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Yukawa-Hooke Equasion

  1. Jan 23, 2004 #1

    Hooke's Law:
    [tex]W(x) = - \frac{kx^2}{2}[/tex]
    k - spring force constant

    Yukawa Potential:
    [tex]U(r) = - f^2 \frac{e^- \frac{(r/r_0)}{}}{r}[/tex]
    f - interaction strength
    r0 = 1.5*10^-15 m

    [tex]U(r) = W(r)[/tex]

    Yukawa-Hooke Equasion:
    [tex]-f^2 \frac{e^- \frac{(r/r_0)}{}}{r} = -\frac{kr^2}{2}[/tex]

    [tex]f^2 = \frac{kr^3}{2e^- \frac{(r/r_0)}{}}[/tex]

    [tex]f = \sqrt{ \frac{kr^3}{2e^- \frac{(r/r_0)}{}}}[/tex]

    [tex]r = \sqrt[3]{ \frac{2f^2 e^- \frac{(r/r_0)}{}}{k}}[/tex]

    [tex]E(r) = U(r) + W(r)[/tex]
    [tex]E(r) = -f^2 \frac{e^- \frac{(r/r_0)}{}}{r} - \frac{kr^2}{2}[/tex]

    Yukawa Meson Mass-Energy Spectrum:
    [tex]\pi ^o (135 Mev) -> \eta ^o (548.8 Mev)[/tex]
    r1 = 1.461 Fm -> .359 Fm

    [tex]E(r) = W(r)[/tex]

    [tex]- \frac{\hbar c}{r_1} = - \frac{kr_1 ^2}{2}[/tex]

    [tex]k = \frac{2 \hbar c}{r_1 ^3}[/tex]

    [tex]E(r) = U(r)[/tex]
    [tex]- \frac{\hbar c}{r_1} = -f^2 \frac{e^- \frac{(r_1/r_0)}{}}{r_1}[/tex]

    [tex]\hbar c = f^2 e^- \frac{(r_1/r_0)}{}[/tex]

    [tex]f = \sqrt{ \frac{\hbar c}{{e^- \frac{(r_1/r_0)}{} }}[/tex]

    How effective is the Yukawa-Hooke Equasion at emulating a Nuclear Force Mediator?

    What is the depth of such an equasion?

  2. jcsd
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