Watch for typos. This is typed 2003 from handwritten text of Lec. 2.4 Scaling [and exponent relations]

Transcription

1 Watch for typos. This is typed 2003 from handwritten text of Lec. 2.4 Scaling [and exponent relations] The scaling form idea is one of the ideas from this course that you re likeliest to use someday, I d wager. It s important for the RG lectures to come (I ll show that the RG implies a scaling form, and then refer you back to this handout for all the consequences of that.. But we write scaling forms (e.g. in analyzing simulations in many situations where we don t know how to write the R.G., or maybe none exists. It s not worth putting this thicket of algebra on the board just to be copied, so I m mainly leaving this topic to this reading. I call it Lec. 2.4 since properly it should have been a separate lecture. As it did historically, the scaling notion comes after the mean-field lectures (these inspire the idea of scaling forms and before the R.G. Other notes (i the scaling form idea was introduced in 1965 by Ben Widom who is still a chemistry professor at Cornell; (iii the best reference on scaling forms is probably chapter 11 of H. Eugene Stanley s book Introduction to phase transitions and critical phenomena (1971; this appeared a few months before the R.G. which immediately made it obsolete. There will be two developments: first for scaling of the thermodynamic functions (free energy, but also magnetization and susceptibility which are just its derivatives. Then, for scaling of correlation functions. These are the same two tracks we followed originally in the ideas of spontaneous symmetry breaking and long-range order respecively (Lec. 2.1; mean-field theory (Lec ; and later in the R.G. itself. O. Overview The basic idea of critical exponents meant quantities that were (a priori arbitrary functions of one variable namely t, the scaled temperature could be reduced to just one arbitrary constant, times a known (universal power law: e.g. m(t (const( t β Now, we ll take the next step: analogously, you can replace functions of two variables by functions of one variable. Equivalently, the functional dependences have universal shapes. Example from Lec. 2.3D (Landau theory Start from the Landau free energy density f L (m; t = atm bm4 hm ( The minimum of (* defined m, the spontaneous magnetization within the Landau theory. Taking f L / m = 0 gives 2atm + 2bm 3 = h. ( 1

2 Divide (** by t 3/2 : That is, 3 2at + 2b = t 1/2 t 1/2 h t 3/2. m(± t, h = t 1/2 m ± t 3/2 where m ± (... are two scaling functions (one for the high side and one for the low side of T c ; we think of such functions as having dimensionless arguments and values. Here, m ± (u are defined as the roots of the cubic equation (1 ±2a m + 2b m 3 = u. (2 In other words, m(h has the same, unique shape at any temperature t > 0, beginning linear in h and ending as h 1/3, which is expressed by m + (u; similarly, m (u expresses the shape at T < T c, which starts as a constant and ends as h 1/3. (Below, section 2.4 B [scaling plots] develops this idea. We could also plug back into eq. (* to express the free energy density f(t, h f L (m(t, h; t: f(t, h t = a + t 1/2 2 b t 1/2 t 3/2 t 1/2 (3 Or where f(t, h t 2 = f ± (h/ t 3/2, f ± (u = a m ± (u b m ±(u 4 u m ± (u (4 Here f ± (u is another pair of scaling functions. A. Scaling and scaling tricks Widom s ansatz (1965 was to say the same functional structure holds beyond meanfield theory, except with non-mean-field exponents: f(± t, h = t m f± (h/ t (5 where we called the exponents m and. From here on, most of the story is simply the applicatoin of three mathematical tricks, to put the scaling form into whatever form we need. Trick #1. Insert a limiting case We knew that f(± t, h = 0 t 2 α. (5 2

3 [That came (Lec. 2.2 A by definition of the exponent α for C(t, plus the thermodynamic identities relating that to derivatives of the free energy. Important note: here f( refers only to the singular part of free energy, elsewhere called f sing (.] Eq. (5 says f(t, h does not go to zero or diverge as h 0. Comparing (5 to (5, the only way both can be true is if the h 0 behavior of (5 is f(± t, 0 t m f± (0 with f ± (0 a finite constant. This requires identifying m = 2 α This allows us to rewrite the ansatz in the new form f(± t, h = t 2 α f± (h/ t (6 This is our basic starting point. It has two unknown exponents in it. Trick #2. Differentiate (or integrate. We have the basic identity * m(t, h f(t, h, (6+ h So plug (6 into (6+: we get Thus m(± t, h = h f(± t, h = t 2 α [ f ] ±. t m(± t, h = t 2 α m ± t : (7 a new scaling relation for magnetization, with scaling function m( obtained by differentiating f(. Next, we repeat Trick 1: set h 0 in (7, we get m(± t, h = t 2 α m ± (0. (7 On the other hand, we can compare with what we know physically (recall Lec. 2.2: { m(t = 0 for t > 0 m(t ( t β for t < 0. (7 * Reminder: obtained by writing out Z(t, h and finding m = Z 1 β Z/ h, e.g. as done in Lec

4 Thus the behavior must be m + (0 = 0, m (0 = const, and we must identify the exponents in (7 and (7 : i.e., finally, β = 2 α. (8 One more time. We know susceptibility is Go back to (7 plug into (8+; i.e. differentiate : we see χ(± t, h m h. (8+ χ(± t, h t 2 α 2 m ± t. Again, set h 0 (only after we ve differentiated! and we see χ(± t t 2 α 2 m ±(0. On the other hand, definition of critical exponent γ says χ(± t t γ, hence Combining (8 and (9, we conclude: γ = 2 α 2 (9 = β + γ; (10 α + 2β + γ = 2. (11 [Eq. (11 was one of the exponent relations on the table of exponents in Lec. 2.3.] Trick #3. Change of scaling variable Assume h > 0 w.l.o.g here. Eq. (7 says m(± t, h h β/ ( t β ([ h 1/ ] m±. h 1/ t Thus, we have two factors which are functions of the same ratio, and we can absorb the prefactor by defining another scaling function m 1 (: ( t m(± t, h h β/ m 1 h 1/ (12 where m 1 (u { u β m + (1/ u for u > 0; u β m (1/ u for u < 0. (13 Compare our starting point (7 to (12: in effect, we turned the scaling function inside out and switched the roles of t and h. 4

5 Next, deploy trick #1 by setting t 0 in (12. We get m(t = 0, h h β/ m 1 (0; (13 a on the other hand, from the definition of critical exponent δ we have m(t = 0, h h 1/δ (13 b Hence, it must be that δ = β = 1 + γ β, (14 another of the exponent relations. A final note Note the basic ansatz (6 is equivalent to another form, f(λt, λ h λ 2 α = f(t, h (15 for any rescaling ratio λ > 0. An f( satisfying Eq. (15 is called a generalized homogeneous function (a homogeneous function is the special case = 1. Proof. (a Eq (6 (15 by straight substitution. (b. Assume (15, then choose λ = 1/ t in (15. We get f(± t, h = f(±1, h/ t (1/ t 2 α t 2 α f± (h/ t,, provided we define f ± (u f(±1, u. Thus (15 (6, QED. 5