http://www.deonto-ethics.org/quantic/index.php?action=history&feed=atom&title=YukawaYukawa - Historique des versions2024-11-22T13:38:09ZHistorique pour cette page sur le wikiMediaWiki 1.25.2http://www.deonto-ethics.org/quantic/index.php?title=Yukawa&diff=121&oldid=prevJacques Lavau : Page créée avec « [QUOTE=bjschaeffer;1579003]The Yukawa potential is a solution of the Klein-Gordon equation <tex>(\nabla^{2}-\frac{1}{c^{2}}\frac{\partial^2}{\partial t^{2}})U(r,t)= \left... »2018-06-19T19:52:54Z<p>Page créée avec « [QUOTE=bjschaeffer;1579003]The Yukawa potential is a solution of the Klein-Gordon equation <tex>(\nabla^{2}-\frac{1}{c^{2}}\frac{\partial^2}{\partial t^{2}})U(r,t)= \left... »</p>
<p><b>Nouvelle page</b></p><div>[QUOTE=bjschaeffer;1579003]The Yukawa potential is a solution of the Klein-Gordon equation<br />
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<tex>(\nabla^{2}-\frac{1}{c^{2}}\frac{\partial^2}{\partial t^{2}})U(r,t)= \left( \frac{2\pi m_0 c}{h}\right)^2 U(r,t)</tex><br />
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Assuming U independent of time, it may be written<br />
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<tex>(\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r} \right)U(r)= \left( \frac{2\pi}{\lambda_C}\right)^2 U(r)</tex><br />
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where <tex>\lambda_C</tex> is the Compton wavelength of the particle. The simplest solution is then :<br />
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<tex>U(r)=\frac{U_0 e^{2\pi r/\lambda_C}}{r}</tex><br />
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However, this solution is not physical: it is wrong to suppress the time derivative. We have to solve the complete equation with a periodical but stationary potential:<br />
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<tex>U(r)=U_0\frac{e^{2\pi r/\lambda_C-i\omega t}}{r}</tex><br />
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The complete equation is:<br />
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<tex>(\nabla^{2}-\frac{1}{c^{2}}\frac{\partial^2}{\partial t^{2}})U(r,t)= \left( \frac{2\pi m_0 c}{h}\right)^2 U(r,t)</tex><br />
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Replacing U we get after simplyfying the time dependent exponential :<br />
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<tex>\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r} \right)U(r)= - \left(\frac{\omega^2}{c^{2}} -\left( \frac{2\pi}{\lambda_C}\right)^2 \right)U(r)</tex><br />
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The solution is almost the same as above :<br />
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<tex>U(r)=\frac{U_0}{r} e^\sqrt{\left( \frac{2\pi}{\lambda_C}\right)^2-{\frac{\omega^2}{c^{2}}}</tex><br />
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There is a physical solution if :<br />
<tex>\frac{2\pi}{\lambda_C}<{\frac{\omega}{c}</tex><br />
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That is when the solution is imaginary and tending to zero at infinity:[/QUOTE]<br />
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[QUOTE=ophase;1550210]Here's the original proof by Yukawa<br />
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Yukawa potential <tex>U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{-r/a}}{r}</tex><br />
gs: Yukawa constant<br />
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Yukawa proposed that nuclear force has to be like elektromagnetic force. So the potential above need to satisfy second green equation with a source term:<br />
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<tex>(\nabla^{2}-\frac{1}{a^{2}})U(r)=g^{2}_{s}\delta(r)</tex> <br />
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Yukawa generalized the equation for non-static states. <br />
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<tex>(\nabla^{2}-\frac{d^{2}}{c^{2}dt^{2}}-\frac{1}{a^{2}})U(r,t)=0</tex> (*)<br />
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This equation is also relativistical invariant. Then Yukawa quantized the potential:<br />
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<tex>U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{ipr/\hbar-iEt/\hbar}}{r}</tex><br />
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Now we put that potential expression in the second green equation (*) and we get:<br />
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<tex>\frac{E^{2}}{c^{2}\hbar^{2}}=\frac{p^{2}}{\hbar^{2}}+\frac{1}{a^{2}}</tex><br />
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<tex>E^2 =p^2c^2+\frac{c^{2}\hbar^{2}}{a^{2}}</tex><br />
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Here the last term should be the mass term:<br />
<tex>m^{2}_{u}c^{4}=\frac{c^{2}\hbar^{2}}{a^{2}}</tex><br />
If we assume a=2 fm, then the exchange particle mass is mu= 100 MeV. <br />
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In 1947 Pion discovered at 140 MeV and it's proved that there is no meson in the nucleus according to Yukawa theory.<br />
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I don't know the rest of that story. Probably someone made a correction about the calculation above. Any ideas??[/QUOTE]</div>Jacques Lavau