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We define the energy as $$E = I_F + I_K + I_V,$$ where, $$I_F [A]= \frac{1}{2} \int d^Dx \operatorname{tr} F^2_{ij},$$ $F_{ij}$ represents the electromagnetic force. $$I_K [\phi,A]= \frac{1}{2} \int d^Dx (D_j \phi)^\dagger (D_j \phi),$$ $$I_V [\phi]= \int d^Dx V(\phi),$$ Given a solution $\bar \phi(x)$, $\bar A_j(x)$, we define scaled fields $$f_\lambda(x)= \bar\phi(\lambda x),$$ $$g_{j\lambda}(x)= \lambda \bar A_j(\lambda x).$$ Proceeding as in the pure scalar case, we find that \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] \tag{1} \end{align} which is stationary at $\lambda = 1$ if

I just don't understand how $$I_F [A]=\lambda^{4-D} I_F [\bar A],$$ is in the equation (1)

The post is slightly related with this: link

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    $\begingroup$ Yang-Mills theory? $\endgroup$
    – Josh
    Jun 12, 2013 at 12:07
  • $\begingroup$ sorry my bad , yup. $\endgroup$
    – user52950
    Jun 12, 2013 at 12:08
  • $\begingroup$ I do not understand a thing: if $I_F[g_\lambda]=\int d^Dx tr F^2(g_\lambda)$, with $F_{ij}(g_\lambda)(x)=\partial_i \lambda \bar{A}_j(\lambda x)-\partial_j \lambda \bar{A}_i(\lambda x)$, then in $tr F^2$ I can extract immediatly 2 powers of $\lambda$ and with the change of variables $\lambda x=y$ we get $-D$ powers, and so $2-D$ in total. Did I misunderstand your notation? $\endgroup$
    – Avitus
    Jun 13, 2013 at 10:59
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    $\begingroup$ @Avitus: Did you take into account the derivatives? You should be able to extract 4 powers of $\lambda$. $\endgroup$ Jun 13, 2013 at 11:14
  • $\begingroup$ @Raskolnikov you are right; now I see it. I messed up with notation :-) $\endgroup$
    – Avitus
    Jun 13, 2013 at 12:19

1 Answer 1

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So,

$$F_{ij}(g_\lambda)(x)=\frac{\partial \lambda \bar{A}_j(\lambda x)}{\partial x_i} -\frac{\partial \lambda \bar{A}_i(\lambda x)}{\partial x_j}$$

becomes with the change of variables $y=\lambda x$

$$F_{ij}(g_\lambda)(x)=\frac{\partial \lambda^2 \bar{A}_j(\lambda x)}{\partial \lambda x_i} -\frac{\partial \lambda^2 \bar{A}_i(\lambda x)}{\partial \lambda x_j} = \lambda^2 \left(\frac{\partial\bar{A}_j(y)}{\partial y_i} -\frac{\partial \bar{A}_i(y)}{\partial y_j}\right)$$

which implies that

$$I_F[g_\lambda]=\int d^Dx \; \mbox{tr } F^2(g_\lambda) = \lambda^{4-D}\int d^Dy \; \mbox{tr } F^2(A) \; .$$

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