Instantaneous-boost effective action for a sine-Gordon kink ($L_{\mathrm{eff}}=-8\sqrt{1-{\dot{X}}^2}-2\pi Xf(t)$)

Substitute the instantaneous Sine-Gordon kink profile ${\theta }_K(\gamma (x-X))$ and prescribed velocity $-\gamma \dot{X}{\theta }_K^{\prime }$ with $\gamma =(1-{\dot{X}}^2{)}^{-1/2}$ into the field density. For $A_1=0$, $A_0=xf(t)$, the integrals $\int ({\theta }_K^{\prime }{)}^2=8$ and $\int {\theta }_K^{\prime }=2\pi$ give the displayed collective-coordinate effective Lagrangian. Its equation is $d(8\gamma \dot{X})/dt=-2\pi f$. This is relativistic particle motion with mass $8$ and topological coupling charge $2\pi$. When $\gamma$ varies, the prescribed velocity omits a width-change contribution; acceleration correction for an instantaneous boosted kink describes the exact chain-rule distinction.

The actual derivative of ${\theta }_K(\gamma (t)(x-X(t)))$ is $[\dot{\gamma }(x-X)-\gamma \dot{X}]{\theta }_K^{\prime }$. The omitted contribution is first order in acceleration. Its cross term with the translation velocity integrates to zero because $\int y({\theta }_K^{\prime }(y){)}^{2}dy=0$. The square adds ${\dot{\gamma }}^2\int y^2({\theta }_K^{\prime }{)}^{2}dy/(2{\gamma }^3)={\pi }^2{\dot{\gamma }}^2/(3{\gamma }^3)$ to the spatially integrated density, since ${\theta }_K^{\prime }=2\mathrm{sech}y$ and $\int y^2({\theta }_K^{\prime }{)}^{2}dy=2{\pi }^2/3$. Thus the instantaneous-boost effective action for a sine-Gordon kink is an adiabatic truncation rather than the exact action restricted to that accelerating configuration-space profile.