Talk:Polarizability: Difference between revisions
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The following short scientific paper was appended to the article some time ago. While it's interesting work it is simply not appropriate for an encyclopedia. Perhaps the conclusions merit reference in the text? |
The following short scientific paper was appended to the article some time ago. While it's interesting work it is simply not appropriate for an encyclopedia. Perhaps the conclusions merit reference in the text? |
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:{{u|Mschuma3}}, [[WP:TPG|article talk pages should not be used by editors as platforms for their personal views on a subject]]. I have removed the excessively long text. [[User:ToBeFree|~ ToBeFree]] ([[User talk:ToBeFree|talk]]) 21:57, 13 July 2019 (UTC) |
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<pre> |
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{{Expert-talk}} |
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{{physics|class=start|importance=high}} |
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=Polarizability of the nucleon = |
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===Preface=== |
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The polarizabilities belong to the fundamental structure constants of the nucleon, in addition to the mass, the electric charge, the spin and the magnetic moment. The proposal to measure the polarizabilities dates back to the 1950th. Two experimental options were considered (i) Compton scattering by the proton and (ii) the scattering of slow neutrons in the Coulomb field of heavy nuclei. The idea was that the nucleon with its pion cloud obtains an electric dipole moment under the action of an electric field vector which is proportional to the electric polarizability. After the discovery of the photoexcitation of the Δ resonance it became obvious that the nucleon also should have a strong paramagnetic polarizability, because of a virtual spin-flip transition of one of the constituent quarks due to the magnetic field vector provided by a real photon in a Compton scattering experiment. However, experiments showed that this expected strong paramagnetism is not observed. Apparently a strong diamagnetism exists which compensates the expected strong paramagnetism. Though this explanation is straightforward, it remained unknown how it may be understood in terms of the structure of the nucleon. A solution of this problem was found very recently when it was shown that the diamagnetism is a property of the structure of the constituent quarks. In retrospect this is not a surprise, because constituent quarks generate their mass mainly through interactions with the QCD vacuum via the exchange of a σ meson. This mechanims is predicted by the linear σ model on the quark level (QLLσM) which also predicts the mass of the σ meson to be mσ=666 MeV. The σ meson has the capability of interacting with two photons being in parallel planes of linear polarization. We will show in the following that the σ meson as part of the constituent quark structure, therefore, provides the largest part of the electric polarizability and the total diamagnetic polarizability. |
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==Definition of electromagnetic polarizabilities== |
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A nucleon in an electric field ''' E ''' and a magnetic field '''H''' obtains an electric dipole moment '''d ''' and magnetic dipole moment '''m''' given by.<ref name=ref1>M. Schumacher, Prog. Part. Nucl. Phys. 55 (2005) 567, |
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arXiv:hep-ph/0501167.</ref> |
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:<math label=e1> |
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{\mathbf d}\,\,=4\pi\,\alpha \,{\mathbf E} |
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</math> |
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:<math> |
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{\mathbf m}=4\pi\,\beta \,{\mathbf H} |
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</math> |
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in a unit system where the electric charge <math>e</math> is given by <math>e^2/4\pi=\alpha_e=1/137.04</math>. The proportionality constants |
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<math>\alpha</math> and <math>\beta</math> are denoted as the electric and magnetic polarizabilities, respectively. These polarizabilities may be understood |
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as a measure of the response of the nucleon structure to the fields provided by a real or virtual photon and it is evident that we need a second photon to measure the polarizabilities. This may be expressed through the relations |
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:<math>\delta W=-\frac12 \,4\pi\,\alpha\,{\mathbf E}^2-\frac12 \,4\pi\,\beta\,{\mathbf H}^2</math> |
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where <math>\delta W</math> is the energy change in the electromagnetic field due to the presence of the nucleon in the field. The definition implies that the polarizabilities are measured in units of a volume, i.e. in units of fm<math>^3</math> (1 fm=<math>10^{-15}</math> m). |
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==Modes of two-photon reactions and experimental methods== |
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Static electric fields of sufficient strength are provided by the Coulomb field of heavy nuclei. Therefore, the electric polarizability of the neutron |
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can be measured by scattering slow neutrons in the electric field '''E''' of a Pb nucleus. The neutron has no electric charge. Therefore, |
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two simultaneously interacting electric field vectors (two virtual photons) are required to produce a deflection of the neutron. Then the electric polarizability can be obtained from the differential cross section measured at a small |
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deflection angle. A further possibility is provided by Compton scattering of real photons by the nucleon, where during the scattering process two electric and two magnetic field vectors simultaneously interact with the nucleon. |
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In the following we discuss the experimental options we have to measure the polarizabilities of the nucleon. As outlined above two photons are needed which simultaneously interact with the electrically charged parts of the nucleon. |
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These photons may be in parallel or perpendicular planes of linear polarization and in these two modes measure the |
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polarizabilities <math>\alpha</math>, <math>\beta</math> or spinpolarizabilities <math>\gamma</math>, respectively. |
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The spinpolarizability is nonzero only for particles having a spin. |
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In total the experimental options discussed above provide us with 6 combinations of two electric and magnetic field vectors. These are described in the following two equations: |
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* Photons in parallel planes of linear polarization |
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:<math>(\text{case}\, 1)\,\, \alpha: \,\,\,\,{\mathbf E}\uparrow\uparrow {\mathbf E}'\quad\quad (\text{case}\, 2)\,\,\beta: \, |
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{\mathbf H} \rightarrow\rightarrow {\mathbf H}'\quad\, (\text{case}\,\, 3)\,\, -\beta: \,{\mathbf H}\rightarrow\leftarrow {\mathbf H}'</math> |
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* Photons in perpendicular planes of linear polarization |
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:<math>(\text{case}\,\,4)\,\, \gamma_E: {\mathbf E}\uparrow\rightarrow {\mathbf E}'\quad \,\,(\text{case}\,\,5)\,\,\gamma_H: {\mathbf H} \rightarrow\downarrow {\mathbf H}'\quad \,\,(\text{case}\,\,6)\,\, -\gamma_H: {\mathbf H}\rightarrow\uparrow {\mathbf H}'</math> |
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Case (1) corresponds to the measurement of the electric polarizability <math>\alpha</math> via two parallel electric field vectors '''E'''. These parallel electric field vectors may either be provided as longitudinal photons by the Coulomb field of a heavy nucleus, or by Compton scattering in the forward |
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direction or by reflecting the photon by 180°. Real photons simultaneously provide transvers electric '''E''' and magnetic '''H''' field vectors. This means that in a Compton scattering experiment linear combinations of electric and magnetic polarizabilities and linear combinations of electric and magnetic spinpolarizabilities are measured. |
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The combination of case (1) and case (2) measures <math>\alpha+\beta</math> and is observed in forward-direction Compton scattering. The combination |
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of case (1) and case (3) measures <math>\alpha-\beta</math> and is observed in backward-direction Compton scattering.The combination of case (4) and case (5) |
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measures <math>\gamma_0\equiv \gamma_E+\gamma_H</math> and is observed in forward-direction Compton scattering. |
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The combination of case (4) and case (6) |
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measures <math>\gamma_\pi\equiv \gamma_E-\gamma_H</math> and is observed in backward-direction Compton scattering. |
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Compton scattering experiments exactly in the forward direction and exactly in the backward direction are not possible |
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from a technical point of view. Therefore, the respective quantities have to be extracted from Compton scattering experiments carried out at intermediate angles. |
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== Experimental results == |
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The experimental polarizabilities of the proton (p) and the neutron (n) may be summarized as follows<ref name=ref1 /> |
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<ref name=ref2>M. Schumacher, Nucl. Phys. A 826 (2009) 131, arXiv:0905.4363 [hep-ph].</ref><ref name=ref3>M. Schumacher, M.I. Levchuk. Nucl. Phys. A 858 (2011) 48, arXiv:1104.3721 [hep-ph].</ref> |
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:<math>\alpha_p=12.0\pm 0.6,\quad \beta_p=1.9\mp 0.6, \quad \alpha_n=12.5\pm 1.7, \quad \beta_n=2.7\mp 1.8\,\text{ in units of}\,\, 10^{-4}\,{\rm fm}^3</math>. |
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The experimental spinpolarizabilities of the proton (p) and neutron (n) are |
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:<math>\gamma^{(p)}_\pi=-36.4\pm 1.5, \quad \gamma^{(n)}_\pi=58.6\pm 4.0\,\text{ in units of}\,\, 10^{-4}\,{\rm fm}^4</math>. |
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The experimental polarizabilities of the proton have been obtained as an average from a larger number of Compton scattering experiments. The experimental electric polarizability of the neutron is the average of an experiment on electromagnetic scattering of a neutron in the Coulomb field of a Pb nucleus and a Compton scattering experiment on a quasifree neutron, i.e. a neutron |
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separated from a deuteron during the scattering process. The two results are (see <ref name=ref1 />) |
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:<math>\alpha_n=12.6\pm 2.5</math> from electromagnetic scattering of a slow neutron in the electric field of a Pb nucleus, and |
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:<math>\alpha_n=12.5\pm 2.3</math> from quasifree Compton scattering by a neutron initially bound in the deuteron. |
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: The average given above is obtained from these two numbers. |
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Furthermore, there are experiments at the University of Lund (Sweden) where the electric polarizability of the neutron is determined through Compton scattering by the deuteron. |
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== Calculation of polarizabilities == |
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Recently great progress has been made in disentangling the total photoabsorption cross section into parts |
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separated by the spin, the isospin and the parity of the intermediate state, using the meson photoproduction amplitudes of Drechsel et al.<ref name=ref4>D. Drechsel, S.S. Kamalov, L. Tiator, Eur. Phys. J. A 34 (2007) 69, arXiv:0710.0306 [nucl.-th].</ref> The spin of the intermediate state may be <math>s=1/2</math> or <math>s=3/2</math> depending on the spin directions of the photon and the nucleon in the initial state. |
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The parity change during the transion from the ground state to the intermediate state is <math>\Delta P= \text{yes}</math> for the multipoles <math>E1,\,M2,\cdots</math> and <math>\Delta P= \text{no}</math> for the multipoles |
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<math>M1,\,E2,\cdots</math>. Calculating the respective partial cross sections from photo-meson data, the following |
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sum rules can be evaluated: |
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:<math> \alpha+\beta=\frac{1}{2\pi^2} \int^\infty_{\omega_0}\frac{\sigma_{\rm tot}(\omega)}{\omega^2}d\omega</math>, |
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:<math> \alpha-\beta=\frac{1}{2\pi^2}\int^\infty_{\omega_0}\sqrt{1+\frac{2\omega}{m}}\left[\sigma(\omega,E1,M2,\cdots)-\sigma(\omega,M1,E2,\cdots)\right]\frac{d\omega}{\omega^2} +(\alpha-\beta)^t</math>, |
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:<math> \gamma_0=-\frac{1}{4\pi^2}\int^\infty_{\omega_0}\frac{\sigma_{3/2}(\omega)-\sigma_{1/2}(\omega)}{\omega^3}d\omega</math>, |
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:<math>\gamma_\pi=\frac{1}{4\pi^2}\int^\infty_{\omega_0}\sqrt{1+\frac{2\omega}{m}}\left(1+\frac{\omega}{m}\right)\sum_n P_n[\sigma^n_{3/2}(\omega)-\sigma^n_{1/2}(\omega)]\frac{d\omega}{\omega^3}+\gamma^t_\pi</math>, |
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:<math>P_n=-1\,\text{for}\,E1,M2,\cdots\,\text{multipoles and}\, P_n=+1\,\text{for} \,M1,E2,\cdots\,\text{multipoles}</math>. |
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:<math>(\alpha-\beta)^t=\frac{1}{2 \pi}\left[\frac{g_{\sigma NN}{ M}(\sigma\to \gamma\gamma)}{ m^2_\sigma} |
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+\frac{g_{f_0 NN}{ M}(f_0\to \gamma\gamma)}{ m^2_{f_0}} |
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+\frac{g_{a_0 NN}{ M}(a_0\to \gamma\gamma)}{ m^2_{a_0}}\tau_3\right]</math>, |
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:<math>\gamma^t_\pi=\frac{1}{2\pi m}\left[ |
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\frac{g_{\pi NN}{ M}(\pi^0\to \gamma\gamma)}{ m^2_{\pi^0}}\tau_3 |
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+\frac{g_{\eta NN}{ M}(f_0\to \gamma\gamma)}{m^2_\eta} |
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+\frac{g_{\eta' NN}{ M}(a_0\to \gamma\gamma)}{m^2_{\eta'}}\right]</math>. |
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where <math>\omega</math> is the photon energy in the lab frame. |
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The sum rules for <math>\alpha+\beta</math> and <math>\gamma_0</math> depend on nucleon-structure degrees of freedom only, whereas the sum rules for <math>\alpha-\beta</math> and <math>\gamma_\pi</math> have to be supplemented by the quantities <math>(\alpha-\beta)^t</math> and <math>\gamma^t_\pi</math>, respectively. These are <math>t</math>-channel |
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contributions which may be interpreted as contributions of scalar and pseudoscalar mesons being parts of the constituent-quark structure. The sum rule for |
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<math>\alpha+\beta</math> depends on the total photoabsorption cross section and, therefore, does not require a disentangling with respect to quantum numbers. |
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The sum rule for <math>\alpha-\beta</math> requires a disentangling with respect to the parity change of the transition. The sum rule for <math>\gamma_0</math> |
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requires a disentangling with respect to the spin of the intermediate state. The sum rule for <math>\gamma_\pi</math> requires a disentangling with respect to |
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spin and parity change. |
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The <math>t</math>-channel contributions depend on those scalar and pseudoscalar mesons which (i) are part of the structure of the constituent quarks and (ii) |
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are capable of coupling to two photons. These are the mesons <math>\sigma(600)</math>, <math>f_0(980)</math> and <math>a_0(980)</math> |
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in case of <math>(\alpha-\beta)^t</math>, and the mesons <math>\pi^0</math>, <math>\eta</math> and <math>\eta'</math> in case of <math>\gamma^t_\pi</math>. |
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The contributions are dominated by the <math>\sigma</math> and the <math>\pi^0</math> whereas the other mesons only lead to small corrections. |
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==Results of calculation== |
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The results of the calculation are summarized in the following eight equations <ref name=ref2 /><ref name=ref3 />: |
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:<math>\alpha_p=\,\,\,+4.5\,\text{(nucleon)}+7.6\,\text{(const. quark)}=+12.1</math> |
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:<math>\beta_p=\,\,\,+9.4\,\text{(nucleon)}-7.6\,\text{(const. quark)}=\,\,\,+1.8</math> |
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:<math>\alpha_n=\,\,\,+5.1\,\text{(nucleon)}+7.6\,\text{(const. quark)}=+12.7</math> |
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:<math>\beta_n=+10.1\,\text{(nucleon)}-7.6\,\text{(const. quark)}=\,\,\,+2.5\,\,\text{in units of}\,\,10^{-4}{\rm fm}^3</math> |
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:<math>\gamma^{(p)}_0=\,\,\,-0.58\pm 0.20\,\text{(nucleon)}</math> |
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:<math>\gamma^{(n)}_0=\,\,\,+0.38\pm 0.22\,\text{(nucleon)}</math> |
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:<math>\gamma^{(p)}_\pi=\,\,\,+8.5\,\text{(nucleon)}-45.1\,\,\text{(const. quark)}=-36.6</math> |
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:<math>\gamma^{(n)}_\pi=+10.0\,\text{(nucleon)}+48.3\,\,\text{(const. quark)}=+58.3\,\,\text{in units of}\,\,10^{-4}{\rm fm}^4</math> |
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The electric polarizabilities <math>\alpha_p</math> and <math>\alpha_n</math> are dominated by a smaller component due to the pion cloud (nucleon) |
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and a larger component due to the <math>\sigma</math> meson as part of the constituent-quark structure (const. quark). The magnetic polarizabilities |
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<math>\beta_p</math> and <math>\beta_n</math> have a large paramagnetic part due to the spin structure of the nucleon (nucleon) and an only slightly smaller |
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diamagnetic part due to the <math>\sigma</math> meson as part of the constituent-quark structure (const. quark). The contributions of the <math>\sigma</math> |
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meson are not supplemented by corrections due to the <math>f_0(980)</math> and <math>a_0(980)</math> mesons because these corrections are small. <ref name=ref2 /> |
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<ref name=ref3 /><ref name=ref5>M. Schumacher, Eur. Phys. J. C 67 (2010) 283, arXiv:1001.0500 [hep-ph].</ref><ref name=ref6>M. |
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Schumacher, Journal of Physics G: Nucl. Part. Phys. 38 (2011) 083001, arXiv:1106.1015 [hep-ph].</ref><ref name=ref7>M. Schumacher, M.D. Scadron, Fortschr. Phys. 61, 703 (2013); arXiv:1301.1567 [hep-ph]</ref><ref name=ref8>M. Schumacher, arXiv:1307.2215 [hep-ph]</ref><ref name=ref9>M. Schumacher, Ann. Phys. (Berlin) 526, 215 (2014); arXiv:1403.7804 [hep-ph]</ref> |
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The spinpolarizabilities <math>\gamma_0^{(p)}</math> and <math>\gamma_0^{(n)}</math> are dominated by destructively interfering components from the pion cloud |
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and the spin structure of the nucleon. The different signs obtained for the proton and the neutron are due to this destructive interference. |
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The spinpolarizabilities <math>\gamma^{(p)}_\pi</math> and <math>\gamma^{(n)}_\pi</math> have a minor component due to the structure of the nucleon (nucleon) |
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and a major component due to the pseudoscalar mesons <math>\pi^0</math>, <math>\eta</math> and <math>\eta'</math> as structure components of the constituent quarks |
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(const. quark). |
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The agreement with the experimental data is excellent in all eight cases. |
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== Summary == |
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In the foregoing we have shown that the polarizabilities of the nucleon are well understood. Differing from previous belief the mesonic structure of |
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the constituent quark is essential for the sizes and the general properties of the polarizabilities. |
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== Linear polarizability == |
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In the definition of the polarizability the Induced dipole is proportional to the electric field. Shouldn't it be mentioned in the article that this assumes low enough electric fields such that only the linear term that contribute to the induced dipole. I believe the article should also mention higher order processes or refer to them, which each have their own non-linear susceptibility. [[User:Eranus|Eranus]] ([[User talk:Eranus|talk]]) 09:15, 25 February 2014 (UTC) |
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Answer: The polarizabilities are dependent on the photon energy <math>\omega</math>. The numbers given above refer to the case |
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<math>\omega \rightarrow 0 </math>. Details may be found in references [7] and [8] given below. |
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</pre> |
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{{collapse bottom}} |
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== Tendencies == |
== Tendencies == |
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Revision as of 21:57, 13 July 2019
| This topic is in need of attention from an expert on the subject. The section or sections that need attention may be noted in a message below. |
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Magnetic Polarizability
There was an article on this subject that was deleted. I have found a link to this subject in an ibox in the Neutron article, and there are probably other links in other articles. So I added this subject to this article as a section, however I am not an expert on the subject and did the best I could, which was probably wanting. The section definitely needs the massaging from an expert who can accurately display this subject for readers in terms they will understand.
— Paine (Ellsworth's Climax) 02:20, 26 May 2010 (UTC)
Should Polarizability be split into two articles?
I would like to make some changes to this article's description of electric polarizability. But before doing so, I wanted to raise the question of whether this article should be split into two: one for electric polarizability and one for magnetic polarizability. This would parallel Wikipedia's treatment of electric susceptibility and magnetic susceptibility. In addition, there are mixed electric-magnetic polarizabilities that might eventually be described in yet a third article.
If it seems advisable to split this article into two, the next question would be "What should happen to the Polarizability article?" In the fields of chemical/molecular physics and physical chemistry, a researcher who refers to "polarizability" without further qualification usually means the electric dipole-electric dipole polarizability α. So an argument could be made for the term "polarizability" to redirect readers to "electric polarizability", with a hatnote placed at the top of the electric polarizability article that refers readers to the magnetic polarizability article. Alternatively, the Polarizability article could become a very short article with links to longer articles that discuss electric polarizabilities, magnetic polarizabilities, mixed electric-magnetic polarizabilities, and hyperpolarizabilities in more detail.
– Rhinde (talk) 18:02, 3 August 2011 (UTC)
- I like your idea of creating longer articles for each topic, and having this article just give an overview. Christopher King (talk) 20:26, 4 August 2011 (UTC)
- I support splitting this article. I propose to split it into 2: "polarizability" (containing the present content about the electric polarizability of molecules, plus links to magnetic polarizability and to the other part) and an article labeled "nuclear polarizability" (with the present rich material about the nuclear polarizabilities). Since there is no specific article on "electric polarizability", we must keep the present contents, and since most scientists referring to polarizability mean usually electric polarizability, I would avoid creating a third page about "electric polarizability". I do now a little pre-cleanup, removing some residual mixup of molecular and nuclear stuff, but someone should then do the split. Who does it? Rhinde? Nicola.Manini (talk) 12:07, 14 November 2011 (UTC)
- OK, since nobody volounteers on this task, I can do it myself one of the next days, as soon as I have one spare hour. Nicola.Manini (talk) 10:47, 28 February 2012 (UTC)
Alkanes are the most polarizable molecules. Are they?
This statement, at face value, has a good chance to be wrong. An alkane composed of, say 30 or 50 atoms has no chance of being even remotely as polarizable as, e.g., a protein containing 10000 or 100000 atoms. It is quite possible that the polarizability per atom or maybe per unit volume of alkanes is record high, but this must be clarified, because, as it stands, this statement refers to the polarizability per molecule. So, even though this statement is taken from an organic chemistry textbook, it must be either corrected or removed. I suggest who wrote it checks and corrects it, but I'll come back in a few weeks, and revise it myself if nobody does. Thanx! Nicola.Manini (talk) 09:18, 9 November 2011 (UTC)
Expunged paper
The following short scientific paper was appended to the article some time ago. While it's interesting work it is simply not appropriate for an encyclopedia. Perhaps the conclusions merit reference in the text?
- Mschuma3, article talk pages should not be used by editors as platforms for their personal views on a subject. I have removed the excessively long text. ~ ToBeFree (talk) 21:57, 13 July 2019 (UTC)
Tendencies
In the "Tendencies" paragraph, it states that
- Generally, polarizability increases as the volume occupied by electrons increases. In atoms, this occurs because larger atoms have more loosely held electrons in contrast to smaller atoms with tightly bound electrons. On rows of the periodic table, polarizability therefore increases from left to right.
However, that is incorrect; the the largest atom within the period is the alkali metal and the smallest is the noble gas, as periodic table page lists under trends. Should this be corrected to say "polarizability increases from right to left," or did I just not understand this correctly? KingisNitro (talk) 15:24, 4 February 2018 (UTC)