Lecture Plan: 1) Cosmic Ray acceleration- accelerated spectrum, - PowerPoint PPT Presentation

Lecture Plan: 1) Cosmic Ray acceleration- accelerated spectrum, efficient accelerators, nuclei friendly PROBLEMS 2) Cosmic Ray proton + nuclei interaction rates in extragalactic radiation fields PROBLEMS 3) Cosmic Ray propagation through Galactic and extragalactic magnetic fields Andrew Taylor

When E -2 and when not? Andrew Taylor

Strong shock wave Shock Acceleration propagating at supersonic velocity (sound speed depends on temperature) u 1 u 2 upstream downstream E 0 1 , µ 0 E 1 , µ 1 1 ✓ 1 + β µ 1 ◆ E 2 = E 1 Andrew Taylor E 0 1 , µ 0 1 + β µ 2 E 2 , µ 2 2

Fermi Acceleration (more) Energy Number ∆ E = 4v 3c = 4 ∆ N = − 4v 3c = − 4 (energy gain) (advection 3 β 3 β downstream) E N ✓ ◆ ✓ ◆ 1 + 4 1 − 4 E 1 = 3 β E 0 N 1 = 3 β N 0 ◆ n ◆ n ✓ ✓ 1 + 4 1 + 4 3 β 3 β N n = N 0 E n = E 0 So n~1/ β crossings are needed SNRs have v sh ~10 3 km s -1 Andrew Taylor before the particle population is so β ~10 -2 significantly altered

Fermi Acceleration (more) Energy Number β ~10 -2 Andrew Taylor

Fermi Acceleration (more) So, ◆ n ✓ 1 − 4 β / 3 ∆ N ∆ E = N 0 1 + 4 β / 3 E 0 ≈ N 0 ( 1 + 4 β / 3 ) − 2n E 0 ≈ N 0 E 0 E − 2 Andrew Taylor

Stochastic Acceleration/Propagation D xx D pp ≈ β 2 scat p 2 β scat β scat β scat β scat β scat β scat Andrew Taylor

x Random Walks γ ( t + 1 ) = t ! t Z ∞ x t e − x dx γ ( t + 1 ) = 0 f ( x , t ) = γ ( t + 1 ) / [ γ ([ t − x ] / 2 + 1 ) γ ([ x + t ] / 2 + 1 )] / ( 2 t ) e − x 2 / ( 2t ) f ( x , t ) ≈ [ π / ( t / 2 )] 1 / 2 Andrew Taylor

Random Walks dN Spatial spread: dx ∝ e − x 2 / 4D xx t dN dx ∝ e − x 2 / 4c 2 t scat t Momentum spread: ∆ E ∝ β E dN dp ∝ e − (ln p ) 2 / 4 ( D pp / p 2 ) t dN dp ∝ e − (ln p ) 2 / 4 ( t / t acc ) Andrew Taylor

Gamma-Ray Probes of Particle Acceleration –Flaring Blazars (Mrk 501) No energy losses astro-ph/1107.1879, Tramacere et al. Andrew Taylor 10

Fermi (Second Order) Acceleration  � ∂ f p τ esc ( p ) + Q f ∂ t = r p · ( D pp r p f ) � τ loss ( p ) f � p 2 Acceleration Radiative Escape Source term Losses Andrew Taylor

Stochastic Particle Acceleration- Random Walk Result (Spatial) r · ( D xx r f ) = δ ( r ) ∂ 2 f ∂ f ∂ r 2 + 2 ∂ r = δ ( r ) r f = r − α − α ( − α − 1 ) − 2 α = 0 α ( α − 1 ) = 0 Andrew Taylor

Stochastic Particle Acceleration- Random Walk Result (Momentum)  � ∂ f p τ esc ( p ) + Q f ∂ t = r p · ( D pp r p f ) � τ loss ( p ) f � p 2 Delta injection Steady state No losses D pp ∝ p q ∂ 2 f ∂ p 2 + ( 2 + q ) ∂ p − 4 τ acc ∂ f f p 2 = δ ( p ) τ esc p For f = p − α and q = 2 α 2 − 3 α − 4 τ acc = 0 Andrew Taylor τ esc

Stochastic Particle Acceleration- Random Walk Result (Momentum) α 2 − 3 α − 4 τ acc = 0 τ esc ◆ 1 / 2 ✓ 4 τ acc α = 3 + 9 2 ± 4 τ esc τ acc = 1 τ esc f = dN d 3 p = p − 4 Andrew Taylor

Fermi (First Order) Acceleration Time t acc = E ∆ t cycle ∆ E cycle Transport of particles in each region is dictated by competition between diffusion and advection downstream upstream t di ff = R 2 R t adv = v adv D xx Balancing these timescales D xx Andrew Taylor t resid = ( c β sh ) 2

Fermi (First Order) Acceleration Time t acc = E ∆ t cycle ∆ E cycle D xx t resid = ( c β sh ) 2 However, during the time it takes advection to dominate over diffusion, the particle will have crossed the shock times 1 / β D xx ∆ t cycle = ( c 2 β sh ) Andrew Taylor

Fermi (First Order) Acceleration Time t acc = E ∆ t cycle ∆ E cycle E 0 1 , µ 0 D xx E 1 , µ 1 1 ∆ t cycle = ( c 2 β sh ) E 0 1 , µ 0 E 2 , µ 2 2 ∆ E cycle = E β sh ✓ 1 + β µ 1 ◆ E 2 = E 1 1 + β µ 2 ( c β sh ) 2 = t scat D xx t acc = β 2 sh Andrew Taylor

Fermi (Second Order) Acceleration Time t acc = E ∆ t scat ∆ E scat ∆ E scat = E β 2 scat t acc = t scat β 2 scat Andrew Taylor

Efficient Accelerators … .what means efficient? Andrew Taylor

Particle Acceleration in AGN R 2 t acc = η R lar t esc . = c β 2 η cR lar R lar = β AM Hillas (1984) Maximum energy η R (Hillas criterion) ✓ ◆ ✓ 1 mG ◆ E R lar ( E , B ) = 10 pc 10 EeV B Andrew Taylor

Compactness of UHECR Sources: Proton/Nuclei Synchrotron Losses AM Hillas (1984) assumed in above plot η ≈ 1 Andrew Taylor

Particle Acceleration with Cooling ✓ m e t acc = η R lar ◆ 9 t cool = t lar E sync c β 2 8 πα γ ≈ 9 4 η − 1 β 2 m e E sync γ α Maximum synchrotron energy tells us how efficient accelerator is! η < 10 3 Andrew Taylor 22

Emission Site? Cen A ~2 kpc Where are the misaligned (X)HBLs? Hardcastle et al. (1103.1744) η < 10 3 Andrew Taylor 23

Future Probes- Cutoff Region total 1 10 2 bg 10 1 0.1 E e dN/dE e bg dN/dE γ 10 0 0.01 E γ 10 -1 0.001 10 -3 10 -2 10 -1 10 0 10 1 0.001 0.01 0.1 1 10 bg E e E γ B crit = 4 × 10 13 G ✓ B ◆ 10 3 E sync = Γ 2 m e total e γ B crit 10 2 10 1 ◆ dN E γ dN/dE Z ✓ E γ ✓ E γ ◆ dN dN 10 0 E γ E e dE e = E 2 E 2 dE γ tot dE γ dE e 10 -1 e e 10 -2 10 -3 Andrew Taylor Possibility to probe cutoff region 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 0 10 1 E γ 24

Nuclei Friendly Accelerators Andrew Taylor

UHECR Air Showers Andrew Taylor

UHECR Air Showers Andrew Taylor

Composition Measurements by the PAO Andrew Taylor

Nuclei Transmutation Within their Source 0 1500 3000 4500 Time [Myr] Andrew Taylor

IMPLICATIONS for UHECR Sources Andrew Taylor

IMPLICATIONS for UHECR Sources Photo-disintegration threshold: , where Since, , ergo.... Andrew Taylor A similar expression holds for TeV photon transparency

IMPLICATIONS for UHECR Sources Since, Only heavily sub-Eddington power objects need apply! IF magnetic + photon luminosity are in equipartition: Requiring, to ensure safe passage. Andrew Taylor OVERALL MESSAGE: Compact Sources Disfavoured

Are there Any Candidate Sources Left? Hillas Diagram Accelerators of Iron nuclei LHC Andrew Taylor

Are there Any Candidate Sources Left? Andrew Taylor

Example Candidate UHECR Source (a Nuclei Friendly Environment) Diagram taken from Ferrari -1998 Stochastic Acceleration in Radio Lobes: for General PROBLEM for Large Accelerators- ACCELERATION TIME Andrew Taylor

Can Centaurus A's Radio Lobes Accelerate UHECR? ( δ B / B 0 ) 2 = 1 (me 1D 3D Yes, but requires: β A = 1 B β A > 0 . 1 p 4 π m p n p c where Andrew Taylor astro-ph/0903.1259, O’Sullivan et al. 36

Diffusion Coefficient • From resonant scattering between particles and magnetic field perturbations With Larmor radius R L B 0 + δB(k) resonance for k ~ R L -1 P ( k ) ∝ k − q Probability to scatter off resonant B 2 ⌧ � D xx R L mode within Larmor period 0 = R L = ( δ B ( k )) 2 kP ( k ) β ∝ p 2 − q Since p 2 ∼ D xx • Bohm -> q=1 β 2 D pp scat • Kolmogorov -> q=5/3 Andrew Taylor • Kraichnan -> q=3/2 D pp ∝ p q • Hard-sphere -> q=2

Particle Transport Equation • Cut-offs arise naturally in the general solution of the transport equation for particles  � ∂ f p τ esc ( p ) + Q f ∂ t = r p · ( D pp r p f ) � τ loss ( p ) f � p 2 Acceleration Radiative Escape Source term Losses Andrew Taylor

Cut-off Shape • Interplay of acceleration and cooling defines the value of the cut-off of the primary particles: dN e − ( E e / E max ) β e β e = 2 − q − r ∝ E − Γ e dE e • In the following, demonstrations for this result will be shown for the case of stochastic acceleration scenarios. However, in reality, this result is more general, holding also for shock acceleration scenarios. [see Schlickeisser et al. 1985, Zirakashvili et al. 2007, Stawarz et al. 2008] Andrew Taylor

A Simple Case- q=1, only escape • Bohm diffusion ( q=1 ) + only escape results in simple exponential cutoff. • Some simplifications to the transport equation:  � ∂ f p τ esc ( p ) + Q f ∂ t = r p · ( D pp r p f ) � τ loss ( p ) f � p 2 Delta injection Steady state No losses Andrew Taylor

A Simple Case (II)- q=1, only escape • Rearranging the terms (and explicitly stating the dependences from p of the parameters): ✓ ◆ ∂ ∂ f 1 p f τ esc ( p ) ∝ p − 1 p 2 D 0 τ esc ( p ) = δ ( p ) , − p 2 ∂ p p 0 ∂ p ∂ 2 f ✓ ◆ ∂ f ∂ p 2 + 3 1 f = δ ( p ) ∂ p − p D 0 τ 0 Cutoff comes from f ∝ Ae − p / p τ balancing 1 st and 3 rd term Recall generally, β e = 2 − q − r q = 1 , r = 0 , → β e = 1 Andrew Taylor (Note- energy losses for the case will not alter this result) r = 0

Intuitive Insights into Cut-off Shape Origin Consider the steady-state case of diffusion (constant diffusion coefficient) of particles into an absorbing medium f r · ( D xx r f ) � τ ( x ) = δ ( r ) τ ( x ) = τ ∗ ( x / x ∗ ) 2 f ∝ const . For f ∝ e − x / x τ For τ ( x ) = τ ∗ For f ∝ e − ( x / x τ ) 2 τ ( x ) = τ ∗ ( x / x ∗ ) − 2 Andrew Taylor

End of First Lecture Andrew Taylor