6
$\begingroup$

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?

Thanks!

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.