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Use metric signature $(+---)$, as required by the sign of the scalar product in this question. For a massive particle, the four-momentum isFor a photon of positive frequency $\nu$ and direction $e$, the Planck constant gives $E=h\nu$ andA future-pointing timelike four-vector $P=(P^0,p)$ has $P^0>|p|$ and $P^0>0$ in this signature. The ordinary Euclidean Cauchy-Schwarz inequality yieldsThis is stronger than the requested nonnegative inequality. For an electron and a positron of rest mass $m>0$, their future-pointing timelike four-momenta have $P_1^2=P_2^2=m^2{c}^2$. Thuswhich cannot equal the null norm of a single photon. Four-momentum conservation forbids spontaneous photon decay into an electron-positron pair in vacuum. An external body or field able to exchange momentum would be a different system.

For $u>0$, take the initial electron direction as the longitudinal axis and put $\beta =u/c$. Incoming four-momentum conservation givesThe transverse components must cancel. In the noncollinear case the two transverse directions are opposite, and ${\nu }_1\sin {\theta }_1={\nu }_2\sin {\theta }_2$. Hence the ratio of longitudinal momentum times $c$ to energy isWriting $s=({\theta }_1+{\theta }_2)/2$ and $d=({\theta }_1-{\theta }_2)/2$, the last ratio is $\cos s/\cos d$. It also equals $(1+\cos ({\theta }_1+{\theta }_2))/(\cos {\theta }_1+\cos {\theta }_2)$ whenever that denominator is nonzero. Since ${\gamma }^2{\beta }^2={\gamma }^2-1$, the angular relation for two-photon electron-positron annihilation isThe positive root follows from $u>0$. For noncollinear emission the positive momentum-to-energy ratio ensures $\cos s>0$, and $|d|<\pi /2$ ensures the stated denominator is positive.

The printed quotient needs a qualification for exactly collinear photons. The allowed configuration ${\theta }_1=0$, ${\theta }_2=\pi$ has positive frequencieswhich obey both conservation equations, but the printed left side is $0/0$. Thus the quotient holds for noncollinear angles, or as its appropriate noncollinear limit. The undivided identityremains true in the collinear case, where both sides vanish. The limit $u=0$ similarly requires care because the initial direction is then undefined and the photons are antiparallel.

For $0<v<c$, the Lorentz transformation with coincident origins isFor two simultaneous events in $S$, $\Delta t=0$, so $\Delta t^{\prime }=-\gamma v\Delta x/c^2$. Distinct spatial positions therefore generally give different times in the boosted frame: this is relativity of simultaneity.

Choose the emission events at $(t,x)=(0,0)$ and $(0,d)$. After emission, the photon world lines are $x=ct$ and $x=d+ct$. Applying the Lorentz transformation to either givesThus the two photons move at speed $c$ in $S^{\prime }$ on parallel lines whose intercepts differ by $\gamma (1+v/c)d$. Measuring their positions at the same $t^{\prime }$ after both emissions gives the photon separation under a collinear Lorentz boost:This separation is constant. It is not the contracted distance between the stationary sources: the two emission events are not simultaneous in $S^{\prime }$, and one photon has already moved when the other is emitted.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.