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: enter image description here

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: enter image description here

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.

enter image description here

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:

enter image description here

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:

enter image description here

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


  • $\begingroup$ Permalink: doi.org/10.1016/0550-3213(84)90287-6 $\endgroup$ – Qmechanic Jan 27 at 0:07
  • $\begingroup$ 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. $\endgroup$ – user7077252 Feb 22 at 18:16
  • 1
    $\begingroup$ @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 $\partial\beta/\partial\lambda$. $\endgroup$ – Henry Deith Feb 23 at 18:22
  • $\begingroup$ Where $\epsilon$ is small, meaning that $\epsilon \to 0 $. Thank you so much for your help. $\endgroup$ – 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: enter image description here

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.

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