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In the general formulation of renormalization group in "statistical mechanics" by P.K.Pathria, each point in parameter space is represented by a vector $\vec{K}$ and the transformed vector would be given by $$\vec{K}'=R(\vec{K})$$ If we look at what happens when it gets very close to the fixed point $\vec{K}^*$, then we would have $$\vec{K}'=\vec{K}^*+\vec{k}'=R(\vec{K}^*+\vec{k})+\mathcal{O}(k_i^2) $$ where both $\vec{k}'$ and $\vec{k}$ will be very small.

Now the problem is that in this book, the author directly puts $$R(\vec{K}^*+\vec{k})=R(\vec{K}^*)+R(\vec{k})=\vec{K}^*+R(\vec{k})$$ and then conclude that the two deviations are related by $$\vec{k}'=R(\vec{k})$$ The step $R(\vec{K}^*+\vec{k})=R(\vec{K}^*)+R(\vec{k})$ looks obviously problematic to me, or did I miss anything?

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It turns out to be a mistake in this version of the book (2nd edition), I checked the latest version (3rd edition) and it has been corrected there. Naturally, the corrected one would be $$R(\vec{K}^*+\vec{k})=R(\vec{K}^*)+R'(\vec{K}^*)\vec{k}+...$$ which after linearization simply becomes $\vec{K}^*+R'(\vec{K}^*)\vec{k}$, and then we have a linearized relation between $\vec{k}'$ and $\vec{k}$, which is $$\vec{k}'=R'(\vec{K}^*)\vec{k}$$

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You missed the fact that this operator $R$ is a linear operator, at least for sufficiently small $\vec{k}$. This is a linearization of the renormalization group transformation around the fixed point $\vec{K}^*$. Let $T$ be a renormalization group transformation. If you expand around the fixed point $\vec{K}^*$ you get:

$$ T(\vec{K}^*+\vec{k})=T(\vec{K}^*)+[T^{(1)}(\vec{K}^*)]\vec{k}+\mathcal{O}(k_i^2) $$

Where $[T^{(1)}(\vec{K}^*)]$ is a linear operator acting on $\vec{k}$. Actually, making $R=T$ until first order in $k$ give us: $$ R(\vec{K}^*+\vec{k})=T(\vec{K}^*)+[T^{(1)}(\vec{K}^*)]\vec{k} $$ and $$ R(\vec{K}^*+\vec{0})=T(\vec{K}^*)=\vec{K}^* $$ $$ [T^{(1)}(\vec{K}^*)]\vec{k}=R(\vec{K}^*+\vec{k})-\vec{K}^*=(\vec{K}^*+\vec{k}')-\vec{K}^*=\vec{k}' $$ Then $R$ is an affine transformation and $R(\vec{K}^*+...)-\vec{K}^*$ is a linear transformation. So, what the author realy mean is: $$ \vec{k}'=R(\vec{K}^*+\vec{k})-\vec{K}^*=[T^{(1)}(\vec{K}^*)]\vec{k} $$

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  • $\begingroup$ yes, u r right, but this is exactly the same answer I gave above. The original problem was due to a mistake in the old version of the book, which I realised after checking the latest version. $\endgroup$ – M. Zeng Jan 21 '16 at 5:34
  • $\begingroup$ Yes. Is a problem of the book. But when you read this a lot of time in a lot of places you naturally overcome this little mistakes. $\endgroup$ – Nogueira Jan 21 '16 at 6:37

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