the amplitude

𝐴𝑒−

𝑡𝛼𝑤

Aw for the different planes can be written as:

𝛼𝑥 = ℐ𝑥𝑃𝛾

2𝐸, 𝛼𝑦 = ℐ𝑦

𝑃𝛾

2𝐸, 𝛼𝐸 = ℐ𝐸

𝑃𝛾

2𝐸,

Where

ℐ𝑥 = 1 − 𝒟, ℐ𝑦 = 1, ℐ𝐸 = 2 + 𝒟

-> Robinson Theorem ℐ𝑖 = 4, 𝑜𝑟 ℐ𝑥 + ℐ𝐸 = 3

The correspondent damping time constant will be:

𝜏𝑥 =2𝐸

ℐ𝑥 𝑃𝛾, 𝜏𝑦 =

2𝐸

ℐ𝑦 𝑃𝛾, 𝜏𝐸 =

2𝐸

ℐ𝐸 𝑃𝛾,

Can we go to zero emittance?

• NO • Why?• Photons are stochastically emitted in a very short

time ~r/cg

• This is much shorter than the revolution period. So emissions are instantaneous.. In this Dt the particle makes a discontinuous energy jump. Emissions are independent so the obey the Poisson distribution. These emissions perturb the particle orbit adding a random noise spread, so increasing the average oscillations.

• Equilibrium is attained when the quantum fluctuation rate equals the radiation damping

Quantum excitation

• In the energy domain a particle with DE in respect to the nominal energy undergoes to synchrotron oscillations with amplitude A.

• If at t1 a photon with energy u is emitted DE will be

∆𝐸 = 𝐴0𝑒𝑗𝜔(𝑡−𝑡0) − 𝑢𝑒𝑗𝜔 𝑡−𝑡0 = A𝑒𝑗𝜔 𝑡−𝑡0

with 𝐴2 = 𝐴02+ 𝑢2 − 2𝐴0 u cos𝜔 𝑡 − 𝑡0

Since the emission is independent t is equally distributed in time and in average the oscillatory term will disappear giving:

δ𝐴2 = 𝐴2 − 𝐴02

𝑡= 𝑢2

So we will have:

𝑑𝐴2

𝑑𝑡=

𝑑 𝐴2

𝑑𝑡= 𝒩𝑢2 with =emission rate

Equilibrium energy spread• Let’s add to the last equation the damping term as seen in the damping part:

𝑑 𝐴2

𝑑𝑡= −2

𝐴2

𝜏𝐸+𝒩𝑢

The stationary state solution will be 𝐴2 =1

2𝒩𝜏𝐸𝑢

2 so for a monoenergetic emission

𝜎𝐸2 =

𝐴2

2=1

4𝒩𝜏𝐸𝑢

2

• To take into account a continuous spectrum 𝜎𝐸2 =

1

4𝐺𝐸𝜏𝐸

• Where

a) 𝐺𝐸 = 𝒩 𝑢2𝑠

= average emission on the ring in all the spectrum

b) 𝒩 = 0∞𝑛 𝑢 𝑑𝑢 = rate of the emitted photons in all the spectrum

c) 𝒩 𝑢2 = 0∞𝑢2𝑛 𝑢 𝑑𝑢 =

𝑑 𝐴2

𝑑𝑡= Average amplitude growth rate by photons emitted in all the spectrum

Remember (slide on Q fluctuation) 𝑢2 =3𝐶𝑢𝐶𝛾

4𝜋

ℏ𝑐2

𝑚𝑐2 3

𝐸7

𝜌3If we insert 𝑃𝛾𝑑𝑒𝑠𝑖𝑔𝑛 𝑜𝑟𝑏𝑖𝑡

=𝑐 𝐶𝛾

2𝜋

𝐸4

𝜌2=

𝑃𝛾1

𝜌2𝜌2

We have 𝒩 𝑢2 𝑑𝑒𝑠𝑖𝑔𝑛 𝑜𝑟𝑏𝑖𝑡 =3ℏ𝑐

2𝐶𝑢

𝛾3

𝜌3

𝑃𝛾1

𝜌2

𝑤𝑖𝑡ℎ 𝐶𝑢 =55

24 3

So taking (a) and 𝜏𝐸 =2𝐸

ℐ𝐸 𝑃𝛾

𝐺𝐸 =3ℏ𝑐

2𝐶𝑢𝛾

3𝑃𝛾1𝜌2

1

𝜌3

𝜎𝐸2 =

3ℏ𝛾4𝑚𝑐3

4ℐ𝐸

𝐶𝑢1𝜌2

1

𝜌3

𝜎𝐸𝐸

2

=𝐶𝑞

4ℐ𝐸

𝛾2

1𝜌2

1

𝜌3

𝑤𝑖𝑡ℎ 𝐶𝑞 =3𝐶𝑢ℏ

4𝑚𝑐= 3.83 10−13 𝑚

For an isomagnetic ring :

𝜎𝐸

𝐸

2=

𝛾2𝐶𝑞

ρℐ𝐸

yields 𝜎𝐸

𝐸= 0.62 10−6

𝛾

ℐ𝐸𝜌[𝑚]

Equilibrium emittance• Same procedure:

• δ𝑥𝛽 = − D s𝑢

𝐸, 𝛿𝑥′𝛽 = − D′ s

𝑢

𝐸+

𝑢

𝐸𝜃 (𝜃 < 1 𝛾)

• The second term in the angular kick is negligible

• Plugging in the Courant Snyder invariant to find the average amplitude oscillations (complex)

• 1) δ 𝐴2 = ℋ𝑢

𝐸

2with

2) ℋ =1

𝛽𝑥𝐷2 + 𝛽𝑥𝐷

′ −1

2𝛽′𝑥𝐷

2

• Considering (1) the growth rate is obtained by replacing u2 by 𝒩 𝑢2 and averaging on the accelerator circumference

𝐺𝑥 =𝑑 𝐴2

𝑑𝑡=

1

2𝜋𝑅𝐸2 𝒩 𝑢2 ℋds =

ℋ𝒩 𝑢2𝑠

𝐸2

• Adding the RF damping term we always get a damped equation:

𝑑 𝐴2

𝑑𝑡= −2

𝐴2

𝜏x+ 𝐺𝑥 with solution -> 𝐴2 =

1

2𝐺𝑥𝜏𝑥 and 𝜎𝑥𝛽𝑥

2 =1

2𝛽𝑥 𝐴2

• Taking into account 𝒩 𝑢2 𝑑𝑒𝑠𝑖𝑔𝑛 𝑜𝑟𝑏𝑖𝑡 we finally get :

• 𝐺𝑥 =𝐶𝑞𝑐𝑟0

𝜌2𝛾5

1

𝜌2

ℋ

|𝜌|3

• 𝜖𝑥 =𝜎𝑥𝛽𝑥

2

𝛽𝑥=

1

4𝐺𝑥𝜏𝑥 =

𝐶𝑞

ℐ𝑥

𝛾2

1

𝜌2

ℋ

|𝜌|3

• 𝐶𝑞 =3𝐶𝑢ℏ

4𝑚𝑐= 3.83 10−13 𝑚

• This is called natural emittance. For isomagnetic rings:

• 𝜖𝑥 =𝐶𝑞

ℐ𝑥

𝛾2

ρℋ 𝑑𝑖𝑝𝑜𝑙𝑒𝑠 =

ℐ𝐸

ℐ𝑥

𝜎𝐸

𝐸

2ℋ 𝑑𝑖𝑝𝑜𝑙𝑒𝑠

Vertical plane

• δy𝛽 = 0, 𝛿y′𝛽=

𝑢

𝐸𝜃 𝑦

• δ 𝐴2 =𝑢

𝐸

2𝜃 𝑦

2𝛽𝑦

• 𝜎𝑦𝛽𝑦2 =

1

2𝜏𝑦𝛽𝑦𝐺𝑦

• 𝐺𝑦 =𝒩 𝑢2𝜃 𝑦

2 𝛽𝑦 𝑠

𝐸2~

𝒩 𝑢2 𝜃 𝑦2 𝛽𝑦 𝑠

𝐸2~

𝒩 𝑢2 𝛽𝑦

𝛾2𝐸2

• Since 𝒩 𝑢2 = 𝐺𝐸yields 𝜎𝑦

2

𝜎𝐸2=

𝜏𝑦𝛽𝑦𝐺𝑦

𝜏𝐸𝐺𝐸~

ℐ𝐸 𝛽𝑦

ℐ𝑦𝛾2𝐸2and using ℐ𝑦=1

we get

• 𝜎𝑦2~

𝐶𝑞 𝛽𝑦

𝜌, 𝜖𝑦~

𝐶𝑞 𝛽𝑦

𝜌

Application -> laser cooling (or heating?)

• We insert a focalized laser to impinge on the focalized electron beam. For quasi head on collisions energetic photons are emitted by Compton backscattering effect

• 1st assumption : the laser acts like an undulator with magnetic field 𝐵𝑤 =2 2

𝑐𝑍0𝐼 (Z0=(ce0)-1 =377

-> free space impedance and I = laser intensity).

This emits photons with: 𝑃𝛾 =32𝜋

3𝑟𝑒𝛾

2𝐼 , 𝑟𝑒 = classical electron radius

• 2nd assumption: the collision beam laser are located in a non dispersive region (without D the beam sizes are smaller ...the minimum beam size increase the rate of production ->luminosity ...next course)

• 3rd assumption: laser pulse length short in respect the Rayleigh length ZR (distance of divergence). Same for the electron beam (no hourglass effect). In this way we suppose that the collision between the laser pulse and the electron beam is represented by the crossing of two cylinder with constant radius, neglecting the natural divergence in the IP.

• We gat back to the previous method : we have to define the damping rate and the quantum excitation. Their equilibrium will provide the equilibrium emittances and energy spread.

Damping

• Consider a laser with 𝐸𝐿 = 𝐼 𝑑𝑥 𝑑𝑦𝑑𝑧

𝑐

• For an electron travelling in z the emitted power will be:

𝑃𝛾𝑑𝑧

𝑐=

32𝜋

3𝑟𝑒𝛾

2 𝐸𝐿

𝑍𝑅𝐿

with L= laser wavelength (from lasers 𝑍𝑅𝐿

4𝜋= 𝜎𝑤

2. The waist is the location with minimum transverse sizes…).

• This is the electron energy loss DE!!!

• So we can provide the number of collisions (turns if 1 collision per turn) necessary to damp all the beam energy

𝑛𝑑 =𝐸

∆𝐸=1.6 105𝐿 𝜇𝑚 𝑍𝑅[𝑚𝑚]

𝐸𝐿 𝐽 𝐸[𝑀𝑒𝑉]

• The average damping rate will be :

𝜏𝐿 =1

𝑛𝑑𝑇0= −

1

𝜖𝑛,𝑥,𝑦

𝑑𝑦

𝑑𝑥=

∆𝐸𝐸𝑇0

Quantum excitation

• Zero dispersion region

• Emission in a straight session of un undulator

• Compton is a kinematical effect. The most efficient in boost the photon energy. The photon are emitted in a Spectrum with an energy cut off

wc.𝑑𝑁𝛾

𝑑𝜔=

1

ℏ𝜔

𝑑𝐸𝛾

𝑑𝜔=

3 ∆𝐸

ℏ𝜔𝑐2 1 − 2𝛶 + 2𝛶2 ,

with 𝜔𝑐 = 4𝛾2𝜔𝐿, 𝛶 =𝜔

𝜔𝑐and wL = laser frequency

Kinematic -> energy – angle emitted photon relationship

𝜔𝛾 =𝜔𝐶

1+𝛾2𝜃2

with q = emission angle

Emittance

• The diffusion recoil of the particle after the photon emission is

𝛿𝜑 = ℏ𝜔𝜃

𝐸

Integrating on the full photon spectrum and projecting on the two planes we obtain the emittancediffusion:

∆𝜖𝑛,𝑥\y =𝛾𝛽𝐼𝑃𝑥,𝑦

2 0

𝜔𝑐 𝛿𝜑 2

2

𝑑𝑁𝛾

𝑑𝜔𝑑𝜔 =

3

10

𝑐

𝐿

∆𝐸

𝐸𝛽𝐼𝑃𝑥,𝑦

With c= compton wavelength = h/mc = 2.43 10-12 [m]

• So the average quantum excitation rate will be

𝑑𝜖

𝑑𝑡=

3

10

𝑐

𝐿

∆𝐸

𝐸

𝛽𝐼𝑃𝑥,𝑦

𝑇0

The balance between the Q.Excitation and the damping rate will give :

∈𝑛,𝑥\y 𝑒𝑞𝑢𝑖𝑙.=𝑑𝜖

𝑑𝑡=

3

10

𝑐

𝐿𝛽𝐼𝑃𝑥,𝑦

Energy spread

• In this case the quantum excitation rate

is 𝑑 𝜎𝐸

2

𝑑𝑡=

1

𝑇0 0𝜔𝑐 ℏ𝜔 2 𝑑𝑁𝛾

𝑑𝜔𝑑𝜔 =

7

10ℏ𝜔𝑐

∆𝐸

𝑇0

• Balancing with the damping rate:

• 𝜎δ =𝜎𝐸

𝐸 𝑒𝑞𝑢𝑖𝑙=

7

10

𝑐

𝐿𝛾

aE• To get the parameter as a function of the accelerators parameters we have to calculate W:

𝑈(radiated energy) = 𝑃𝛾𝑑𝑡 = 𝑃𝛾𝑑𝑡

𝑑𝑠𝑑𝑠 =

1

𝑐𝑃𝛾(1 +

𝑥

𝜌)𝑑𝑠 =

1

𝑐𝑃𝛾 1 +

𝐷∆𝐸

𝜌𝐸0𝑑𝑠

• So:

1)𝑊 =𝑑𝑈

𝑑𝐸=

1

𝑐

𝑑𝑃𝛾

𝑑𝐸+𝐷𝑃𝛾

𝜌𝐸𝐸0

𝑑𝑠,

Taking 𝑃𝛾~E2B2

𝑑𝑃𝛾

𝑑𝐸 𝐸0

= 2𝑃𝛾

𝐸0+ 2

𝑃𝛾

𝐵

𝑑𝐵

𝑑𝐸= 2

𝑃𝛾

𝐸0+ 2

𝑃𝛾

𝐵

𝑑𝑥

𝑑𝐸

𝑑𝐵

𝑑𝑥= 2

𝑃𝛾

𝐸0+ 2

𝑃𝛾𝐷

𝐵𝐸0

𝑑𝐵

𝑑𝑥

• Substituting in (1)

𝑑𝑈

𝑑𝐸=

1

𝑐2𝑃𝛾

𝐸0+ 2

𝑃𝛾𝐷

𝐵𝐸0

𝑑𝐵

𝑑𝑥+𝐷𝑃𝛾

𝜌𝐸𝐸0

𝑑𝑠 =𝑈0𝐸0

2 +1

𝑐𝑈0 𝐷𝑃𝛾

2

𝐵

𝑑𝐵

𝑑𝑥+1

𝜌𝐸0

𝑑𝑠

So : 𝛼𝐸=𝑊

2𝑇0=

𝑈0

𝑇0𝐸2 + 𝒟 where 𝒟 is the partition number

𝒟 =1

𝑐𝑈0 𝐷𝑃𝛾

2

𝐵

𝑑𝐵

𝑑𝑥+

1

𝜌 𝐸0

𝑑𝑠 = 𝐷

𝜌2K(s) +

1

𝜌2𝑑𝑠

𝑑𝑠

𝜌2

−1

, 𝑤𝑖𝑡ℎ 𝐾 𝑠 =1

𝐵𝜌

𝑑𝐵

𝑑𝑥(Qpole gradient)

For an isomagnetic ring: 𝒟 =1

2𝜋 𝐷 2K(s) +

1

𝜌2 𝐷𝑖𝑝𝑜𝑙𝑒𝑑𝑠 and if the isomagnetic ring has separated functions magnets :

𝒟 =1

2𝜋𝜌 D(𝑠)

𝜌𝑑𝑠 =

𝛼𝑐𝑅

𝜌

So for separate functions machines 𝛼𝐸 =𝑃𝛾

𝐸=

𝑈0

𝐸𝑇0

Coupling

• Linear coupling. Coupling coefficient κ• Errors, solenoids, skew Qpoles…• Quantum excitation and damping is

shared between the two planes• If we take into account that, since also

in the vertical plane we feel the dispersion, 𝜖𝑥 + 𝜖𝑦=𝜖𝑛𝑎𝑡𝑢𝑟𝑎𝑙 ->

• 𝜖𝑥 =1

1+κ𝜖𝑛𝑎𝑡𝑢𝑟𝑎𝑙 , 𝜖𝑦 =

κ1+κ

𝜖𝑛𝑎𝑡𝑢𝑟𝑎𝑙