According to this wikipedia article in the $\phi^4$ section, the equation

$$\frac{1}{c^2}\frac{∂^2}{∂t^2}\phi(x,t)-\sum_i\frac{∂^2}{∂x_i^2}\phi(x,t)+g\ \phi(x,t)^3=0,$$

in 4 dimensions is invariant under

$$x\rightarrow \lambda\ x,$$ $$t\rightarrow \lambda\ t,$$ $$\phi\rightarrow \lambda^{-1}\ \phi.$$

I have a problem seeing this, which might be because I don't really know what to do with the derivatives. If I just replace $\phi(x,t)$ by $\lambda^{-1}\ \phi(\lambda x,\lambda t)$, then I get

$$\frac{1}{c^2}\lambda^2\lambda^{-1}\frac{∂^2}{∂(\lambda\ t)^2}\phi(\lambda x,\lambda t)-\lambda^2\lambda^{-1}\sum_i\frac{∂^2}{∂(\lambda\ x_i)^2}\phi(\lambda x,\lambda t)+\lambda^{-3} g\ \phi(\lambda x,\lambda t)^3=0,$$

which doesn't work out.

If I compare with the last section then this seems to be what one should do. Because In that case

$$x\rightarrow \lambda\ x,$$ $$t\rightarrow \lambda\ t,$$ $$\rho\rightarrow \lambda\ \rho,$$ $$u\rightarrow u,$$ and therefore the naviar stokes equation with $$...+\rho\ u\ \nabla u+\mu\ \nabla^2 u+...$$ is such, that its invariant if the derivative produces an additional $\lambda$, the way I did above.

Sidenote: is scale invariance $\phi(x)=\phi^{-\Delta}(\lambda x)$ of fields something that should more acurately be called self similarity?

Also, I don't understand the last sentence in the $\phi^4$ paragraph, which says that the dimensionlessness of the coupling $g$ is "the main point". Why is it a problem to also scale the coupling? However, I never understood why dimensionless couplings are prefered anyway. If the coupling is somehow unique, then why does it make sense to search for a process dependend coupling in renormalization theory?


1 Answer 1


The reason you're not finding scale invariance in your equation is that you've inserted a factor of $\lambda^2$ in the first two terms, which as far as I can see has no reason to be there.

I'm guessing you put it in because of your uncertainty as to how to dilate derivatives. Transforming a derivative is nothing complicated, though. You simply make the transformation on the variable that is being differentiated with respect to, treat it as a differential in the denominator, and simplify. For example:

$$\frac{\partial}{\partial x_i} \to \frac{\partial}{\partial(\lambda x_i)} = \frac{\partial}{\lambda\partial x_i} = \lambda^{-1}\frac{\partial}{\partial x_i}$$

because $\mathrm{d}(\lambda x_i) = \lambda\mathrm{d}x_i$.

When you do this, in each of the first two terms you get $\lambda^{-1}$ from the transformation of the field and $\lambda^{-2}$ from the transformation of the derivative, for an overall scaling factor of $\lambda^{-3}$. And in the last term, you get $\lambda^{-3}$ from the fields alone. All those factors of $\lambda^{-3}$ then cancel out.

The reason dimensionless couplings are preferred (in certain applications) is that they don't set any scale for the theory. If your coupling $g$ had a dimension of $L^{-1}$, for example, then you could establish a characteristic length scale for that theory as $g^{-1}$. But being scale invariant means that there should be no preferred length scale; you can change any scale $x \to \lambda x$ without altering the theory. If you have a dimensionful coupling, that is obviously not the case, because you can distinguish between $x$ and $\lambda x$ by measuring them in units of $g^{-1}$. You can get around that by having $g$ change under a scale transformation as well, but then it's no longer a coupling constant, which makes things more complicated.

  • $\begingroup$ I thought of the product rule when substituting $\frac{∂}{∂x}$ with $\lambda\frac{∂}{∂x}$. Of course I thought about doing it like you did, but as I pointed out in the question, I don't see how that procedure fits with the other scaling example. If I consider these two terms, then it seems like $\nabla$ should scale like $\rho$. Also what exactly does "which makes things more complicated" mean? Does it mean that one practically does this? And why is $g^{-1}$ the length scale, doesn't that depend on many things? $\endgroup$
    – Nikolaj-K
    Commented Jan 29, 2012 at 11:21
  • $\begingroup$ No need for the product rule here. I think the dynamic viscosity has to scale as $\mu\to\lambda^{2}\mu$, otherwise something is wrong with the argument about scale invariance of the Navier-Stokes equations. Changing the coupling constant under a scale transformation gets into aspects of renormalization which I don't understand as well as I should. And what are these many things you're saying the length scale depends on? If $g$ has units of $L^{-1}$, then $g^{-1}$ is a length, simple as that. $\endgroup$
    – David Z
    Commented Jan 29, 2012 at 21:13
  • 2
    $\begingroup$ @Nick Kidman and David Zaslavsky: There was an error on wikipedia in the scaling law for $\rho$, which is now corrected. $\endgroup$
    – Qmechanic
    Commented Jan 29, 2012 at 23:20
  • $\begingroup$ Ah, that makes more sense now. $\endgroup$
    – David Z
    Commented Jan 29, 2012 at 23:53

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