Derivation of the Proca equation from the Proca Lagrangian

[timewalker]

How to show the Proca equation by using the given Proca Lagrangian?
Surely, I know the Euler-Lagrange equation, but I can't solve this differentiation!(TT)

The given Proca lagrangian is,
\mathcal{L}= -\frac{1}{16\pi}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})+ \frac{1}{8 \pi} (\frac{mc}{\hbar})^2 A^{\nu} A_{\nu}

and the Euler-Lagrangian equation is,
\partial_{\mu}(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A^{\nu})}) = \frac{\partial \mathcal{L}}{\partial A^\nu}

At first, I just tried to solve

\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A^{\nu})}= \frac{\partial}{\partial(\partial_{\nu}A^{\mu})}(-\frac{1}{16 \pi}(\partial^{\mu}A^{\nu}-\partial{^\nu}A^{\mu})(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})+\cdots)

but I think I am misunderstand and not very well to handle these indices. So I think I can understand if I can see correct solving procedure. Please help me :(

 

[torus]

Hi,
you need to raise and lower the indices so they match your derivative-operator, i.e. write

\partial^\mu A^\nu = g^{\mu \alpha} \partial_\alpha A^\nu

then you can use
\frac{\partial}{\partial (\partial_\alpha A^\beta)} \partial_\mu A^\nu = \delta^\alpha_\mu \delta^\nu_\beta

Hope this helps,

torus

 

[timewalker]

Thank you so much! After I see your reply, I thought a little bit and I got right answer! :)
Let me finish this post. :D

Now we have the Proca Lagrangian given

\mathcal{L}=-\frac{1}{16 \pi} (\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu} )(\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu} ) + \frac{1}{8 \pi} (\frac{mc}{\hbar})^2 A_{\nu} A^{\nu}

Here we use the index lowering/raising as 'torus' said,

\partial^{\mu} A^{\nu} = g^{\mu \alpha} \partial_{\alpha} A^{\nu}
\partial_{\mu} A_{\nu} = g_{\nu \gamma} \partial_{\mu} A^{\gamma}

then we have the Lagrangian in a modified form.

\mathcal{L}=-\frac{1}{16 \pi} (g^{\mu \alpha} \partial_{\alpha} A^{\nu} - g^{\nu \beta} \partial_{\beta} A^{\mu} )(g_{\nu \gamma} \partial_{\mu} A^{\gamma} - g_{\mu \delta} \partial_{\nu} A^{\delta} ) + \frac{1}{8 \pi} (\frac{mc}{\hbar})^2 A_{\nu} A^{\nu}

Now expand the parenthesis in the first term.

(g^{\mu \alpha} \partial_{\alpha} A^{\nu} - g^{\nu \beta} \partial_{\beta} A^{\mu} )(g_{\nu \gamma} \partial_{\mu} A^{\gamma} - g_{\mu \delta} \partial_{\nu} A^{\delta} ) :: Let this be (*).
<br /> =g^{\mu \alpha} g_{\nu \gamma}(\partial_{\alpha} A^{\nu})(\partial_{\mu}A^{\gamma})<br /> -g^{\mu \alpha} g_{\mu \delta}(\partial_{\alpha} A^{\nu})(\partial_{\nu} A^{\delta})<br /> -g^{\nu \beta} g_{\nu \gamma} (\partial_{\beta} A^{\mu})(\partial_{\mu} A^{\gamma}) +g^{\nu \beta} g_{\mu \delta} (\partial_{\beta} A^{\mu})(\partial_{\nu} A^{\delta})

Now we calculate \frac{\partial \mathcal{L}}{\partial (\partial_{\rho} A^{\sigma})} to get the Euler-Lagrange equation that \partial_{\rho} ( \frac{\partial \mathcal{L}}{\partial (\partial_{\rho} A^{\sigma})}) = \frac{\partial \mathcal{L}}{\partial A^{\sigma}}.

\frac{\partial \mathcal{L}}{\partial (\partial_{\rho} A^{\sigma})}
=-\frac{1}{16 \pi}\frac{\partial(*) }{\partial (\partial_{\rho} A^{\sigma})} +0

Using the product rule of the differentiation and \frac{\partial A^{i}}{\partial A^{j}}=\delta_{i}^{j}, \frac{\partial(*) }{\partial (\partial_{\rho} A^{\sigma})} is,

\frac{\partial(*) }{\partial (\partial_{\rho} A^{\sigma})} = 4\partial^{\rho} A_{\sigma} - 4 \partial_{\sigma} A^{\rho}

Therefore

\frac{\partial \mathcal{L}}{\partial (\partial_{\rho} A^{\sigma})} = -\frac{1}{4 \pi} (\partial^{\rho} A_{\sigma} - \partial_{\sigma} A^{\rho} )

and, using \frac{\partial \mathcal{L}}{\partial A^{\sigma}} = \frac{1}{4 \pi} (\frac{mc}{\hbar})^2 A^{\sigma}, the Euler-Lagrange equation yields

\partial_{\mu} (\partial^{\mu} A_{\nu} - \partial_{\nu} A^{\mu} ) + (\frac{mc}{\hbar})^2 A^{\nu} = 0 Q.E.D.

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P.S. Is there any difference between taking the Proca equation by solving
\partial_{\mu}(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu})}) = \frac{\partial \mathcal{L}}{\partial A_{\nu}}
and
\partial_{\mu}(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A^{\nu})}) = \frac{\partial \mathcal{L}}{\partial A^{\nu}}
??

Actually my textbook(D.J. Griffiths, Introduction to Elementary Particles, 2nd Edition, Chap. 10.2 Example 3) supposed the vector field A^{\mu} but solved the first one. I can't agree with that so I asked about the second one. Is there any problem?

 

[torus]

No, there is no difference as they are connected by just raising/lowering the index nu.

 

[mrmokri]

thanx timewalker