Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 308 / 1

Solution

Write

V (ϕ) = \frac{1}{2} (q^{2} - ϕ^{2})^{2} (p^{2} - ϕ^{2})^{2}, q > p > 0.

The four vacua are

- q, - p, p, q

. Hence there are three adjacent scalar-field kink sectors,

(- q, - p)

(- p, p)

and

(p, q)

, together with their three reversed antikinks. With

W^{'} (ϕ) = (q^{2} - ϕ^{2}) (p^{2} - ϕ^{2}), W (ϕ) = p^{2} q^{2} ϕ - \frac{p ^{2} + q ^{2}}{3} ϕ^{3} + \frac{1}{5} ϕ^{5},

E = ∣ W (ϕ_{+}) - W (ϕ_{-}) ∣

. The central kink obeys

ϕ^{'} = W^{'} (ϕ)

and has

E_{c} = \frac{4 p ^{3}}{15} (5 q^{2} - p^{2}) .

The two outer kinks obey

ϕ^{'} = - W^{'} (ϕ)

and have equal energy

E_{o} = \frac{2}{15} (q - p)^{3} (p^{2} + 3 pq + q^{2}) .

For the kink passing through zero, choose

ϕ (0) = 0

. Its profile is determined implicitly by

\frac{1}{q ^{2} - p ^{2}} [\frac{1}{p} artanh \frac{ϕ}{p} - \frac{1}{q} artanh \frac{ϕ}{q}] = x .

q = p

, the central equation becomes

ϕ^{'} = (p^{2} - ϕ^{2})^{2}

, so

\frac{ϕ}{2 p ^{2} ( p ^{2} - ϕ ^{2} )} + \frac{1}{2 p ^{3}} artanh \frac{ϕ}{p} = x .

For

q = p + ε

E_{o} = \frac{2}{3} p^{2} ε^{3} + O (ε^{4}), E_{c} = \frac{16}{15} p^{5} + O (ε),

so the leading ratio is

E_{o} : E_{c} : E_{o} = 1 : \frac{8}{5} (p / ε)^{3} : 1 .

There are no discussions about this article yet.

There are currently no matching articles.