Get An Introduction to Modern Astrophysics PDF

By Bradley W. Carroll, Dale A. Ostlie

ISBN-10: 0321442849

ISBN-13: 9780321442840

An creation to fashionable Astrophysics, moment version has been completely revised to mirror the dramatic adjustments and developments in astrophysics that experience happened during the last decade. the second one version of this market-leading e-book has been up-to-date to incorporate the most recent effects from correct fields of astrophysics and advances in our theoretical figuring out of astrophysical phenomena. Designed for sophomore-level astrophysics for astronomy and physics majors,An creation to fashionable Astrophysics is now provided in spinoff types: creation to fashionable Stellar Astrophysics, moment variation and creation to trendy Galactic Astrophysics and Cosmology, moment variation. The middle textual content is geared for classes in stellar constitution and evolution, whereas the briefer books supply extra insurance of the sunlight method, galactic and extragalactic astronomy, dynamics, evolution, and cosmology.

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THE TIME-INDEPENDENT SCHRODINGER EQUATION 53 V(x) x Hψ0 = − 2 2 2 a d2 ψ0 − sech2 (ax)ψ0 2m dx2 m 2 = Aa2 − sech(ax) tanh2 (ax) + sech3 (ax) − 2 2 a A sech3 (ax) m 2m 2 2 a A = − sech(ax) tanh2 (ax) + sech3 (ax) − 2 sech3 (ax) 2m 2 2 a =− A sech(ax) tanh2 (ax) + sech2 (ax) . 2m 1 sinh2 θ + 1 sinh2 θ But (tanh2 θ + sech2 θ) = + = = 1, so 2 2 cosh θ cosh θ cosh2 θ =− 1 = |A|2 2 2 a ψ0 . QED 2m ∞ −∞ Evidently E = − sech2 (ax)dx = |A|2 1 tanh(ax) a 2 2 a . 2m ∞ = −∞ 2 2 |A| =⇒ A = a a . 2 ψ(x) x (c) dψk A = (ik − a tanh ax)ik − a2 sech2 ax eikx .

36 Start with Eq. 22: ψ(x) = A sin kx + B cos kx. This time the boundary conditions are ψ(a) = ψ(−a) = 0: A sin ka + B cos ka = 0; −A sin ka + B cos ka = 0. Subtract : A sin ka = 0 ⇒ ka = jπ or A = 0, Add : B cos ka = 0 ⇒ ka = (j − 12 )π or B = 0, (where j = 1, 2, 3, . . ). If B = 0 (so A = 0), k = jπ/a. In this case let n ≡ 2j (so n is an even √ integer); then k = nπ/2a, a ψ = A sin(nπx/2a). Normalizing: 1 = |A|2 −a sin2 (nπx/2a) dx = |A|2 /2 ⇒ A = 2. If A = 0 (so B = 0), k = (j − 12 )π/a. In this case let n ≡ 2j − 1 (n is an odd integer); again k = nπ/2a, √ a ψ = B cos(nπx/2a).

Dt V = 1 dV mω 2 x2 =⇒ = mω 2 x. 2 dx dV dp = −mω 2 x = −mω 2 a cos ωt = , so Ehrenfest’s theorem is satisfied. 50 (a) (E + 12 mv 2 ) ∂Ψ mα ∂ = − 2 |x − vt| − i Ψ; ∂t ∂t ∂ |x − vt| = ∂t −v, if x − vt > 0 v, if x − vt < 0 . We can write this in terms of the θ-function (Eq. 143): 2θ(z) − 1 = i ∂Ψ = ∂t i 1, if z > 0 −1, if z < 0 mαv , so ∂ |x − vt| = −v[2θ(x − vt) − 1]. ∂t 1 [2θ(x − vt) − 1] + E + mv 2 Ψ. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.

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An Introduction to Modern Astrophysics by Bradley W. Carroll, Dale A. Ostlie


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