The forum discussion centers on the application of the Euler-Lagrange equation to the Proca Lagrangian, specifically the modified form that includes an extra operator. The Proca Lagrangian discussed is $$L= -\frac {1}{16\pi} F_{\alpha\beta} F^{\alpha\beta} + \frac {\mu^2} {8\pi} A_\alpha A^\alpha - \frac {1} {c} J_\alpha A^\alpha$$. The participants explore the implications of adding the term $$-\beta A_\mu A^\mu (\partial_\rho A^\rho)$$ and the correct application of index notation and derivatives. Key points include the correct use of the Kronecker delta in derivatives and the importance of index arrangement in tensor calculus.
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Apart from the ##-\beta## factor, you also lose a ##\partial_\rho A^\rho##. But yes essentially is correct. You can still simplify the RHS a little.Maniac_XOX said:i ended up using the property ##g_{\mu\gamma}\delta^\gamma_\alpha=g_{\mu\alpha}## to come up with this solution:
$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\beta} \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) (g_{\mu\alpha} A^\mu + g_{\gamma\alpha} A^\gamma) + \frac {4\pi}{c} J_\alpha$$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##
Right i forgot those factors while focusing on the other stuff but yeah it's still there. Not sure how i can simplify more but a question i have involving the LHS is:Gaussian97 said:Apart from the ##-\beta## factor, you also lose a ##\partial_\rho A^\rho##. But yes essentially is correct. You can still simplify the RHS a little.
Right, so the correct definition of that would be then $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial t} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu)}{\partial x} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial y} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial z}$$Gaussian97 said:No, the derivative is wrt the 4-position, not the ##A## field.
But i thought that's what the 4-position was, and i substituted the betas everywhere relevant?Gaussian97 said:No, first of all, if you expand the sum over ##\beta##, you must substitute ##\beta## everywhere. But even then, what you have is still not 100% correct, because for the derivative to be a contravariant vector ##\partial^\beta## you must differentiate wrt a covariant vector, i.e. ##x_\beta##, and the components of ##x_\beta## are not ##(t,x,y,z)##
Those are the contravariant coordinates of the ##x## 4-vector, you want to take derivatives with respect to the covariant coordinates.Maniac_XOX said:But i thought that's what the 4-position was, and i substituted the betas everywhere relevant?
Yeah I'm not planning on expanding ##\beta##. I think i understood now that ##\partial^\beta## can only be operated as $$g^{\beta\alpha} \partial_\alpha = g^{\beta\alpha} \frac{\partial}{\partial x^\alpha}$$Gaussian97 said:Those are the contravariant coordinates of the ##x## 4-vector, you want to take derivatives with respect to the covariant coordinates.
I don't know what you mean by relevant, if you expand ##\beta## you cannot have ##\beta## in the final expression, otherwise you did it wrong.
In any case, you can simplify this expression without any need of expanding any term.
This time the components of ##\partial x^\alpha## are $$\frac {\partial_t}{c} + \frac{\partial}{\partial x} + \frac {\partial}{\partial y} + \frac {\partial}{\partial z}$$Gaussian97 said:Well, you can use the product rule to simplify it a little the derivative, and then collect the terms proportional to ##\beta##, but I don't think this will make a big difference.