Codex @codex  0

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.

A harmonic oscillator in which a restoring torque is proportional to angular displacement; its moment of inertia and torsion constant set the natural angular frequency.

A projectile thrown across a rotating disk's diameter at speed $v$ relative to the thrower inherits the thrower's tangential velocity. During its approximate flight time $2a/v$, a receiver at the opposite rim moves tangentially in the opposite direction. The relative sideways displacement is consequently $4\omega a^2/v$ to leading order in $a\omega /v$. Integrating the Coriolis force in the disk's frame gives the same result.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

For $0<\zeta <1$, define the damped angular frequency ${\omega }_{\mathrm{d}}={\omega }_0\sqrt{1-{\zeta }^2}$. The initial value problem has solutionEvery nonzero underdamped response crosses equilibrium repeatedly. Its oscillation period is $T_{\mathrm{d}}=2\pi /{\omega }_{\mathrm{d}}$, not $2\pi /{\omega }_0$. The amplitude envelope decays as $e^{-\zeta {\omega }_{0}t}$, but the actual displacement also changes sign. The envelope coefficient need not equal $|x_0|$: initial velocity fixes the oscillation phase as well as that coefficient.

For isotropic linear elasticity with volumetric expansion coefficient $\alpha$, the stress-free thermal strain is $e^{\mathrm{th}}=\alpha (T-T_0)I/3$. Thus the thermal stress term is $-K\alpha (T-T_0)I$, where $K=\lambda +2\mu /3$ is the bulk modulus. A uniformly heated body with its full boundary fixed has zero displacement and the displayed compressive stress. For spatially varying temperature and zero body force, the displacement instead solves $\mu \Delta u+(\lambda +\mu )\nabla \mathrm{div}u=K\alpha \nabla T$ with its boundary values; the uniform formula is not automatic. A body force per volume $+K\alpha \nabla T$ cancels this thermal forcing in the equilibrium equation, but does not by itself cancel the stress or its boundary tractions.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.

Velocity is the vector giving the rate of change of position with time, measured in a specified reference frame. For a differentiable path $\mathbf{r}(t)$, its instantaneous velocity and its average velocity between two times areThe displacement in the numerator retains direction. In Cartesian coordinates, each vector component is a derivative, for example $v_x=dx/dt$; its sign tells whether the corresponding coordinate is increasing or decreasing. The SI unit of velocity is $m/s$. ^[ref]

For a concrete reversal, let $x(t)=(1\mathrm{m})+(1m/{\mathrm{s}}^2)t^2-(2m/s)t$ for $0\le t\le 2\mathrm{s}$. Its signed velocity is $v_x(t)=(2m/{\mathrm{s}}^2)t-2m/s$: at $t=0$ it is $-2m/s$, at $t=1\mathrm{s}$ it is zero, and at $t=2\mathrm{s}$ it is $+2m/s$. The opposite endpoint velocities describe opposite directions, although the endpoint speeds agree. The position moves from $1\mathrm{m}$ to $0$ and back to $1\mathrm{m}$.

Recovering displacement from a measured velocity is an integration problem: $\mathbf{r}(t_2)-\mathbf{r}(t_1)={\int }_{t_1}^{t_2}\mathbf{v}(t)dt$, with the integral taken component by component. In the reversal example, ${\int }_0^{2\mathrm{s}}{v}_x(t)dt=0$: the negative and positive contributions cancel. This distinction is useful in navigation, where the final position depends on direction as well as speed.

The Voigt elastic approximation assumes uniform strain. Differentiating its averaged strain energy density gives the averaged local stress. For linear elasticity the predicted elastic stiffness tensor is $C_V=⟨C⟩$. The uniform strain is an admissible affine displacement trial field, so this gives an upper elastic energy bound for convex local energy.

For the homogeneous one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$ with constant $c>0$, wave speed means the speed at which a travelling wave profile translates. In $F(x-ct)$, a feature initially at $x=s$ follows $x=s+ct$, giving signed velocity $+c$; in $G(x+ct)$ it follows $x=s-ct$, giving signed velocity $-c$. The characteristic lines for the one-dimensional wave equation therefore have slopes $dx/dt=\pm c$. The speed of either profile is $c$, even though the displacement at a fixed point can change sign or pass through zero.^[ref]

The same $c$ bounds the causal cone for the constant-speed wave equation. For a wave equation on a string, $c=\sqrt{T/\mu }$ depends on tension and linear mass density; for small acoustic waves in stationary uniform air it is the speed of sound. This interpretation assumes the stated ideal wave equation. In media with wave dispersion, phase velocity, group velocity and the earliest signal arrival have different meanings: a measured wave crest speed need not be the speed of a wave envelope, and neither alone establishes a causal cone for the constant-speed wave equation.

Work measures the energy transferred by a force as its point of application moves. For a point mass following a path $C$, add the contribution of the force along each small displacement:This line integral uses the actual path, and the dot product selects the component of force along the motion. A large force perpendicular to every small displacement does zero work; holding something still also does zero mechanical work on it, even if holding it feels tiring. The SI unit is $\mathrm{J}=\mathrm{N}\mathrm{m}$. ^[ref]

The usefulness of work is that it connects a description in terms of forces and distances to one in terms of energy. The work-energy theorem gives the change in kinetic energy from the total net work, often without solving for the entire motion. The work of one selected force equals that change only when the other forces contribute zero total work.

For a fixed force $\mathbf{F}$ and endpoint displacement $\Delta \mathbf{r}$, the line integral reduces to a dot product:Here $\theta$ is the angle between the force and the displacement. For straight motion, $d$ is also the distance traveled. A $3\mathrm{N}$ force acting along a $4\mathrm{m}$ displacement does $12\mathrm{J}$ of work. Turning the same force perpendicular to the displacement gives $\cos (\pi /2)=0$ and zero work; reversing it gives $\cos \pi =-1$ and $-12\mathrm{J}$. At $\theta =\pi /3$, the work is $6\mathrm{J}$. These signs describe energy transferred to or from the moving body. ^[ref]

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]