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Choose the source motion along $+x$, write $\beta =v/c$, and let primes denote its instantaneous rest frame. A photon with rest-frame energy $E^{\prime }$ and direction ${\theta }^{\prime }$ has $p_x^{\prime }=E^{\prime }\cos {\theta }^{\prime }/c$ and transverse momentum $E^{\prime }\sin {\theta }^{\prime }/c$. A Lorentz boost to the laboratory givesSince $|\mathbf{p}|=E/c$, the relativistic aberration relation and its inverse areThe mapping is monotone. The boundary of the rest-frame forward hemisphere, $\cos {\theta }^{\prime }=0$, therefore maps to $\cos \theta =\beta$. An isotropic photon count assigns exactly half the photons to that hemisphere, provingThe photon count inside a relativistic beaming cone concerns an emitted ensemble and counts photons, not their total energy or the arrival-time-weighted power at a detector. The laboratory photon energies depend on direction, so “half the photons” need not mean half the radiated energy. At large $\gamma$ the angular scale is ${\theta }_{1/2}\simeq 1/\gamma$.

Use the metric $\eta =\mathrm{diag}(-1,1,1,1)$ and four-velocity $u^{\mu }=\gamma (c,\mathbf{v})$. Its derivative with respect to proper time is the four-acceleration. Sincea purely transverse laboratory acceleration has $d\gamma /dt=0$. With $d/d\tau =\gamma d/dt$,The invariant proper acceleration is the magnitude measured in the instantaneous rest frame, $a_0^2=A^{\mu }{A}_{\mu }$. Consequently,For longitudinal acceleration the factor is instead ${\gamma }^3$ multiplying the laboratory acceleration. The distinction matters for circular storage-ring motion, where the acceleration is centripetal and transverse even though its magnitude is nonzero.

For the source's definition $s^{\alpha }=\gamma (\mathbf{s}\cdot \mathbf{v}/c,\mathbf{s})$, direct contraction with $u^{\alpha }=\gamma (c,\mathbf{v})$ givesIt remains zero under the torque-free transport law. Differentiate the contraction, using the product rule and $u^2=-c^2$:Thus an initially spatial spin in the instantaneous rest frame remains orthogonal to the velocity. This is the orthogonal-spin specialization of Fermi-Walker transport. The sign is tied to the negative timelike norm; using the opposite metric sign without changing the transport formula would spoil the cancellation.

First note that the supplied spin equation preserves $s\cdot u=0$: its electromagnetic bracket has contraction $u_a{F}^{a\lambda }{s}_{\lambda }+(u^2/c^2)F^{\mu \nu }{u}_{\mu }{s}_{\nu }=0$, and the remaining transport term cancels $s\cdot \dot{u}$ as in i.

For an explicitly electromagnetic acceleration, set ${\dot{u}}^a=(q/m)F^{a\lambda }{u}_{\lambda }$ and define $C=F^{\mu \nu }{u}_{\mu }{s}_{\nu }$. Antisymmetry gives $s\cdot \dot{u}=-(q/m)C$, so the spin equation can be writtenContract with $2s_a$:The first term vanishes because a symmetric product of the two spin covectors contracts an antisymmetric tensor; the second vanishes by spin–velocity orthogonality. In the instantaneous rest frame $s^0=0$, so $s^a{s}_a$ is precisely the squared proper spatial-spin magnitude. It remains constant for arbitrary $g$ under this transport, though its direction can precess. In this equation $g$ is the dimensionless g-factor; the dimensional gyromagnetic ratio includes $gq/(2m)$.

Use the same event pair already made simultaneous in the rear instantaneous rest frame. Its proper spatial separation is the Minkowski interval length, soThe inequality selects the front rocket rather than the equally long backward extension. Equivalently the Lorentz-transformed spatial difference obeys $\gamma [h(t_B)-f(t_A)-v(t_B-t_A)]=L$. This is not the condition $h(t)-f(t)=L$ at equal laboratory times, which would describe the stretching arrangement in a.