I am studying the textbook *An introduction to Quantum Field Theory* by Peskin and Schroeder, and am not able to understand a seemly important and straightforward result on running coupling constant. It is probably a stupid one but I just cannot figure it out.

After some calculations, one obtains the expression for the running coupling in $\phi^4$ theory, namely, Eq.(12.82) on P.422 of *Chapter* 12 which reads

$$\bar{\lambda}(p)=\frac{\lambda}{1-(3\lambda/16\pi^2)\log(p/M)}.$$

Owing to Eq.(12.69) which gives the following relation between the Coleman's hydrodynamic-bacteriological equation and the Callan-Symanzik equation for $\phi^4$ theory

$$\log(p/M) \leftrightarrow t,$$

$$\lambda \leftrightarrow x,$$

$$-\beta(\lambda) \leftrightarrow v(x),$$

we have

$$\bar{\lambda}(\log(p/M);\lambda) \leftrightarrow \bar{x}(t;x).$$

Now, according to the interpretation above Eq.(12.70), $\bar{x}$ is the initial position (at $t=0$) of the element which is at $x$ at time $t$. To be more specific, one emphasizes that $\bar{x}(t;x)$ is not the position at time $t$. Similarly, I understand that $\bar{\lambda}(p)\equiv\bar{\lambda}(\log(p/M);\lambda)$ is *not* the coupling constant at $p$. $\bar{\lambda}$ is the value of the coupling constant at renormalization energy scale $p=M$, which evolves to $\lambda$ at scale $p$.

If the above understanding is correct, Eq.(12.82) cannot be interpreted as written in the textbook

*The running coupling constant goes to zero at a logarithmic rate as $p\rightarrow 0$.*

This is because, again, $\bar{\lambda}$ itself corresponds to the coupling fixed at $M$, it does not run.

On the contrary, the inverse function $\lambda(\log(p/M);\bar{\lambda})$ shall describe the running coupling, as analogically $x(t;\bar{x})$ will do. In fact, similarly to Eq.(12.70) (to be more explicit, I changed $d$ to $\partial$)

$$\frac{\partial}{\partial t'}\bar{x}(t';x)=-v(\bar{x}),$$

which states that if an element arrives at a given $x$ but $\Delta t$ later, its departure position at initial time $t=0$ should be shifted to the left (indicated by the negative sign) by an amount $v(\bar{x})\Delta t$, where $v(\bar{x})$ is the fluid velocity at the initial position. On the other hand, one has

$$\frac{\partial}{\partial t'}x(t';\bar{x})=v(x).$$

The above equation states that for an element starts at $\bar{x}$ at $t=0$, its position is shifted to the right (corresponding to the $+$ sign before the velocity) by an amount $v(x)\Delta t$ , if we let it evolve for $\Delta t$ longer. Therefore, it seems the solution of $\lambda(\log(p/M);\bar{\lambda})$ is simply (12.82) by exchanging $\lambda$ with $\bar{\lambda}$ and adding an extra $(-1)$ before $\log(p/M)$. But then the conclusion will be completely the opposite?! What am I missing? Many thanks for the explanation.

**Edit**

After some thinking, I try to refine my doubts as follows:

1. Why $p\rightarrow 0$ is interpreted as *going to small energy scale*? Imagine that one would like to study the running coupling of $\phi^4$ theory at a (small) scale $p$? Intuitively, I understand that this can be achieved experimentally by carrying out a collision between few (two to two, to be exact) $\phi$s with energy roughly at $E\sim p$, and then measuring the cross section of the process. In this context, the coupling we talk about ($\bar{\lambda}$) should be evaluated at the corresponding momentum $M$ (not $p$).

2. If we want to study the coupling as a function of energy scale for *the same theory*, the renormalization condition should be something given in the first place, and it should not be modified afterwards. For instance, one can study $\lambda$ as a function of $p$ for given $\bar{\lambda}$ and $M$. I understand, as a matter of fact, that $\bar{\lambda}$ indeed can be viewed as a function of $p$. However, if one evaluates $\bar{\lambda}$ while varies $p$ for given $\lambda$ and $M$, it seems to me that we are no longer talking about the *same* theory. (c.f. in fact, fix $\lambda$ and $p$ and study $\bar\lambda$ as a function of $M$ is also ok, but not mix variables from different pairs.)

**P.S.** If you do not have the time to write an answer, please consider leaving a brief comment below to drop a hint.