Momentum Shell RG für Ising-Modell

Question

Momentum Shell RG für Ising-Modell

Physik
Renormierung
Quantenfeldtheorie
Statistische-mechanik

Artem Alexandrov

Ich studiere das Buch "Introduction to Functional RG" von Kopietz. Nachdem die Autoren alle erforderlichen Konzepte eingeführt hatten, erhielten sie die folgenden RG-Flussgleichungen:

\partial_{l} {\bar{r}}_{l} = 2 {\bar{r}}_{l} + \frac{1}{2} \frac{{\bar{u}}_{l}}{1 + {\bar{r}}_{l}};;

\partial_{l} {\bar{u}}_{l} = (4 - - D.) {\bar{u}}_{l} - - \frac{3}{2} \frac{{\bar{u}}_{l}^{2}}{(1 + {\bar{r}}_{l})^{2}},

wo

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} / (c_{0} Λ_{0}^{2})

{\bar{r}}_{l} = r_{l} /. (c_{0} Λ_{0}^{2})

und

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = K_{D} u_{l} / (c_{0}^{2} Λ_{0}^{4 - D})

{\bar{u}}_{l} = {K.}_{D.} u_{l} /. (c_{0}^{2} Λ_{0}^{4 - - D.})

und

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T - T_{c}) / (a^{2} T_{c})

r_{l} = (T. - - {T.}_{c}) /. ({ein}^{2} {T.}_{c})

,

c_{0} = 1 / (2 D)

c_{0} = 1 / (2 D)

c_{0} = 1 / (2 D)

c_{0} = 1 / (2 D)

c_{0} = 1 / (2 D)

c_{0} = 1 / (2 D)

c_{0} = 1 /. (2 D.)

,

u_{l} = 2 a^{D - 4}

u_{l} = 2 a^{D - 4}

u_{l} = 2 a^{D - 4}

u_{l} = 2 a^{D - 4}

u_{l} = 2 a^{D - 4}

u_{l} = 2 a^{D - 4}

u_{l} = 2 a^{D - 4}

u_{l} = 2 a^{D - 4}

u_{l} = 2 {ein}^{D. - - 4}

(wie ich es verstehe). Ich möchte die ungefähre kritische Temperatur für das 3D-Ising-Modell (nicht das mittlere Feld) ermitteln. Was soll ich mit Flussgleichungen machen?

Erstens kann man, wie im Buch beschrieben, den Wilson-Fisher-Punkt (WF-Punkt) betrachten und Strömungsgleichungen in der Nähe des WF-Punkts linearisieren. Aber es macht für mich keinen Sinn, wie man eine ungefähre Angabe erhält $T_{c}$ $T_{c}$ $T_{c}$ $T_{c}$ ${T.}_{c}$ dh wie man eine Temperaturverschiebung findet.

Zweitens habe ich diese Notizen gefunden (S. 33-36), aber ich verstehe nicht, wie der Autor nur Verschiebung definiert $t_{c}$ $t_{c}$ $t_{c}$ $t_{c}$ $t_{c}$ (Gleichungen 178 und 179).

Antworten (1)

Momentum Shell RG für Ising-Modell

Artem Alexandrov · Answer 1

Entschuldigung für die Dummy-Frage, es war einfach. Schreiben Sie einfach auf $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = (T_{c} - T_{c 0}) / (a^{2} T_{c 0})$ $r = ({T.}_{c} - - {T.}_{c 0}) /. ({ein}^{2} {T.}_{c 0})$ , wo $T_{c 0} = 2 D J$ $T_{c 0} = 2 D J$ $T_{c 0} = 2 D J$ $T_{c 0} = 2 D J$ $T_{c 0} = 2 D J$ $T_{c 0} = 2 D J$ ${T.}_{c 0} = 2 D. J.$ ist die kritische MFA-Temperatur. Dies ergibt den folgenden Ausdruck für das 3D-Modell:

{T.}_{c} = {T.}_{c 0} (1 - - \frac{1}{6 π})

, wo

ϵ = 4 - D = 1

ϵ = 4 - D = 1

ϵ = 4 - - D. = 1

. Es gibt

T_{c} = 5.68 J

T_{c} = 5.68 J

T_{c} = 5.68 J

T_{c} = 5.68 J

T_{c} = 5.68 J

T_{c} = 5.68 J

{T.}_{c} = 5.68 J.

aber es scheint seltsam. Zum Beispiel kann man im 2D-Fall erhalten

T_{c} = 3.68 J

T_{c} = 3.68 J

T_{c} = 3.68 J

T_{c} = 3.68 J

T_{c} = 3.68 J

T_{c} = 3.68 J

{T.}_{c} = 3.68 J.

Das ist so weit vom genauen Ergebnis entfernt

T_{c}^{exact} = 2.27 J

T_{c}^{exact} = 2.27 J

T_{c}^{exact} = 2.27 J

T_{c}^{exact} = 2.27 J

T_{c}^{exact} = 2.27 J

T_{c}^{exact} = 2.27 J

T_{c}^{exact} = 2.27 J

T_{c}^{exact} = 2.27 J

{T.}_{c}^{genau} = 2.27 J.

. Ist es richtig?...

Momentum Shell RG für Ising-Modell

Artem Alexandrov

Antworten (1)

Artem Alexandrov

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