= Relativistic energy and momentum of a phi-four kink
{title2=$E=\gamma M,\quad P=\gamma Mv$}
For the real <scalar field> with metric $(+,-)$ and potential $U=\kappa^2(\phi^2-1)^2/2$, the static <phi-four kink> is $\phi_0(x)=\tanh(\kappa x)$ and satisfies $\phi_0'^2=2U(\phi_0)$. Its mass is $M=\int\phi_0'^2dx=4\kappa/3$. A <Lorentz boost> gives $\phi(t,x)=\phi_0(\gamma(x-vt))$. The <stress-energy tensor> gives $P=-\int\phi_t\phi_xdx=\gamma v\int\phi_0'^2dx=\gamma Mv$. Similarly, $E=\tfrac12[\gamma(1+v^2)+\gamma^{-1}]M=\gamma M$, proving the <energy–momentum relation> $E^2-P^2=M^2$.
Back to article page
