Derivation of the Proca equation from the Proca Lagrangian


by timewalker
Tags: formulation, lagrangian, proca equation
timewalker
timewalker is offline
#1
Jul5-11, 06:42 AM
P: 15
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,
[itex]\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}[/itex]

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

At first, I just tried to solve

[itex]\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)[/itex]

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 :(
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torus
torus is offline
#2
Jul5-11, 07:13 AM
P: 21
Hi,
you need to raise and lower the indices so they match your derivative-operator, i.e. write

[itex]\partial^\mu A^\nu = g^{\mu \alpha} \partial_\alpha A^\nu[/itex]

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

Hope this helps,

torus
timewalker
timewalker is offline
#3
Jul5-11, 11:06 AM
P: 15
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

[itex]\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}[/itex]

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

[itex]\partial^{\mu} A^{\nu} = g^{\mu \alpha} \partial_{\alpha} A^{\nu} [/itex]
[itex]\partial_{\mu} A_{\nu} = g_{\nu \gamma} \partial_{\mu} A^{\gamma} [/itex]

then we have the Lagrangian in a modified form.

[itex]\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}[/itex]

Now expand the parenthesis in the first term.

[itex](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} )[/itex] :: Let this be (*).
[itex]
=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})[/itex]

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

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

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

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

Therefore

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

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

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

=======================================================================

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

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

torus
torus is offline
#4
Jul6-11, 06:10 AM
P: 21

Derivation of the Proca equation from the Proca Lagrangian


No, there is no difference as they are connected by just raising/lowering the index nu.
mrmokri
mrmokri is offline
#5
Nov30-12, 10:22 AM
P: 1
thanx timewalker


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