Lecture 21 April 15, 2010

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1 Lecture 21 April 15, 2010 We found that the power radiated by a relativistic particle is given by Liénard, P = 2 q 2 [ 3 c γ6 β ) 2 β ] β ) 2. This is an issue for high-energy accelerators. There are two main types, linear circular. In a linear accelerator the direction of β is constant so β β P = 2e2 3c γ6 β) 2 = 2e2 3c = 2e 2 3m 2 c 3 ) de 2. dx γ) 2 β 2 = 2e2 3m 2 c 5 de dt ) / 2 ) dx 2 cdt So the power radiated is independent of the energy depends only on the rate of energy change.

2 The ratio of power lost to power input, de/dt P de/dt = 2e2 1 de 3m 2 c 3 v dx 2 r e de 3 mc 2 dx, where r e = e 2 /mc 2 is the classical radius of the electron, 2.82 fm. As we are unlikely to increase the energy of an electron by its rest energy in a distance of its tiny classical radius, energy loss in a linac is negligible.

3 Circular Acclerators In a synchrotron, particles are in circular orbit with the energy changing slowly but the direction of the momentum changing rapidly, β = ω β β, so P = 2 q 2 [ 3 c γ6 β) 2 β = 2 q 2 c 3 ρ 2 γ4 β 4, β) 2 ] = 2 3 q 2 c γ6 ω 2 β 2 [ 1 β 2] where ρ is the orbit radius we used cβ = ωρ. The energy loss per revolution δe is the integral of this over an orbit, δt = 2π/ω = 2πρ/cβ, or δe = 4π ] 1 3 q2 β 3 γ [ρ 4 where the expression in braces is given rather than a simple 1/ρ in case you design your accelerator to have magnets not completely covering the circumference 1. 1 See footnote lecture 15 p. 2, slide 4. ρ 2

4 Energy loss in synchrotrons For ultrarelativistic particles, β 1, δe E 4 /ρ, with the proportionality constant m/gev 4 for electrons m/gev 4 for protons. For Lep, an electron beam of roughly 80 GeV a radius of about 4 km, the electrons lose nearly a GeV per turn! This is why the ring is as big as it is. For the LHC, which will have 7 TeV protons at the same radius, I get a loss of only 4 KeV per turn, so energy loss is not the crucial issue for proton synchrotrons, but bending radius is.

5 Angular Distribution We derived he complete expression for F αβ in covariant form F αβ = q U ρ x ρ r ρ 1) τ)) [ d x rτ)) α U β τ) x rτ)) β U α ] τ) dτ U µ x µ r µ τ)) but it is often useful to have the expression more explicitly in three dimensional language. Using R as the 3-vector from r α τ 0 ) to x α, with magnitude R direction ˆn, we have R α := x α r α τ 0 ) = R, Rˆn), U α τ 0 ) = γc, γc β) du α dτ = γ du α dt = γ 4 cβ β, cγ 2 β + cγ 4 β β β ), where we used γ = γ 3 β β, we underst that β, γ are to be evaluated at the retarded time τ 0. β τ 0.

6 In??) we then have dx α r α τ)/dτ = U α U ρ x ρ r ρ τ 0 )) = Rγc1 ˆn β), but d dτ U x r) = U 2 du α + x r) α dτ = c 2 + R cγ 4 β β ˆn βcγ 2 ˆn βcγ 4 β β ). Thus F αβ = q R 3 γ 3 c 3 1 ˆn β) 3 [R α du β dτ du α Rβ dτ R α U β R β U α ) c 2 + R ) Rcγ1 ˆn β) { cγ 4 β β ˆn βcγ 2 ˆn βcγ 4 β β}) ].

7 E For the electric field, E = F i0 ê i we have E = q R 3 γ 3 c 3 1 ˆn [ β) 3 R nγ 4 cβ β Rcγ 2 β + cγ 4 β β β) ) Rcγ1 ˆn β) γcr n Rγcβ) c 2 + R { cγ 4 β β ˆn βcγ 2 ˆn βcγ 4 β β }) ] Some algebra spelled out in the lecture notes gives qˆn β) E = R 2 γ 2 1 ˆn β) + q ˆn ˆn β) ) β 3 Rc 1 ˆn β). 3

8 B For the magnetic field, qɛ ijk B i = 1 2 ɛ ijkf jk = R 3 γ 3 c 3 1 ˆn [ β) 3 Rn j cγ 2 βk + Rn j cγ 4 β ββ k )Rcγ1 ˆn β) Rn j γcβ k c 2 +Rcγ 4 β β1 ˆn β) ˆn )] βcγ 2 qˆn β) = i R 2 γ 2 1 ˆn β) 3 q [ Rc1 ˆn β) ˆn ] β1 ˆn β) + ˆn βˆn β 3 i so B = ˆn E. as it should be for a radiation-zone field.

9 Power Flux Thus we can derive the expression for the power radiated towards the observer, the flux being given by the Poynting vector ˆn S = c 4π E2. At large distances this is ˆn S = ret q2 4πcR 2 [ ˆn ˆn β ) 1 ˆn β ] β ret ) 3 2. For ultrarelativistic β 1 particle, near forward direction, ˆn β β) 6. 1 flux received 1 ˆn β) 5. But the power radiated is only 1 ˆn Why?

10 Energy per Unit Area The power/unit area ˆn S received during [t, t + t] is ret determined by retarded times [t e, t e + t e ] corresponding to emission τ 0 s. Light received at t + t had to travel a distance ˆn v t e less than the light received at t, so t = 1 ˆn β) t e. Total energy emitted = total energy received, so power emitted is 1 ˆn β) times the power received. Natural to express things in terms of the emission time, t e, with t = t e + Rt e )/c

11 The energy per unit area we receive is E/A = dt ˆn S = dt ˆn S ret = dt ˆn S 1 ˆn β ). t t ) d dt t + Rt ) c So the expression which determines the energy distribution is [ ˆn ˆn β ) ]) 2 β ) 5. 2) dp da = q2 4πcR 2 1 ˆn β

12 Linear acceleration Let us consider two important special cases. The first has the acceleration in the same direction as the motion, β β. Then the numerator is ˆn ˆn β) 2 = sin 2 θ v 2 /c 2, dp da = q2 v 2 4πc 3 R 2 sin 2 θ 1 β cos θ) 5. For β close to 1 this is very strongly peeked in the forward direction. The maximum intensity is when so d dx 1 x 2 1 βx) 5 ) x=cos θ x = cos θ max = = 0 = 2x 1 βx) 5 + 5β1 x2 ) 1 βx) β 2 1 With β = 1 γ 2 1 1/2γ 2 ), x 1 1 8γ 2 θ max 1/2γ. 3β.

13 For such small angles, with θ << 1 but without taking γθ small, the intensity is dp da = 8q2 v 2 γθ) 2 πc 3 R 2 γ8 1 + γ 2 θ 2 ) 5. As an example, consider the linear accelerator at SLAC, which accelerates electrons to 50 GeV over a distance of 3 km. At the end, γ f = 50GeV/0.511MeV 10 5,, as the travel has been vitually at the speed of light, t = 10 5 s. Assuming the energy gain per meter is constant, m e c 2 dγ dt = m ec 2 γ 3 β β = m e c 2 γ f / t, so the final value of β is 1/γ 2 f t = 10 5 /s. The angle of maximum intensity is θ max = 1/200, 000 rad = 4.1 seconds of arc, the power per sterradian from this single electron at that angle is π e 2 β2 γ 8 = W, c just from one electron. Why so much?

14 Total Linac Power Note this is not the power, it is the power/sterradian. The total power is P 2πR 2 θ dθ dp 0 da = 16q2 v 2 c 3 γ 8 γθ) 2 θ dθ γ 2 θ 2 ) 5 = 8q2 v 2 c 3 γ 6 u du 1 + u) 5 = 2q2 3c 3 v2 γ 6. 0

15 Circular Accelerators Another important special case is a circular storage ring, where the acceleration is perpendicular to the velocity. Taking β in the z direction β in the x, using the usual spherical angles for ˆn = sin θ cos φ, sin θ sin φ, cos θ), we may evaluate the numerator of??) as ˆn [ˆn β) ) 2 ) β] = ˆn βˆn β) [ˆn ˆn β)] 2 β = ˆn β) 2 ˆn β) 2 21 β cos θ)ˆn β) βˆn β) +1 β cos θ) 2 β) 2 = ˆn β) 2 1 2β cos θ + β 2 21 β cos θ)) +1 β cos θ) 2 β) 2 = [ sin θ cos φ) 2 γ 2 ) + 1 β cos θ) 2] β) 2 dp so dω = e2 v) 2 [ 4πc 3 1 β cos θ) 3 1 sin2 θ cos 2 ] φ 1 β cos θ) 2. Again this is strongly peaked in the forward direction.

16 If we take θ << 1, but keeping γθ to all orders, so 1 β cos θ 1 + γ 2 θ 2 )/2γ 2, we have dp dω 2e2 γ 6 v) 2 [ πc γ 2 θ 2 ) 3 1 4γ2 θ 2 cos 2 ] φ 1 + γ 2 θ) 2. The total power radiated in all directions is, from Liénard, P = 2 e 2 3 c 3 v) 2 γ 4, as β) 2 β β) 2 = β) 2 1 β 2 ) = γ 2 β) 2. But do not be mislead into thinking this is weaker than in the case with β β, where we had γ 6 instead of γ 4, because it is very hard to accelerate in the direction of β. A 4-force F in the direction of β produces d dτ mcβγ = F = mc βγ + β 2 γ 3 ) = mc βγ 3 = β = F mcγ 3, while a force in the transverse direction has mcγ β = F, or β = F/mcγ. So the β) 2 is likely to be γ 4 bigger in the transverse case.

17 In particular, at the LHC, with 7 TeV protons travelling at roughly c around a 4.3 km radius circle have β = ω β = / s, 10 9 times bigger than the electrons at SLAC, even though their γ is a factor of 13 smaller than the γ of the electrons. Note that for a given size ring, with ultrarelativistic particles travelling at essentially c, the angular velocity therefore v is fixed, so the power radiated is proportional to γ 4 or, for a fixed kind of particle, to E 4. This becomes a serious problem at large energies, especially for electrons as the power radiated is independent of mass for fixed γ).

18 Frequency of radiation Consider a particle in ultrarelativistic circular motion, with radius ρ. As its radiation is essentially confined to a direction δθ = 1/γ, the arc of the circle during which it irradiates a given distant observer is of length d = ρ/γ, which it does in a time δt = ρ/γv. This pulse of light has its leading edge travelling towards the observer a distance D = cρ/γv during this time, while the trailing edge of the pulse is emitted at d, so the pulse has a length D d = ρ/γ)β 1 1) ρ/2γ 3. Thus the duration of the received pulse is ρ/2cγ 3 which means it contains frequencies up to ω c c/ρ)γ 3. Thus synchrotrons are a good source of X-rays.

r,t r R Z j ³ 0 1 4π² 0 r,t) = 4π

r,t r R Z j ³ 0 1 4π² 0 r,t) = 4π 5.4 Lienard-Wiechert Potential and Consequent Fields 5.4.1 Potential and Fields (chapter 10) Lienard-Wiechert potential In the previous section, we studied the radiation from an electric dipole, a λ/2