http://www.deonto-ethics.org/quantic/index.php?title=Yukawa&feed=atom&action=history Yukawa - Historique des versions 2024-05-17T03:54:49Z Historique pour cette page sur le wiki MediaWiki 1.25.2 http://www.deonto-ethics.org/quantic/index.php?title=Yukawa&diff=121&oldid=prev Jacques 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 &lt;tex&gt;(\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 /> <br /> &lt;tex&gt;(\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)&lt;/tex&gt;<br /> <br /> Assuming U independent of time, it may be written<br /> <br /> &lt;tex&gt;(\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)&lt;/tex&gt;<br /> <br /> where &lt;tex&gt;\lambda_C&lt;/tex&gt; is the Compton wavelength of the particle. The simplest solution is then :<br /> <br /> &lt;tex&gt;U(r)=\frac{U_0 e^{2\pi r/\lambda_C}}{r}&lt;/tex&gt;<br /> <br /> 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 /> <br /> &lt;tex&gt;U(r)=U_0\frac{e^{2\pi r/\lambda_C-i\omega t}}{r}&lt;/tex&gt;<br /> <br /> The complete equation is:<br /> <br /> &lt;tex&gt;(\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)&lt;/tex&gt;<br /> <br /> Replacing U we get after simplyfying the time dependent exponential :<br /> <br /> &lt;tex&gt;\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)&lt;/tex&gt;<br /> <br /> The solution is almost the same as above :<br /> <br /> &lt;tex&gt;U(r)=\frac{U_0}{r} e^\sqrt{\left( \frac{2\pi}{\lambda_C}\right)^2-{\frac{\omega^2}{c^{2}}}&lt;/tex&gt;<br /> <br /> There is a physical solution if :<br /> &lt;tex&gt;\frac{2\pi}{\lambda_C}&lt;{\frac{\omega}{c}&lt;/tex&gt;<br /> <br /> That is when the solution is imaginary and tending to zero at infinity:[/QUOTE]<br /> <br /> [QUOTE=ophase;1550210]Here&#039;s the original proof by Yukawa<br /> <br /> Yukawa potential &lt;tex&gt;U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{-r/a}}{r}&lt;/tex&gt;<br /> gs: Yukawa constant<br /> <br /> 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 /> <br /> &lt;tex&gt;(\nabla^{2}-\frac{1}{a^{2}})U(r)=g^{2}_{s}\delta(r)&lt;/tex&gt; <br /> <br /> Yukawa generalized the equation for non-static states. <br /> <br /> &lt;tex&gt;(\nabla^{2}-\frac{d^{2}}{c^{2}dt^{2}}-\frac{1}{a^{2}})U(r,t)=0&lt;/tex&gt; (*)<br /> <br /> This equation is also relativistical invariant. Then Yukawa quantized the potential:<br /> <br /> &lt;tex&gt;U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{ipr/\hbar-iEt/\hbar}}{r}&lt;/tex&gt;<br /> <br /> Now we put that potential expression in the second green equation (*) and we get:<br /> <br /> &lt;tex&gt;\frac{E^{2}}{c^{2}\hbar^{2}}=\frac{p^{2}}{\hbar^{2}}+\frac{1}{a^{2}}&lt;/tex&gt;<br /> <br /> &lt;tex&gt;E^2 =p^2c^2+\frac{c^{2}\hbar^{2}}{a^{2}}&lt;/tex&gt;<br /> <br /> Here the last term should be the mass term:<br /> &lt;tex&gt;m^{2}_{u}c^{4}=\frac{c^{2}\hbar^{2}}{a^{2}}&lt;/tex&gt;<br /> If we assume a=2 fm, then the exchange particle mass is mu= 100 MeV. <br /> <br /> In 1947 Pion discovered at 140 MeV and it&#039;s proved that there is no meson in the nucleus according to Yukawa theory.<br /> <br /> I don&#039;t know the rest of that story. Probably someone made a correction about the calculation above. Any ideas??[/QUOTE]</div> Jacques Lavau