---
schema_version: 1
id: PHYS-2026-07-28-01
date: 2026-07-28
updated_at: 2026-07-28
title: One static fixed point, two critical clocks
summary: "Why the same equilibrium phi-four theory can produce Model-A and Model-B response poles, different relaxation laws, and different dynamical critical exponents."
language: en
entry_kind: daily
status: published
level: graduate-advanced
user_difficulty: unrated
domains:
  - statistical-mechanics
  - quantum-field-theory
  - condensed-matter
estimated_minutes: 55
---

## Key claim

**A static RG fixed point does not determine critical relaxation by itself.**

Two systems can share the same equilibrium $\phi^4$ theory and static susceptibility while conservation changes their response poles and dynamical exponents.

## Theme

**Static Wilson–Fisher universality versus dynamical critical slowing.**

## Guiding question

How can two systems have

$$
\chi(\mathbf q,0)\sim q^{-2+\eta}
$$

yet exhibit different low-frequency response functions and different dynamical exponents $z$?

The equilibrium functional fixes static probabilities and correlations. A retarded response also needs an equation of motion, its conservation laws, kinetic coefficients, and noise.

## Setup

Consider a classical scalar order parameter in $d=4-\epsilon$ spatial dimensions:

$$
\mathcal F[\phi;h]
=
\int\dd^dx
\left[
\frac12(\nabla\phi)^2
+\frac r2\phi^2
+\frac{u}{4!}\phi^4
-h\phi
\right].
$$

For $\epsilon>0$ and $u>0$, tuning $r$ to the critical surface leads to the Wilson–Fisher fixed point.

The equilibrium measure is

$$
P_{\rm eq}[\phi]
=
\frac1Z
\exp\left[-\frac{\mathcal F[\phi]}{T}\right],
$$

where $k_B=1$.

This measure contains no rule for how $\phi$ relaxes.

Start with nonconserved relaxational dynamics, Model A:

$$
\partial_t\phi(\mathbf x,t)
=
-\Gamma
\frac{\delta\mathcal F}{\delta\phi(\mathbf x,t)}
+\zeta(\mathbf x,t).
$$

The white noise satisfies

$$
\left\langle
\zeta(\mathbf x,t)
\zeta(\mathbf x',t')
\right\rangle
=
2\Gamma T\,
\delta^d(\mathbf x-\mathbf x')
\delta(t-t').
$$

The drift and noise amplitude obey detailed balance and make $P_{\rm eq}$ stationary.

At criticality, dynamic scaling takes the form

$$
\chi^R(q,\omega)
=
q^{-2+\eta}
\Phi\left(\frac{\omega}{q^z}\right),
$$

up to nonuniversal metric factors.

The static exponent $\eta$ does not determine $z$.

## Analysis

### Derivation: derive the Model-A pole

Set $u=0$. In Fourier space,

$$
\left[
-\ii\omega+\Gamma(r+q^2)
\right]
\phi(\mathbf q,\omega)
=
\Gamma h(\mathbf q,\omega)
+\zeta(\mathbf q,\omega).
$$

The retarded susceptibility is

$$
\chi_A^R(q,\omega)
=
\frac{
\delta\langle\phi(\mathbf q,\omega)\rangle
}{
\delta h(\mathbf q,\omega)
}
=
\frac{\Gamma}{
\Gamma(r+q^2)-\ii\omega
}.
$$

Equivalently,

$$
\chi_A^R(q,\omega)
=
\frac1{
r+q^2-\ii\omega/\Gamma
}.
$$

For $\omega>0$, its dissipative part is

$$
\chi_A''(q,\omega)
=
\frac{
\Gamma\omega
}{
\Gamma^2(r+q^2)^2+\omega^2
}.
$$

The pole lies at

$$
\omega_*=-\ii\Gamma(r+q^2),
$$

so

$$
\tau_q
=
\frac1{\Gamma(r+q^2)}.
$$

At the Gaussian critical point $r=0$,

$$
\tau_q\sim q^{-2},
\qquad
z_A^{\rm Gaussian}=2.
$$

The equilibrium correlation spectrum is

$$
C_A(q,\omega)
=
\frac{
2\Gamma T
}{
\omega^2+\Gamma^2(r+q^2)^2
}.
$$

It obeys the classical fluctuation–dissipation relation

$$
C_A(q,\omega)
=
\frac{2T}{\omega}
\chi_A''(q,\omega).
$$

The correlation function and dissipative response therefore contain the same relaxation rate.

### Scope and assumptions: conservation changes the clock

The functional $\mathcal F[\phi]$ does not select Model A.

Suppose $\phi$ is a conserved density. Local conservation requires

$$
\partial_t\phi+\nabla\cdot\mathbf j=0.
$$

Choose the dissipative current

$$
\mathbf j
=
-\Gamma\nabla
\frac{\delta\mathcal F}{\delta\phi}
+\boldsymbol\xi.
$$

Its noise obeys

$$
\left\langle
\xi_i(\mathbf x,t)
\xi_j(\mathbf x',t')
\right\rangle
=
2\Gamma T\,
\delta_{ij}
\delta^d(\mathbf x-\mathbf x')
\delta(t-t').
$$

The order parameter then follows Model-B dynamics:

$$
\partial_t\phi
=
\Gamma\nabla^2
\frac{\delta\mathcal F}{\delta\phi}
+\zeta_B,
\qquad
\zeta_B=-\nabla\cdot\boldsymbol\xi.
$$

The conserved noise vanishes at zero momentum:

$$
\left\langle
\zeta_B(\mathbf q,t)
\zeta_B(-\mathbf q,t')
\right\rangle
=
2\Gamma Tq^2\delta(t-t').
$$

In the Gaussian theory,

$$
\chi_B^R(q,\omega)
=
\frac{
\Gamma q^2
}{
\Gamma q^2(r+q^2)-\ii\omega
}.
$$

Its static limit matches Model A:

$$
\chi_B^R(q,0)
=
\frac1{r+q^2}.
$$

Its pole does not:

$$
\omega_*
=
-\ii\Gamma q^2(r+q^2).
$$

At $r=0$,

$$
\tau_q\sim q^{-4},
\qquad
z_B^{\rm Gaussian}=4.
$$

The extra $q^2$ comes from local conservation. At $q=0$, the total conserved order parameter cannot relax.

At the interacting fixed point,

$$
z_A=2+O(\epsilon^2),
$$

while equilibrium Model B obeys

$$
z_B=4-\eta.
$$

The Model-B result assumes detailed balance and no additional slow field coupled to the order parameter.

### Physical interpretation: separate restoring force from mobility

Write both linearized dynamics as

$$
\left[
-\ii\omega
+\mathcal K(q)(r+q^2)
\right]\phi
=
\mathcal K(q)h+\zeta.
$$

The static inverse susceptibility

$$
r+q^2
$$

is the thermodynamic restoring force.

The kinetic kernel is

$$
\mathcal K_A(q)=\Gamma
$$

for Model A and

$$
\mathcal K_B(q)=\Gamma q^2
$$

for Model B.

The static functional fixes the restoring force. Conservation fixes the small-$q$ structure of the kinetic kernel, which then sets the response pole.

The correspondence can be summarized as follows:

| Structure | Model A | Model B |
| --- | --- | --- |
| Order parameter | Nonconserved | Conserved |
| Kinetic kernel | $\Gamma$ | $\Gamma q^2$ |
| Gaussian rate at $r=0$ | $\Gamma q^2$ | $\Gamma q^4$ |
| Gaussian exponent | $z=2$ | $z=4$ |
| Static susceptibility | $q^{-2}$ | $q^{-2}$ |

This correspondence has a clear boundary.

Coupling the order parameter to conserved energy produces Model C. Coupling a conserved order parameter to momentum density leads toward Model H.

Colored noise, memory kernels, driving, or nonreciprocal couplings can also invalidate the equilibrium fluctuation–dissipation mapping.

## False claim to diagnose

> Once the static RG flow reaches the Wilson–Fisher fixed point, the low-frequency susceptibility is fixed uniquely because all microscopic kinetic coefficients are irrelevant.

The claim confuses numerical kinetic coefficients with the structure of the kinetic operator.

Within one dynamical universality class, $\Gamma$ sets a nonuniversal time scale. But conservation changes $\mathcal K(q)$ from a constant to a quantity that vanishes as $q^2$.

That power of $q$ survives coarse-graining because it expresses a conservation law.

A defensible statement is:

> Static universality fixes equilibrium scaling. Dynamic universality additionally requires the slow variables, conservation laws, and noise structure.

## What follows — and what does not

Four statements that sound similar have different logical status:

| Statement | Status |
| --- | --- |
| Given $\mathcal F$ and detailed balance, the equilibrium measure fixes static correlations. | Equilibrium statement |
| Given a linear Langevin equation, its kinetic kernel fixes the Gaussian response pole. | Exact within the Gaussian dynamics |
| Model A and Model B can share the Wilson–Fisher static fixed point. | RG classification |
| They must share the same $z$. | False |

The static fixed point classifies equal-time long-distance fluctuations. The dynamical fixed point classifies how the slow variables approach equilibrium.

## Exercise

Use the Gaussian functional

$$
\mathcal F_0
=
\frac12
\int\dd^dx
\left[
(\nabla\phi)^2+r\phi^2
\right]
-\int\dd^dx\,h\phi.
$$

1. Derive $\chi_A^R(q,\omega)$ from Model-A dynamics.
2. Find the frequency at which $\chi_A''(q,\omega)$ is maximal.
3. At $r=0$, put $\chi_A^R$ into the form

   $$
   \chi_A^R(q,\omega)
   =
   q^{-2}
   \Phi_A\left(
   \frac{\omega}{\Gamma q^2}
   \right)
   $$

   and read off $z_A$.

4. Repeat the scaling analysis for Model B.
5. Verify that both models have the same $\chi^R(q,0)$.
6. Derive the factor $q^2$ in the Model-B noise covariance from $\zeta_B=-\nabla\cdot\boldsymbol\xi$.

<details>
<summary>Hint 1</summary>

For

$$
f(\omega)=\frac{\omega}{a^2+\omega^2},
$$

solve $\dd f/\dd\omega=0$.

</details>

<details>
<summary>Hint 2</summary>

Read $z$ from the dimensionless combination of $q$ and $\omega$ in the scaling function.

</details>

**Oral check 1.** Can two systems have identical equilibrium susceptibilities but parametrically different relaxation times?

**Oral check 2.** Which factor in Model B records local conservation?

<details class="solution">
<summary>Solution</summary>

For Model A,

$$
\chi_A^R(q,\omega)
=
\frac{\Gamma}{
\Gamma(r+q^2)-\ii\omega
},
$$

and

$$
\chi_A''(q,\omega)
=
\frac{
\Gamma\omega
}{
\Gamma^2(r+q^2)^2+\omega^2
}.
$$

Differentiating with respect to $\omega$ gives

$$
\omega_{\rm peak}
=
\Gamma(r+q^2).
$$

At $r=0$,

$$
\chi_A^R(q,\omega)
=
q^{-2}
\frac1{
1-\ii\omega/(\Gamma q^2)
}.
$$

Thus

$$
z_A^{\rm Gaussian}=2.
$$

For Model B,

$$
\chi_B^R(q,\omega)
=
\frac{
\Gamma q^2
}{
\Gamma q^2(r+q^2)-\ii\omega
}.
$$

Equivalently,

$$
\chi_B^R(q,\omega)
=
\frac1{
r+q^2-\ii\omega/(\Gamma q^2)
}.
$$

At $r=0$,

$$
\chi_B^R(q,\omega)
=
q^{-2}
\frac1{
1-\ii\omega/(\Gamma q^4)
}.
$$

Therefore,

$$
z_B^{\rm Gaussian}=4.
$$

Setting $\omega=0$ gives

$$
\chi_A^R(q,0)
=
\chi_B^R(q,0)
=
\frac1{r+q^2}.
$$

Finally,

$$
\zeta_B(\mathbf q,t)
=
-\ii\mathbf q\cdot
\boldsymbol\xi(\mathbf q,t).
$$

Using the current-noise covariance,

$$
\left\langle
\zeta_B(\mathbf q,t)
\zeta_B(-\mathbf q,t')
\right\rangle
=
2\Gamma Tq^2\delta(t-t').
$$

The same $q^2$ that suppresses uniform noise also slows the conserved mode.

</details>

## Check your understanding

Which part of $\chi^R(q,\omega)$ follows from the static $\phi^4$ fixed point, and which part requires a dynamical universality class?

Answer by naming one static datum and one dynamical datum.

You may also reply with “deeper,” “too easy,” “too hard,” or your derivation.

## Further Reading

- [Theory of dynamic critical phenomena](https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.49.435)
- [Model-A dynamical exponent through fourth order in the epsilon expansion](https://arxiv.org/abs/0808.1347)
- [Dynamic scaling of order-parameter fluctuations in Model B](https://journals.aps.org/prd/abstract/10.1103/PhysRevD.108.074004)

## Connections and next step

- Conceptual primitive: static fixed point versus dynamical fixed point.
- Fields connected: static RG, Langevin dynamics, and linear response.
- Toy system: scalar $\phi^4$ theory with Model-A and Model-B dynamics.
- Decisive structure: $\mathcal K_A(q)=\Gamma$ versus $\mathcal K_B(q)=\Gamma q^2$.
- Assumption challenged: static universality uniquely fixes critical dissipation.
- Next step: dynamic RG and renormalization of $\Gamma$.
- Avoid repeating soon: basic analytic continuation and isolated quasiparticle decay.