= Solution
Use metric $\eta=\operatorname{diag}(1,-1,-1,-1)$ and common spacetime origins. A <four-vector> transforms as $A'=\Lambda A$ under a <Lorentz transformation>. A <Lorentz invariant> is a scalar unchanged by that transformation. Since $\Lambda^T\eta\Lambda=\eta$,
$$
\boxed{A'\cdot B'=A^T\Lambda^T\eta\Lambda B=A^T\eta B=A\cdot B.}
$$
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,
$$
\phi=\omega t-\mathbf k\cdot\mathbf r=K_\mu X^\mu,
\qquad K_\mu=\partial_\mu\phi.
$$
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,
$$
\boxed{\omega'=\gamma(\omega-vk_x),\quad
k'_x=\gamma(k_x-v\omega/c^2),\quad k'_y=k_y,\quad k'_z=k_z.}
$$
For vacuum <light>, $k_x=(\omega/c)\cos\theta$, giving
$$
\boxed{\omega'=\gamma\omega(1-\beta\cos\theta).}
$$
A <photon> has <four-momentum> $p^\mu=\hbar K^\mu$, with $K^2=0$. For two <photons> separated by angle $\alpha$,
$$
(p_1+p_2)^2=2\hbar^2K_1\cdot K_2
=\frac{2\hbar^2\omega_1\omega_2}{c^2}(1-\cos\alpha).
$$
In the <centre-of-momentum frame> the least possible pair <energy> is $2m_ec^2$, when both final particles are at rest. Thus the threshold has $(p_1+p_2)^2=4m_e^2c^2$; above it, excess <energy> supplies their <kinetic energy>. Therefore
$$
\boxed{\omega_{2,\rm th}=
\frac{2m_e^2c^4}{\hbar^2\omega_1(1-\cos\alpha)}.}
$$
The 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>:
$$
\boxed{E_{2,\min}=\frac{(m_ec^2)^2}{E_1}\quad(\alpha=\pi).}
$$
The <blackbody-peak photon pair-production estimate>, using the frequency-spectrum blackbody peak, $E_1=2.82144k_BT=7.29\times10^{-4}\,\mathrm{eV}$ at $3\,\mathrm K$, gives \b[$E_{2,\min}\simeq3.58\times10^{14}\,\mathrm{eV}$], or $358\,\mathrm{TeV}$. If “peak” means the wavelength-spectrum peak, $E_1=4.96511k_BT$ instead gives $2.03\times10^{14}\,\mathrm{eV}$. 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.
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