A Calculation in Perturbative QCD by Muller

[MathematicalPhysicist]

TL;DR Summary

Derivation of of Eq(31) on page 16 from Eq(23) which is on page 16.

First let's define as on page 13 of the book the perturbative coefficients of the hard scattering cross section $H_{ab}$ by:
$$(22) \ \ \ \ H_{ab}=H_{ab}^{0}+\frac{\alpha_s}{\pi}H_{ab}^{(1)}+\mathcal{O}(\alpha_s^2)$$

Now, on page 16 it's written that the RG equation for $H_{ab}$ is:
$$(31) \ \ \ \ \mu \frac{d}{d\mu} H_{ab}(x_A,x_B,Q;\frac{\mu}{Q},\alpha_s(\mu))=$$
$$=-\sum_c \int_{x_A}^1d \zeta_A P_{c/a}(\zeta_A,\alpha_s(\mu))H_{cb}(\frac{x_A}{\zeta_A},x_B,Q;\frac{\mu}{Q},\alpha_s(\mu))$$
$$-\sum_d\int_{x_B}^1 d\zeta_BP_{d/b}(\zeta_B,\alpha_s(\mu))H_{ad}(x_A,\frac{x_B}{\zeta_B},Q;\mu/Q,\alpha_s(\mu)).$$

Here $P_{c/a}(\xi,\alpha_s(\mu))$ is the all orders Altarelli-Parisi kernel. It has a perturbative expansion:
$$(32) \ \ \ \ P_{c/a}(\xi,\alpha_s(\mu))=\frac{\alpha_s(\mu)}{\pi}P_{c/a}^{(1)}(\xi)+\ldots$$
where $P_{c/a}^{(1)}(\xi)$ is the function that appears in equation (23).
$$(23) \ \ \ \ f_{a/b}(x;\epsilon)=\delta_{ab}\delta(1-x)-\frac{1}{2\epsilon}\frac{\alpha_s}{\pi}P_{a/b}^{(1)}(x)+\mathcal{O}(\alpha_s^2)$$

Thus at lowest order the renormalization group equation (31) is a simple consequence of differentiating eq. (23).
With respect to what does he take a derivative?
Can show me explicitly the calculation?

Thanks!

Forgive my idioticity I meant equation (23) is on page 13 of course.

 

[nrqed]

Summary:: Derivation of of Eq(31) on page 16 from Eq(23) which is on page 16.

First let's define as on page 13 of the book the perturbative coefficients of the hard scattering cross section $H_{ab}$ by:
$$(22) \ \ \ \ H_{ab}=H_{ab}^{0}+\frac{\alpha_s}{\pi}H_{ab}^{(1)}+\mathcal{O}(\alpha_s^2)$$

Now, on page 16 it's written that the RG equation for $H_{ab}$ is:
$$(31) \ \ \ \ \mu \frac{d}{d\mu} H_{ab}(x_A,x_B,Q;\frac{\mu}{Q},\alpha_s(\mu))=$$
$$=-\sum_c \int_{x_A}^1d \zeta_A P_{c/a}(\zeta_A,\alpha_s(\mu))H_{cb}(\frac{x_A}{\zeta_A},x_B,Q;\frac{\mu}{Q},\alpha_s(\mu))$$
$$-\sum_d\int_{x_B}^1 d\zeta_BP_{d/b}(\zeta_B,\alpha_s(\mu))H_{ad}(x_A,\frac{x_B}{\zeta_B},Q;\mu/Q,\alpha_s(\mu)).$$

Here $P_{c/a}(\xi,\alpha_s(\mu))$ is the all orders Altarelli-Parisi kernel. It has a perturbative expansion:
$$(32) \ \ \ \ P_{c/a}(\xi,\alpha_s(\mu))=\frac{\alpha_s(\mu)}{\pi}P_{c/a}^{(1)}(\xi)+\ldots$$
where $P_{c/a}^{(1)}(\xi)$ is the function that appears in equation (23).
$$(23) \ \ \ \ f_{a/b}(x;\epsilon)=\delta_{ab}\delta(1-x)-\frac{1}{2\epsilon}\frac{\alpha_s}{\pi}P_{a/b}^{(1)}(x)+\mathcal{O}(\alpha_s^2)$$

Thus at lowest order the renormalization group equation (31) is a simple consequence of differentiating eq. (23).
With respect to what does he take a derivative?
Can show me explicitly the calculation?

Thanks!

Forgive my idioticity I meant equation (23) is on page 13 of course.

They don't mean Eq.23, obviously. They mean Eq.(19). And the derivative is mu d/dmu. But they do that in several steps, working in the lowest in the string coupling constant expansion.

 

[MathematicalPhysicist]

@nrqed I will come back to your comment after I'll return to read the book.
(in the meantime I returned on reading Schutz's book in chapter 10, so after I will finish reading this chapter I'll read your comment and let you know if I need some more guidance).

Cheers! thanks!

 

[MathematicalPhysicist]

They don't mean Eq.23, obviously. They mean Eq.(19). And the derivative is mu d/dmu. But they do that in several steps, working in the lowest in the string coupling constant expansion.

Hi @nrqed can you elaborate on the several steps which they did in the book?
BTW, you do refer to the strong coupling constant, right? there's no string theory in this book as far as I can tell.

 

[MathematicalPhysicist]

Never mind, I see their approach.
I now understand their calculation.

Thanks!

 

[MathematicalPhysicist]

In fact it seems to be a simple matter of derivative of $d/d\ln \mu$... hahaha

 

[nrqed]

Never mind, I see their approach.
I now understand their calculation.

Thanks!

Good ! Good job.

(and yes, I meant "strong coupling constant" and not "string coupling constant". I typed the "i" instead of the ket next to it, the "o", by mistake. It is funny that it led to a mistake that was still physically meaningful, by pure luck!)

 

[MathematicalPhysicist]

Good ! Good job.

(and yes, I meant "strong coupling constant" and not "string coupling constant". I typed the "i" instead of the ket next to it, the "o", by mistake. It is funny that it led to a mistake that was still physically meaningful, by pure luck!)

Some might say luck, but is it really?
Do you mean instead of 'k' since I don't understand what 'ket' has to do with it?
I mean ket is from the bra and ket of Dirac's notation in QM, but what does that have to do with this post?
Anyway thanks!
We make mistakes all the time, part of being human after all.