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 $T$^{[1]}, 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

^{[2]}, 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:

$2-\alpha ={\frac {d}{y_{t}}}\quad \qquad \Delta ={\frac {y_{h}}{y_{t}}}$

- ↑ We have already stated that considering $K$ as a coupling constant is equivalent to considering $T$ as such.
- ↑ 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.