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For $V=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, $m^2<0$, the vacuum manifold is a sphere of radius $v=\sqrt{-m^2/\lambda }$. At a chosen vacuum the Hessian matrix is $2\lambda v^{2}n{n}^T$, where $n$ is its unit direction. The radial scalar has mass squared $2\lambda v^2$; the $n-1$ angular modes are physical Goldstone bosons when the symmetry is global.

For the two-component case, write $\mathbf{\varphi }=(v+\rho )(\cos \theta ,\sin \theta )$. The kinetic term becomes $\frac{1}{2}({\partial }_{\mu }\rho {)}^2+\frac{1}{2}(v+\rho {)}^2({\partial }_{\mu }\theta {)}^2$. At quadratic order the canonically normalized angular field is $\pi =v\theta$, with no mass term. Its Euler-Lagrange equation is the wave equation, so ${\omega }^2=|\mathbf{k}{|}^2$ in units where the propagation speed is one. The radial field instead obeys ${\omega }^2=|\mathbf{k}{|}^2+2\lambda v^2$. Thus a common rotation costs no energy, whereas a slowly varying rotation propagates as a collective wave. With the selecting source of continuous vacuum selection by an external source, use $\pi =w\theta$ and obtain ${\omega }^2=|\mathbf{k}{|}^2+h/w$. This elementary dispersion relation separates a Goldstone boson, a pseudo-Goldstone boson and a massive radial excitation within one example.

For equality-constrained optimization, restrict the Hessian matrix $H$ of the optimization Lagrangian to feasible tangent directions. If the columns of $Z$ form a basis of the nullspace of the constraint Jacobian matrix $A$, the resulting quadratic form has matrix $Z^{T}HZ$. Positive definiteness gives positive quadratic curvature in every nonzero tangent direction. Nonsingularity of this reduced matrix, together with full row rank of $A$, characterizes nonsingularity of the corresponding equality-constrained saddle-point matrix.

A Lagrangian is regular when its velocity Hessian matrix is invertible. The inverse function theorem then makes the fiber derivative $p_i=L_{v^i}$ locally invertible, allowing a local Legendre transform in mechanics to a Hamiltonian. If that fiber derivative is globally invertible, the Lagrangian is called a hyperregular Lagrangian. Regularity alone does not assert global invertibility, and singular Lagrangians may give constrained rather than ordinary unconstrained Hamilton's equations.

A saddle point of a scalar function is a stationary point with nearby values both greater and less than its value. An indefinite Hessian matrix proves this by choosing directions with opposite signs of the quadratic term. A degenerate Hessian matrix may require higher-order terms: $x^2-y^4$ has a saddle point of a scalar function at the origin despite a zero quadratic coefficient in the $y$ direction. This is distinct from a saddle equilibrium of a dynamical system.

The scalar mass matrix is the Hessian matrix of the scalar potential at a vacuum. Its eigenvalues are the tree-level squared masses; broken symmetry directions are zero eigenvectors before a Higgs mechanism or explicit symmetry breaking removes the corresponding massless modes.

For twice continuously differentiable coordinate functions $u,v$ and an outer scalar function $f$, apply the multivariable chain rule twice along an original coordinate $s$. The first three terms differentiate the outer function, while the last two differentiate the coordinate coefficients. They cannot be omitted for a nonlinear coordinate map. In matrix notation this is the Hessian matrix formula $H_{f\circ g}=J_g^T{H}_f{J}_g+f_u{H}_u+f_v{H}_v$ for $g=(u,v)$. The extra terms vanish for an affine map.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.

Let $f$ be a twice continuously differentiable function on an open domain stable under positive scaling, with $f(\lambda x)={\lambda }^{n}f(x)$. Apply the chain rule twice to the scalar function $\lambda \mapsto f(\lambda x)$ and set $\lambda =1$. The result isIn two variables, this reads $x^2{f}_{xx}+2xy{f}_{xy}+y^2{f}_{yy}=n(n-1)f$. In matrix notation it is $x^T({\nabla }^{2}f)x=n(n-1)f$, where ${\nabla }^{2}f$ is the Hessian matrix. Equivalently, the radial differential operator $D={\sum }_i{x}_i{\partial }_i$ obeys $D^{2}f=Df+{\sum }_{i,j}{x}_i{x}_j{\partial }_i{\partial }_{j}f$; substituting $Df=nf$ explains the factor $n(n-1)$.

The gradient of this polynomial vanishes exactly when $x=0,\pm \sqrt{a/2}$ and $y=0,\pm \sqrt{b/2}$, giving nine stationary points. Its diagonal Hessian matrix has entries $12x^2-2a$ and $2b-12y^2$. The second-derivative test therefore gives two strict local maxima at $(0,\pm \sqrt{b/2})$, with value $b^2/4$, and two strict local minima at $(\pm \sqrt{a/2},0)$, with value $-a^2/4$. The origin and the four points $(\pm \sqrt{a/2},\pm \sqrt{b/2})$ are saddle points of a scalar function, with values zero and $(b^2-a^2)/4$, respectively. Every stationary point is nondegenerate.

Completing the squares organizes the contour plot:The level set through the four outer saddle points of a scalar function factors asIt is a circle together with a hyperbola when $a\ne b$; for $a=b$, the latter factor's complementary zero set is the pair of lines $y=\pm x$. Neither the local maxima nor the local minima are global, since the restrictions to the two coordinate axes are unbounded in opposite directions. The case $a=1$, $b=4$ gives four outer saddle values $15/4$ and an exact circle-hyperbola contour crossing at all four outer saddle points of a scalar function.

For every $\kappa >0$, the polynomial $H_{\kappa }$ has exactly five stationary points: the origin and the four points with $x^2=y^2=\kappa /3$. Its gradient factors asIf a coordinate vanishes at a stationary point, both do. Otherwise put $u=x^2$, $v=y^2$ to obtain $u^2=(\kappa /3)v$ and $v^2=(\kappa /3)u$. Dividing gives $u=v$ and then $u=v=\kappa /3$.

At the origin, the Hessian matrix vanishes. Axis restrictions give $H_{\kappa }(t,0)=t^6>0$, while diagonal restrictions give $H_{\kappa }(t,\pm t)=2t^6-\kappa t^4<0$ for small nonzero $t$. Thus the origin is a degenerate saddle point of a scalar function. Its leading quartic term is nonpositive, showing why the sign of the first nonzero homogeneous term alone need not settle the problem when that term vanishes along some directions.

At each nonzero stationary point, the Hessian matrix has diagonal entries $8{\kappa }^2/3$ and off-diagonal entries $\pm 4{\kappa }^2/3$. Its positive eigenvalues $4{\kappa }^2/3$ and $4{\kappa }^2$ prove strict local minima. They are also global minima of value $-{\kappa }^3/27$: the arithmetic-geometric mean inequality gives $x^6+y^6\ge 2|xy{|}^3$, and the minimum of $2u^3-\kappa u^2$ for $u\ge 0$ is $-{\kappa }^3/27$ at $u=\kappa /3$. Equality requires $|x|=|y|$.

At a nondegenerate critical point placed at the origin, integral Taylor theorem gives $f(x)-f(0)=x^{T}A(x)x$, where $A(x)$ is a smooth symmetric matrix and $A(0)=\frac{1}{2}\mathrm{Hess}f(0)$. A constant linear change makes $A(0)$ diagonal with entries $\pm 1$. Successive completion of squares, with the Schur complement at each step, produces a smooth invertible triangular matrix $B(x)$ such that $x^{T}A(x)x={\sum }_i{ϵ}_i(B(x)x{)}_i^2$. The pivots keep their nonzero signs near zero. The map $x\mapsto B(x)x$ has invertible derivative at zero, so the inverse function theorem makes it a coordinate system. The signs are the Hessian matrix signature by Sylvester's law of inertia.

For a graph $z=x+i\nabla f$ in ${\mathbb{C}}^3$, symmetry of the Hessian matrix makes the graph Lagrangian. Pulling back $\Omega =d{z}^1\wedge d{z}^2\wedge d{z}^3$ gives $\det (I+i{D}^{2}f)$. Its imaginary part is $\mathrm{tr}{D}^{2}f-\det D^{2}f$, while its real part is $1-{\sigma }_2(D^{2}f)$. Vanishing of the imaginary part and positivity of the real part give the displayed equation and oriented branch. The equation is nonlinear even though its linearization near a flat graph is the Laplace equation.

// Target: string-theory.bigb

At an equilibrium, a positive-definite Hessian matrix of the potential gives stable small oscillations. If a continuous symmetry supplies zero modes, positive definiteness on the directions transverse to the symmetry gives stability of the relative configuration.

For a twice continuously differentiable real function, define the Newton map in optimization $N(p)=p-H(p{)}^{-1}\nabla f(p)$ wherever its Hessian matrix $H$ is invertible. Every stationary point in this domain is a fixed point, because its gradient is zero. This includes a local maximum and a saddle point of a scalar function, as well as a local minimum. Thus Newton's method in optimization needs a separate minimum classification or a method for ensuring descent when it is used for minimization problem.

For example, $f(x,y)=(x+y^2{)}^2+(x^2+y{)}^2$ has a stationary point at $(-1/2,-1/2)$, with Hessian matrixIts eigenvalues are $-1$ and $7$, so it is a saddle point of a scalar function, but starting Newton's method in optimization there gives a constant sequence. Its value is $1/8$, whereas the global minimum zero is attained at $(0,0)$ and $(-1,-1)$.

The gradient vanishes only when $y=0$ and $\tan x=-2x$. On each tangent branch, $\tan x+2x$ increases strictly from negative to positive infinity, so it has exactly one root $x_s$. With $z_s=e^{-x_s^2}\cos x_s$, the Hessian matrix is diagonal with entries $-(4x_s^2+3)z_s$ and $-2z_s$. Its determinant is positive, so every stationary point is a strict maximum when $z_s>0$ or a strict minimum when $z_s<0$. Signs alternate between successive tangent branches. This gives infinitely many nondegenerate extrema under a decaying Gaussian function envelope.

Suppose $h$ is differentiable and $F^{\prime }>0$ on its range. The multivariable chain rule shows that $F\circ h$ has exactly the same stationary points as $h$. Since $F$ is strictly increasing, the type of every local maximum, local minimum, and saddle point is preserved. If both functions are twice differentiable, then at a stationary pointso the Hessian matrix signature is unchanged. The hypothesis $F^{\prime }>0$ matters for the converse about stationary points: a strictly increasing function such as $F(s)=s^3$ can have a zero derivative and create new stationary points in a composition.

Subtracting the quadratic Taylor polynomial at the rescaling centre makes the function, gradient and Hessian matrix vanish there. Dividing by $A{r}^{2+\mu }$ preserves the normalized Hölder oscillation of the Hessian matrix while turning a small forcing-to-oscillation ratio into a harmonic limiting equation. The three vanished Taylor coefficients provide local bounds needed by the Arzelà-Ascoli theorem.

At an equilibrium ${\mathbf{r}}_{\ast }$, the total force vanishes. A small transverse displacement must produce a transverse restoring force; a small axial displacement must also restore the bead. Downstream of a tight focus, the backward optical gradient force can oppose the forward optical scattering force. Equality at one point is insufficient: the total axial force must decrease as the bead moves downstream through that point. The equilibrium can therefore lie downstream of the intensity maximum.

For uncoupled overdamped coordinates, ${\zeta }_i\dot{\delta r_i}=-k_i\delta r_i$ with ${\zeta }_i>0$ decays only when each $k_i>0$. More generally, linearizing gives $\dot{\delta \mathbf{r}}=MJ\delta \mathbf{r}$, where $M$ is the hydrodynamic mobility matrix and $J_{ij}=\partial F_i/\partial r_j$ at equilibrium. All eigenvalues of $MJ$ must have negative real parts for linear asymptotic stability. For a conservative force with symmetric positive-definite mobility, a positive-definite Hessian matrix of the potential energy supplies this condition. A general optical force need not possess a global potential energy.

Increasing numerical aperture admits more strongly converging light rays and a steeper axial variation of the field. But a larger nominal numerical aperture alone does not establish trapping: illumination of the objective pupil, optical aberration, particle properties and the actual in-liquid focus also matter. A harmonic optical trap describes local confinement after this stability question has been answered; it does not predict capture from arbitrary positions or an infinite holding force.

The acceleration tidal tensor is $T_{jk}={\partial }_k{g}_j=-{\partial }_j{\partial }_k\Phi$ for gravitational acceleration $\mathbf{g}=-\nabla \Phi$. It describes differential acceleration $\Delta \mathbf{g}\simeq T\Delta \mathbf{r}$. Some authors instead call the positive Hessian matrix ${\partial }_j{\partial }_k\Phi$ the tidal tensor, reversing this sign convention. For a point mass and the positive potential convention $\mathbf{g}=\nabla \psi$, the tensor is $Gm(3\mathbf{R}{\mathbf{R}}^{\mathsf{T}}/R^5-1/R^3)$.

The tidal tensor is the Jacobian matrix of the Newtonian gravitational field, rather than the gravitational acceleration itself. Its entries have units ${\mathrm{s}}^{-2}$: multiplying by a displacement produces an acceleration. For a smooth Newtonian gravitational potential, the tidal tensor is a symmetric matrix, since mixed partial derivatives commute. A spatially uniform gravitational acceleration has zero tidal tensor, however large that acceleration is.

Tidal torque theory explains the early angular momentum of a protohalo through the misalignment of its mass second-moment tensor $I_{ij}$ and the Hessian matrix $T_{ij}={\partial }_i{\partial }_j\Phi$ of the surrounding peculiar gravitational potential. At leading order,