$O(3)$ sigma model for lumps

Question

I'm studying the $O(3)$ $\sigma$-model related to lumps through chapter 6 of Manton's book.

There appears that $$\mathcal{L} = (1/4)\partial _{\mu}\phi \cdot \partial ^{\mu}\phi + \nu (1-\phi \cdot \phi)$$ (note: where $\phi = (\phi _{1},\phi _{2},\phi _{3})$ and the model is in (d+1) dimensional Minkowski space-time) and that the resulting E-L equation is, after eliminating $\nu$, $\partial _{\mu}\partial ^{\mu}\phi + (\partial _{\mu}\phi \cdot \partial ^{\mu}\phi)$ $\phi$ = 0. Could someone derive the expression for $\nu$ please? I'm seriously stuck on it.

On the other hand, where is derived $(\partial _{i}\phi \pm \epsilon _{ij}\phi \times \partial_{j}\phi) \cdot (\partial _{i}\phi \pm \epsilon _{ik}\phi \times \partial_{k}\phi) > 0$ from?

Answer 1

You start with the Lagrangian $$ \mathcal L = \frac 1 4 \partial_\mu\vec \phi \cdot \partial^\mu\vec \phi + \nu(1 - \vec \phi \cdot \vec \phi), $$ where $\vec \phi = (\phi_1, \phi_2, \phi_3)^T$ is the $O(3)$ triplet and $\nu$ is a Lagrange-multiplier to fix the constraint $|\vec\phi| = 1$. Using the EL-equation we arrive at $$ \frac 1 2 \partial_\mu\partial^\mu \vec\phi +2\nu\vec\phi = 0. \implies \nu = -\frac 1 4 \partial_\mu\partial^\mu \vec\phi \cdot \vec\phi $$ From $|\vec\phi| = 1$ we know (by repeated differentiation) $$ 0 = \partial_\mu \vec\phi \cdot \vec\phi + \vec\phi \cdot\partial_\mu \vec\phi \implies 2 \partial_\mu \partial^\mu\vec\phi \cdot \vec\phi + 2\partial_\mu\vec \phi \cdot \partial^\mu\vec \phi= 0 $$ $$ \partial_\mu \partial^\mu\vec\phi \cdot \vec\phi = -\partial_\mu\vec \phi \cdot \partial^\mu\vec \phi. $$ So inserting the above into $\nu$ $$ \nu = \frac 1 4 \partial_\mu\vec \phi \cdot \partial^\mu\vec \phi $$ and $\nu$ into the EoM $$ \frac 1 2 \partial_\mu\partial^\mu \vec\phi + \frac 2 4 (\partial_\mu\vec \phi \cdot \partial^\mu\vec \phi )\vec\phi = 0. $$ which results in $$ \partial_\mu\partial^\mu \vec\phi + (\partial_\mu\vec \phi \cdot \partial^\mu\vec \phi )\vec\phi = 0. $$ The full answer to your second question is given in the answer by mike stone.

Answer 2

The field $\nu(x,t) $ is a Lagrange multiplier enforcing the constraint ${\boldsymbol \phi}\cdot \boldsymbol \phi=1$. It takes whatever value it needs to do this. To find an expression for it, first find the EL equation for $\boldsymbol\phi$ including an arbitrary $\nu(x,t) $ Take the dot product with ${\boldsymbol \phi}$, use ${\boldsymbol \phi}\cdot \boldsymbol \phi=1$, and read off the expression for $\nu $.

As for your last question: the dot product of any real three vector with itself is $\ge 0$.