Differentiating the potential gives
$$
U'(\phi)=2(1-\phi^2)(\epsilon-\phi).
$$
Thus $\phi=\pm1$ are stationary points, and
$$
U''(1)=4(1-\epsilon)>0,
\qquad
U''(-1)=4(1+\epsilon)>0
$$
because $|\epsilon|<1$. Both are local minima. Their energies are
$$
U(1)=\frac{4\epsilon}{3},
\qquad
U(-1)=-\frac{4\epsilon}{3}.
$$

For $\epsilon=0$, the static energy can be completed to a square:
$$
E=\int dx\left\{\frac12[\phi'-(1-\phi^2)]^2
+\phi'(1-\phi^2)\right\}.
$$
The increasing <scalar-field kink> therefore obeys the first-order <Bogomolny equation>
$$
\boxed{\phi'=1-\phi^2}.
$$
With center $X$, its solution and energy are
$$
\boxed{\phi_K(x)=\tanh(x-X)},
\qquad
\boxed{E_K=\int_{-1}^{1}(1-\phi^2)\,d\phi=\frac43}.
$$
The antikink uses the opposite sign.

For small positive $\epsilon$, the true vacuum $\phi=-1$ lies below the false vacuum $\phi=1$ by
$$
\Delta U=U(1)-U(-1)=\frac{8\epsilon}{3}.
$$
This pressure exerts force $\Delta U$ on a kink with $-1$ on its left and $+1$ on its right. Dividing by its leading mass $4/3$ gives acceleration toward the false-vacuum side:
$$
\boxed{\ddot X=2\epsilon+O(\epsilon^2)}.
$$

An antikink followed by a kink encloses a region of the lower vacuum while approaching $\phi=1$ at both infinities. Vacuum pressure pushes the pair apart, whereas their attraction pulls them together. At a static separation $s$,
$$
\frac{8\epsilon}{3}=32e^{-2s},
$$
so
$$
\boxed{s\simeq\frac12\log\frac{12}{\epsilon}}.
$$
This estimate is self-consistent for $\epsilon\ll1$, when the two soliton cores are well separated.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 308 / 1 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-308/1/solution