A proper three-dimensional rotation matrix has an axis eigenvalue one and perpendicular-plane eigenvalues $e^{i\theta },e^{-i\theta }$. Their sum gives the displayed trace identity. It determines the angle cosine but does not by itself choose the oriented axis or angle sign.

Apply $R_1$ to a column coordinate vector, then apply $R_2$ in the same fixed basis. The function composition isThus the rightmost matrix acts first. The product is again a rotation matrix, since $(R_2{R}_1{)}^T(R_2{R}_1)=I$ and $\det (R_2{R}_1)=1$. For planar rotations about the same origin, the sine addition formula and cosine addition formula giveFor a concrete composition of rotation matrices, first rotate $(2,0{)}^T$ by $\pi /3$ radians, reaching $(1,\sqrt{3}{)}^T$, then by $\pi /6$ radians:The total angle is $\pi /2$. Reversing these two planar rotations gives the same answer, but noncommuting rotations about fixed axes show why one cannot assume that property for rotations in three dimensions.^[ref]

Rotate an orthonormal basis counterclockwise by $\theta$ while keeping the vector $v$ fixed. The new basis vectors, written in the old basis, are the columns of $P=R_{\theta }$. The change-of-basis matrix convention therefore givesFor example, the fixed vector with old coordinate vector $(2,0{)}^T$ has new coordinate vector $(1,-\sqrt{3}{)}^T$ after the basis turns by $\pi /3$ radians. Its negative second coordinate means that the arrow lies below the new first axis. By contrast, an active planar rotation through that same angle moves the arrow to old coordinates $(1,\sqrt{3}{)}^T$. The sign difference follows from which object moves. The passive component formulas are $x^{\prime }=x\cos \theta +y\sin \theta$, $y^{\prime }=-x\sin \theta +y\cos \theta$.^[ref]

If coordinate vectors are instead written as rows, transposing the active equation gives $(v^{\prime }{)}^T=v^T{R}_{\theta }^T$. Switching to rows changes the side and the transpose; it does not by itself turn an active planar rotation into passive rotation of coordinates.

For homogeneous, incompressible, inviscid flow with constant rotation vector and negligible inertia, taking the curl of $2\mathbf{\Omega }\times \mathbf{u}=-\nabla p$ gives $(\mathbf{\Omega }\cdot \nabla )\mathbf{u}=0$. Thus a geostrophic flow is independent of distance along the rotation axis. Body forces modify the relation by their curl; it applies unchanged in unforced exterior regions. A nontrivial exterior column cannot both obey this leading balance and decay along the rotation axis.

For a convective star, the activity convention compares its rotation period with a convective turnover time. It differs by fixed factors from the fluid Rossby number estimated from speed, length and angular velocity. Small ${\mathrm{Ro}}_{\ast }$ denotes strong rotational influence on convection. Magnetic activity usually rises as this ratio decreases in the unsaturated regime, but levels off in a saturated rapid-rotation regime.

A vertex whose parent in a rooted tree is the specified vertex.

For a nonroot vertex, its adjacent predecessor on the unique graph path from the root of a tree. The root of a tree has no parent.

The distinguished vertex that specifies a rooted tree. Its choice determines the parent in a rooted tree and children in a rooted tree relationships.

With $m$ the characteristic-prime-free part of $n$, all roots of unity over a finite field lie in ${\mathbb{F}}_{q^s}$ exactly when $m\mid q^s-1$. This follows from the cyclicity of the multiplicative group of a finite field. For $m=1$ the least positive degree is one. A primitive root of order $m$ has the same field degree, because its Frobenius orbit closes exactly when $m\mid q^s-1$.

A round body of radius $R_0$ in a fixed cylinder of radius $R_1$ has center of mass path radius $a=R_1-R_0$. With orbital angle $\theta$ and spin angular velocity $\Omega$ both positive about the same axis, rolling without slipping gives $a\dot{\theta }+R_0\Omega =0$. For moment of inertia $I=\gamma m{R}_0^2$, its kinetic energy is $\frac{1}{2}m{a}^2(1+\gamma ){\dot{\theta }}^2$. Release from rest at angle $\alpha$ gives, by conservation of energy, a descent time $\sqrt{1+\gamma }$ times that of frictionless sliding along the same center of mass path. The spin sign reverses if its positive direction is chosen opposite to the orbital angle.

For a uniform solid cylinder, take $s$ positive uphill and $\omega$ positive for uphill rolling without slipping. The velocity of the contact point is $w=\dot{s}-a\omega$. Uphill kinetic friction gives $\ddot{s}=-g(\sin \alpha -\mu \cos \alpha )$ and $a\dot{\omega }=-2\mu g\cos \alpha$. Starting from $\dot{s}(0)=a\omega (0)$, one obtains $w=-gt(\sin \alpha -3\mu \cos \alpha )$. Thus $\tan \alpha >3\mu$ makes the assumed uphill kinetic friction self-consistent for the entire ascent and descent. Without slip the required uphill static friction is $mg\sin \alpha /3$; it is available exactly when $\tan \alpha \le 3\mu$ under the same limiting coefficient.

Let the flat face make angle $\theta$ with horizontal while the rounded face contacts a horizontal plane. The centroid has height $a-(3a/8)\cos \theta$ and contact distance squared $b^2=a^2+(3a/8{)}^2-2a(3a/8)\cos \theta$. For rolling without slipping, its speed is $b|\dot{\theta }|$. Combining translation and rotation gives kinetic energy $\frac{1}{2}(I_G+m{b}^2){\dot{\theta }}^2$. Conservation from rest at $\theta =\pi /2$ gives the displayed result until the flat-face/contact geometry changes.

For a rigid body rolling from rest down a straight incline with fixed rolling radius $R$ and axial moment of inertia $I$, translation and rotation give $(M+I/R^2)\ddot{s}=Mg\sin \theta$. A smooth slide has $\ddot{s}=g\sin \theta$. Travel over the same distance from rest therefore gives the displayed time ratio. Static friction must suffice for rolling without slipping; its contact point is at rest, so this ideal friction does no work. A uniform solid cylinder has $I=M{R}^2/2$ and ratio $\sqrt{3/2}$.

A body ejecting mass at constant rate $Q>0$ and relative exhaust speed $V>0$, against quadratic drag $\alpha u^2$, has remaining mass $m=m_0-Qt$. A short-interval momentum balance gives $m\dot{u}=QV-\alpha u^2$. From rest, separation of variables givesFor final mass $m_f>0$, this formula applies until $t=(m_0-m_f)/Q$. The zero-drag limit is the rocket equation $u_f=V\log (m_0/m_f)$. Two dimensionless variables are $\alpha V/Q$ and $m_0/m_f$.

For fuel mass rate $s>0$ and backward relative exhaust speed $u$, a rocket sliding on a horizontal plane has remaining mass $q(t)$ and satisfies the displayed rocket equation with kinetic friction. Integration over a sliding interval gives $v(t)-v(t_0)=u\log [q(t_0)/q(t)]-\mu g(t-t_0)$. From initial rest, immediate forward sliding additionally requires thrust to overcome static friction. If both friction coefficients equal $\mu$, motion starts only when $su>\mu gq(t)$. An algebraically negative sliding speed is not the physical solution at rest.

Rocket thrust is the propulsive force associated with ejecting rocket propellant. In the ideal rocket equation with no pressure correction, $T=u(-\dot{M})$, where $\dot{M}=dM/dt<0$ is the signed rate of change of rocket mass and $-\dot{M}$ is the positive mass loss rate. For $u=2000m/s$ and $\dot{M}=-5kg/s$, $T=10000\mathrm{N}$. With no other force, the acceleration is $T/M$: initially $10m/{\mathrm{s}}^2$ for $M=1000\mathrm{kg}$, rising as mass falls if this rocket thrust stays constant.

A real exhaust outlet can also contribute a pressure term: $T=u(-\dot{M})+A_e(p_e-p_a)$, with exit area $A_e$, exit pressure $p_e$ and ambient pressure $p_a$. The ideal formula here assumes that term vanishes; it must not silently absorb this correction while still interpreting $u$ as the actual relative rocket exhaust speed. ^[ref]

The rocket mass ratio compares the initial and final total rocket masses for a burn. It is dimensionless, and the ideal rocket equation becomes $\Delta v=u\ln R$. For the same fixed rocket exhaust speed, two successive burns with rocket mass ratios $R_1$ and $R_2$ have combined rocket mass ratio $R_1{R}_2$ and total velocity gain $u\ln R_1+u\ln R_2$.

Rocket propellant is the onboard material expelled to produce rocket thrust. A chemical rocket normally carries both fuel and oxidizer; both contribute to its mass and to the expelled material. In the ideal rocket equation, $M(t)=M_{\mathrm{dry}}+M_{\mathrm{propellant}}(t)$, with fixed rocket dry mass and decreasing remaining rocket propellant. ^[ref]

The rocket exhaust speed is the magnitude of the emitted material's velocity relative to the instantaneous rocket. In the ideal rearward-ejection convention of the rocket equation, a rocket with inertial velocity $v$ emits material with inertial velocity $v-u$ to first order. A constant $u$ does not make $v-u$ constant.

A valuative criterion tests a morphism of schemes $f:X\to Y$ by lifting maps from a valuation ring. Given a valuation ring $R$ with fraction field $K$, a map $\mathrm{Spec}R\to Y$, and a compatible map $\mathrm{Spec}K\to X$, ask whether the latter extends to $\mathrm{Spec}R\to X$. For finite-type morphisms between Noetherian schemes, uniqueness of every such lift characterizes separatedness; existence together with uniqueness characterizes proper morphisms. The test treats the generic point as a moving point and asks whether it has a unique limiting position. A projective morphism satisfies the existence-and-uniqueness property; an open immersion can fail existence when the limit leaves the open set.