I Laws of Motion for New Lagrangian: Partial Differential Equations

[Maniac_XOX]

TL;DR Summary

Is it possible to simplify this complicated equation??

$$\partial^\beta F_{\beta\alpha} +\partial^\beta A_\mu A^\mu \delta^\alpha_\sigma \delta^\rho_\beta+\mu^2 A_\alpha = 2A_\mu (\partial_\rho A^\rho) +\frac {4\pi}{c}J_\alpha$$

 

[Ibix]

You don't have the same free indices on each of your terms. There are three free indices on the second one on the left and you have a free $\mu$ instead of $\alpha$ in the first term on the right. Correct your typos and we may be able to help.

 

[Maniac_XOX]

You don't have the same free indices on each of your terms. There are three free indices on the second one on the left and you have a free $\mu$ instead of $\alpha$ in the first term on the right. Correct your typos and we may be able to help.

Hmm you're right, the complicated part where i most def went wrong was finding the euler lagrange equation $$\partial^\beta \frac{\partial L}{\partial(\partial^\beta A^\alpha)}=\frac {\partial L}{\partial A^\alpha}$$
to the following section of the lagrangina i used: $-\beta A_\mu A^\mu (\partial_\rho A^\rho)$

Need help finding the answer to this.

I found that $$\partial^\beta \frac{\partial L}{\partial(\partial^\beta A^\alpha)}=\partial^\beta(-\beta A_\mu A^\mu \delta^\alpha_\sigma \delta^\rho_\beta)$$

and also that $$\frac {\partial L}{\partial A^\alpha}= -2\beta A_\alpha (\partial_\rho A^\rho) $$

Could u help me out?

 

[Orodruin]

I found that ∂β∂L∂(∂βAα)=∂β(−βAμAμδσαδβρ)

This is quite obviously wrong as you do not have the same free indices on both sides. Can you find your error and/or tell us how you arrive at this?

 

[Maniac_XOX]

This is quite obviously wrong as you do not have the same free indices on both sides. Can you find your error and/or tell us how you arrive at this?

hello yeah i realized actually, I've come to the right equation being $$\partial^\beta(A_\mu A^\mu g_{\rho\sigma} \delta^\sigma_\beta \delta^\rho_\alpha)=(\partial_\rho A^\rho) g_{\mu\gamma} (\delta_\alpha^\gamma A^\mu + A^\gamma \delta^\mu_\alpha)$$ from euler lagrange $$\partial^\beta \frac{\partial L}{\partial(\partial^\beta A^\alpha)}=\frac {\partial L}{\partial A^\alpha}$$
When put inside the whole complete equation it becomes:
$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\sigma} \delta^\sigma_\beta \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) g_{\mu\gamma} (\delta_\alpha^\gamma A^\mu + A^\gamma \delta^\mu_\alpha) + \frac {4\pi}{c} J_\alpha$$
Can you help me simplify this?

 

[Orodruin]

I would start by using the summation properties of the Kronecker delta.

 

[Ibix]

I'd move the $g_{\mu\lambda}$ inside the bracket before following Orodruin's advice.

 

[Maniac_XOX]

I'd move the $g_{\mu\lambda}$ inside the bracket before following Orodruin's advice.

I would start by using the summation properties of the Kronecker delta.

Okay so I've tried lowering the indices on the right hand side and using the property $$g_{\mu\gamma}g^{\mu\nu}=\delta^\nu_\gamma$$
therefore on the right hand side I get
$$(\partial_\rho A^\rho) (\delta_\gamma^\nu \delta_\alpha^\gamma A_\nu + \delta_\gamma^\lambda \delta^\mu_\alpha A_\lambda) + \frac {4\pi}{c} J_\alpha$$
which should be able to simplify as:
$$(\partial_\rho A^\rho) (\delta_\alpha^\nu A_\nu + \delta_\gamma^\lambda \delta^\mu_\alpha A_\lambda) + \frac {4\pi}{c} J_\alpha$$

Is this correct? If so how do i go from there?

 

[Ibix]

The point is that $g_{\mu\gamma}\delta^\gamma_\alpha=g_{\mu\alpha}$.

 

[Maniac_XOX]

The point is that $g_{\mu\gamma}\delta^\gamma_\alpha=g_{\mu\alpha}$.

OH i get it, so i should be getting this$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\beta} \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) (g_{\mu\alpha} A^\mu + g_{\gamma\alpha} A^\gamma) + \frac {4\pi}{c} J_\alpha$$which finally becomes$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\alpha\beta}) + \mu^2 A_\alpha = (A_\alpha + A_\alpha) + \frac {4\pi}{c} J_\alpha$$
so then the right hand side becomes $2A_\alpha + \frac {4\pi}{c} J_\alpha$ ?

 

[Ibix]

You seem to have lost a $\partial_\rho A^\rho$ on the right hand side. Otherwise, yes.

 

[Maniac_XOX]

You seem to have lost a $\partial_\rho A^\rho$ on the right hand side. Otherwise, yes.

Right i forgot that and also $-\beta$ factors on both sides while focusing on the other stuff. A question i have involving the LHS is:
Does $\partial^\beta(-\beta \times g_{\alpha\beta}A_\mu A^\mu)$
mean the same as $$\frac {\partial (-\beta \times g_{\alpha\beta}A_\mu A^\mu)}{\partial A^\beta}$$ ? Cuz then wouldn't that bit simplify to $-2\beta A_\alpha$ as well?