Derivation of the Proca equation from the Proca Lagrangian

In summary, the given Proca Lagrangian can be used to obtain the Proca equation, which can be solved by using the Euler-Lagrange equation. The index raising and lowering process is crucial in the solution process. There is no difference between solving for A^{\mu} or A_{\mu}, as they are connected by just raising or lowering the index \nu.
  • #1
timewalker
15
1
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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  • #2
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
 
  • #3
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?
 
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  • #4
No, there is no difference as they are connected by just raising/lowering the index nu.
 
  • #5
thanx timewalker
 

FAQ: Derivation of the Proca equation from the Proca Lagrangian

1. What is the Proca equation and what is its significance in physics?

The Proca equation is a relativistic wave equation that describes the behavior of massive particles with spin 1, such as the photon. It is derived from the Proca Lagrangian, which is a mathematical expression that describes the dynamics of such particles. The Proca equation is significant in physics because it helps us understand how these particles interact with other particles and fields in the universe.

2. How is the Proca equation derived from the Proca Lagrangian?

The Proca equation is derived from the Proca Lagrangian using the principle of least action. This principle states that the path a particle takes between two points is the one that minimizes the action, which is the integral of the Lagrangian over time. By varying the action with respect to the particle's position and momentum, we can obtain the Proca equation.

3. What are the assumptions made in the derivation of the Proca equation from the Proca Lagrangian?

The derivation of the Proca equation from the Proca Lagrangian assumes that the particle is a massive boson with spin 1, that it interacts with an external electromagnetic field, and that it follows the principle of least action. It also assumes that the particle's mass is constant and that the field is described by a vector potential.

4. How does the Proca equation differ from the Klein-Gordon equation?

The Proca equation differs from the Klein-Gordon equation in that it includes terms for the particle's mass and spin, whereas the Klein-Gordon equation only describes the behavior of spinless particles. The Proca equation also incorporates the effects of an external electromagnetic field, while the Klein-Gordon equation does not.

5. What are some applications of the Proca equation in physics?

The Proca equation has many applications in physics, including in the study of electromagnetic interactions, such as the behavior of photons and other spin 1 particles. It is also used in the theory of quantum electrodynamics, which describes the interactions between charged particles and electromagnetic fields. Other applications include the study of superconductivity and the behavior of particles in the early universe.

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