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[QUOTE=bjschaeffer;1579003]The Yukawa potential is a solution of the Klein-Gordon equation

LaTeX: (\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)

Assuming U independent of time, it may be written

LaTeX: (\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)

where LaTeX: \lambda_C is the Compton wavelength of the particle. The simplest solution is then :

LaTeX: U(r)=\frac{U_0 e^{2\pi r/\lambda_C}}{r}

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:

LaTeX: U(r)=U_0\frac{e^{2\pi r/\lambda_C-i\omega t}}{r}

The complete equation is:

LaTeX: (\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)

Replacing U we get after simplyfying the time dependent exponential :

LaTeX: \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)

The solution is almost the same as above :

LaTeX: U(r)=\frac{U_0}{r} e^\sqrt{\left( \frac{2\pi}{\lambda_C}\right)^2-{\frac{\omega^2}{c^{2}}}

There is a physical solution if : LaTeX: \frac{2\pi}{\lambda_C}<{\frac{\omega}{c}

That is when the solution is imaginary and tending to zero at infinity:[/QUOTE]

[QUOTE=ophase;1550210]Here's the original proof by Yukawa

Yukawa potential LaTeX: U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{-r/a}}{r} gs: Yukawa constant

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:

LaTeX: (\nabla^{2}-\frac{1}{a^{2}})U(r)=g^{2}_{s}\delta(r)

Yukawa generalized the equation for non-static states.

LaTeX: (\nabla^{2}-\frac{d^{2}}{c^{2}dt^{2}}-\frac{1}{a^{2}})U(r,t)=0 (*)

This equation is also relativistical invariant. Then Yukawa quantized the potential:

LaTeX: U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{ipr/\hbar-iEt/\hbar}}{r}

Now we put that potential expression in the second green equation (*) and we get:

LaTeX: \frac{E^{2}}{c^{2}\hbar^{2}}=\frac{p^{2}}{\hbar^{2}}+\frac{1}{a^{2}}

LaTeX: E^2 =p^2c^2+\frac{c^{2}\hbar^{2}}{a^{2}}

Here the last term should be the mass term: LaTeX: m^{2}_{u}c^{4}=\frac{c^{2}\hbar^{2}}{a^{2}} If we assume a=2 fm, then the exchange particle mass is mu= 100 MeV.

In 1947 Pion discovered at 140 MeV and it's proved that there is no meson in the nucleus according to Yukawa theory.

I don't know the rest of that story. Probably someone made a correction about the calculation above. Any ideas??[/QUOTE]