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Constant velocity depends on the reference frame. For nonrotating axes and a moving origin of laboratory position $\mathbf{R}(t)$, subtracting that origin givesIf $\mathbf{R}(t)={\mathbf{R}}_0+\mathbf{U}t$ with constant velocity $\mathbf{U}$, the Galilean transformation leaves acceleration unchanged. The cart moving at $1.5m/s$ in the laboratory is at rest in a frame moving right at $1.5m/s$. Both descriptions obey Newton's first law.

Instead choose an origin with $R_x(t)=\frac{1}{2}A{t}^2$, where the stipulated constant $A=0.50m/{\mathrm{s}}^2$ points right. The same cart hasAt $t=3\mathrm{s}$, the cart is instantaneously at rest relative to this accelerating origin, but its relative acceleration is nonzero. In this translating accelerated reference frame, Newton's second law must include the effective inertial force $-mA=-1.0\mathrm{N}$. This term accounts for the accelerating coordinates; the actual external forces on the cart still sum to zero. No rotation-dependent terms are present because the axes remain parallel to the laboratory axes.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

In thermal-time units, taking the Prandtl number large removes momentum inertia at leading order. The Stokes flow momentum equation then determines velocity instantaneously from the temperature perturbation, while the nonlinear heat equation retains its time derivative. Small Prandtl number does not generally give this model. An exact contrasting shear in a free-slip layer is $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$: it decays slowly as the Prandtl number tends to zero but is excluded by the instantaneous Stokes equation.

A fluid displacement measured from an initial position must vanish at the initial time. Integrating a harmonic velocity therefore introduces phase-dependent constants. Mean-orbit labels absorb those constants and leave purely oscillatory displacements. The instantaneous displacement correction to velocity can differ under the two reference conventions, but its period mean gives the same leading Stokes drift.

Using the extensional screening length of a slumping column, an initially uniform column of length $\Lambda$ obeys $W_{ZZ}=1+W$, $W(0)=0$ and $W_Z(\Lambda )=0$. Thus $W=\cosh (\Lambda -Z)/\cosh \Lambda -1$ and its initial area-fraction growth is ${\varphi }_T={\varphi }_0\sinh (\Lambda -Z)/\cosh \Lambda$. For $\Lambda \gg 1$, the velocity is $e^{-Z}-1$ outside exponentially small end corrections; for $\Lambda \ll 1$, it is $Z^2/2-\Lambda Z$. The first limit balances excess weight with annular hydraulic drag, the second with extensional viscous stress.

Instantaneous power is power at a particular moment. Consider a point mass with mass $m=2\mathrm{kg}$, initially at rest, acted on by a constant net force $F=3\mathrm{N}$ along a straight path. Newton's second law gives constant acceleration $a=F/m=1.5m/{\mathrm{s}}^2$, hence velocity $v(t)=at$ andThe instantaneous power starts at zero even though the force is nonzero, because the initial velocity is zero. At distance $4\mathrm{m}$, the kinetic energy example below gives speed $\sqrt{12}m/s$, so the final instantaneous power is $3\sqrt{12}\mathrm{W}\approx 10.3923\mathrm{W}$. Negative instantaneous power means the selected force is removing energy at that moment.

An integral adds up contributions of a function across a domain. For a continuous function $f$ on $[a,b]$, the Riemann integral is the common limit of its Riemann sums as the largest subinterval width tends to zero:Each product combines a value with the width over which it acts. When $f$ is nonnegative, the result is the area beneath its graph; values below the horizontal axis contribute with the opposite sign. The signed area under a graph can therefore vanish even when the graph encloses nonzero geometric area.

For a velocity $v(t)=1+2t$ on $0\le t\le 3$, use $n$ equal time steps. The left and right Riemann sums areBecause the function is increasing, they bound the integral. With six steps they give $10.5$ and $13.5$; with sixty steps they give $11.85$ and $12.15$. Both approach the same total, $12$. This is accumulation made exact by refining finite measurements.

An integral retains units: velocity times time gives displacement, and a probability density function integrated over an interval gives a probability. Weighting a density by a function gives its expectation. The same operation serves different subjects because it expresses a common pattern, value multiplied by extent, then accumulated.

The fundamental theorem of calculus often evaluates a definite integral by finding an antiderivative, but a finite sum or a numerical approximation is still useful when no convenient antiderivative is available. An indefinite integral denotes a family of antiderivatives; a definite integral denotes an accumulated value between specified endpoints. More general constructions, including Lebesgue integration, extend the class of functions and domains that can be handled.

For a fixed collection of two constant-mass bodies in an inertial frame, separate their mutual forces from the resultant external force on each:Adding the two Newton's second law equations and using Newton's third law givesThis is the proof of internal-force cancellation in system momentum. For more bodies, each internal force cancels with its partner once both appear in the sum. A boundary that excludes one interacting body leaves its force in the external force sum. A massive connecting spring must also be included, or its own changing momentum accounted for; its two endpoint forces are not automatically one third-law interaction pair between the carts.

Zero resultant external force gives momentum conservation; complete absence of external forces is sufficient but unnecessary. For constant total mass $M$, the centre of mass has velocity $\mathbf{V}=\mathbf{P}/M$ and obeys $Md\mathbf{V}/dt={\mathbf{F}}_{\mathrm{ext}}$. Internal interactions can move the parts apart while leaving the centre of mass moving uniformly.

Additional external forces do not invalidate the identified mutual pair. At the $4\mathrm{N}$ instant above, suppose a hand also exerts $+3\mathrm{N}$ on $A$. The mutual forces remain $-4\mathrm{N}$ and $+4\mathrm{N}$, but the net forces become $-1\mathrm{N}$ on $A$ and $+4\mathrm{N}$ on $B$. Consequently $a_A=-1m/{\mathrm{s}}^2$, $a_B=+2m/{\mathrm{s}}^2$, $dP/dt=+3\mathrm{N}$, and the centre of mass has acceleration $+1m/{\mathrm{s}}^2$. Pair equality by itself does not promise constant total momentum.

At a step in buoyancy frequency without a jump in background mass density, vertical velocity and fluid pressure are continuous. For a fixed horizontal wavenumber and angular frequency, these are continuity of $W$ and $W^{\prime }$. They determine reflection and transmission between propagating internal gravity waves and evanescent waves.

A steady galactic distribution function depending only on relative energy $E=\psi -v^2/2$ is isotropic and has zero mean velocity. If its one-dimensional velocity dispersion is constant, the Spherical Jeans equation reduces to ${\sigma }_0^2{\rho }^{\prime }=\rho {\psi }^{\prime }$, yielding the displayed density. This is a tracer relation in a specified gravitational potential, not a requirement that the tracer generate that potential. With no escape-energy cutoff its local velocity distribution is Maxwell-Boltzmann velocity distribution.

Without an escape-energy cutoff, a galactic distribution function proportional to $\exp [(\Psi -v^2/2)/{\sigma }^2]$ gives $\rho \propto e^{\Psi /{\sigma }^2}$ and a constant one-dimensional velocity dispersion. Since $\Psi ={\Phi }_0-\Phi$, differentiation gives the displayed hydrostatic relation. An ideal gas at constant temperature has pressure $P=\rho k_{B}T/m_0$, so its hydrostatic equilibrium equation is identical at the displayed dispersion-temperature identification. Coupling either relation to the same Poisson equation for Newtonian gravity, with matched central density and boundary data, therefore gives the same density structure. Truncating the stellar velocities at escape changes this equivalence.

Starting from rest, compose $n$ positive collinear Lorentz boosts, each with the same finite rapidity $\alpha >0$. Each step has relative velocity $w=c\tanh \alpha$, measured in the preceding rest inertial frame. The relativistic velocity-addition formula givesThese are equal increments of rapidity, not equal increments of velocity measured in the original inertial frame. They can describe a chain of relatively moving inertial frames or successive idealized changes of a massive particle's motion; no particular acceleration schedule is assumed. ^[ref].

For a relative velocity $w=c/2$ at each step,The first four velocities are $u_0=0$, $u_1=c/2$, $u_2=4c/5$, and $u_3=13c/14$. At step eight, $u_8/c=3280/3281\approx 0.99969521$. The rapidities grow by the same amount at each step, while the successive velocity increments become smaller.

A short derivation applies the hyperbolic-function identity for $\tanh (a+b)$ to the recurrence, giving ${\theta }_{n+1}={\theta }_n+\alpha$. Substituting $\tanh x=(e^{2x}-1)/(e^{2x}+1)$ and $e^{2\alpha }=3$ gives the displayed closed expression. In particular,for every finite nonnegative integer $n$. The limit is $u_n\to c$ as $n\to \infty$, but no finite sequence reaches the speed of light. This illustrates the beauty of rapidity: unbounded additive parameters coexist with bounded physical velocities. Replacing every step by a negative rapidity gives the corresponding approach to $-c$.