# Derivation of the Proca equation from the Proca Lagrangian

1. Jul 5, 2011

### 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 :(

Last edited: Jul 5, 2011
2. Jul 5, 2011

### 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

3. Jul 5, 2011

### 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 (*).
$=g^{\mu \alpha} g_{\nu \gamma}(\partial_{\alpha} A^{\nu})(\partial_{\mu}A^{\gamma}) -g^{\mu \alpha} g_{\mu \delta}(\partial_{\alpha} A^{\nu})(\partial_{\nu} A^{\delta}) -g^{\nu \beta} g_{\nu \gamma} (\partial_{\beta} A^{\mu})(\partial_{\mu} A^{\gamma}) +g^{\nu \beta} g_{\mu \delta} (\partial_{\beta} A^{\mu})(\partial_{\nu} A^{\delta})$

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?

Last edited: Jul 5, 2011
4. Jul 6, 2011

### torus

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

5. Nov 30, 2012

### mrmokri

thanx timewalker