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On a one-dimensional uniform finite element mesh of spacing $h$, let $w$ be an affine function and ${\varphi }_i$ the interior piecewise-linear hat functions. The weighted mass matrix has entriesand zero entries for nonadjacent indices. The diagonal product ${\varphi }_i^2$ is symmetric about $x_i$ and integrates to $2h/3$; the neighbouring product is symmetric about $(x_i+x_{i+1})/2$ and integrates to $h/6$. The odd part of an affine function integrates to zero around each centre, giving the formulas. This evaluates a variable reaction term exactly without substituting a constant coefficient or a quadrature approximation.

A breathing mode is a collective normal mode in which a structure expands and contracts radially while preserving its overall symmetry. For a uniform scaling pattern, the displacement of point $i$ is proportional to its position relative to the center of mass, as displayed above. In a central spring framework this pattern is a normal mode only when applying the stiffness matrix returns a multiple of the same pattern with the appropriate mass matrix; radial motion alone does not establish an eigenvector.

For point masses joined by unstretched central springs obeying Hooke's law, let $n_{ij}$ be the equilibrium unit vector along an edge and $u_i$ the small displacement of its endpoint. The leading extension is $n_{ij}\cdot (u_j-u_i)$, so its quadratic elastic potential energy is $k_{ij}[n_{ij}\cdot (u_j-u_i){]}^2/2$. Consequently the block stiffness matrix has off-diagonal blocks $K_{ij}=-k_{ij}{n}_{ij}{n}_{ij}^T$ and diagonal blocks $K_{ii}={\sum }_j{k}_{ij}{n}_{ij}{n}_{ij}^T$, with absent edges contributing zero. It is symmetric and positive semidefinite. The normal modes solve $Kq={\omega }^{2}Mq$, where $M$ is the mass matrix. Rigid translations and infinitesimal rotations are rigid-motion zero modes of a spring framework. A prestressed spring can add transverse quadratic terms, so the unstretched assumption matters.

Coupled normal-contact penalty modes are small oscillations generated by several simultaneously compressed penalty contacts acting through shared body inertia. A displacement at one contact can accelerate another because both load the same translating or rotating bodies. Separate scalar effective mass in a normal-contact penalty mode estimates do not generally give the coupled frequencies.

Freeze a contact configuration, use independent generalized displacement perturbations $q$, positive-definite mass matrix $M$, and gap perturbations $\xi =Jq$. Let $K$ be a positive diagonal matrix of normal penalty stiffnesses. After subtracting equilibrium loads, the restoring generalized force is $-J^{\mathsf{T}}KJq$, soThe contact-Jacobian force/velocity pairing and stiff compliant-contact model come from the primary multibody formulation; the local modal reduction here follows by substitution.^[ref]

Although $AK$ need not be symmetric, it is similar to the symmetric positive-semidefinite matrix $K^{1/2}A{K}^{1/2}$. Set $z=K^{1/2}\xi$ to obtainIts positive eigenvalues are squared angular frequencies. A zero eigenvalue may describe redundant gap coordinates outside the realizable displacement subspace; it should not automatically be interpreted as an extra physical oscillation. Free generalized motions satisfying $Jq=0$ are a separate possibility.

For an original planar example, a rigid bar has vertical translation $y$, small rotation $\theta$, mass $m$, center-of-mass moment of inertia $I$, and two vertical contacts at horizontal offsets $-a,+a$. ThenWith equal contact stiffness $k$, the equal-gap mode is vertical translation and the opposite-gap mode is rocking. Their angular frequencies areTake $m=2\mathrm{kg}$, $I=0.005\mathrm{kg}{\mathrm{m}}^2$, $a=0.10\mathrm{m}$ and $k=10^{5}N/m$. Then $A$ has diagonal entries $2.5\mathrm{k}{\mathrm{g}}^{-1}$ and off-diagonal entries $-1.5\mathrm{k}{\mathrm{g}}^{-1}$. Its eigenvalues are $1$ and $4\mathrm{k}{\mathrm{g}}^{-1}$, giving $316.2$ and $632.5rad/s$. Treating the contacts independently would assign each $500rad/s$ and miss the faster rocking mode.

Contact-mode timestep stability depends on the largest active frequency and on the integrator. For the undamped central-difference recurrence, one mode has amplification equationThe distinct oscillatory roots lie on the unit circle for $0<h\omega <2$; the endpoints need separate treatment. The example therefore requires $h<2/632.5\approx 3.162\mathrm{ms}$ for this linear stability condition. Resolving the waveform accurately normally requires a smaller step. This is not a stability rule for every integration method.

The derivation applies while both contacts remain compressed and geometry is locally constant. A sufficiently large rocking amplitude can unload one contact, changing the active model. A proposed check preloads both supports, applies small symmetric and antisymmetric perturbations, and compares measured or simulated frequencies with the two predictions. Repeat with smaller amplitudes and smaller timesteps to separate linearization error, contact opening and numerical error.

The damped quadratic eigenvalue problem finds exponential motions of a linear mechanical model with mass matrix $M$, damping matrix $C$ and stiffness matrix $K$. Substituting $u(t)=r{e}^{\lambda t}$ into $M\ddot{u}+C\dot{u}+Ku=0$ givesThe dependence on $\lambda$ is quadratic, so solving the undamped problem $Kr={\omega }^{2}Mr$ alone does not find the damped decay rates and mode shapes.^[ref]

A useful linearization sets $z=(r,\lambda r{)}^T$ and rewrites the problem asThis generalized eigenvalue problem has twice the displacement dimension. With invertible $M$, it becomes an ordinary first-order state eigenproblem by multiplication by the block mass inverse. Singular $M$ requires more care and cannot be handled by silently using $M^{-1}$.

For an original one-coordinate example in consistent units, take $M=1$, $C=2$ and $K=5$. Then ${\lambda }^2+2\lambda +5=0$, giving $\lambda =-1\pm 2i$. Real free motions have the formThe decay rate is $1$ and damped oscillation frequency is $2$, while the undamped natural frequency is $\sqrt{5}$. Increasing $C$ to6 instead gives real roots $-1,-5$, showing that the quadratic problem also describes nonoscillatory decay.

For real symmetric positive-definite $M,K$ and positive-semidefinite $C$, an oscillatory eigenpair satisfies a useful energy check. Multiply by the conjugate transpose $r^{\ast }$ and write $m=r^{\ast }Mr>0$, $c=r^{\ast }Cr\ge 0$, $k=r^{\ast }Kr>0$. The imaginary part of $m{\lambda }^2+c\lambda +k=0$ gives $\mathrm{Re}\lambda =-c/(2m)\le 0$ when $\mathrm{Im}\lambda \ne 0$. In a multi-coordinate system, complex eigenvectors can encode relative phase differences between coordinates; nonclassical damping generally prevents replacing them by the undamped real mode shapes.

A damping matrix $C$ maps generalized velocity to the velocity-dependent resisting-force term in a linear structural dynamics model:Here $M$ is the mass matrix and $K$ is the stiffness matrix. The sign convention places $C\dot{u}$ on the left, so the actual resisting force is $-C\dot{u}$. For pure translational coordinates, entries have units $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$; mixed rotational and translational coordinates require their corresponding generalized units.

Consider an original two-coordinate model with three viscous dampers: one from each coordinate to a fixed support, and a third between them. With positive coefficients $c_1,c_2,c_{12}$,For $\dot{u}=(v_1,v_2{)}^T$, the removed mechanical power isThis makes $C$ a positive semidefinite matrix. In particular, the connecting damper dissipates nothing when both coordinates move at the same speed. Choosing $c_1=c_2=10$ and $c_{12}=5$ in newton-seconds per metre gives $C=\left(\begin{array}{cc}15& -5\\ -5& 15\end{array}\right)$. Speeds $(1,0)\mathrm{m}{\mathrm{s}}^{-1}$ dissipate15watts, while $(1,1)\mathrm{m}{\mathrm{s}}^{-1}$ dissipate20watts.

For symmetric constant $M,K$, multiplying the equation by ${\dot{u}}^T$ gives an energy balance:The matrix therefore encodes where dissipation occurs and which relative motions activate it.

Rayleigh damping approximates $C$ by a weighted combination of the mass and stiffness matrices. OpenSees exposes separate coefficients for mass and different stiffness states, and cautions that this model can give incorrect results in some nonlinear concentrated-plasticity analyses.^[ref] A convenient matrix prescription must still represent the intended dissipation mechanism.

The effective mass in a normal-contact penalty mode is the inertia seen by one specified relative normal displacement, rather than automatically the total mass of either body. A contact away from a center of mass can excite rotation as well as translation, changing that inertia and the normal contact penalty time scale.

Use independent generalized velocities $v$, a positive-definite mass matrix $M$, and a normal gap $g$ whose local velocity is $\dot{g}=Jv$. A scalar normal force $\lambda$ produces generalized force $J^{\mathsf{T}}\lambda$. The primary multibody formulation gives this contact-Jacobian velocity/force pairing explicitly.^[ref] At a frozen configuration, the part of gap acceleration caused by that force isIf the scalar $a=J{M}^{-1}{J}^{\mathsf{T}}$ is positive, defineThe definition uses a physical normal gap measured in metres and a matching normal force measured in newtons. Arbitrarily rescaling the gap coordinate changes the associated inertia. Constraints must first be accounted for by independent coordinates or the appropriate constrained response.

A short normal impulse $P$ gives $\Delta v=M^{-1}{J}^{\mathsf{T}}P$, hence $\Delta \dot{g}=P/m_{\mathrm{eff}}$. This supplies a direct interpretation: less effective mass means a larger relative normal-velocity change for the same impulse. At a fixed local configuration, a compressed linear penalty spring of stiffness $k$ gives the small-motion estimateIt omits changing contact geometry, other driving forces and damping; the source separately explains why stiff compliant contact challenges numerical integration.^[ref]

For two freely translating particles, the relative gap sees $a=1/m_1+1/m_2$, so $m_{\mathrm{eff}}=m_1{m}_2/(m_1+m_2)$. This is the reduced mass. Equal $2\mathrm{kg}$ particles give $m_{\mathrm{eff}}=1\mathrm{kg}$; replacing one by a fixed wall gives $2\mathrm{kg}$ instead.

For a planar rigid body above a fixed plane, let its mass be $m$, rotational moment of inertia about its center be $I$, and the contact normal be vertical. A contact a horizontal distance $r$ from the center hasFor $m=2\mathrm{kg}$, $I=0.020\mathrm{kg}{\mathrm{m}}^2$ and $r=0.10\mathrm{m}$, $m_{\mathrm{eff}}=1\mathrm{kg}$. At the center it is $2\mathrm{kg}$. With $k=10^{5}N/m$, the off-center angular frequency is $316.2rad/s$, compared with $223.6rad/s$ at the center. Using the full mass in both cases misses this rotational response.

Several simultaneous contacts require more than separate scalar replacements. Their matrix $A=J{M}^{-1}{J}^{\mathsf{T}}$ generally has off-diagonal terms: a force at one contact changes another contact’s acceleration. With frozen geometry and positive diagonal stiffness matrix $K$, compressed gap perturbations satisfy $\ddot{\xi }+AK\xi =0$; its nonzero modal squared frequencies are those of the symmetric matrix $K^{1/2}A{K}^{1/2}$. Coupled normal-contact penalty modes can therefore be faster than estimates treating each contact independently. A proposed verification applies a small known impulse or displacement perturbation at a stated configuration and checks relative contact motion, followed by a timestep convergence study.

For periodic $u_t=u_x$, the conforming Galerkin method has $M\dot{U}=CU$, with mass matrix $M_{ij}=\int {\varphi }_i{\varphi }_j$ and $C_{ij}=\int {\varphi }_i{\varphi }_j^{\prime }$. Integration by parts and periodicity give $C^T=-C$. ThereforeThe conserved quantity is the squared L2 norm of the finite element function. On a uniform one-dimensional mesh of spacing $h$, the piecewise-linear hat functions give $M_{ii}=2h/3$, $M_{i,i\pm 1}=h/6$, $C_{i,i+1}=1/2$ and $C_{i,i-1}=-1/2$.

Without fixed gauge charges or interactions distinguishing particles from antiparticles, $N$ free Dirac fields give $2N$ left-handed Weyl fields after charge conjugation of the right-handed fields. Their massless kinetic term has unitary group $U(2N)$. Equal nonzero Dirac masses have equal positive singular values and can be written as $2N$ equal Majorana mass terms; their preserving group is the orthogonal group $O(2N)$. Explicitly, in a basis with symmetric mass matrix $m1$, invariance requires $V^{†}V=V^{T}V=1$, hence real $V$. Restricting to transformations preserving a chosen common fermion number gives the usual $U(N{)}_L\times U(N{)}_R$ at zero mass and $U(N)$ at nonzero mass.

A finite element mesh partitions a domain into simple regions called elements, such as intervals, triangles or tetrahedra. In a conforming partition, distinct element interiors are disjoint, their closures cover the domain closure, and any nonempty intersection is a common face, edge or vertex of both elements. Chen's author notes give this compatibility condition for simplicial meshes.^[ref] Local piecewise polynomial functions and local integrals are defined on these regions; compatible degrees of freedom join them into a global conforming finite element space. The assembled stiffness matrix and mass matrix then inherit the local overlap pattern.

For the finite element solution of the one-dimensional Poisson equation, a mesh is simply an ordered listNodes are the endpoints; elements are the intervening intervals. Four equal elements have five nodes but only three unknown nodal values after imposing the zero Dirichlet boundary conditions. A nonuniform partition such as $0,1/8,1/4,1/2,1$ has element lengths $1/8,1/8,1/4,1/2$ and $h=1/2$, so counting elements alone does not measure the largest unresolved scale.

Refining a finite element mesh replaces some or all elements by smaller ones. For continuous piecewise linear functions in one dimension, retaining all old nodes gives nested conforming finite element spaces $V_h\subset V_{h/2}$ when every interval is bisected. The old approximation remains available, and the new interior nodes allow additional changes of slope. With a symmetric coercive bilinear form and exact integration, best approximation from Galerkin orthogonality means that the energy norm error cannot increase under such nested refinement.

For the finite element solution of the one-dimensional Poisson equation, uniform refinement from four to eight elements increases the unknown count from three to seven. The exact formulas give maximum errors $0.0078125$ and $0.001953125$, and energy norm errors approximately $0.0721688$ and $0.0360844$. Refinement near a region of strong curvature can spend unknowns where the linear interpolation error matters most; convergence in a general problem still needs approximation and regularity assumptions such as those in the finite element interpolation estimate.

The Hessian of potential energy at a mechanical equilibrium, expressed in specified displacement coordinates. It determines restoring forces to first order. Combining it with the mass matrix gives the dynamical matrix whose eigenvalues are squared mode angular frequencies.

For a uniform piecewise-linear Galerkin heat discretization, the consistent mass matrix is $M=(h/6)\mathrm{tridiag}(1,4,1)$ and the stiffness matrix is $K=h^{-1}\mathrm{tridiag}(-1,2,-1)$. The generalized eigenvalues are $6(1-\cos \theta )/[h^2(2+\cos \theta )]$. Applying forward Euler method requires $k\lambda \le 2$, giving the displayed uniform bound for $\mu =k/h^2$. For $J$ intervals with Dirichlet endpoints the exact finite-grid upper bound is $(2-\cos (\pi /J))/[3(1+\cos (\pi /J))]$, which approaches $1/6$. Replacing the consistent mass matrix by a lumped one changes the stability restriction.

In a Hermitian neutral-kaon mass matrix, independently choosing square-root signs of $z$ and $z^{\ast }$ can interchange the eigenvalue labels. Choose $z=r{e}^{i\phi }$, $a=\sqrt{r}{e}^{i\phi /2}$ and $b=\sqrt{r}{e}^{-i\phi /2}$, so $ab=r>0$, continuously from the desired CP-conserving limit. Then $(a,-b)$ is the lower-mass eigenvector. At a degenerate off-diagonal zero the mass eigenbasis is not fixed by diagonalization alone.

The L2 inner product is the inner product $⟨f,g⟩=\int f\overline{g}d\mu$ on L2 space. Its induced norm is the L2 norm, and the mass matrix represents it in a finite element basis.

A mass eigenstate is a quantum state with a definite mass. For a self-adjoint operator $M$ representing this quantum observable, it satisfies $M|n⟩=m_n|n⟩$; a measurement of that observable in the normalized state gives $m_n$ with probability one.

In a flavor eigenstate basis, the matrix representing the mass observable can have off-diagonal entries. Diagonalizing that matrix identifies the mass eigenstates as linear combinations of the flavor eigenstates. For a finite-dimensional self-adjoint operator, these states can be chosen to form an orthonormal eigenbasis. Propagation of unstable particles can instead be modeled by an effective complex matrix; its eigenvectors need not have the orthogonality of a self-adjoint observable. The numerical finite-element mass matrix is a different construction: an L2 inner product matrix used in discretization.

For a finite conservative linear mechanical system $M\ddot{q}+Kq=0$ with real symmetric positive definite mass matrix $M$ and real symmetric stiffness matrix $K$, the normal modes satisfy $K{u}_i={\omega }_i^{2}M{u}_i$. Taking the scalar product with $u_j$ and using symmetry in the two equations gives $({\omega }_i^2-{\omega }_j^2)u_i^{T}M{u}_j=0$. Distinct-frequency modes are therefore orthogonal in the inner product defined by $M$, even if the dynamical matrix $M^{-1}K$ is not symmetric in the ordinary Euclidean inner product. Within a degenerate mode space one can choose an $M$-orthogonal basis.

Mechanical dynamic condensation eliminates selected coordinates from a frequency-dependent mechanical response while retaining their effect on the remaining coordinates. It is exact algebraic elimination at the chosen transform variable when the eliminated block is invertible. It is not equivalent to clamping those coordinates. Starting from the partitioned mechanical dynamic-stiffness matrix,solve the internal equation as ${\mathbf{X}}_i={\mathbf{D}}_{ii}^{-1}({\mathbf{F}}_i-{\mathbf{D}}_{ir}{\mathbf{X}}_r)$. Substitution givesThe effective matrix is a Schur complement; the right side is the condensed mechanical load vector. Both matter when the internal coordinate is externally forced. The coupled dynamic equations and measurement boundary conditions are the starting model, rather than independent scalar stiffness reciprocals.^[ref]

For the two-mass example at $\omega =5rad/s$, take $D_{11}=D_{22}=175N/m$ and $D_{12}=D_{21}=-100N/m$. With $F_1=0$ and $F_2=1\mathrm{N}$, eliminating coordinate2 gives effective stiffness $117.8571N/m$ and effective load $100/175\mathrm{N}$. Hence $X_1=100/20625\mathrm{m}\approx 4.848\mathrm{mm}$. Omitting the load correction would incorrectly predict no response at coordinate1.

For constant original mass matrix, damping and mechanical stiffness matrix, the condensed matrix is generally a rational function of frequency, not simply a smaller constant mass–damping–stiffness triple. A low-frequency approximation requires its own justification. The elimination can also become singular at an internal-block frequency even when the complete system matrix remains invertible. For example, $\left(\begin{array}{cc}0& -k_c\\ -k_c& 0\end{array}\right)$ has invertible full matrix for $k_c\ne 0$, but its lower-right scalar block cannot be inverted. One must then retain that coordinate or choose a different elimination.

A mechanical dynamic-stiffness matrix maps displacement coordinates to applied forces in a linear structural dynamics model at a specified transform variable or frequency. Starting with constant mass matrix $\mathbf{M}$, damping matrix $\mathbf{C}$ and mechanical stiffness matrix $\mathbf{K}$,zero initial conditions give $\mathbf{F}(s)=\mathbf{D}(s)\mathbf{X}(s)$ with $\mathbf{D}=s^2\mathbf{M}+s\mathbf{C}+\mathbf{K}$. For the $e^{i\omega t}$ convention, $\mathbf{D}(i\omega )=\mathbf{K}-{\omega }^2\mathbf{M}+i\omega \mathbf{C}$. Its matrix inverse, when it exists, is the complete mechanical receptance matrix. Models and measured responses must use matching coordinates and boundary conditions.^[ref]

Consider two $1\mathrm{kg}$ masses, each grounded by a $100N/m$ spring and joined by another $100N/m$ spring, with no damping. At angular frequency $5rad/s$,A force amplitude $F_1=1\mathrm{N}$ with $F_2=0$ gives $X_1=175/20625\mathrm{m}\approx 8.485\mathrm{mm}$ and $X_2=100/20625\mathrm{m}\approx 4.848\mathrm{mm}$. Clamping coordinate2 instead gives $X_1=1/175\mathrm{m}\approx 5.714\mathrm{mm}$, because the clamp supplies a reaction force.

Partitioning retained and internal coordinates gives mechanical dynamic condensation. With ${\mathbf{F}}_i=0$ and invertible ${\mathbf{D}}_{ii}$,The effective matrix is a Schur complement, generally frequency dependent even when the original mass matrix, damping and mechanical stiffness matrix are constant. In the example its scalar value is $175-100^2/175\approx 117.857N/m$, rather than175. A reduced force-free internal coordinate and a clamped coordinate describe different experiments. At a singular block, that particular elimination is unavailable; it does not establish that the full mechanical system lacks a valid solution.

With CPT symmetry, a Hermitian kaon mass matrix is $M=\left(\begin{array}{cc}{r}_0& z\\ z^{\ast }& r_0\end{array}\right)$. Its eigenvalues are $r_0\pm |z|$. For correlated roots $a^2=z$, $b^2=z^{\ast }$ and $ab=|z|$, the normalized lower and upper eigenvectors are $(a,-b)/\sqrt{2|z|}$ and $(a,b)/\sqrt{2|z|}$. They are orthogonal. This is a mass-only result; including decay requires the generally non-Hermitian effective matrix $M-i\Gamma /2$.

Nonclassical damping occurs when a linear mechanical system's damping matrix cannot be diagonalized together with its mass matrix and stiffness matrix by an undamped normal mode transformation. Consequently the damping terms couple the modal equations even though their inertia and stiffness terms are separated. This can matter when a localized dissipative element is attached to a structure whose elastic and inertial responses are distributed differently.^[ref]

For $M\ddot{u}+C\dot{u}+Ku=0$, take a complete mass-normalized undamped mode matrix $\Phi$ with ${\Phi }^{T}M\Phi =I$ and ${\Phi }^{T}K\Phi ={\Omega }^2$. Then $u=\Phi q$ givesWith distinct undamped natural frequencies, a nonzero off-diagonal $D_{ij}$ shows modal damping coupling. If a frequency is repeated, first allow a rotation of the modes within its repeated-frequency subspace: an off-diagonal entry in one arbitrary basis is not by itself proof of nonclassical damping. For symmetric $M,K,C$ with $M$ positive definite, the basis-independent condition for classical decoupling isFailure of this condition establishes nonclassical damping under those assumptions.^[ref]

For an original example in consistent normalized units, letThe damping acts only on the first physical coordinate. The undamped mode vectors $(1,1)/\sqrt{2}$ and $(1,-1)/\sqrt{2}$ giveThus an initial modal state $q=(1,0{)}^T$, $\dot{q}=(1,0{)}^T$ has ${\ddot{q}}_2=-0.2$. Dropping the off-diagonal damping terms would incorrectly predict zero initial acceleration in the second modal coordinate. Nevertheless ${\dot{u}}^{T}C\dot{u}=0.4{\dot{u}}_1^2\ge 0$: passive dissipation does not require modal decoupling.

The damped quadratic eigenvalue problem follows from $u(t)=r{e}^{\lambda t}$:In oscillatory cases its complex vectors can encode different phases at different physical coordinates. One can solve the coupled equations directly or use an appropriate damped-mode method; assigning separate scalar decay equations to the undamped modes is an additional approximation, not a consequence of positive dissipation.