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Yukawa potential

  1. May 1, 2016 #1
    1. The problem statement, all variables and given/known data

    (a) Calculate the equations of motion for a massive vector ##A_{\mu}## from the Lagrangian

    ##\mathcal{L}=-\frac{1}{4}F_{\mu\nu}^{2}+\frac{1}{2}m^{2}A_{\mu}^{2}-A_{\mu}J_{\mu},##

    where ##F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}##. Assuming ##\partial_{\mu}J_{\mu}=0##, use the equations to find a constraint on ##A_{\mu}##.

    (b) For ##J_\mu## the current of a point charge, show that the equation of motion for ##A_0## reduces to

    ##A_{0}(r)=\frac{e}{4\pi^{2}ir}\ \int\limits_{-\infty}^{\infty}\ \frac{k\ dk}{k^{2}+m^{2}} e^{ikr}.##

    (c) Evaluate this integral with contour integration to get an explicit form for ##A_{0}(r)##.

    (d) Show that as ##m \rightarrow 0## you reproduce the Coulomb potential.

    (e) In 1935 Yukawa speculated that this potential might explain what holds protons together in the nucleus. What qualitative features does this Yukawa potential have, compared to a Coulomb potential, that make it a good candidate for the force between protons? What value for ##m## might be appropriate (in ##\text{MeV}##)?

    (f) Plug the constraint on ##A_\mu## that you found in part (a) back into the Lagrangian, simplify, then rederive the equations of motion. Can you still find the constraint? What is acting as a Lagrange multiplier in the Lagrangian given in part (a)?

    2. Relevant equations

    3. The attempt at a solution

    (a) ##\mathcal{L}=-\frac{1}{4}F_{\mu\nu}^{2}+\frac{1}{2}m^{2}A_{\mu}^{2}-A_{\mu}J_{\mu}##

    ##=-\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^{2}+\frac{1}{2}m^{2}A_{\mu}^{2}-A_{\mu}J_{\mu}##

    ##=-\frac{1}{2}(\partial_{\mu}A_{\nu})^{2}+\frac{1}{2}(\partial_{\mu}A_{\mu})^{2}+\frac{1}{2}m^{2}A_{\mu}^{2}-A_{\mu}J_{\mu},##

    where the last step is valid only up to an integration by parts.

    Now,

    ##\partial_{\mu}\frac{\partial(\partial_{\rho}A_{\sigma})^{2}}{\partial(\partial_{\mu}A_{\nu})}=\partial_{\mu}\Big[2(\partial_{\rho}A_{\sigma})\frac{\partial(\partial_{\rho}A_{\sigma})}{\partial(\partial_{\mu}A_{\nu})}\Big]=\partial_{\mu}[2(\partial_{\rho}A_{\sigma})g_{\rho\mu}g_{\sigma\nu}]=2\partial_{\mu}(\partial_{\mu}A_{\nu})## and

    ##\partial_{\mu}\frac{\partial(\partial_{\alpha}A_{\alpha})^{2}}{\partial(\partial_{\mu}A_{\nu})}=\partial_{\mu}\Big[2(\partial_{\alpha}A_{\alpha})\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})}g_{\beta\gamma}\Big]=\partial_{\mu}[2(\partial_{\alpha}A_{\alpha})g_{\beta\mu}g_{\gamma\nu}g_{\beta\gamma}]=2\partial_{\nu}(\partial_{\alpha}A_{\alpha})##.

    Then, the Euler-Lagrange equations ##\frac{\partial \mathcal{L}}{\partial A_{\nu}}-\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}A_{\nu})}\Big)## imply

    ##m^{2}A_{\nu}-J_{\nu}-\partial_{\mu}(-\partial_{\mu}A_{\nu})-\partial_{\nu}(\partial_{\mu}A_{\mu})=0##

    so that

    ##\partial_{\mu}F_{\mu\nu}=J_{\nu}-m^{2}A_{\nu}##.

    Assuming ##\partial_{\nu}J_{\nu}=0##, we have

    ##\partial_{\mu}F_{\mu\nu}=J_{\nu}-m^{2}A_{\nu}##

    ##\implies \partial_{\nu}\partial_{\mu}F_{\mu\nu}=\partial_{\nu}J_{\nu}-m^{2}\partial_{\nu}A_{\nu}##

    ##\implies \partial_{\nu}\partial_{\mu}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})=\partial_{\nu}J_{\nu}-m^{2}\partial_{\nu}A_{\nu}##

    ##\implies \partial_{\nu}(\partial_{\mu}\partial_{\mu}A_{\nu})-\partial_{\mu}(\partial_{\nu}\partial_{\nu}A_{\mu})=-m^{2}\partial_{\nu}A_{\nu}##

    ##\implies m^{2}\partial_{\nu}A_{\nu}=0##

    so that, if ##m \neq 0##, then ##\partial_{\nu}A_{\nu}=0##.

    Is my working correct so far?
     
  2. jcsd
  3. May 6, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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