Yes, your working is correct so far. Keep going with the rest of the problem!

In summary, the Lagrangian for a massive vector ##A_{\mu}## can be written as ##\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}##. By using the Euler-Lagrange equations, it can be shown that ##\partial_{\mu}F_{\mu\nu}=J_{\nu}-m^{2}A
  • #1
spaghetti3451
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Homework Statement



(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)?

Homework Equations



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?
 

1. What does it mean when someone says "Yes, your working is correct so far. Keep going with the rest of the problem!"?

When someone says this, it means that they have reviewed your work so far and believe that you are on the right track. They are encouraging you to continue with the rest of the problem and trust in your approach.

2. Why is it important to receive confirmation that my working is correct so far?

It is important to receive confirmation because it gives you confidence in your problem-solving skills and reassurance that you are not making any major mistakes. It also helps to avoid wasting time and effort on incorrect solutions.

3. What if I am not sure if my working is correct so far?

If you are not sure, it is always a good idea to double-check your work and ask for feedback from someone else. You can also consult with a teacher or reference materials to confirm your approach.

4. What should I do if I receive confirmation that my working is correct so far?

If you receive confirmation, you should continue with the rest of the problem using the same approach. It is important to stay focused and not get overconfident, as there may still be potential mistakes in the remaining steps.

5. Is it possible for my working to be correct so far but still lead to the wrong answer?

Yes, it is possible. Sometimes, even if your working is correct so far, a small mistake or miscalculation in the later steps can lead to an incorrect answer. That is why it is important to always check and recheck your work to ensure accuracy.

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