# Yukawa

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

$(\nabla^{2}-\frac{1}{c^{2}}\frac{\partial^2}{\partial t^{2}})U(r,t)= \left( \frac{2\pi m_0 c}{h}\right)^2 U(r,t)$

Assuming U independent of time, it may be written

$(\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r} \right)U(r)= \left( \frac{2\pi}{\lambda_C}\right)^2 U(r)$

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

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

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:

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

The complete equation is:

$(\nabla^{2}-\frac{1}{c^{2}}\frac{\partial^2}{\partial t^{2}})U(r,t)= \left( \frac{2\pi m_0 c}{h}\right)^2 U(r,t)$

Replacing U we get after simplyfying the time dependent exponential :

$\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r} \right)U(r)= - \left(\frac{\omega^2}{c^{2}} -\left( \frac{2\pi}{\lambda_C}\right)^2 \right)U(r)$

The solution is almost the same as above :

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

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

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 $U(r)=\frac{-g^{2}_{s}}{4\pi}\frac{e^{-r/a}}{r}$ 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:

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

Yukawa generalized the equation for non-static states.

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

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

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

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

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

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

Here the last term should be the mass term: $m^{2}_{u}c^{4}=\frac{c^{2}\hbar^{2}}{a^{2}}$ 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]