 Lorentz boost of a Q-ball

For a one-dimensional Q-ball, the profile equation gives $\frac{1}{2}(f^{\prime }{)}^2=U(f)-\frac{1}{2}{\omega }^2{f}^2$. Its rest energy is therefore $M=\int [(f^{\prime }{)}^2+{\omega }^2{f}^2]$. 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 $\int [\frac{1}{2}|{\varphi }_t{|}^2+\frac{1}{2}|{\varphi }_x{|}^2+U]$ and momentum $-\int \mathrm{Re}({\bar{\varphi }}_t{\varphi }_x)$. Substitution and $dx=d\xi /\gamma$ give the displayed values and $E^2-P^2=M^2$.