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I have a couple of questions regarding Peskin and Schroeder's derivation of the solution to the Callan-Symanzik equation. First of all, they claim that using $$\int_\lambda^\bar{\lambda}\frac{d\lambda'}{\beta(\lambda')} = \int_{p'=M}^{p'=p}d\log(p'/\:M),\qquad(1)$$ one arrives at $$\left(p\frac{\partial}{\partial p}-\beta(\lambda)\frac{\partial}{\partial\lambda}\right)\bar{\lambda}=0,\qquad(2)$$ which I unfortunately do not see. On another note, they then claim that using this relation, one can prove that the function $$G^{(2)}(p,\lambda) = \hat{\mathcal{G}}(\bar{\lambda}(p;\lambda)){\cdot}\exp\left(-\int_{p'=M}^{p'=p}d\log(p'/\:M)\cdot2[1-\gamma(\bar{\lambda}(p';\lambda))]\right),\qquad(3)$$ where $\bar{\lambda}(p';\lambda)$ $${d\over d\log(p/M)}\bar{\lambda}(p;\lambda)=\beta(\bar{\lambda}),\qquad\bar{\lambda}(M;\lambda) = \lambda\qquad(4)$$ solves the Callan-Symanzik Equation$$\left[\frac{\partial}{\partial p}-\beta(\lambda)\frac{\partial}{\partial\lambda}+2(1-\gamma(\lambda))\right]G^{(2)}(p)=0.\qquad(5)$$ I cannot prove this either, and I am sure it is because I am taking derivatives improperly, but any help would be much appreciated.

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  • $\begingroup$ I can hand-wavingly prove that (2) is true with a PLUS sign between the two terms. Maybe it is a typo in the book. I could check this if I knew how to prove that (3) was a solution of (5) using (2), but alas, I cannot. $\endgroup$ – Nick Murphy Oct 12 '15 at 13:33
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The eqs. do check out, although you are right, it is very easy to botch the derivatives.

To derive Eq.(2) from Eq.(1) take into account that $\bar{\lambda}(M, \lambda) =\lambda$ and rewrite the latter so as to clarify the integral limits and the right hand side integrand: $$ \int_{\bar{\lambda}(M, \lambda)}^{\bar{\lambda}(p, \lambda)}{\frac{d\lambda'}{\beta(\lambda')}} = \int_{M}^{p}{\frac{dp'}{p'}} \tag{1a} $$ Now take the derivative w.r.t. $p$ on both sides:
$$ \frac{\partial \bar{\lambda}}{\partial p}(p, \lambda) \frac{1}{\beta(\bar{\lambda}(p, \lambda))} = \frac{1}{p} $$ and rearrange as $$ p \frac{\partial \bar{\lambda}}{\partial p}(p, \lambda) = \beta(\bar{\lambda}(p, \lambda)) \tag{2a} $$ The above looks deceptively close to the desired identity, but it is not really, since $\beta(\bar{\lambda}(p, \lambda)) \neq \beta(\lambda)$. To bring in $\beta(\lambda)$, apply the same procedure to the integral identity w.r.t. $\lambda$ and obtain: $$ \frac{\partial \bar{\lambda}}{\partial \lambda}(p, \lambda) \frac{1}{\beta(\bar{\lambda}(p, \lambda))} - \frac{\partial \bar{\lambda}}{\partial \lambda}(M, \lambda) \frac{1}{\beta(\bar{\lambda}(M, \lambda))} = 0 $$ But since $\bar{\lambda}(M, \lambda) =\lambda$, we have $(\partial \bar{\lambda}/\partial \lambda)(M, \lambda) = 1$ and $\beta(\bar{\lambda}(M, \lambda)) = \beta(\lambda)$, and we can rearrange again into $$ \beta(\bar{\lambda}(p, \lambda)) = \beta(\lambda) \frac{\partial \bar{\lambda}}{\partial \lambda}(p, \lambda) \tag{2b} $$ Obviously, substituting (2b) into (2a) produces Eq.(2): $$ p \frac{\partial \bar{\lambda}}{\partial p}(p, \lambda) - \beta(\lambda) \frac{\partial \bar{\lambda}}{\partial \lambda}(p, \lambda) = 0 $$

To check that $G^{(2)}(p,\lambda)$ in Eq.(3) satisfies the Callan-Symanzik Eq.(5):

Use identity (1a) in Eq.(3) to switch the integration over $p$ to an integration over $\lambda$: $$ G^{(2)}(p,\lambda) = \hat{\mathcal{G}}(\bar{\lambda}(p,\lambda))\exp\left(-\int_{\bar{\lambda}(M, \lambda)}^{\bar{\lambda}(p, \lambda)}{\frac{d\lambda'}{\beta(\lambda')}2[1-\gamma(\lambda')]}\right) $$ Now we can take the derivatives w.r.t. $p$ and $\lambda$. For the derivative w.r.t. $p$ we obtain $$ \frac{\partial G^{(2)}}{\partial p}(p,\lambda) = \frac{d \hat{\mathcal{G}}}{d \bar{\lambda}}(\bar{\lambda}(p,\lambda)) \frac{\partial \bar{\lambda}}{\partial p}(p, \lambda) \exp\left(-\int_{\bar{\lambda}(M, \lambda)}^{\bar{\lambda}(p, \lambda)}{\frac{d\lambda'}{\beta(\lambda')}2[1-\gamma(\lambda')]}\right) - \\ -\; \frac{\partial \bar{\lambda}}{\partial p}(p, \lambda) \frac{2[1-\gamma(\bar{\lambda}(p,\lambda))]}{\beta(\bar{\lambda}(p,\lambda))} G^{(2)}(p,\lambda) $$ But from (2a) we have $(\partial \bar{\lambda}/\partial p)(p, \lambda) = \beta(\bar{\lambda}(p,\lambda))/p$, which substituted above yields, after slight rearragement, $$ p \frac{\partial G^{(2)}}{\partial p}(p,\lambda) + 2[1-\gamma(\bar{\lambda}(p,\lambda))] G^{(2)}(p,\lambda) = \\ = \beta(\bar{\lambda}(p,\lambda))\frac{d \hat{\mathcal{G}}}{d \bar{\lambda}}(\bar{\lambda}(p,\lambda)) \exp\left(-\int_{\bar{\lambda}(M, \lambda)}^{\bar{\lambda}(p, \lambda)}{\frac{d\lambda'}{\beta(\lambda')}2[1-\gamma(\lambda')]}\right) \tag{5a} $$ Again, it looks deceptively like we already have 2 terms of the Callan-Symanzik equation, but in fact $\gamma(\bar{\lambda}(p,\lambda)) \neq \gamma(\lambda)$. Let's take the derivative w.r.t. $\lambda$: $$ \frac{\partial G^{(2)}}{\partial \lambda}(p,\lambda) = \frac{d \hat{\mathcal{G}}}{d \bar{\lambda}}(\bar{\lambda}(p,\lambda)) \frac{\partial \bar{\lambda}}{\partial \lambda}(p, \lambda) \exp\left(-\int_{\bar{\lambda}(M, \lambda)}^{\bar{\lambda}(p, \lambda)}{\frac{d\lambda'}{\beta(\lambda')}2[1-\gamma(\lambda')]}\right) - \\ -\; \frac{\partial \bar{\lambda}}{\partial \lambda}(p, \lambda) \frac{2[1-\gamma(\bar{\lambda}(p,\lambda))]}{\beta(\bar{\lambda}(p,\lambda))} G^{(2)}(p,\lambda) + \frac{\partial \bar{\lambda}}{\partial \lambda}(M, \lambda) \frac{2[1-\gamma(\bar{\lambda}(M,\lambda))]}{\beta(\bar{\lambda}(M,\lambda))} G^{(2)}(p,\lambda) $$ Using again the fact that $\bar{\lambda}(M, \lambda) =\lambda$, $(\partial \bar{\lambda}/\partial \lambda)(M, \lambda) = 1$, and $\beta(\bar{\lambda}(M, \lambda)) = \beta(\lambda)$, as well as (2b) in the form $(\partial \bar{\lambda}/\partial p)(p, \lambda) = \beta(\bar{\lambda}(p,\lambda))/\beta(\lambda)$, we have $$ \beta(\bar{\lambda}(p,\lambda))\frac{d \hat{\mathcal{G}}}{d \bar{\lambda}}(\bar{\lambda}(p,\lambda)) \exp\left(-\int_{\bar{\lambda}(M, \lambda)}^{\bar{\lambda}(p, \lambda)}{\frac{d\lambda'}{\beta(\lambda')}2[1-\gamma(\lambda')]}\right) = \\ = \beta(\lambda) \frac{\partial G^{(2)}}{\partial \lambda}(p,\lambda) + 2[1-\gamma(\bar{\lambda}(p,\lambda))] G^{(2)}(p,\lambda) - 2[1-\gamma(\lambda)]G^{(2)}(p,\lambda) \tag{5b} $$ Substituting (5b) into the r.h.s. of (5a) gives $$ p \frac{\partial G^{(2)}}{\partial p}(p,\lambda) + 2[1-\gamma(\bar{\lambda}(p,\lambda))] G^{(2)}(p,\lambda) = \\ = \beta(\lambda) \frac{\partial G^{(2)}}{\partial \lambda}(p,\lambda) + 2[1-\gamma(\bar{\lambda}(p,\lambda))] G^{(2)}(p,\lambda) - 2[1-\gamma(\lambda)]G^{(2)}(p,\lambda) $$ and finally, the Callan-Symanzik Eq.(5), $$ p \frac{\partial G^{(2)}}{\partial p}(p,\lambda)\; - \;\beta(\lambda) \frac{\partial G^{(2)}}{\partial \lambda}(p,\lambda)\;+\;2[1-\gamma(\lambda)]G^{(2)}(p,\lambda) = 0 $$

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    $\begingroup$ Ahh thank you so much! I now see the mistake. I was only differentiating the integral w.r.t lambda_bar(p,lambda) when taking the derivative of the integral w.r.t to lambda, and not including the derivative with respect to lambda(M,lambda). Thanks again for taking the time, I was still stuck on this! $\endgroup$ – Nick Murphy Oct 17 '15 at 18:12
  • $\begingroup$ Welcome, and good luck! $\endgroup$ – udrv Oct 17 '15 at 19:28
  • $\begingroup$ @udrv Would you please take a look at this question, any comment will be really appreciated. :) physics.stackexchange.com/questions/358353/… $\endgroup$ – gamebm Oct 3 '17 at 13:51
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I have a complementary derivation to udrv's answer, it is based on the hydro-bacteriological analogy discussed in the textbook by Peskin and Schroeder. To me, it helps to understand the physical content of Eq.(12.75) from an a somewhat different aspect. The relation between the hydro-bacteriological environment and running coupling constant is as follows $$\log(p/M) \leftrightarrow t,$$ $$\lambda \leftrightarrow x,$$ $$-\beta(\lambda) \leftrightarrow v(x),$$ and $$\bar{\lambda}(\log(p/M);\lambda) \leftrightarrow \bar{x}(t;x).$$

In the textbook, Eq.(12.70) (to be more explicit, I changed $d$ to $\partial$) reads $$\frac{\partial}{\partial t'}\bar{x}(t';x)=-v(\bar{x}).\;\;\;\qquad{(1)}$$ It 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.

Though not written out explicitly, one also has the following relation concerning the spatial derivative $$\frac{\partial}{\partial x}\bar{x}(t';x)=\frac{v(\bar{x})}{v(x)}.\;\;\;\qquad{(2)}$$ The above equation states if an element arrives at $x+\Delta x$ (instead of $x$) at a given instant $t$, its initial position $\bar x$ also should be shifted to the right, namely, it becomes $\bar x+\Delta\bar x$. Intuitively, the ratio of them is given by $\frac{\Delta\bar x}{\Delta x}=\frac{v(\bar x)}{v(x)}$.

The above two equations give immediately $$\frac{\partial}{\partial t'}\bar{x}(t';x)+v(x)\frac{\partial}{\partial x}\bar{x}(t';x)=0.\;\;\;\qquad{(3)}$$

Using the matching relation, it is readily to find that the integral version of Eq.(1) corresponding to Eq.(12.74) and Eq.(3) gives, Eq.(12.75), the desired result.

We note Eq.(2) corresponds to Eq.(2b) in udrv's derivation. In this context, the above derivation shows that it is not necessary to integrate Eq.(12.70) first and then carry out partial derivatives.

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