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.