= Solution
Use $\hbar=c=1$ in the field-theory calculations. Work with signature $(+,-)$ and $a\ne0$. Since $x=(x_++x_-)$ and $t=(x_+-x_-)$, $\partial_+=\partial_x+\partial_t$, $\partial_-=\partial_x-\partial_t$, and the <Sine-Gordon equation> is $\phi_{+-}=\sin\phi$. Introduce $w=(\rho+\kappa)/2$ and $d=(\rho-\kappa)/2$. The specified <Sine-Gordon Bäcklund transformation> becomes $d_+=a\sin w$, $w_-=a^{-1}\sin d$. Its compatibility gives
$$
d_{+-}=\cos w\sin d,\qquad w_{+-}=\cos d\sin w.
$$
Adding and subtracting these equations establishes
$$
\boxed{\rho_{+-}=\sin(w+d)=\sin\rho,\qquad \kappa_{+-}=\sin(w-d)=\sin\kappa.}
$$
Thus a compatible transformed field satisfies the same <Sine-Gordon equation>. Smoothness is needed to interchange mixed derivatives. The statement cannot literally include $a=0$, since one defining equation contains $1/a$.
\b[Generating the <kink>.] With zero seed, the two first-order equations give $\rho_+=2a\sin(\rho/2)$ and $\rho_-=2a^{-1}\sin(\rho/2)$. On a nonconstant branch put $p=\tan(\rho/4)$; then $p_+=ap$, $p_-=a^{-1}p$. Integration gives
$$
\boxed{\rho=4\arctan\left[C\exp\left(\tfrac12(a+a^{-1})x+\tfrac12(a-a^{-1})t\right)\right].}
$$
For $a>0,C>0$, this is the <Sine-Gordon kink>, after absorbing $C$ in the center position. Its speed and Lorentz factor obey
$$
\boxed{v=\frac{1-a^2}{1+a^2},\qquad (1-v^2)^{-1/2}=\tfrac12(a+a^{-1}),\qquad a=e^{-\eta},\ v=\tanh\eta.}
$$
Changing the sign of $a$ or of the exponential coefficient selects the corresponding <antikink> orientation; $|v|<1$ whenever $a$ is finite and nonzero. Constant vacuum branches, omitted by division by $\sin(\rho/2)$, can be added separately.
The continuously parametrized <Bäcklund transformation> does more than produce this one solution. Successive compatible transformations and <Bianchi permutability for sine-Gordon Bäcklund transformations> construct <multisoliton solutions>; expansion of the generating relations gives the <local conserved-charge hierarchy of sine-Gordon theory>. Together with its <Lax pair> and <inverse scattering transform>, this is the structure behind <classical integrability>, elastic scattering and the absence of generic radiative energy loss in <soliton> collisions. Existence of a transformation alone is not a proof that every generated <conserved charge> is independent and in involution; those are additional properties of the integrable system.
\b[Transforming the static <kink> at $a=1$.] The seed obeys $\kappa_x=2\operatorname{sech}x$, $\sin(\kappa/2)=\operatorname{sech}x$, $\cos(\kappa/2)=-\tanh x$. Adding and subtracting the two <Sine-Gordon Bäcklund transformation> equations yields
$$
\rho_x=-2\tanh x\sin(\rho/2),\qquad \rho_t=2\operatorname{sech}x[1+\cos(\rho/2)].
$$
Again set $p=\tan(\rho/4)$. These simplify to $p_x=-p\tanh x$, $p_t=\operatorname{sech}x$. The first gives $p=f(t)/\cosh x$, and the second gives $f'=1$. Therefore
$$
\boxed{\rho(x,t)=4\arctan\frac{t-t_0}{\cosh x}.}
$$
Continuous inverse-tangent branches may be chosen without creating artificial jumps. At spatial infinity the field approaches the same vacuum, so its net <topological charge> is zero. For large $|t-t_0|$ there are two well-separated transitions near $x_\pm=\pm\operatorname{arcosh}|t-t_0|$, with speeds tending to zero. The two transitions exchange their <kink>/<antikink> orientations as they pass through the collision, while the field remains smooth.
This is the <Sine-Gordon threshold kink-antikink solution>, a <separatrix> between finite-speed scattering and the bounded <Sine-Gordon breather> motion, rather than a finite-period breather. In fact it is the $v\to0$ limit of the scattering field in question 3. At $t=t_0$, $\rho=0$ and $\rho_t=4\operatorname{sech}x$, so the reduced energy is $\int\rho_t^2dx/2=16$, exactly twice the reduced single-kink energy $8$.
For <mass> comparisons below, the original PDF really prints $m^2/\beta$. Restoring physical coordinates $X=x/m$, $T=t/m$ gives the angular-field action $\beta^{-1}\int dtdx[\tfrac12(\partial\phi)^2+\cos\phi-1]$ and literal <kink> <mass> $8m/\beta$, for $\beta>0$. The usual <Sine-Gordon theory> instead has prefactor $m^2/\beta^2$ and <mass> $8m/\beta^2$. The equation of motion and every field above are unchanged by this overall normalization; the quantum scattering comparison in question 3 is not.
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