Full-mass unpolarized formal pair-production coefficient

By [F2], the spin sums reduce the squared-current factors to $L_e^{ab}=\frac{1}{4}\mathrm{Tr}[(\mathrm{̸}{p}_2-m){\gamma }^a(\mathrm{̸}{p}_1+m){\gamma }^b],L_{\mu ,ab}=\mathrm{Tr}[(\mathrm{̸}{k}_1+M){\gamma }_a(\mathrm{̸}{k}_2-M){\gamma }_b]\mathrm{.}$ The odd traces vanish. Expanding the even terms with [F2] and using $p_1\cdot p_2+m^2=s\mathrm{/}2$ and $k_1\cdot k_2+M^2=s\mathrm{/}2$ gives $L_e^{ab}=p_2^a{p}_1^b+p_2^b{p}_1^a-\frac{s}{2}{h}^{ab},L_{\mu ,ab}=4(k_{1a}{k}_{2b}+k_{1b}{k}_{2a}-\frac{s}{2}{h}_{ab})\mathrm{.}$ These mass-shell dot products follow from $(p_1+p_2{)}^2=(k_1+k_2{)}^2=s$.

Contracting the tensors gives $L_e^{ab}{L}_{\mu ,ab}=4\{2[(p_2\cdot k_1)(p_1\cdot k_2)+(p_2\cdot k_2)(p_1\cdot k_1)]+s^2-s(p_1\cdot p_2+k_1\cdot k_2)\}\mathrm{.}$ The two cross-metric terms contribute $-s(p_1\cdot p_2+k_1\cdot k_2)$, while the metric-metric term contributes $s^2$; substituting $p_1\cdot p_2=(s-2m^2)\mathrm{/}2$ and $k_1\cdot k_2=(s-2M^2)\mathrm{/}2$ reduces the last two terms to $s(m^2+M^2)$.

The finite matrix hypothesis [L1] supplies coefficientwise inverse and phase-covariance identities only for its specified finite graph kernel. For the separate amplitude in [A1], the spinor equations from [F2] give $\bar{v}(p_2)\mathrm{̸}{p}_2=-m\bar{v}(p_2)$ and $\mathrm{̸}{p}_{1}u(p_1)=mu(p_1)$, so their sum contracted with the incoming current is zero. Likewise $\bar{u}(k_1)\mathrm{̸}{k}_1=M\bar{u}(k_1)$ and $\mathrm{̸}{k}_{2}v(k_2)=-Mv(k_2)$, so the outgoing current is transverse to $P=k_1+k_2$. A longitudinal propagator term is therefore multiplied by two zero contractions. No broader tree-current claim is taken from [L1].

In the center-of-momentum frame, the equal-mass positive-shell formulas of [F3] give $\mathrm{\mid }{\mathbf{p}}_1^{\ast }\mathrm{\mid }=\sqrt{s}{\beta }_e\mathrm{/}2$ and $\mathrm{\mid }{\mathbf{k}}_1^{\ast }\mathrm{\mid }=\sqrt{s}{\beta }_{\mu }\mathrm{/}2$. By the definition of $\theta$ in the Statement, the four cross-dot-products in step 2.1 are $s\mathrm{/}4\pm \mathrm{\mid }{\mathbf{p}}_1^{\ast }\mathrm{\mid }\mathrm{\mid }{\mathbf{k}}_1^{\ast }\mathrm{\mid }\cos \theta$, with each sign occurring twice. Substitution into step 2.1 and multiplication by $g^4\mathrm{/}{s}^2$ gives $\overline{\mathrm{\mid }\mathcal{M}{\mathrm{\mid }}^2}=g^4[3-{\beta }_e^2-{\beta }_{\mu }^2+{\beta }_e^2{\beta }_{\mu }^2{\cos }^2\theta ]$. Each basis spinor ${\xi }_{\pm ,s}$ is dimensionless, so $u_s$ and $v_s$ have units ${\mathrm{J}}^{1\mathrm{/}2}$; each current has units $\mathrm{J}$, and $g^2\mathrm{/}s$ has units ${\mathrm{J}}^{-2}$. Thus $\mathcal{M}$ is dimensionless as required by the coefficient map. For each outgoing direction $\mathbf{n}\in S^2$, the rank-two projections $P_{+}$ and $P_{-}$ of [F2] depend continuously on $\mathbf{n}$, since the outgoing energies are positive. For either projection field $P(\mathbf{n})$, the sets on which each fixed pair of columns $P(\mathbf{n})e_j,P(\mathbf{n})e_k$ is linearly independent are open and cover $S^2$: a rank-two matrix has two linearly independent columns. Order the six pairs and partition $S^2$ by the first pair whose columns are independent. On each Borel piece, Gram--Schmidt gives a Borel orthonormal frame of the range. Thus the outgoing spinors can be chosen Borel in $\mathbf{n}$; the incoming spinors are fixed. The spin-resolved amplitudes are Borel, and their finite spin-label sums are Borel. The trace formula already computed makes the averaged squared amplitude continuous. [F2] The strict threshold gives positive ${\beta }_e,{\beta }_{\mu }$, and [F3] gives incoming flux $F=2s{\beta }_e>0$.

For distinguishable final particles, [F3] gives $d{\mathrm{\Phi }}_2\mathrm{/}d\mathrm{\Omega }={\beta }_{\mu }\mathrm{/}(32{\pi }^2)$ and $F=2s{\beta }_e$. Substituting the squared amplitude from step 3.1 into [F3] yields the displayed differential coefficient after using $g^4=(4\pi \alpha {)}^2$. To integrate, choose the polar axis along ${\mathbf{p}}_1^{\ast }$ and write the outgoing unit direction as $X(\theta ,\phi )=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$. The phase-space angular measure is the chart surface measure on $S^2$ and is invariant under rotations by [F3]. This chart has area density $\sin \theta$; the omitted meridian and poles have zero chart area. Thus ${\int }_{S^2}d\mathrm{\Omega }=2\pi {\int }_0^{\pi }\sin \theta d\theta =4\pi$ and ${\int }_{S^2}{\cos }^2\theta d\mathrm{\Omega }=2\pi {\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =4\pi \mathrm{/}3$. This gives the displayed total coefficient.

For $\delta =4m^2\mathrm{/}s\le 1\mathrm{/}2$, set $B_0=2-{\beta }_{\mu }^2+{\beta }_{\mu }^2{\cos }^2\theta$ and $Q=1-{\beta }_{\mu }^2{\cos }^2\theta$. Then $B_0\ge 1$, $0\le Q\le 1$, and the ratio of the full coefficient to its $m=0$ value is $R=(B_0+\delta Q)\mathrm{/}(B_0\sqrt{1-\delta })\ge 1$. Also $R-1=\frac{1}{\sqrt{1-\delta }}-1+\frac{\delta Q}{B_0\sqrt{1-\delta }}\le 2\sqrt{2}\delta \le 4\delta ,$ because $1\mathrm{/}\sqrt{1-\delta }-1=\delta \mathrm{/}(\sqrt{1-\delta }(1+\sqrt{1-\delta }))\le \sqrt{2}\delta$ for $0\le \delta \le 1\mathrm{/}2$, and $B_0\sqrt{1-\delta }\ge 1\mathrm{/}\sqrt{2}$. This compares only the two formal tree coefficients.