- #1
neorich
- 20
- 1
Hi All,
I am trying to derive the equation of motion from a Lagrangian with a gauge fixing term, and I think get quite close to the result, but am missing some steps somewhere. These indices and notation is new to me, so please feel free to bring any mistakes to my attention.
So the Lagrangian is:
[tex]
{\cal L} = -\,\frac{1}{4}\,F_{\mu\nu}F^{\mu\nu} +\, j_{\mu}A^{\mu} -\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2
[/tex]
And the equation of motion I have to derive is:
[tex]
\left(\Box g^{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial^\mu \partial^\nu \right) A_\nu = j^{\mu}
[/tex]
using the Euler-Lagrange Equation:
[tex]
\partial_{\mu}\,(\frac{\partial{\cal L}}{\partial\,(\partial_{\mu}A_{\nu})}) -\, \frac{\partial{\cal L}}{\partial\,A_{\nu}}=0
[/tex]
So the first term in the E-L equation will only act on the derivatives in the Lagrangian, and the second term will only act in the non-derivatives, so I get:
[tex]
\frac{\partial{(-\,\frac{1}{4}\,F_{\mu\nu}F^{\mu\nu})}}{\partial\,(\partial_{\mu}A_{\nu})}=-F_{\mu\nu}=\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu}
[/tex]
[tex]
\frac{\partial{(-\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2)}}{\partial\,(\partial_{\mu}A_{\nu})}=-\,\frac{1}{\xi}\,g^{\mu\nu}(\partial_{\mu}A_{\nu})
[/tex]
[tex]
\frac{\partial{(j_{\mu}A^{\mu}})}{\partial\,A_{\nu}}=\frac{\partial{(j_{\mu}g^{\mu\nu}A_{\nu}})}{\partial\,A_{\nu}}=g^{\mu\nu}j_{\mu}=j^{\nu}
[/tex]
Now, I am not entirely sure about my use of gmunu's here. Anyway, combining these, I get:
[tex]
\partial_{\mu}(\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu})-\frac{1}{\xi}g^{\mu\nu}\partial_{\mu}\partial_{\mu}A_{\nu}-j^{\nu}=0
[/tex]
Can you please tell me if this is correct so far, and if so, can you please direct me as to how to obtain the box times the metric, and the two derivatives with upper indices on the right of the left hand side on the equation of motion?
Thanks for your help,
Regards
neorich.
I am trying to derive the equation of motion from a Lagrangian with a gauge fixing term, and I think get quite close to the result, but am missing some steps somewhere. These indices and notation is new to me, so please feel free to bring any mistakes to my attention.
So the Lagrangian is:
[tex]
{\cal L} = -\,\frac{1}{4}\,F_{\mu\nu}F^{\mu\nu} +\, j_{\mu}A^{\mu} -\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2
[/tex]
And the equation of motion I have to derive is:
[tex]
\left(\Box g^{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial^\mu \partial^\nu \right) A_\nu = j^{\mu}
[/tex]
using the Euler-Lagrange Equation:
[tex]
\partial_{\mu}\,(\frac{\partial{\cal L}}{\partial\,(\partial_{\mu}A_{\nu})}) -\, \frac{\partial{\cal L}}{\partial\,A_{\nu}}=0
[/tex]
So the first term in the E-L equation will only act on the derivatives in the Lagrangian, and the second term will only act in the non-derivatives, so I get:
[tex]
\frac{\partial{(-\,\frac{1}{4}\,F_{\mu\nu}F^{\mu\nu})}}{\partial\,(\partial_{\mu}A_{\nu})}=-F_{\mu\nu}=\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu}
[/tex]
[tex]
\frac{\partial{(-\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu})^2)}}{\partial\,(\partial_{\mu}A_{\nu})}=-\,\frac{1}{\xi}\,g^{\mu\nu}(\partial_{\mu}A_{\nu})
[/tex]
[tex]
\frac{\partial{(j_{\mu}A^{\mu}})}{\partial\,A_{\nu}}=\frac{\partial{(j_{\mu}g^{\mu\nu}A_{\nu}})}{\partial\,A_{\nu}}=g^{\mu\nu}j_{\mu}=j^{\nu}
[/tex]
Now, I am not entirely sure about my use of gmunu's here. Anyway, combining these, I get:
[tex]
\partial_{\mu}(\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu})-\frac{1}{\xi}g^{\mu\nu}\partial_{\mu}\partial_{\mu}A_{\nu}-j^{\nu}=0
[/tex]
Can you please tell me if this is correct so far, and if so, can you please direct me as to how to obtain the box times the metric, and the two derivatives with upper indices on the right of the left hand side on the equation of motion?
Thanks for your help,
Regards
neorich.