Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 configura- tions, 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)?

share|improve this question
Related: math.stackexchange.com/q/476497/11127 –  Qmechanic Aug 30 '13 at 15:06

2 Answers 2

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)

share|improve this answer
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 $\frac{dE}{d \lambda}$ @ $\lambda = 1$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.