One static fixed point, two critical clocks

Key claim

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

Two systems can share the same equilibrium ${\varphi }^4$ 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

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 $d=4-ϵ$ spatial dimensions:

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

The equilibrium measure is

where $k_B=1$.

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

Start with nonconserved relaxational dynamics, Model A:

The white noise satisfies

The drift and noise amplitude obey detailed balance and make $P_{\mathrm{e}\mathrm{q}}$ stationary.

At criticality, dynamic scaling takes the form

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,

The retarded susceptibility is

Equivalently,

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

The pole lies at

so

At the Gaussian critical point $r=0$,

The equilibrium correlation spectrum is

It obeys the classical fluctuation–dissipation relation

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

Scope and assumptions: conservation changes the clock

The functional $\mathcal{F}[\varphi ]$ does not select Model A.

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

Choose the dissipative current

Its noise obeys

The order parameter then follows Model-B dynamics:

The conserved noise vanishes at zero momentum:

In the Gaussian theory,

Its static limit matches Model A:

Its pole does not:

At $r=0$,

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

At the interacting fixed point,

while equilibrium Model B obeys

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

The static inverse susceptibility

is the thermodynamic restoring force.

The kinetic kernel is

for Model A and

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	$\mathrm{\Gamma }$	$\mathrm{\Gamma }{q}^2$
Gaussian rate at $r=0$	$\mathrm{\Gamma }{q}^2$	$\mathrm{\Gamma }{q}^4$
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, $\mathrm{\Gamma }$ 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

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

   $${\chi }_A^R(q,\omega )=q^{-2}{\mathrm{\Phi }}_A\left(\frac{\omega }{\mathrm{\Gamma }{q}^2}\right)$$Copy LaTeX

   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 ${\zeta }_B=-\mathrm{\nabla }\cdot \mathit{\xi }$.

Hint 1

For

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

solve $\mathrm{d}f/\mathrm{d}\omega =0$.

Hint 2

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

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?

Solution

For Model A,

$${\chi }_A^R(q,\omega )=\frac{\mathrm{\Gamma }}{\mathrm{\Gamma }(r+q^2)-\mathrm{i}\omega },$$Copy LaTeX

and

$${\chi }_A^{″}(q,\omega )=\frac{\mathrm{\Gamma }\omega }{{\mathrm{\Gamma }}^2(r+q^2{)}^2+{\omega }^2}.$$Copy LaTeX

Differentiating with respect to $\omega$ gives

$${\omega }_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}=\mathrm{\Gamma }(r+q^2).$$Copy LaTeX

At $r=0$,

$${\chi }_A^R(q,\omega )=q^{-2}\frac{1}{1-\mathrm{i}\omega /(\mathrm{\Gamma }{q}^2)}.$$Copy LaTeX

Thus

$$z_A^{\mathrm{G}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{a}\mathrm{n}}=2.$$Copy LaTeX

For Model B,

$${\chi }_B^R(q,\omega )=\frac{\mathrm{\Gamma }{q}^2}{\mathrm{\Gamma }{q}^2(r+q^2)-\mathrm{i}\omega }.$$Copy LaTeX

Equivalently,

$${\chi }_B^R(q,\omega )=\frac{1}{r+q^2-\mathrm{i}\omega /(\mathrm{\Gamma }{q}^2)}.$$Copy LaTeX

At $r=0$,

$${\chi }_B^R(q,\omega )=q^{-2}\frac{1}{1-\mathrm{i}\omega /(\mathrm{\Gamma }{q}^4)}.$$Copy LaTeX

Therefore,

$$z_B^{\mathrm{G}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{a}\mathrm{n}}=4.$$Copy LaTeX

Setting $\omega =0$ gives

$${\chi }_A^R(q,0)={\chi }_B^R(q,0)=\frac{1}{r+q^2}.$$Copy LaTeX

Finally,

$${\zeta }_B(\mathbf{q},t)=-\mathrm{i}\mathbf{q}\cdot \mathit{\xi }(\mathbf{q},t).$$Copy LaTeX

Using the current-noise covariance,

$$⟨{\zeta }_B(\mathbf{q},t){\zeta }_B(-\mathbf{q},t^{\prime })⟩=2\mathrm{\Gamma }T{q}^2\delta (t-t^{\prime }).$$Copy LaTeX

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

Check your understanding

Which part of ${\chi }^R(q,\omega )$ follows from the static ${\varphi }^4$ 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
• Model-A dynamical exponent through fourth order in the epsilon expansion
• Dynamic scaling of order-parameter fluctuations in Model B

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 ${\varphi }^4$ theory with Model-A and Model-B dynamics.
• Decisive structure: ${\mathcal{K}}_A(q)=\mathrm{\Gamma }$ versus ${\mathcal{K}}_B(q)=\mathrm{\Gamma }{q}^2$.
• Assumption challenged: static universality uniquely fixes critical dissipation.
• Next step: dynamic RG and renormalization of $\mathrm{\Gamma }$.
• Avoid repeating soon: basic analytic continuation and isolated quasiparticle decay.