Help with the Derrick scaling argument and topological solitons

[JackHolmes]

TL;DR Summary

Derrick scaling argument and topological soliton energy bound seem to contradict each other?

I have been reading Manton & Sutcliffe for some time now and can't quite wrap my head around something.

If you take the Hopf invariant N of a topological soliton ϕ then its Skyrme-Faddeev energy (which I hope I've gotten right up to some constants)

E=∫∂_iϕ⋅∂_iϕ+(∂_iϕ×∂_jϕ)⋅(∂_iϕ×∂_jϕ) d³x​

satisfies the bound E>c|N|^3/4 (as per page 423). This seems like a really nice result. (BTW ϕ here is a vector field in ℝ³)

However, using the idea behind Derrick's theorem, I can rescale x→μx for some μ∈ℝ. Then the energy for the re-scaled soliton is something like E(μ(x))∽μE(x). Re-scaling space doesn't affect the Hopf invariant though (since it's a topological property) so you can make E arbitrarily small without changing N just by re-scaling the topological soliton. So I don't see how the inequality can hold since E can be made arbitrarily small.

What is wrong with my intuition here?

 

[TeethWhitener]

Topological field theory isn’t really in my wheelhouse, but I think your answer comes on page 84-85 of Manton and Sutcliffe, where it explains that Derrick’s theorem doesn’t apply for theories with terms higher than quadratic in the derivatives of the field. Maybe someone with more experience in topology can give you a better answer.

Edit: this is, in fact, the correct explanation. Manton and Sutcliffe apply Derrick’s theorem to baby skyrmions on page 152-153 (in 2 dimensions, but an analogous argument goes through for 3D).

 

[TeethWhitener]

For posterity, in case folks with similar questions don’t have access to the book:
The scaling argument considers a finite scalar field $\phi(\mathbf{x})$ in $\mathbb{R}^d$ which is scaled by a factor of $\mu>0$ to give $\phi(\mu\mathbf{x})$. The scaled energy of the field is a function of $\mu$ in general. In particular, the energy for a field with a scalar potential will be:
$$\begin{align*}
E(\phi(\mu\mathbf{x})) &=\int{\left[\mathbf{\nabla}\phi(\mu\mathbf{x})\cdot\mathbf{\nabla}\phi(\mu\mathbf{x}) + U(\phi(\mu\mathbf{x}))\right]d^dx} \\
&=\int{\left[\mu^{2-d}\mathbf{\nabla}\phi(\mu\mathbf{x})\cdot\mathbf{\nabla}\phi(\mu\mathbf{x}) + \mu^{-d}U(\phi(\mu\mathbf{x}))\right]d^d(\mu x)}
\end{align*}$$
So for $d\geq2$, the energy will, in general, decrease monotonically with increasing $\mu$. Derrick’s theorem states that, since the scaled energy function has no stationary point, the only static solution of the field equation is the trivial (vacuum) solution. Therefore, there does not exist a stable topological soliton.

However, for the baby Skyrme model:
$$
E(\phi)=\int{\left[\mathbf{\nabla}\phi\cdot\mathbf{\nabla}\phi + (\mathbf{\nabla}\phi\times\mathbf{\nabla}\phi)\cdot(\mathbf{\nabla}\phi\times\mathbf{\nabla}\phi)\right]d^dx}
$$
Calling the first term of the integral $E_2$ and the second term $E_4$, we see that, in $\mathbb{R}^3$:
$$E=\mu^{-1}E_2+\mu E_4$$
which has a minimum at finite $\mu$, meaning that topologically non-trivial solutions to the field equation are not specifically ruled out by Derrick’s theorem.