4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

Hint: Change order of integration

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

because

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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