Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 47 / 2 / i

Solution

Take

r > 0

and

m > 0

, the positive-energy sigma-model regime, and keep

θ = 0

for the energy calculation. For a static field the energy is

E = \frac{r}{4} \int_{R} [ϕ_{x}^{2} + sin^{2} ϕ α_{x}^{2} + m^{2} sin^{2} ϕ] d x .

For either sign

s = \pm 1

, complete the square:

E = \frac{r}{4} \int [(ϕ_{x} - s m sin ϕ)^{2} + sin^{2} ϕ α_{x}^{2}] d x + \frac{r m s}{2} \int ϕ_{x} sin ϕ d x = \frac{r}{4} \int [(ϕ_{x} - s m sin ϕ)^{2} + sin^{2} ϕ α_{x}^{2}] d x + r m s T .

Here

\int ϕ_{x} sin ϕ d x = - [cos ϕ]_{- \infty}^{+ \infty} = 2 T

. Choose

s

to have the sign of

T

. The Bogomolny bound is therefore

E \geq r m ∣ T ∣.

Saturation requires

ϕ_{x} = s m sin ϕ

and

sin ϕ α_{x} = 0

. Finite-energy vacua have

sin ϕ = 0

, so their endpoint cosine values are

\pm 1

and this particular topological charge is

T \in {- 1, 0, 1}

r

would instead make the derivative energy unbounded below, so positivity is an essential physical convention.

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