By Avi

Winner of the 2003 Newbery Medal, Avi's action-packed event, Crispin, transports us again to 14th-century England, the place a boy stands accused of homicide.

**Read or Download Crispin: The Cross of Lead (Crispin, Book 1) PDF**

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**Extra resources for Crispin: The Cross of Lead (Crispin, Book 1)**

**Example text**

E. suﬃciently weak magnetic ﬁelds, we obtain κ ,⊥ (ω) α sin θ eB m 3 2πλ 1 dν (1−ν 2 )−1/2 e−[4/(λ(1−ν × 2 ))] 1− ν2 3 , 1 ν2 + 2 6 0 = eB 1 α sin θ 2 m 3 2πλ ∞ 2 1 + 2 3u 3u du (u−1)−1/2 × , 2 1 − 2 3u 6u 1 = ⊥ eB −4/λ 1 α sin θ e 2 m 3 2 1 2 , 1 4 . 109) ⊥ These ﬁndings coincide with the expressions found in [158, 117, 74, 14, 116, 135, 166]. 109) describe the asymptotic absorptive behavior of the magnetic ﬁeld. Both absorption coeﬃcients tend to zero for λ → 0 or λ → ∞. In between, a maximum exists at λ 16 for κ /ωL and at λ 19 for κ⊥ /ωL .

30 2. 83) in the limit |E| = |B| and E · B = 0. A diﬀerent representation of Π µν for the case of crossed ﬁelds, in terms of integrals over Airy functions, can be found in [139]; the corresponding formulas for the functions Πi are particularly useful for studying the analytical properties of Π µν in the complex k 2 plane. Finally, we apply these ﬁndings to the question of light propagation in a crossed-ﬁelds background; for convenience, we conﬁne ourselves to the softphoton limit ω m and consider weak ﬁelds only.

Since only small values of z are relevant for the integral, it suﬃces to approximate the functions N ,⊥ by the ﬁrst terms of their Taylor expansions: 1 1 (1 − ν 2 ) 1 − ν 2 z 2 , 2 3 1 1 2 1 N⊥ = (1 − ν 2 ) + ν z2. 107) 0 has been employed. 106) demonstrates that the absorption coeﬃcients are proportional to the Larmor frequency ωL = eB/m. 104). In the various limits of λ, one can make use of the usual asymptotic expansions of the Bessel function. e. 69)⊥ . e. suﬃciently weak magnetic ﬁelds, we obtain κ ,⊥ (ω) α sin θ eB m 3 2πλ 1 dν (1−ν 2 )−1/2 e−[4/(λ(1−ν × 2 ))] 1− ν2 3 , 1 ν2 + 2 6 0 = eB 1 α sin θ 2 m 3 2πλ ∞ 2 1 + 2 3u 3u du (u−1)−1/2 × , 2 1 − 2 3u 6u 1 = ⊥ eB −4/λ 1 α sin θ e 2 m 3 2 1 2 , 1 4 .