= Solution
<Goldstone's theorem> states that a spontaneously broken continuous internal <global symmetry in field theory> of a local relativistic <quantum field theory> produces massless <Goldstone bosons>. For a <Lorentz invariance> <vacuum> breaking $G$ to its stabilizer $H$, there are $\dim G-\dim H$ independent symmetry-required massless directions. Additional accidentally massless modes are possible. A gauge redundancy does not obey this physical-state counting, and nonrelativistic systems need not have one independent mode per broken generator.
For the classical proof, consider real <scalar fields> $\phi_i$ with a positive canonical <kinetic term> and an invariant smooth <scalar potential> $V$. An infinitesimal generator $T^a$ satisfies
$$
\frac{\partial V}{\partial\phi_i}(T^a\phi)_i=0.
$$
Let $v$ be a stationary <vacuum> and differentiate this identity with respect to $\phi_j$. At $v$, the term proportional to $\partial_iV$ vanishes, leaving
$$
\boxed{\mathsf M^2_{ji}(T^av)_i=0,\qquad \mathsf M^2_{ji}=\left.\frac{\partial^2V}{\partial\phi_j\partial\phi_i}\right|_v.}
$$
The <Hessian matrix> $\mathsf M^2$ is the <scalar mass matrix>. An unbroken generator has $T^av=0$; broken generators span the tangent space of the <vacuum manifold> $G/H$. Each independent nonzero tangent vector is a zero <eigenvector> of $\mathsf M^2$. The quadratic <Euler-Lagrange equations> in these directions are $\Box\pi^a=0$, giving the massless <dispersion relation> $\omega^2=\mathbf p^2$. This proves the classical result and explains why the <Goldstone bosons> describe motion along the degenerate minima rather than against a restoring curvature.
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