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High Energy Astrophysics Dr. Adam Ingram Lecture 3 Synchrotron Radiation Introduction • Electrons Accelerated to Ultra-Relativistic Energies at Shock Fronts (E.G

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High Energy Astrophysics Dr. Adam Ingram Lecture 3 Synchrotron Radiation Introduction • Electrons Accelerated to Ultra-Relativistic Energies at Shock Fronts (E.G High Energy Astrophysics Dr. Adam Ingram Lecture 3 Synchrotron Radiation Introduction • Electrons accelerated to ultra-relativistic energies at shock fronts (e.g. jet lobes, supernova remnants) will spiral around B-field, resulting in synchrotron radiation. • Observe strong radio emission from such regions. • e.g. AGN/XRB jet lobes, SN remnants, galaxies (sum of radio emission from SN remnants & XRBs) Radio galaxy Cygnus A at 5 GHz (VLA: Carilli and Barthel 1996, A&A Reviews) Evidence for synchrotron • Smooth, featureless broadband spectrum over many orders of magnitude in frequency; • Power-law spectrum (will address turn-over next time); • High degree of linear polarisation. Conway et al (1963) Synchrotron radiation • Radiation released by relativistic electrons spiralling around magnetic field lines. Synchrotron radiation • Radiation released by relativistic electrons spiralling around magnetic field lines. • Motion is helical: …velocity parallel to B-field v∥ = constant …velocity perpendicular to B-field v⊥ = circular electron: mass m, charge e v⊥ B-field moving out of screen moving out of screen v∥ Synchrotron radiation • Radiation released by relativistic electrons spiralling around magnetic field lines. • Motion is helical: …velocity parallel to B-field v∥ = constant …velocity perpendicular to B-field v⊥ = circular Pitch angle = angle between v and B = α electron: mass m, charge e v⊥ B-field moving α out of screen moving out of screen v∥ Acceleration of electron • Lorentz force: F = q(ℰ + v × B) = − e(ℰ + v × B) Electron has -ve charge Electric field vector Electric charge (E is reserved for electron energy) a v⊥ B-field moving out of screen moving out of screen v∥ Acceleration of electron • Lorentz force: F = q(ℰ + v × B) = − e(ℰ + v × B) • Static B-field ⟹ F = − ev × B a v⊥ B-field moving out of screen moving out of screen v∥ Acceleration of electron • Lorentz force: F = q(ℰ + v × B) = − e(ℰ + v × B) • Static B-field ⟹ F = − ev × B d • Therefore equation of motion: (γmv) = − ev × B dt a v⊥ B-field moving out of screen moving out of screen v∥ Acceleration of electron • Lorentz force: F = q(ℰ + v × B) = − e(ℰ + v × B) • Static B-field ⟹ F = − ev × B d • Therefore equation of motion: (γmv) = − ev × B dt • Therefore acceleration: evB sin α a = − r̂ γm a v⊥ B-field moving out of screen moving out of screen v∥ Gyroradius • Calculate orbital radius of electron by setting a equal to centripetal acceleration: evB sin α v2 v2 sin2 α a = = ⊥ = γm r r a r v⊥ B-field moving out of screen moving out of screen v∥ Gyroradius • Calculate orbital radius of electron by setting a equal to centripetal acceleration: evB sin α v2 v2 sin2 α a = = ⊥ = γm r r • Therefore gyroradius: γmv sin α r = g eB a rg v⊥ B-field moving out of screen moving out of screen v∥ Gyroradius • Calculate orbital radius of electron by setting a equal to centripetal acceleration: evB sin α v2 v2 sin2 α a = = ⊥ = γm r r • Therefore gyroradius: γmv sin α r = g eB • Gyro angular frequency: v⊥ eB ωg = = a rg γm rg v⊥ B-field moving out of screen moving out of screen v∥ Gyroradius • Calculate orbital radius of electron by setting a equal to centripetal acceleration: evB sin α v2 v2 sin2 α a = = ⊥ = γm r r • Therefore gyroradius: γmv sin α r = g eB • Gyro angular frequency: v⊥ eB ωg = = a rg γm • Gyrofrequency: rg v⊥ eB νg = = v⊥ 2πrg 2πγm B-field moving out of screen moving out of screen v∥ Gyroradius • Calculate orbital radius of electron by setting a equal to centripetal acceleration: evB sin α v2 v2 sin2 α a = = ⊥ = γm r r • Therefore gyroradius: γmv sin α r = g eB • Gyro angular frequency: v⊥ eB ωg = = a rg γm • Gyrofrequency: rg v⊥ eB νg = = v⊥ 2πrg 2πγm B-field moving • Gyroperiod: out of screen 1 2πγm tg = = νg eB moving out of screen v∥ Radiation generated Electron at rest From Coulomb’s law, electric field is: e ℰ = r 2 4πϵ0r ℰθ = 0 Thomson’s reasoning Radiation generated Now give electron velocity Δv in time interval Δt From Coulomb’s law, electric field is: e ℰ = r 2 4πϵ0r ℰθ = 0 Thomson’s reasoning Radiation generated After time t: • Electron has travelled distance cΔt Δv t • Electric field is radial centred on electron in sphere of radius ct ct • Field outside of this sphere Δv hasn’t adjusted yet a = Δt • Kink in field in shell of radius Δv . t cΔt where inner and outer fields join up • This kink corresponds to non- zero ℰθ Thomson’s reasoning Radiation generated At later time t: Pulse of non-zero has • ℰθ moved out Thomson’s reasoning Radiation generated After time t: Pulse of non-zero has • ℰθ tΔ v moved out cΔt sin θ • Pulse strength (in electron rest frame) depends on angle to acceleration θ: θ ℰ tΔv sin θ ar Δv . t θ = = sin θ 2 ℰr cΔt c Thomson’s reasoning Power radiated (non-relativistic) e ℰ ar 1 ℰ = θ = sin θ c2 = r 2 2 4πϵ0r ℰr c ϵ0μ0 Energy flow in pulse / time / area at distance r = modulus of Poynting vector: 2 2 2 1/2 e a sin θ S = ϵ /μ ℰ2 = ( 0 0) θ 2 3 2 (4π) ϵ0c r Therefore power radiated into full sphere (remember dA = r 2π sinθ dθ): e2a2 P = 3 6πϵ0c Power radiated e2a2 P = 3 6πϵ0c • We want to deal with ultra-relativistic electrons, so need relativistic limit! Luckily, can use non-relativistic formula in the instantaneous electron rest frame S’ and then use Lorentz invariance of dE/dt to trivially move back to the observer’s frame S (in which the B-field is at rest) Power radiated e2a2 P = 3 6πϵ0c • We want to deal with ultra-relativistic electrons, so need relativistic limit! Luckily, can use non-relativistic formula in the instantaneous electron rest frame S’ and then use Lorentz invariance of dE/dt to trivially move back to the observer’s frame S (in which the B-field is at rest) • This comes about because dE and dt Lorentz transform in the same way: dE = γdE′;$ dt = γdt′$ ∴ (dE/dt) = (dE/dt)′$ Power radiated e2a2 P = 3 6πϵ0c • We want to deal with ultra-relativistic electrons, so need relativistic limit! Luckily, can use non-relativistic formula in the instantaneous electron rest frame S’ and then use Lorentz invariance of dE/dt to trivially move back to the observer’s frame S (in which the B-field is at rest) • This comes about because dE and dt Lorentz transform in the same way: dE = γdE′;$ dt = γdt′$ ∴ (dE/dt) = (dE/dt)′$ Electron restframe, S’: Observer restframe, S’: $ z z′ v B′$ B α′$ v α v x′$ x Power radiated In S frame: • Bx = B cos α; By = 0; Bz = B sin α ℰ = 0 Power radiated In S frame: • Bx = B cos α; By = 0; Bz = B sin α ℰ = 0 • In S’ frame: 2 e a′$ P′$ = 3 6πϵ0c F′$ = ma′$ = e(ℰ′$ + v′$ × B′)$ Power radiated In S frame: • Bx = B cos α; By = 0; Bz = B sin α ℰ = 0 • In S’ frame: 2 e a′$ P′$ = 3 6πϵ0c F′$ = ma′$ = e(ℰ′$ + v′$ × B′)$ • By design of the rest frame, v’=0, therefore: ma′$ = eℰ′$ Power radiated In S frame: • Bx = B cos α; By = 0; Bz = B sin α ℰ = 0 • In S’ frame: 2 e a′$ P′$ = 3 6πϵ0c F′$ = ma′$ = e(ℰ′$ + v′$ × B′)$ • By design of the rest frame, v’=0, therefore: ma′$ = eℰ′$ • Lorentz transforms to get electric field in S’: ℰx′$ = ℰx ℰy′$ = γ(ℰy − vBz) ℰz′$ = γ(ℰz − vBy) Power radiated In S frame: • Bx = B cos α; By = 0; Bz = B sin α ℰ = 0 • In S’ frame: 2 e a′$ P′$ = 3 6πϵ0c F′$ = ma′$ = e(ℰ′$ + v′$ × B′)$ • By design of the rest frame, v’=0, therefore: ma′$ = eℰ′$ • Lorentz transforms to get electric field in S’: ℰx′$ = ℰx = 0 ℰy′$ = γ(ℰy − vBz) = − γvBz = − γvB sin α ℰz′$ = γ(ℰz − vBy) = 0 Power radiated • Therefore: eγvB sin α a′$ = − m Power radiated • Therefore: eγvB sin α a′$ = − • Therefore: m 2 4 2 2 2 2 e a′$ e γ B v sin α P = P′$ = = 3 3 2 6πϵ0c 6πϵc me Power radiated • Therefore: eγvB sin α a′$ = − • Therefore: m 2 4 2 2 2 2 e a′$ e γ B v sin α P = P′$ = = 3 3 2 6πϵ0c 6πϵc me • Re-arrange: 1 B2 2 = energy density of magnetic field c = Umag = ϵ0μ0 2μ0 e4 σ = T 2 4 2 = Thomson cross-section 6πϵ0c m 2 v ⟹ P = 2σ cU γ2 sin2 α T mag ( c ) Power radiated • Therefore: eγvB sin α a′$ = − • Therefore: m 2 4 2 2 2 2 e a′$ e γ B v sin α P = P′$ = = 3 3 2 6πϵ0c 6πϵc me • Re-arrange: 1 B2 2 = energy density of magnetic field c = Umag = ϵ0μ0 2μ0 e4 σ = T 2 4 2 = Thomson cross-section 6πϵ0c m 2 v ⟹ P = 2σ cU γ2 sin2 α T mag ( c ) i.e. Tighter helixes with the same v radiate more (because more of velocity is in circular motion) Power radiated 2 v P = 2σ cU γ2 sin2 α T mag ( c ) • Average over isotropic distribution of pitch angles: 2 4 v ∴ ⟨P⟩ = σ cU γ2 3 T mag ( c ) i.e. For a population of electrons travelling in random initial directions but all with the same speed (and therefore the same Lorentz factor and same energy). Spectrum γ=1: gyroradiation Linearly polarised ) t ( ϕ 2 cos 10 1 ∝ − ) 0123 t ( P t ϕ a rg v⊥ θ B-field moving out of screen Spectrum γ=1: gyroradiation ] ) Linearly polarised t ( ϕ 2 t [ g cos + 1 10 1 − ∝ 0123 ) t ( P t ϕ a rg v⊥ θ B-field moving out of screen Spectrum γ=1: gyroradiation ] ) Linearly polarised t ( ϕ 2 t [ g cos + 1 10 1 − ∝ 0123 ) t ( P t ϕ Fourier transform to get spectrum: a rg v⊥ θ 2νg Fν B-field moving out of screen ν Spectrum γ>>1: Relativistic beaming tg Linearly polarised ) t ( P 10 1 − 0123 t ϕ a rg v⊥ θ B-field moving out of screen Spectrum γ>>1: Relativistic beaming tg Linearly polarised ) t ( P 10 1 − 0123 t ϕ Fourier transform to get spectrum: a rg v⊥ θ 2νg νc Fν B-field moving out of screen ν Spectrum γ>>1: Relativistic beaming tg Linearly polarised ) t ( P Δt 10 1 − 0123 t ϕ Fourier transform to get spectrum: a rg v⊥ θ 2νg νc ∼ 1/Δt Fν B-field moving out of screen ν Spectrum γ>>1: Relativistic beaming Relativistic aberration formula: cos ϕ′$ + v/c cos ϕ = 1 + (v/c)cos ϕ′$ ϕ a rg v⊥ θ Observer’s rest frame: ϕ B-field moving out of screen Electron’s rest frame: ϕ′$ Relativistic aberration Directly
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