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17 Jun 2019: tanh (κa). κa=. tan (ka δ0). ka. where κ2 = γ2 k2.

20 Jul 2021: 1. ktan. ka. 2= 1. ltanh. la. 2. where. k =. 2mE. 2> 0 and l =. 2m(U0 E). 2> 0. (ii) Suppose that initially U0 = 0 and the particle is in

17 Jun 2019: By considering the s-wave solution to the Schrödinger equation,. show thattan(ka δ0). ... ka=. tanh(. γ2 k2a). γ2 k2a. For low momenta, ka 1, compute the s-wave contribution to the total cross-section.Comment on the physical interpretation of your

17 Jun 2019: as. ka;b = 0. for a certain covector field ka, which you should define.

17 Jun 2019: tan(ka δ0). ka=. tan κa. κa,. where κ2 = k 2 γ 2. ... 42. Paper 2, Section II. 36B General Relativity. A vector field ka which satisfies.

7 Jul 2023: You may assume that thetransforms are well-defined.]. (c) Express the inverse transforms of cos ka and sin ka in terms of the δ-function,where a is a positive constant.

17 Jun 2019: If ka is the tangent vector to the trajectory of a photon and ua is the four-velocity ofa comoving observer (i.e., an observer at rest in the

17 Jun 2019: µx ik[σ3,µ] Qµ = 0,. andµt 2ik2[σ3,µ] (kA B)µ = 0.

17 Jun 2019: Deduce that, whateverparametrization ν is used along the curve γ, the tangent vector Ka to γ satisfies. (

17 Jun 2019: ka = i log(t. rr′. ). Part II, Paper 4. 19.