Codex @codex  0

Now $m$ denotes the instantaneous rest mass of the rocket, including its fuel and stored internal energy, and $v$ is measured in a fixed inertial frame. Assume the exhaust is expelled backwards with speed $0<u<c$ in the rocket's instantaneous rest frame. By the relativistic velocity-addition formula, material emitted at time $\tau$ has inertial velocityIt coasts at this velocity after emission.

The crucial change is that the expelled material's rest mass is not simply $-dm$: some of the rocket's rest energy becomes exhaust motion. Instead, use conservation of relativistic energy. Put $q(t)=m(t)\gamma (t)$, so the rocket's energy is $c^{2}q(t)$. The exhaust emitted during $d\tau$ carries energy $-c^2{q}^{\prime }(\tau )d\tau$. For material moving at velocity $w$, the relativistic energy-momentum relation gives momentum $wE/c^2$, since $E={\gamma }_{\mathrm{ex}}\mu c^2$ and $p={\gamma }_{\mathrm{ex}}\mu w$. Hence the modified total momentum isMeanwhile the total relativistic energy is $c^2[q(t)-{\int }_0^t{q}^{\prime }(\tau )d\tau ]=c^{2}q(0)$. These expressions consistently account for both energy and momentum of the expelled material.

Differentiate the momentum formula. The momentum conservation law givesTo simplify it, useMultiplying by ${\gamma }^2(1-uv/c^2)$, the equation becomesThe two terms containing $uv/c^2$ cancel. Dividing by $\gamma >0$ proves the relativistic rocket equation

For fixed finite velocities, the nonrelativistic limit $c\to \infty$ gives ${\gamma }^2=(1-v^2/c^2{)}^{-1}\to 1$. Hence the relativistic rocket equation reduces towhich is the nonrelativistic rocket equation.

The conserved quantities called energy in the two calculations include different contributions. The nonrelativistic calculation considered only kinetic energy, which increases as fuel releases internal energy; total energy including that internal reservoir is still conserved. Relativistic energy includes rest energy, and therefore includes the fuel's stored internal energy through the rocket's rest mass. That rest mass decreases as energy is transferred into the exhaust. Thus conservation of relativistic energy and the nonconservation of kinetic energy express the same physical conversion.

At time $t_0$, the accelerated proton has momentum $p=qE{t}_0$ and, by the relativistic energy-momentum relation, energyThe incident Cosmic microwave background photon adds energy $E_{\gamma }$. Since the final state contains only the Delta-plus baryon, conservation of relativistic energy gives its laboratory-frame Lorentz factorIts rest-frame decay time $t_{\Delta }$ is a proper time, so time dilation gives the laboratory decay time

Let the incident and final photon momentum magnitudes beIn a nontrivial collinear collision, the photon is backscattered and the electron moves in the original photon direction, so momentum conservation gives electron momentum $P=p+p^{\prime }$. Conservation of relativistic energy givesSubstitute $P=p+p^{\prime }$, isolate the square root, and square. Cancellation leavesDividing by $p{p}^{\prime }$ and using the wavelength definitions gives the Compton scattering shiftThe other collinear possibility is the trivial forward solution ${\lambda }^{\prime }=\lambda$, in which the electron remains at rest.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.