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At a specified frequency with transparent-medium phase index $n>1$, Cherenkov radiation requires $\beta >1/n$. Inserting the threshold speed into relativistic kinetic energy gives $K_{\mathrm{th}}=m{c}^2[(1-n^{-2}{)}^{-1/2}-1]$. Equality is the kinematic limit; the cone angle and emission vanish there. A dispersive material has different thresholds in different spectral intervals. The result uses phase velocity, not group velocity.

The external work equals the change in relativistic kinetic energy, $W=m_e{c}^2({\gamma }_f-{\gamma }_i)$. With $m_e{c}^2=511\mathrm{keV}$,The rest energy cancels between endpoints. At these speeds, the nonrelativistic estimate $\frac{1}{2}{m}_e{c}^2(0.11^2-0.10^2)=0.53655\mathrm{keV}$ is close but slightly smaller.

Use the inertial frame in which the departure point and destination are stationary, with separation $d$. The astronauts' elapsed proper time is $\tau =t/\gamma$, whereas $d=vt$. HenceThe Earth-frame travel time is $t=\sqrt{41}\mathrm{yr}\simeq 6.40\mathrm{yr}$ and the speed is approximately $0.625c$. The spacecraft's relativistic kinetic energy is thereforeThe five years are not the time to use in $v=d/t$ in the departure frame. This is the cruising kinetic energy, with departure and arrival acceleration intervals omitted, rather than a complete propulsion energy budget.

Interpret the percentage relative to the classical kinetic energy, so that $K_{\mathrm{rel}}=1.02K_{\mathrm{cl}}$. Set $\beta =v/c$ and use the Lorentz factor $\gamma =(1-{\beta }^2{)}^{-1/2}$. The relativistic kinetic energy and classical kinetic energy giveFor the positive nonzero speed, insert ${\beta }^2=1-{\gamma }^{-2}$ and cancel $\gamma -1$:It follows thatHere $m{c}^2=938\mathrm{MeV}$ is the proton's rest energy; the resulting relativistic kinetic energy is $12.479\mathrm{MeV}$. The zero-speed equality before cancellation has no meaningful percentage excess because both kinetic energies then vanish.

The relativistic kinetic energy is the total energy minus the rest energy:Here $\gamma$ is the Lorentz factor and $c$ the speed of light. The binomial series, with $b=v/c$, givesFor $v\ll c$, the leading term is the classical kinetic energy $M{v}^2/2$. The ratio of the fourth-order term to that leading term is $3v^2/(4c^2)$. ConsequentlyThe condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly $1\%$.

For the lander, $u/c=35/37$, soThe relativistic kinetic energy isThe rest energy $m{c}^2$ is excluded from kinetic energy.

Identify the nonrelativistic mass with the rest mass, $m=m_0$. Expanding the relativistic kinetic energy in $p^2/(m^2{c}^2)$ givesFirst-order perturbation theory requires $\Delta E_0=⟨0|H^{\prime }|0⟩$. Put $\alpha =\sqrt{m\hslash \omega /2}$, so $p=i\alpha (a^{†}-a)$. The annihilation relations are $a|0⟩=0$ and $⟨0|a^{†}=0$. In particular, $a^{†}|0⟩$ does not vanish. A short calculation retaining the operator ordering givesEquivalently, of the ordered words in $(a^{†}-a{)}^4$, only $a{a}^{†}a{a}^{†}$ and $aa{a}^{†}{a}^{†}$ have nonzero ground-state expectation values, respectively $1$ and $2$. Words with a creation operator on the far left or an annihilation operator on the far right vanish; the remaining unbalanced words also vanish. ThusThe correction is negative because the relativistic kinetic energy is smaller than $p^2/(2m)$ at fixed nonzero momentum. This relativistic oscillator ground-state kinetic correction requires $\hslash \omega \ll m{c}^2$.

For a massive free particle with invariant rest mass $m$, speed $v<c$, and Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$, its total energy is $E=\gamma m{c}^2$. It includes the rest energy $m{c}^2$ and the relativistic kinetic energy $(\gamma -1)m{c}^2$. Here $c$ is the speed of light. At fixed total energy, a larger speed requires a smaller rest mass.

Expanding the relativistic kinetic energy $\sqrt{m^2{c}^4+c^2{p}^2}-m{c}^2$ for $|p|\ll mc$ gives $p^2/(2m)-p^4/(8m^3{c}^2)+\cdots$. The first correction is negative and contributes to atomic fine structure. Its perturbative use requires small momentum compared with mc; the truncated negative quartic term is not a globally stable exact Hamiltonian.

The expansion of the relativistic kinetic energy at fixed rest mass gives the perturbation $-p^4/(8m^3{c}^2)$. In the harmonic-oscillator ground state, $⟨p^4⟩=3(m\hslash \omega /2{)}^2$, proving the displayed first-order energy correction. Its small parameter is $\hslash \omega /(m{c}^2)$. This correction lowers the nonrelativistic energy and does not include other relativistic interactions.