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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Quantum mechanical scattering of photon and nucleus ： ウィキペディア英語版 Quantum mechanical scattering of photon and nucleus In pair production, a photon creates an electron positron pair. In the process of photons scattering in air (e.g. in lightning discharges), the most important interaction is the scattering of photons at the nuclei of atoms or molecules. The full quantum mechanical process of pair production can be described by the quadruply differential cross section give here:〔Bethe, H.A., Heitler, W., 1934. On the stopping of fast particles and on the creation of positive electrons. Proc. Phys. Soc. Lond. 146, 83–112〕 $$ \begin d^4\sigma &= \frac|\mathbf_+||\mathbf_-| \frac\frac\times \\ &\times\left( \\ &+2\hbar^2\omega^2\frac_-^2\sin^2\Theta_-}_-|\cos\Theta_-)} \\ &+2\left.\frac_-|\sin\Theta_+\sin\Theta_-\cos\Phi}_-|\cos\Theta_-)}\left(2E_+^2+2E_-^2-c^2\mathbf^2\right)\right ). \\ \end with $$ \begin d\Omega_+&=\sin\Theta_+\ d\Theta_+,\\ d\Omega_-&=\sin\Theta_-\ d\Theta_-. \end This expression can be derived by using a quantum mechanical symmetry between pair production and Bremsstrahlung. $Z$ is the atomic number, $\backslash alpha\_\backslash approx\; 1/137$ the fine structure constant, $\backslash hbar$ the reduced Planck's constant and $c$ the speed of light. The kinetic energies $E\_$ of the positron and electron relate to their total energies $E\_$ and momenta $\backslash mathbf\_$ via $$ E_=E_+m_e c^2=\sqrt^2 c^2}. Conservation of energy yields $$ \hbar\omega=E_+E_. The momentum $\backslash mathbf$ of the virtual photon between incident photon and nucleus is: $$ \begin -\mathbf^2&=-|\mathbf_+|^2-|\mathbf_-|^2-\left(\frac\omega\right)^2+2|\mathbf_+|\frac \omega\cos\Theta_+ +2|\mathbf_-|\frac \omega\cos\Theta_- \\ &-2|\mathbf_+||\mathbf_-|(\cos\Theta_+\cos\Theta_-+\sin\Theta_+\sin\Theta_-\cos\Phi), \end where the directions are given via: $$ \begin \Theta_+&=\sphericalangle(\mathbf_+,\mathbf),\\ \Theta_-&=\sphericalangle(\mathbf_-,\mathbf),\\ \Phi&=\text (\mathbf_+,\mathbf) \text (\mathbf_-,\mathbf), \end where $\backslash mathbf$ is the momentum of the incident photon. In order to analyse the relation between the photon energy $E\_+$ and the emission angle $\backslash Theta\_+$ between photon and positron, Köhn and Ebert integrated 〔Koehn, C., Ebert, U., Angular distribution of Bremsstrahlung photons and of positrons for calculations of terrestrial gamma-ray flashes and positron beams, Atmos. Res. (2014), vol. 135-136, pp. 432-465〕 the quadruply differential cross section over $\backslash Theta\_-$ and $\backslash Phi$. The double differential cross section is: $$ \begin \frac = \sum\limits_^ I_j \end with $$ \begin I_1&=\frac} \\ &\times \ln\left(\frac_2)^2+4p_+^2p_-^2\sin^2 \Theta_+}(\Delta^_1+\Delta^_2)+\Delta^_1\Delta^_2}_2)^2+4p_+^2p_-^2\sin^2 \Theta_+}(\Delta^_1-\Delta^_2)+\Delta^_1\Delta^_2 }\right) \\ &\times\left(), \\ I_2&=\frac\ln\left( \frac\right), \\ I_3&=\frac_1p_-c)^2+4m^2c^4p_+^2p_-^2\sin^2\Theta_+ }} \\ &\times\ln\Bigg(\Big((E_-+p_-c)(4p_+^2p_-^2\sin^2\Theta_+(E_--p_-c)+(\Delta^_1+\Delta^_2) ((\Delta^_2E_-+\Delta^_1p_-c) \\ &-\sqrt_1p_-c)^2+4m^2c^4p_+^2p_-^2\sin^2\Theta_+}))\Big)\Big((E_--p_-c) (4p_+^2p_-^2\sin^2\Theta_+(-E_--p_-c) \\ &+(\Delta^_1-\Delta^_2) ((\Delta^_2E_-+\Delta^_1p_-c)-\sqrt_1p_-c)^2+4m^2c^4p_+^2p_-^2\sin^2\Theta_+}))\Big)^\Bigg) \\ &\times\left^ \\ &+\Big(\\ &+2\hbar^2\omega^2p_- m^2c^3(\Delta^_2E_-+\Delta^_1p_-c)\Big ) \Big()^ \\ &+\left.\frac_1p_-c)^2-4m^2c^4p_+^2p_-^2\sin^2\Theta_+)(\Delta^_1E_-+\Delta^_2p_-c)}_1p_-c)^2+4m^2c^4p_+^2p_-^2\sin^2\Theta_+)^2}\right], \\ I_4&=\frac_1p_-c)}_1p_-c)^2+4m^2c^4p_+^2p_-^2\sin^2\Theta_+}+\frac_1p_-c)^2}_1p_-c)^2+4m^2c^4p_+^2p_-^2\sin^2\Theta_+)^2}, \\ I_5&=\frac_1)^2-4p_+^2p_-^2\sin^2\Theta_+) ((\Delta^_2E_-+\Delta^_1p_-c)^2+4m^2c^4p_+^2p_-^2\sin^2\Theta_+)} \\ &\times\left\Big]\Big()^\\ &+ \frac_1\Delta^_2 p_-c+2(\Delta^_2)^2E_-+8p_+^2p_-^2\sin^2\Theta_+ E_-)}\\ &-\Big\Big()^\\ &-\left.\frac_1 E_-)}\right], \\ I_6&=-\frac_1)^2-4p_+^2p_-^2\sin^2\Theta_+)} \end and $$ \begin A&=\frac\frac_-|},\\ \Delta^_1&:=-|\mathbf_+|^2-|\mathbf_-|^2-\left(\frac\omega\right) + 2\frac\omega|\mathbf_+|\cos\Theta_+,\\ \Delta^_2&:=2\frac\omega|\mathbf_i|-2|\mathbf_+||\mathbf_-| \cos\Theta_+ + 2. \end This cross section can be applied in Monte Carlo simulations. An analysis of this expression shows that positrons are mainly emitted in the direction of the incident photon. ==References== 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Quantum mechanical scattering of photon and nucleus」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.