 Solution

Use metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and common spacetime origins. A four-vector transforms as $A^{\prime }=\Lambda A$ under a Lorentz transformation. A Lorentz invariant is a scalar unchanged by that transformation. Since ${\Lambda }^T\eta \Lambda =\eta$,The wave phase labels the same stage of a physical oscillation at a spacetime event, independently of the observer's coordinates. Equal-phase wavefronts and interference phase differences are physical; fixing a common phase origin makes the continuous phase a Lorentz scalar. For a plane wave,The gradient of a scalar is a covector; raising its index gives the four-wavevector $K^{\mu }=(\omega /c,\mathbf{k})$. Under an $x$ boost,For vacuum light, $k_x=(\omega /c)\cos \theta$, givingA photon has four-momentum $p^{\mu }=\hslash K^{\mu }$, with $K^2=0$. For two photons separated by angle $\alpha$,In the centre-of-momentum frame the least possible pair energy is $2m_e{c}^2$, when both final particles are at rest. Thus the threshold has $(p_1+p_2{)}^2=4m_e^2{c}^2$; above it, excess energy supplies their kinetic energy. ThereforeThe kinematic equality is a limiting threshold; nonzero production phase space requires energies above it. For parallel photons no finite threshold is available, whereas head-on incidence minimizes the required energy:The blackbody-peak photon pair-production estimate, using the frequency-spectrum blackbody peak, $E_1=2.82144k_{B}T=7.29\times 10^{-4}\mathrm{eV}$ at $3\mathrm{K}$, gives $E_{2,\min }\simeq 3.58\times 10^{14}\mathrm{eV}$, or $358\mathrm{TeV}$. If “peak” means the wavelength-spectrum peak, $E_1=4.96511k_{B}T$ instead gives $2.03\times 10^{14}\mathrm{eV}$. The spectral convention changes the prefactor, not the conclusion that photons of order a few $10^{14}\mathrm{eV}$ are needed for this idealized peak-photon threshold.