$\mu \rightarrow e\gamma$ in the $R_\xi$ gauge: trouble with momenta and Dirac matrices
My first ever question on stackexchange! Sorry if it is clumsy...
I am trying to follow the computation of $\mu \rightarrow e \gamma$ in <PERSON> and <PERSON> and am confused about the second and third ligns in equation (13.97):
$S_1 \rightarrow \tilde{S}1 = (p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u\mu \right] 2 m_\mu \left[2(1-\alpha_1)^2 + (2\alpha 1-1)\alpha _2\right]\ S_2 \rightarrow \tilde{S}_2 = -k^2(p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u\mu \right] (m_\mu /M^2) \times \left[(3\alpha 2-1)+\left[2\alpha_1^2-\alpha_1+\alpha_2(2\alpha_1-1/2)\right]\right]\ S_3 \rightarrow \tilde{S}_3 = -k^2(p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u\mu \right] (m_\mu /M^2) \left[2\alpha_1^2+\alpha_1+\alpha_2(2\alpha_1-1/2)\right]$
where
$S_1 = \Gamma ^{\mu \nu}N_{\mu \nu}\ S_2 = (k^\lambda \Gamma ^\mu \lambda)(k^\nu N{\mu \nu})/M^2\ S_3 = \left[(k+q)^\lambda \Gamma ^\mu \lambda \right]\left[(k+q)^\nu N{\mu \nu}\right]/M^2$
with
$\Gamma {\alpha \beta} = \left[(2k\cdot \varepsilon) g{\alpha \beta} -(k+2q)\beta \varepsilon \alpha - (k-q)\alpha \varepsilon \beta\right] $
(this being equation (13.81))
and
$N_{\mu \nu} =\bar{u}e(p-q)\gamma \mu ((p+k)\cdot \gamma ) \gamma \nu (1-\gamma _5)u\mu(p)$
(equation (13.88)).
In the calculation $S_i\rightarrow \tilde{S}_i$ we shift the integration variable $k$ to $k-\alpha_1 p - \alpha_2 q$ and discard all terms that are not $\sim (p\cdot \varepsilon)$.
Whereas I managed to obtain the expression for $\tilde{S}1$ (with a minus sign, but apparently that's alright...), I do not understand what to do with the many momenta in the calculation of $\tilde{S}_2$ (and $\tilde{S}_3$): Why do <PERSON> and <PERSON> only get (or keep?) terms $\sim k^2$? I would have thought we would get a whole bunch of terms $\sim m\mu ^2$ from the combinations $\sim p^2$ and $\sim p\cdot q$. (The mass of the electron is neglected here.)
I would be very grateful for some tips... Thank you in advance! 818