Course:Statistical Mechanics/The Renormalization Group/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 T¹ , 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) \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^*) \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:
λ_\ell ^t = \ell ^{y_t} \qquad λ_\ell ^h = \ell ^{yₕ}
Note that we can always do that, it is just a simple definition. In other
words, we are defining yₜ and yₕ as:
y_t = \frac{\ln λ_\ell ^t}{\ln \ell } \qquad y_h = \frac{\ln λ_\ell ^h}{\ln
\ell }
This way we can write:
\begin{pmatrix} t' h'\end{pmatrix} = \begin{pmatrix} λ_\ell ^t & 0 0 &
λ_\ell ^h \end{pmatrix} \begin{pmatrix} t h\end{pmatrix} ⇒ \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 \qquad h^{(n)} = \left (\ell
^{y_h}\right )ⁿ h
and since ingeneral we know that ξ(t',h') = ξ(t,h)/\ell :
ξ(t,h) = \ell ^n ξ(\ell ^{ny_t} t, \ell ^{nyₕ} h )
This is the scaling law of the correlation length. From this we can
determine the critical exponent ν; in fact, setting h=0 and choosing \ell
so that t\ell ^{nyₜ}=b with b a positive real number² , we have:
\ell ^n = \left (\frac{b}{t} \right )^{1/y_t} ⇒ ξ(t) = \left ( \frac{t}{b}
\right )^{-1/y_t} ξ(b,0)
Since in general ξ\sim t^{-ν}, we get:
ν= \frac{1}{yₜ}
This is an extremely important result! In fact, we see that once the RG
transformation R_\ell is known, yₜis straightforward to compute and so we
are actually able to calculate ν and predict its value! We can do even
something more (including giving yₕ 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
^{nyₕ}h)
and choosing \ell so that \ell ^{nyₜ}t = b^{yₜ}, then:
f(t,h) = t^{d/yₜ} b^{-d} f(b^{yₜ},b^{yₕ}h/t^{yₕ/yₜ})
Comparing this to what we have seen in An alternative expression for the
scaling hypothesis we get:
2-α= \frac{d}{yₜ} \qquad Δ= \frac{yₕ}{yₜ}
[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.