# The origins of scaling and critical behaviour

Let us consider a fixed point of the RG flow of a generic system, and assume that it has two relevant directions corresponding to the coupling constants $T$ , the temperature, and $H$ , the external field. We suppose that $T$ and $H$ are transformed under the RG as:

$T'=R_{\ell }^{T}(T,H)\quad \qquad H'=R_{\ell }^{H}(T,H)$ where $R_{\ell }^{T}$ and $R_{\ell }^{H}$ are analytic functions given by the coarse graining procedure. The fixed points $(T^{*},H^{*})$ of the flow will be given by the solutions of:
$T^{*}=R_{\ell }^{T}(T^{*},H^{*})\quad \qquad H^{*}=R_{\ell }^{H}(T^{*},H^{*})$ Linearising the transformation around $(T^{*},H^{*})$ , in terms of the reduced variables $t=(T-T^{*})/T^{*}$ and $h=(H-H^{*})/H^{*}$ we have:
${\begin{pmatrix}t'\\h'\end{pmatrix}}={\boldsymbol {T}}{\begin{pmatrix}t\\h\end{pmatrix}}$ where:
${\boldsymbol {T}}={\begin{pmatrix}\partial R_{\ell }^{T}/\partial T&\partial R_{\ell }^{T}/\partial H\\\partial R_{\ell }^{H}/\partial T&\partial R_{\ell }^{H}/\partial H\end{pmatrix}}_{|T^{*},H^{*}}$ As previously stated we suppose ${\boldsymbol {T}}$ to be diagonalizable. We therefore write its eigenvalues as:
$\lambda _{\ell }^{t}=\ell ^{y_{t}}\quad \qquad \lambda _{\ell }^{h}=\ell ^{y_{h}}$ Note that we can always do that, it is just a simple definition. In other words, we are defining $y_{t}$ and $y_{h}$ as:
$y_{t}={\frac {\ln \lambda _{\ell }^{t}}{\ln \ell }}\quad \qquad y_{h}={\frac {\ln \lambda _{\ell }^{h}}{\ln \ell }}$ This way we can write:
${\begin{pmatrix}t'\\h'\end{pmatrix}}={\begin{pmatrix}\lambda _{\ell }^{t}&0\\0&\lambda _{\ell }^{h}\end{pmatrix}}{\begin{pmatrix}t\\h\end{pmatrix}}\quad \Rightarrow \quad {\begin{pmatrix}t'\\h'\end{pmatrix}}={\begin{pmatrix}\ell ^{y_{t}}t\\\ell ^{y_{h}}h\end{pmatrix}}$ After $n$ iterations we will have:
$t^{(n)}=\left(\ell ^{y_{t}}\right)^{n}t\quad \qquad h^{(n)}=\left(\ell ^{y_{h}}\right)^{n}h$ and since ingeneral we know that $\xi (t',h')=\xi (t,h)/\ell$ :
$\xi (t,h)=\ell ^{n}\xi (\ell ^{ny_{t}}t,\ell ^{ny_{h}}h)$ This is the scaling law of the correlation length. From this we can determine the critical exponent $\nu$ ; in fact, setting $h=0$ and choosing $\ell$ so that $t\ell ^{ny_{t}}=b$ with $b$ a positive real number, we have:
$\ell ^{n}=\left({\frac {b}{t}}\right)^{1/y_{t}}\quad \Rightarrow \quad \xi (t)=\left({\frac {t}{b}}\right)^{-1/y_{t}}\xi (b,0)$ Since in general $\xi \sim t^{-\nu }$ , we get:
$\nu ={\frac {1}{y_{t}}}$ This is an extremely important result! In fact, we see that once the RG transformation $R_{\ell }$ is known, $y_{t}$ is straightforward to compute and so we are actually able to calculate $\nu$ and predict its value! We can do even something more (including giving $y_{h}$ a meaning) from the scaling law of the free energy density. After $n$ iterations of the RG we have:
$f(t,h)=\ell ^{-nd}f(t^{(n)},h^{(n)})=\ell ^{-nd}f(\ell ^{ny_{t}}t,\ell ^{ny_{h}}h)$ and choosing $\ell$ so that $\ell ^{ny_{t}}t=b^{y_{t}}$ , then:
$f(t,h)=t^{d/y_{t}}b^{-d}f(b^{y_{t}},b^{y_{h}}h/t^{y_{h}/y_{t}})$ Comparing this to what we have seen in An alternative expression for the scaling hypothesis we get:

$2-\alpha ={\frac {d}{y_{t}}}\quad \qquad \Delta ={\frac {y_{h}}{y_{t}}}$ 1. We have already stated that considering $K$ as a coupling constant is equivalent to considering $T$ as such.
2. Remember that the value of $\ell$ is not fixed, so we can choose the one we prefer; in this case we are making this choice because $\ell$ does not necessarily have to be an integer.