= Lorentz boost of a Q-ball
{c}
{title2=$E=\gamma M,\quad P=\gamma vM$}
For a one-dimensional <Q-ball>, the profile equation gives $\tfrac12(f')^2=U(f)-\tfrac12\omega^2f^2$. Its rest energy is therefore $M=\int[(f')^2+\omega^2f^2]$. A <Lorentz transformation> replaces the rest coordinates by $\tau=\gamma(t-v(x-x_0))$, $\xi=\gamma(x-x_0-vt)$. The scalar solution $e^{i\omega\tau}f(\xi)$ has energy $\int[\tfrac12|\phi_t|^2+\tfrac12|\phi_x|^2+U]$ and momentum $-\int\operatorname{Re}(\bar\phi_t\phi_x)$. Substitution and $dx=d\xi/\gamma$ give the displayed values and $E^2-P^2=M^2$.
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