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Consider a theory in $D$ spatial dimensions involving one or more scalar fields $\phi_a$, with a Lagrangian density of the form $$L= \frac{1}{2} G_{ab}(\phi) \partial_\mu \phi_a \partial^\mu \phi_b- V(\phi)$$ where the eigenvalues of $G$ are all positive definite for any value of $\phi$, and $V = 0$ at its minima. Any finite energy static solution of the field equations is a stationary point of the potential energy $$E = I_K + I_V ,$$ where $$I_K[\phi]= \frac{1}{2} \int d^Dx G_{ab}(\phi) \partial_j \phi_a \partial_j\phi_b$$ and $$I_V = \int d^Dx V(\phi)$$ are both positive. Since the solution is a stationary point among all configurations, it must, a fortiori, also be a stationary point among any subset of these configurations to which it belongs. Therefore, given a solution $\phi(x)$, consider the one-parameter family of configurations, $$f_\lambda(x)= \bar{\phi}(\lambda x)$$ that are obtained from the solution by rescaling lengths. The potential energy of these configurations is given by \begin{align} E_\lambda &= I_K(f_\lambda) + I_V(f_\lambda)\\ &=\lambda^{2-D} I_K[\bar\phi]+\lambda^{-D} I_V[\bar\phi]\tag{1} \end{align}

$\lambda = 1$ must be a stationary point of $E(\lambda)$, which implies that $$0=(D-2) I_K[\bar\phi]+D I_V[\bar\phi] \tag{2}$$

My problem is, how they got the equation(2) from (1)?

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Related: – Qmechanic Aug 30 '13 at 15:06
$\uparrow$ Who are they? – Qmechanic Mar 16 at 22:09
up vote 4 down vote accepted

I didn't go through all of your equations. However, if you take (1), differentiate it w.r.t $\lambda$ and set $\lambda = 1$, then since $\lambda=1$ is the stationary point $E'(\lambda)|_{\lambda=1} = 0$. This is equation (2)

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Can you look at the equation (1) that how it came from the previous line? – Unlimited Dreamer May 1 '13 at 14:57

You started off stating that the undeformed solution ($\lambda=1$) must be a stationary point. So, if you differentiate the energy of this one-parameter family wrt $\lambda$ you should get zero. So evaluate $\frac{dE}{d \lambda}$ @ $\lambda = 1$.

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