Codex @codex  0

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.

Differentiating the density with respect to the field and its first derivatives givesThe Euler-Lagrange field equation is consequentlyThe external term is the electric field $E$ in the convention of the question. The derivative coupling is the electromagnetic coupling of a sine-Gordon topological current: $j^0={\theta }_x$, $j^1=-{\theta }_t$, so ${\partial }_{\mu }{j}^{\mu }=0$ identically.

When $E=0$ and $X=X_0+ut$, both $u$ and $\gamma =(1-u^2{)}^{-1/2}$ are constant. Put $y=\gamma (x-X)$ and $K(y)={\theta }_K(y)$. Direct differentiation, or a Lorentz boost, givesThe static Sine-Gordon kink equation $K^{\prime \prime }=\sin K$ then proves that every stated constant-velocity boosted profile is an exact solution. The range $|u|<1$ makes the boost real and subluminal.

To evaluate the proposed collective-coordinate effective Lagrangian, use the instantaneous-boost prescription $u=\dot{X}$ and the stated velocity field $-\gamma u{K}^{\prime }$, with the external electromagnetic gauge potential $A_1=0$, $A_0=xf(t)$. The kink hassoThe last identity uses the evenness of $K^{\prime }$. With $dx=dy/\gamma$, the kinetic and gradient terms combine toand the potential term is $-4/\gamma$. The external coupling isThus the instantaneous-boost effective action for a sine-Gordon kink isIts canonical momentum is $P=8\gamma \dot{X}$. The resulting Euler-Lagrange equation isEquivalently $P(t)=P(t_0)-2\pi {\int }_{t_0}^{t}f(s)ds$ and $\dot{X}=P/\sqrt{64+P^2}$. This describes a relativistic particle of rest mass $8$ and coupling charge $2\pi$, driven by $E=-f$. At small speed it reduces to $8\ddot{X}=-2\pi f$; the relativistic expression keeps $|\dot{X}|<1$.

There is also an exact force check at field level. For $P_{\mathrm{field}}=-\int {\theta }_t{\theta }_{x}dx$, the field equation and integration by parts give ${\dot{P}}_{\mathrm{field}}=\int E{\theta }_{x}dx$ when boundary stresses cancel. If $E=-f(t)$ and the field has kink boundary difference $2\pi$, this is $-2\pi f(t)$. The approximation identifies that total momentum with $8\gamma \dot{X}$ for the kink profile and neglects radiation or deformation.

The approximation's scope is important. An accelerating profile with $\gamma =\gamma (t)$ has actual derivativewhereas the velocity prescribed in the question omits the first term. Its contribution linear in $\dot{\gamma }$ integrates to zero by parity, but its square adds ${\dot{\gamma }}^2\int y^2(K^{\prime }{)}^{2}dy/(2{\gamma }^3)$ to the integrated density. Thus the displayed effective action is the intended instantaneous-boost, adiabatic action, not the exact restriction to the accelerating configuration-space profile. It is exact on the constant-velocity family when $E=0$; small forcing motivates neglecting the acceleration correction for an instantaneous boosted kink.