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
Regarding NowIGetToLearnWhatAHeadIs's answer, I try to refine my doubts as follows:
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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 $p$).
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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 pointed out, 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.)
Best Answer
My interpretation is that you are right when you say
However, I don't think you are on the same wavelength (no pun intended) as Peskin and Schroeder (P&S) when you say
I think P&S call $\bar{\lambda}(p)\equiv\bar{\lambda}(\log(p/M);\lambda)$ (actually, they write $\bar{\lambda}(p)\equiv\bar{\lambda}(p;\lambda)$) the running coupling constant, not because of its dependence on $M$, but because of its dependence on $p$ when the coupling constant $\lambda$ at the momentum scale $p$ is held fixed. For example, when they introduce the term "running coupling constant", they say
In the above quote the dependence of $\bar{\lambda}$ on the momentum $p$ is emphasized.
I think if you reread the section with that understanding, everything makes sense. In particular, it makes sense when they say
This makes sense because if you put the scale $p$ way into the infrared by putting $p \to 0$ and you hold the coupling constant $\lambda$ at this scale $p$ fixed, then the coupling constant $\bar{\lambda}$ at the scale $M$ must get very small, because as you integrate out the wavenumbers between $M$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.
If P&S had really meant instead for $\bar{\lambda}$ to be the coupling at $p$ and $\lambda$ to be the coupling at $M$, then you what have the opposite (and therefore incorrect) situation. You would have that as $p\to 0$ with the coupling constant at $M$ fixed, that the coupling constant at $p$ goes to zero, contrary to the fact that $\lambda$ is a relevant parameter. Therefore P&S must really have meant that $\bar{\lambda}$ is the coupling at $M$.