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Following up on a related question. This implies that,

From the reference we can write \begin{align*} E(\lambda) &= I_F [g_\lambda ] + I_K [f_\lambda , g_\lambda ] + I_V [f_\lambda ] \\ & = \lambda^{4-d} I_F [\bar A] + \lambda^{2-d} I_K [\bar \phi,\bar A]+ \lambda^{-d} I_V [\bar \phi]. \end{align*} This must be stationary at $\lambda = 1$, therefore we must have, \begin{equation} 0= (4-d) I_F [\bar A] + (d-2) I_K [\bar \phi,\bar A]+ d I_V [\bar \phi].\tag{1} \end{equation}

I think the equation (1) equates the time derivative to zero. Is it right? If this is true, then why are we taking this derivative and why does it need to be zero?

Source: Erick J. Weinberg, Classical Solutions in Quantum Field Theory: Solitons and Instantons in High Energy Physics (p. 43)

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  • $\begingroup$ Not sure if this matters, but... Where you have $(4-d)$, the book has $(d-4)$. The book claims that, if $(1)$ then the thing is stationary at $\lambda=1$. You state the reverse: if the thing is stationary at $\lambda=1$ then $(1)$. $\endgroup$
    – Řídící
    Commented Sep 8, 2013 at 12:10

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