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I have problems with understanding what the Galilean invariance of KPZ means and how it is connencted to KPZ scale invariance?

How can I see that KPZ is scale invariant?

Why this symmetry impose that the MSR action depends only on the combinantion $\partial_t h- \lambda (\nabla h)^2$? (MSR action is an action from Martin-Siggia-Rose formalism, which depends on so called deterministic and noise variables)

And what is the Ward identity?

I am quite confused after the lecture I have heard...


Ward identity comes in my notes by claiming, that:

"For the full effective action S[h,p] (for the KPZ equation) we may introduce multiplicative renormalization

$\int dt dr p(Z\partial_t h - \lambda Z' (\nabla h)^2)$,

the symmetries implies $Z=Z'$ for all scales (Ward identity)."

Agnieszka
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    You can't just throw abbreviations at us and expect us to know what you're talking about. What is the "MSR action"? Have you tried to just do a scale transform and see the equation is invariant? Why are you talking about "the Ward identity" when this is a general principle of quantum gauge theories, seemingly unrelated to the rest? – ACuriousMind Sep 16 '15 at 12:09
  • Sorry, it is only because I don't know how obvious it is for persons which know a lot about KPZ equation. – Agnieszka Sep 16 '15 at 15:52
  • What do you mean with doing the scale transformation - setting x=bx, with some const b and then see if equation stays the same? – Agnieszka Sep 16 '15 at 16:23
  • What lecture are you talking about? Who is the lecturer? Where is it given? Is it about KPZ equation of statistical physics? – Steven Mathey Sep 22 '15 at 15:57

1 Answers1

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Your question is a bit fuzzy. I will do my best to answer and let you complain about what I did not understand properly. The references that I give are just papers that I know and like. There is much more to read on the topic.

I will use the following sign convention for $\lambda$,

$$ \partial_t h + \frac{\lambda}{2}\left(\vec{\nabla}h\right)^2 = \nu \nabla^2 h + \eta \, . \qquad (1) $$

I assume that you know what every term stands for. All the parameters are positive and $\eta(t,\vec{x})$ is a stochastic noise.

Galilée invariance: KPZ equation is invariant under Galilei transformations. If $h(t,\vec{x})$ is a solution of Eq. (1), then

$$\tilde{h}(t,\vec{x}) = h(t,\vec{x}-t \vec{u}) + \frac{1}{\lambda} \vec{u} \cdot \tilde{x} - \frac{u^2 t}{2 \lambda} \, , \qquad (2)$$

is a solution as well (with $\tilde{\eta}(t,\vec{x}) = \eta(t,\vec{x}-\vec{u}t)$). Note that the last term is quadratic in $\vec{u}$ and is usually not considered since only infinitesimally small values of $\vec{u}$ are needed for the Ward identities. Note that both sides of Eq. (1) are invariant under such a transformation independently. This implies that any function of (in particular) the left-hand-side will be a Galilei scalar.

This transformation is called Galilei invariance because Eq. (1) can be mapped onto Burgers equation,

$$ \partial_t \vec{v} + \left(\vec{v} \cdot \vec{\nabla}\right) \vec{v} = \nu \Delta \vec{v} + \vec{\nabla}\eta \, , \qquad (3)$$

with the definition, $\vec{v} = \lambda \vec{\nabla}h$. Burgers' equation is very similar to Navier-Stokes and describes some kind of hydrodynamic velocity field. The transformation (2) applied to $\vec{v}$ becomes

$$ \tilde{\vec{v}}(t,\vec{x}) = \vec{v}(t,\vec{x}-t\vec{u}) + \vec{u} \, .$$

It is a Galilei transformation.

Scale invariance: There is no easy was to see that Eq. (1) is scale invariant. It takes a lot of work to see it. This can be done using Renormalisation Group (RG) techniques (see e.g. this and more references in it's bibliography) numerical simulations or even experiments. Note that the $d=1$ case has been solved analytically and the scale invariance of Eq. (1) proven mathematically. See e.g. this. Often it is simply just assumed that the stationary state is scale invariant.

In particular, scale invariance of the steady state implies that there are two scaling exponents ($\chi$ and $z$) and a scaling function $g(x)$ such that

$$ \langle \left[h(t+\tau,\vec{x}+\vec{r})-h(t,\vec{x}) \right]^2 \rangle = r^{2\chi} \, g\left(\frac{\tau}{r^z}\right) \, .$$

You can write similar expressions for all the correlation functions. Note that the steady state is invariant under space time translations as well as space rotations, $\left|\vec{r}\right| = r$.

For $d=1$, $\chi$ and $z$ are known. Huge amounts of work go into computing them for higher dimensions. Galilei invariance is a big help here because it relates one exponent to the other,

$$ \chi + z = 2 \, . \qquad (4)$$

Indeed, Galilei invariance insures (in particular) that $\lambda$ is not changed by the renormalisation group. Whatever the level of coarse graining, the two terms of the left-hand-side of Eq. (1) will always be related in the same way. If your system is scale invariant you can rescale space, time and field as

$$ x = \kappa \hat{x} \, , \quad t = \kappa^z \hat{t} \, , \quad h = \kappa^\chi \hat{h} \, .$$

Then the left-hand-side of Eq. (1) is rescaled as

$$ \kappa^{\chi-z} \partial_{\hat{t}} \hat{h} + \kappa^{2\chi-2} \left(\hat{\nabla}\hat{h}\right)^2 = \kappa^{\chi-z} \left[ \partial_{\hat{t}} \hat{h} + \kappa^{\chi+z-2} \left(\hat{\nabla}\hat{h}\right)^2 \right] \, .$$

The whole thing does not change (up to a prefactor that is absorbed into the renormalisation of $\nu$ and $\eta$) only if Eq. (4) is fulfilled.

Advective non-linearity: The left-hand-side of Eq. (1) should be considered as a block because in the hydrodynamic representation (Eq. (3)), it is the advective time derivative,

$$\frac{d \vec{v}(t,\vec{x}(t))}{d t} = \partial_t \vec{v} + \left(\vec{v}\cdot \vec{\nabla}\right) \vec{v} \, .$$

This shows as well why it is Galilei invariance that is responsible for the conservation (under RG transformations) of $\lambda$.

Ward identities: You can find the Ward identities as well as their derivation here.

  • Ok, I will start to analize the papers, thanks! I have one question: by the renormalization theory one often assumes that the scaling functions converge to some constant when the variable goes to infinity, g(x->\infty)->const. Why should this make sense? – Agnieszka Sep 23 '15 at 13:26
  • There are two limits, $g(x\gg 1)$ and $g(x\ll 1)$. One is set to a constant simply as a definition of $\chi$, the other must be a power law. When $\tau= 0$, the correlation function should be a function of $r$ only, hence $g(x\ll 1) = cst$. On the other hand if $r\rightarrow 0$, only $\tau$ should remain, $g(x\gg 1)\sim x^{2\chi/z}$. Be careful about what correlation function you talk about. In the wikipedia page we have the opposite, when the system size is sent to infinity, $W(\infty,t)$ should only depend on $t$. – Steven Mathey Sep 23 '15 at 13:48