Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 5

Varying the real scalar field gives the Euler-Lagrange field equation ${\partial }_t^2\varphi -{\partial }_x^2\varphi +V^{\prime }(\varphi )=0$. Since $V^{\prime }(\varphi )=a\sin (b\varphi )$, the answer isFor $b\ne 0$, introducing $u=b\varphi$ gives $u_{tt}-u_{xx}+ab\sin u=0$, the Sine-Gordon equation. The proposed real localized kink requires $ab>0$; arbitrary real $a,b$ do not guarantee that condition. In what follows put $k=\sqrt{ab}>0$. The displayed potential itself requires $b\ne 0$, unless a limiting prescription is specified.

Allow a freely chosen center $x_0$, set $u=k(x-x_0)$ and $z=e^u$. Direct differentiation of the proposed Sine-Gordon kink givesThe angle identity givesSince $k^2/b=a$, these formulas show $\boxed{{\varphi }^{\prime \prime }=a\sin (b\varphi )}$, precisely the static equation of motion.

For $x\to -\infty$ and $x\to +\infty$, the field approaches $0$ and $2\pi /b$ respectively. These are neighboring minima of the scalar potential when $ab>0$, and ${\varphi }^{\prime }$ vanishes exponentially at either end. The energy is therefore finite and the topological charge $Q=b[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )$ is one. Reversing $x-x_0$ gives the corresponding antikink. If $ab<0$, the printed real exponential profile is unavailable; its formal imaginary scale is not a real finite-energy solution of the stated form.

The Lagrangian density is Lorentz invariant, so a scalar-valued static solution can be composed with a Lorentz transformation. Let $X=\gamma (x-vt-x_0)$, with $\gamma =(1-v^2{)}^{-1/2}$ and $|v|<1$. If $\varphi (x,t)=f(X)$, the chain rule givesThe boosted equation of motion is therefore exactly the static one already verified in solution, andis a solution. The center moves as $x=x_0+vt$: $v$ is the physical velocity in units where the limiting speed is one. The kink width contracts from $k^{-1}$ to $(\gamma k{)}^{-1}$, while its topological charge remains unchanged.

For a static finite-energy field, the Hamiltonian density is $\frac{1}{2}({\varphi }^{\prime }{)}^2+V(\varphi )$. Multiply the static equation of motion by ${\varphi }^{\prime }$ and integrate once:At either vacuum endpoint both terms vanish, so $\frac{1}{2}({\varphi }^{\prime }{)}^2=V$. The total energy density is consequently $({\varphi }^{\prime }{)}^2=4k^2{\mathrm{sech}}^2[k(x-x_0)]/b^2$. Its integral givesFor the usual positive $a,b$ convention this isThe coefficient depends on the dimensionless parameter $b$; it is not a universal numerical constant independent of the potential. In two spacetime dimensions, $[\varphi ]=[b]=0$ and $[a]=2$, so the dimensions are consistent. If both $a$ and $b$ are negative while $ab>0$, the real answer is instead $8\sqrt{|a|}/|b{|}^{3/2}$; the literal factor $\sqrt{a}$ would not be real.

An equivalent positive-parameter calculation substitutes $z=b\varphi$ in $E_0={\int }_0^{2\pi /b}\sqrt{2V}d\varphi$ and uses the provided integral of $\sqrt{1-\cos z}$. The following original plot shows that the localized energy density sits where the field passes between the vacua.