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In Thompson's Modern Particle Physics, in section 6 about electron-positron annihilation, it is stated (p. 152) that

"If the final-state fermion mass is also neglected, (6.63) reduces to the expression for the spin-averaged matrix element squared of (6.25), which was obtained from the helicity amplitudes."

Here:

$$\begin{align}\langle\lvert\mathcal{M}_{fi}\rvert^2\rangle&=2\dfrac{Q_f^2e^4}{(p_1\cdot p_2)^2}\left[(p_1\cdot p_3)(p_2\cdot p_4)+(p_1\cdot p_4)(p_2\cdot p_3)+m_f^2(p_1\cdot p_2)\right]\tag{6.63}\\ \langle\lvert\mathcal{M}_{fi}\rvert^2\rangle&\approx 2e^4\dfrac{(p_1\cdot p_3)^2+(p_1\cdot p_4)^2}{(p_1\cdot p_2)^2}\tag{6.25}\end{align}$$

So how exactly does one show this equality?

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Oh, right, you just need to look at equivalences of Mandelstam variables in the massless limit:

$$\begin{align}t&=(p_1-p_3)^2\approx -2p_1\cdot p_3\\ &=(p_2-p_4)^2\approx -2p_2\cdot p_4\\ u&=(p_1-p_4)^2\approx -2p_1\cdot p_4\\ &=(p_2-p_3)^2\approx -2p_2\cdot p_3\end{align}$$

and the result follows.

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