Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I am trying to get to grips with Altarelli-Parisi-type equations. In chapter 17 of Peskin/Schroeder, they first develop the equations for a similar problem in QED. Equation $(17.123)$ introduces the sum rule $$ \int_0^1 dx ( f_e(x,Q) - f_\bar{e}(x,Q) ) = 1 $$ where $f_e$ and $f_\bar{e}$ are the distribution functions of electrons and antielectrons 'inside' an electron. I'm trying to prove that this is independent of $Q$.

The evolution equations are ($(17.120)$ in Peskin/Schroeder) $$ \frac{d}{d\log Q} f_e(x,Q)= \frac{\alpha}{\pi}\int_x^1 \frac{dz}{z} \left( P_{e\leftarrow e}(z) f_e(\frac{x}{z},Q) + P_{e\leftarrow\gamma}(z)f_\gamma(\frac{x}{z},Q)\right) $$ $$ \frac{d}{d\log Q} f_\bar{e}(x,Q)= \frac{\alpha}{\pi}\int_x^1 \frac{dz}{z} \left( P_{e\leftarrow e}(z) f_\bar{e}(\frac{x}{z},Q) + P_{e\leftarrow\gamma}(z)f_\gamma(\frac{x}{z},Q)\right) $$ where the relevant splitting functions is given by (equation $(17.121)$ in Peskin/Schroeder) $$ P_{e\leftarrow e}(z) = \frac{1+z^2}{(1-z)_+}+\frac{3}{2}\delta(1-z) $$ Using $\frac{d}{d\log Q}$ on $(17.123)$ gives (the part involving $P_{e\leftarrow\gamma}(z)$ cancels): $$ \frac{\alpha}{\pi}\int_0^1 dx \int_x^1 \frac{dz}{z} P_{e\leftarrow e}(z) \left( f_{e}(\frac{x}{z},Q) - f_\bar{e}(\frac{x}{z},Q) \right) $$ Insterting $(17.121)$ and using that $$ \int_0^1 dx \frac{f(x)}{(1-x)_+} = \int_0^1 dx \frac{f(x)-f(1)}{(1-x)} $$ I get $$ \frac{\alpha}{\pi}\int_0^1 dx \int_x^1 \frac{dz}{z} \left(\frac{1+z^2}{(1-z)_+}+\frac{3}{2}\delta(1-z) \right) \left( f_{e}(\frac{x}{z},Q) - f_\bar{e}(\frac{x}{z},Q) \right) $$ $$ = \frac{\alpha}{\pi}\int_0^1 dx \left( \int_x^1 dz \left( \frac{1+z^2}{(z-z^2)}\Delta(\frac{x}{z},Q) -\frac{2}{1-z} \Delta(x,Q) \right) + \frac{3}{2} \Delta(x,Q) \right) $$ This expression is singular and it seems that the singularities in the first two terms should cancle. However, I'm at a loss what to do here. My idea was to extract the singularity in the first term, but that seems like i'm doing it backwards (and I haven't figured out how to do it). Any hint would be appreciated, I'm not looking for full solutions.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Hint: Change order of integration

$$ \int_0^1 dx \int_x^1 \frac{dz}{z} P_{e\leftarrow e}(z) \left( f_{e}(\frac{x}{z},Q) - f_\bar{e}(\frac{x}{z},Q) \right) $$ $$ =\int_0^1 \frac{dz}{z} \int_0^z dx \ P_{e\leftarrow e}(z) \left( f_{e}(\frac{x}{z},Q) - f_\bar{e}(\frac{x}{z},Q) \right) $$ $$ \stackrel{x=zx'}=\int_0^1 dz \ P_{e\leftarrow e}(z) \int_0^1 dx' \left( f_{e}(x',Q) - f_\bar{e}(x',Q) \right) = 0, $$


$$\int_0^1 dz \ P_{e\leftarrow e}(z) = 0, $$

cf. formula

$$ \int_0^1 dz \frac{1+z^2}{(1-z)_+} = -\frac{3}{2}, $$

mentioned between (17.106) and (17.107) in Peskin and Schroeder.

share|improve this answer
That solves my problem. Thank you! –  David M. R. Sep 27 '11 at 12:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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