Question 5.3: The Bremsstrahlung Part of the GLAP Equation The bremsstrahl......

The number of quarks with a specific flavor is given by

N=\int_0^1\mathrm{d}x\;f(x)\;\,.    (1)

One has to show that N does not depend on Q² and consequently not on t. Therefore

\frac{\mathrm{d}}{\mathrm{d}t}N=\int_0^1\mathrm{d}x\frac{\mathrm{d}f}{\mathrm{d}t}=\frac{\alpha_{s}}{2\pi}\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\frac{\mathrm{d}y}{y}f(y)P_{q q}\left(\frac{x}{y}\right)   (2)

must vanish. We exchange the x and y integrations,

\frac{\mathrm{d}}{\mathrm{d}t}N=\frac{\alpha_{s}}{2\pi}\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\frac{\mathrm{d}x}{y}f(y)P_{q q}\left(\frac{x}{y}\right)~,    (3)

and introduce the new variable z = x/y,

\frac{\mathrm{d}}{\mathrm{d}t}N=\frac{\alpha_{s}}{2\pi}\int_0^1\mathrm{d}y\;f(y)\int_0^1\mathrm{d}z P_{q q}(z)=\frac{\alpha_{s}}{2\pi}N\int_{0}^{1}\mathrm{d}z P_{q q}(z)\;.     (4)

By means of (5.49) we have

\int_0^1\mathrm{d}z\,F(z)\left[{\frac{1+z^{2}}{1-z}}\right]_{+}=\int_{0}^{1}\mathrm{d}z\,\left[F(z)-F(1)\right]{\frac{1+z^{2}}{1-z}}    (5.49)

\int_0^1\mathrm{d}z\frac{4}{3}\left(\frac{1+z^{2}}{1-z}\right)_{+}=\int_0^1\mathrm{d}z\left(\frac{4}{3}-\frac{4}{3}\right)\frac{1+z^{2}}{1-z}=0    (5)

and hence

which was to be shown.
It is no coincidence that the right-hand side of (5.58) integrated over x separates into two factors, namely the integrals over y and z. Since QCD is a dimensionless theory, P_{q q} can only depend on ratios of the momenta that occur, i.e., on x/y. Consequently all contributions to the GLAP equation exhibit the same feature. In the next section we shall discuss howthis mathematical structure can be derived from very general assumptions.