# Polchinski's toy model of renormalization group flow: significance of main steps

In "Renormalization and Effective Lagrangians" (Nucl. Phys. B, 231 p269, 1984; preprint), Polchinski begins section 2 with a toy model to demonstrate the renormalization group with a relevant and irrelevant coupling.

He starts with a scalar theory with a 4-point and 6-point interaction. The dimensionless couplings for these interactions are $$\lambda_4$$ and $$\lambda_6 = \Lambda^2 g_6$$ where $$\Lambda$$ is the scale of the theory.

The $$\beta$$-function equations are:

If $$\bar\lambda_4, \bar\lambda_6)$$ are a solution to these equations, then we can examine small deviations from this trajectory: $$\varepsilon_i \equiv \lambda_i - \bar\lambda_i$$.

The equations to $$\mathcal O(\varepsilon)$$ are:

Now here's where I'm a bit confused: Polchinski explains the $$\lambda_6$$ will term affects the relative flow of the $$\lambda_4$$ between nearby trajectories. For example, the following figure: two nearby points A1 and A2 have the same $$\lambda_4$$ but flow to points B1 and B2 which now have quite different $$\lambda_4$$s.

Polchinski then says that even though $$\varepsilon_4$$ is big, there is a point B2' on trajectory 2 that is close to A2. In order to illuminate this, he defines this particular combination of couplings:

Question 1: I don't quite see the motivation for writing this. However, Polchinski says that $$(\xi_4,\xi_6)$$ is a vector that points from B1 to B2', which I think I can see if I think of $$\frac{d\bar\lambda_6/d\Lambda}{d\bar\lambda_4/d\Lambda} = \frac{d\bar\lambda_6}{d\bar\lambda_4} .$$ Is that reasonable? Then I'm thinking of $$\bar\lambda_6$$ as a function of $$bar\lambda_4$$: $$\bar\lambda_6 = \bar\lambda_6(\bar\lambda_4)$$.

In this way, to quote Polchinski above eq (8), we are "subtracting off a multiple of the tangent vector to the trajectory."

Question 2: I'm having a hard time showing that the flow equation for $$\xi_6$$ is:

Is it obvious how this follows from the earlier flow equations?

Thanks!

• Permalink: doi.org/10.1016/0550-3213(84)90287-6 – Qmechanic Jan 27 at 0:07
• I am trying to understand how to obtain the equations 3a and 3b from the equations before those. Do you have any idea of how that was done? I know it is just given to us in the paper but I am trying to get there from the deviation given in the paper. – user7077252 Feb 22 at 18:16
• @user7077252: to derive equation (3a), simply replace $\epsilon_i = \lambda_i - \bar\lambda_i$. Then use equation (2a) in the limit where $\epsilon$ is small. The finite difference between $\beta(\lambda)$ and $\beta(\bar\lambda)$ is approximately $\partial\beta/\partial\lambda$. – Henry Deith Feb 23 at 18:22
• Where $\epsilon$ is small, meaning that $\epsilon \to 0$. Thank you so much for your help. – user7077252 Feb 23 at 19:01

Ah, I think I've sorted it out. I share the key steps here.

For simplicity, let me write $$t = \ln \Lambda$$ so that $$d/dt = \Lambda (d/d\Lambda)$$. I further use the shorthand where $$\dot\lambda = d\lambda/dt$$.

For question 1, I believe an explanation is to annotate Polchinski's figure as follows:

Here the green line is the tangent line of the lower trajectory. The expression for the separation between the trajectories, $$\xi_6$$ is approximated by looking at the vertical separation between B2 and A2, $$\varepsilon_6 =\lambda_6(t+\delta t)-\bar\lambda_6(t+\delta t)$$, minus $$\Delta \varepsilon_6$$, which is the "rise" from the tangent line approximation.

For question 2, one can arrive at this result from simply calculating $$\dot\xi_6$$. It is useful to note that $$\beta_i$$ depends on $$t$$ through $$\lambda_4(t)$$ and $$\lambda_6(t)$$. This means, for example, $$\dot \beta_4 = + \frac{\partial \beta_4}{\partial_4}\dot\lambda_4 + \frac{\partial \beta_4}{\partial_6}\dot\lambda_6$$

Thus the following corollaries may be useful:

(1) $$\frac{\ddot \lambda_4}{\dot\lambda_4} = \frac{d\ln \beta_4}{dt}$$

(2) $$\ddot \lambda_6 = 2\dot\lambda_6 + \frac{\partial \beta_6}{\partial_4}\dot\lambda_4 + \frac{\partial \beta_6}{\partial_6}\dot\lambda_6$$

(3) $$\frac{\partial \beta_4}{\partial_6} = \frac{1}{\dot\lambda_6} \left( \dot\beta_4 - \frac{\partial \beta_4}{\partial_\lambda 4} \dot\lambda_4 \right)$$

Using these relations, I think it is simply a matter of applying the chain rule and grouping terms together.