Planck Pre-Launch Status Papers - APC - Université Paris Diderot ...

C. Rosset et al.: Planck-HFI: polarization calibrationAppendix A: Explicit forms of pointing related functionsWe write the projection of the signal m into a sky map s asŝ = (à T Ã) −1 à T m⎡ ⎤⎤∑ ∑= ⎢⎣ à T d ⎥⎦−1 ⎡⎢⎣ Ãd à T d A ds⎥⎦d⎡∑= ⎢⎣d⎤dà T d ⎥⎦−1 ⎡⎢⎣ Ãdd⎤∑Λ d (γ d ,ɛ d ,ω d )s⎥⎦ .Where A is the pointing matrix. In this expression, Λ d (γ d ,ɛ d ,ω d )isanexplicitfunctionofγ d , ɛ d and ω d .˜g, ˜ρ and ˜α are onlyparameters. If we note t(d) thedatasamplesofdetectord, Λ d reads⎛∑ (1 + γ) (˜ρ d + ɛ ρ )cos2( ψ˜d (t) + ω d )(˜ρ d + ɛ ρ )sin2( ˜⎞ψ d (t) + ω d )Λ d = (1 + γ) ρ˜⎜⎝ d cos 2ψ˜d (t) ρ˜d (˜ρ d + ɛ ρ )cos2ψ˜d (t)cos2( ψ˜d (t) + ω d ) ρ˜d (˜ρ d + ɛ ρ )cos2ψ˜d (t)sin2( ψ˜d (t) + ω d )t(d) (1 + γ) ρ˜d sin 2ψ˜d (t) ρ˜d (˜ρ d + ɛ ρ )sin2ψ˜d (t)cos2( ψ˜d (t) + ω d ) ρ˜d (˜ρ d + ɛ ρ )sin2ψ˜d (t)sin2( ψ˜⎟⎠ .(A.4)d (t) + ω d )Considering small variations around ˜g, ˜ρ and ˜α, wecanwritetheperturbativeexpansiontofirstorderforbothγ ≪ 1, ɛ ≪ 1andω ≪ 1:∆s = ŝ − s⎡∑= ⎢⎣d⎡∑≡ ⎢⎣⎡∑≃ ⎢⎣⎤à T d ⎥⎦−1 ⎡⎢⎣ Ãdd⎤à T d ⎥⎦−1 ⎡⎢⎣ Ãddd⎤ −1 ∑à T d Ãd⎥⎦dd⎤∑à T d A d − à T d Ãd⎥⎦ s⎤∑Λ d (γ d ,ɛ d ,ω d ) − Λ d (0, 0, 0) ⎥⎦ s[ ∂Λd∂γ dγ d + ∂Λ d∂ɛ dɛ d + ∂Λ d∂ω dω d]s. (A.6)Straightforward generalization to second order reads:⎡ ⎤∑−1 ⎡⎤∑ ∑∆s = ⎢⎣ à T ∂Λ dd Ãd⎥⎦⎢⎣ e d + 1 ∑ ∂ 2 Λ d∂e d 2 ∂e d ∂e ′ e d e ′ d⎥⎦ s.ddde∈{γ,ɛ,ω}(e,e ′ )∈{γ,ɛ,ω}Derivatives of Λ d (γ d ,ɛ d ,ω d )withrespecttouncertaintiesofgainγ,polarizationefficiency ɛ and detector orientation ω are given by∣⎛⎞∂Λ ∣∣∣∣(0,0,0) d∑ 1 00= cos 2 ˜ψ∂γ d⎜⎝ d (t)00⎟⎠(A.8)t(d) sin 2 ˜ψ d (t)00∣⎛⎞∂Λ ∣∣∣∣(0,0,0) d∑ 0cos 2 ˜ψ d (t)sin2˜ψ d (t)= 0˜ ρ∂ɛ d⎜⎝ d cos 2 2 ˜ψ d (t) ρ˜d cos 2 ˜ψ d (t)sin2˜ψ d (t) ⎟⎠(A.9)t(d) 0˜ ρ d cos 2 ˜ψ d (t)sin2˜ψ d (t) ρ˜d sin 2 2 ˜ψ d (t)∣⎛⎞∂Λ ∣∣∣∣(0,0,0) d∑ 0−2ρ˜d sin 2 ˜ψ d (t) 2ρ˜d cos 2 ˜ψ d (t)= 0−2ρ˜∂ω d⎜⎝2 d cos 2 ˜ψ d (t)sin2˜ψ d (t)2ρ˜2 d cos 2 2 ˜ψ d (t) ⎟⎠t(d) 0−2ρ˜2 d sin 2 2 ˜ψ d (t) 2ρ˜2 d cos 2 ˜ψ .(A.10)d (t)sin2˜ψ d (t)And second order derivatives reads∂ 2 Λ d∂γd2 = 0∣(0,0,0) ∂ 2 Λ d∂ɛd2 = 0∣(0,0,0) ⎛⎞∂ 2 Λ d∑ 0−4ρ˜d cos 2 ˜ψ d (t) −4ρ˜d sin 2 ˜ψ d (t)∂ω 2 = 0−4ρ˜∣ ⎜⎝2 d cos 2 2 ˜ψ d (t) −4ρ˜2 d cos 2 ˜ψ d (t)sin2˜ψ d (t) ⎟⎠d (0,0,0) t(d) 0−4ρ˜2 d cos 2 ˜ψ d (t)sin2˜ψ d (t)−4ρ˜2 d sin 2 2 ˜ψ d (t)∣ ⎛⎞∂ 2 Λ ∣∣∣∣∣(0,0,0)d∑ 0−2sin2˜ψ d (t)2cos2˜ψ d (t)= 0−4ρ˜∂ɛ d ∂ω d⎜⎝ d cos 2 ˜ψ d (t)sin2˜ψ d (t)4ρ˜d cos 2 2 ˜ψ d (t) ⎟⎠t(d) 0−4ρ˜d sin 2 2 ˜ψ d (t) 4ρ˜d cos 2 ˜ψ d (t)sin2˜ψ d (t)(A.1)(A.2)(A.3)(A.5)(A.7)(A.11)(A.12)(A.13)(A.14)∂ 2 Λ d∂γ d ∂ɛ d∣ ∣∣∣∣∣(0,0,0)=∂2 Λ d∂γ d ∂ω d∣ ∣∣∣∣∣(0,0,0)= 0.(A.15)Page 11 of 12