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Energy–momentum transformations and invariant. Let the primed inertial frame move at signed velocity $u$ along the unprimed $x$ axis. With $\Gamma =(1-u^2/c^2{)}^{-1/2}$, a Lorentz transformation of four-momentum givesUse $p^2=p_x^2+p_y^2+p_z^2$. The longitudinal terms satisfyThe transverse momenta are unchanged, soFor one electron, the relativistic energy-momentum relation identifies this Lorentz invariant as $E_0^2=m_e^2{c}^4$.

Symmetric electron scattering. Assume an isolated elastic collision of the two electrons, with no emitted radiation. The incident electron has total energy $E$ and momentum magnitude $p$, so $p^2{c}^2=E^2-E_0^2$; the target electron initially contributes rest energy $E_0$ and zero momentum. In the nondegenerate configuration with final track angles $+\theta$ and $-\theta$, transverse momentum conservation givesBecause both final particles have the same rest mass, they also have equal total energies. Conservation of energy givesLongitudinal momentum conservation is $p=2p_f\cos \theta$. Combining these expressions, with $E>E_0$, proves the symmetric relativistic equal-mass scattering angle relation:The low-kinetic energy limit has $\theta \to 45^{\circ }$, recovering the familiar right angle between the outgoing tracks in nonrelativistic equal-mass scattering. Relativistic incident energies give $\theta <45^{\circ }$, approaching zero at very high energy.

Accelerating potential. For $\theta =30^{\circ }$, ${\tan }^2\theta =1/3=2E_0/(E+E_0)$, so $E=5E_0$. The acceleration from rest supplies kinetic energy, not total energy. If $e>0$ is the elementary charge, the required accelerating electric potential difference has magnitudeusing electron rest energy $E_0\simeq 0.511\mathrm{MeV}$. The electric potential difference is oriented to accelerate the negative electron; the result above specifies its magnitude.

Nonrelativistic momentum conservation gives $\mathbf{u}={\mathbf{v}}_1+{\mathbf{v}}_2$, while an elastic collision conserves kinetic energy, $u^2=v_1^2+v_2^2$. Squaring the momentum equation yields ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=0$. The two nonzero outgoing velocities are perpendicular. In the degenerate case where one outgoing particle is at rest, its direction is undefined.

For relativistic collisions conserve total four-momentum, with $E=\gamma m{c}^2$, $\mathbf{p}=\gamma m\mathbf{v}$ and $E^2-p^2{c}^2=m^2{c}^4$. Rest energies must be included; for photons $E=pc$.

For the symmetric relativistic equal-mass scattering angle, the incident momentum is $mc\sqrt{\alpha (\alpha +2)}$. Energy conservation gives each final particle $E_f=(1+\alpha /2)m{c}^2$, hence $p_f=(mc/2)\sqrt{\alpha (\alpha +4)}$. Transverse momenta cancel and $2p_f\cos \theta =p_{\mathrm{in}}$, givingThe nonrelativistic limit has $\theta \to 45^{\circ }$, so the outgoing angle is $90^{\circ }$ as above.

For the symmetric two-photon annihilation angle in the specified electron-positron annihilation, each photon has energy $E_{\gamma }=(\alpha +2)m_e{c}^2/2$. Momentum conservation gives $2(E_{\gamma }/c)\cos \varphi =m_{e}c\sqrt{\alpha (\alpha +2)}$, soAs $\alpha \to 0^{+}$ the two photons become back-to-back, transverse to the limiting incident direction; as $\alpha \to \infty$ they are forward collimated. Equal photon energies follow from the specified equal angles, and are not a general property of annihilation in a moving laboratory frame.