For a vertically stratified accretion disk with surface density $\Sigma =\int \rho dz$, the density-weighted mean kinematic viscosity is $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$. In an alpha disk with $\nu =\alpha p/(\rho \Omega )$, this becomes $\bar{\nu }=\alpha (\Sigma \Omega {)}^{-1}\int pdz$.

// Target: astrophysics.bigb

On a connected $n$-dimensional pseudo-Riemannian manifold, evaluation of a Killing vector field and its first derivative is injective by Killing transport. The value has $n$ components and the Killing equation makes the derivative an antisymmetric two-tensor with $n(n-1)/2$ components. Their sum gives the bound. Flat metrics attain it through ${\xi }_a=A_a+B_{ab}{x}^b$ with $B_{ab}=-B_{ba}$. Connectedness is essential for a uniform global bound in terms of $n$: independent fields can be supported on different connected components.

A parallel commuting Dirac current is causal. If timelike, the geometry locally splits as a time line times a three-dimensional Ricci-flat metric; three-dimensional Ricci flatness implies flatness. If null, the stabilizer of a parallel spinor in $\mathrm{Spin}(1,3)$ contains only null rotations, and the local vacuum metric is $d{s}^2=-2dudv+d{x}^2+d{y}^2+H(u,x,y)d{u}^2$ with $H_{xx}+H_{yy}=0$. Constant spinors obeying ${\gamma }^{+}ϵ=0$ are parallel. These are local statements; global quotients must preserve a compatible spin structure.

For constant sectional curvature $-1/a^2$, spin curvature is $[{\nabla }_{\mu },{\nabla }_{\nu }]=-{\gamma }_{\mu \nu }/(2a^2)$. The added Clifford term in $\mathcal{D}$ has the opposite commutator, making $\mathcal{D}$ flat. On a simply connected region, parallel transport of any initial spinor is path independent and constructs solutions of $\mathcal{D}ϵ=0$. Thus Anti-de Sitter spacetime and its universal cover admit the associated Killing spinors locally and, with appropriate global spin data, globally.

A bifurcate Killing horizon consists of two intersecting null Killing horizons of the same Killing vector field, which vanishes on their spacelike intersection, the bifurcation surface. In regular Kruskal coordinates the horizons are $U=0$ and $V=0$, and their intersection is $U=V=0$. The displayed field is tangent and normal to each horizon branch and zero on the intersection.

In an axisymmetric stationary spacetime, the horizon angular velocity is the constant coefficient in the null horizon generator $\chi =k+{\Omega }_{H}m$, where $k$ is the stationary Killing vector normalized at infinity and $m$ generates axial orbits with period $2\pi$. It measures rotation relative to the stationary frame at infinity. For a Kerr black hole, ${\Omega }_H=a/(r_{+}^2+a^2)$. It enters the rotating physical-process first law through the horizon flux of $E-{\Omega }_{H}J$.

The outer-horizon surface gravity is $(r_{+}-r_{-})/(2(r_{+}^2+a^2))$. Hawking radiation or nondegenerate Euclidean black-hole regularity condition gives $T_H=\kappa /(2\pi )$ in units $G=\hslash =c=k_B=1$. The zero-temperature extremal value is the nonextremal limit; conical regularity alone does not fix a Euclidean period at a degenerate horizon.

Differentiate $M=(r_{+}^2+a^2+Q^2)/(2r_{+})$, $J=Ma$ and $A=4\pi (r_{+}^2+a^2)$. Eliminating $d{r}_{+}$ and $da$ gives the displayed relation, with ${\Omega }_H=a/(r_{+}^2+a^2)$ and ${\Phi }_H=Q{r}_{+}/(r_{+}^2+a^2)$. Matching Hawking temperature to this First law of black-hole mechanics fixes the Bekenstein-Hawking entropy as $A/4$.

Lense-Thirring precession is the rotation of orbital angular momentum caused by the gravitational field of a spinning mass. In the secular weak-field picture, the precession vector is parallel to the central spin. Thus an annulus has torque proportional to ${\mathbf{J}}_h\times \delta {\mathbf{J}}_d$, perpendicular to ${\mathbf{J}}_h$. Summing such torques preserves this orthogonality even if their radial coefficients differ. Dissipation and the internal response of a warped accretion disk can supplement precession with an alignment torque.

The nullity of a linear map $T:V\to W$ is $n(T)=\dim \ker T$, the dimension of its kernel. It counts independent input directions that can change without changing the output. For a finite-dimensional vector space $V$, the rank-nullity theorem givesFor an $m\times n$ matrix $A$ of matrix rank $r$, this is $n-r$, using the number of columns, not the number of rows. ^[ref]

The nullity answers a practical question about a linear system: once one input is known to produce the requested output, how much freedom remains? If $T(x_0)=b$, its fiber of a function over $b$ isIndeed, $T(x)=T(x_0)$ exactly when $T(x-x_0)=0$. Thus every nonempty fiber is an affine subspace of dimension $n(T)$, with the same direction space as the kernel. The kernel itself is the fiber over zero. Outputs outside the image of a linear map have empty fibers; the nullity does not guarantee existence, and we assign no solution-space dimension to an empty fiber here.

In the diagram, $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ has matrix $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$, matrix rank one and nullity one. Its kernel has basis $(1,-1)$, and the full solution of $F(x,y)=(c,0)$ is $(x,y)=(c,0)+s(1,-1)$. One free scalar moves the solution along a straight line; the output $(c,1)$ is impossible.

The existing three-input example $T(x,y,z)=(x+y,z)$ instead has rank two and nullity one. Its image of a linear map is all of ${\mathbb{R}}^2$, andEvery output has a solution straight line. In contrast, $U(x,y,z)=(x+y,0)$ has rank one and nullity two, with kernel basis $(1,-1,0),(0,0,1)$. Its reachable output $(a,0)$ has the solution affine subspacea plane, while $(a,b)$ with $b\ne 0$ has no solution. These two linear maps have the same input and output dimensions but different freedom and reachability. Computing the kernel separates what an output determines from what an extra constraint would have to fix. For either example, a basis of the kernel gives one independent free scalar per basis vector, which is why nullity measures freedom rather than the number of solutions.

When a measurement model is a linear map $T$, its kernel records changes that its readout cannot distinguish. For example, the rectangular kernel and solution-fibre example measures only $x+2y$ and $y+z$. Even exact readouts $(3,2)$ leave the whole fiber $u_0+\ker A$ possible. Repeating either readout with greater precision cannot distinguish those inputs: their outputs agree exactly. This ambiguity follows from a nontrivial kernel, rather than from measurement error.

Additional information can remove a null direction of a linear map. In that example, adding a third linear functional that reads $y$ distinguishes $u_0+tv$, because its third readout is $t$. More generally, additional linear functionals $c_1,\dots ,c_k$ give unique recovery of all inputs exactly whenThis follows by applying the trivial kernel criterion to the combined linear map $x\mapsto (Tx,c_1(x),\dots ,c_k(x))$. Merely adding repeated linear equations does not meet this condition. On a finite-dimensional input vector space, each added scalar readout can increase the matrix rank by at most one, so at least $\dim \ker T$ extra independent readouts are needed for recovery of unrestricted inputs.

Restrictions on the possible inputs can also make recovery unique. If the allowed inputs form a set $D$, two allowed inputs are indistinguishable exactly when their difference is in the kernel. Thus $T$ is injective on $D$ precisely when no difference of distinct points of $D$ is a kernel vector. Such a restriction is extra information about the input; the original readout alone determines only its fiber.

Consider the linear map $A:{\mathbb{R}}^3\to {\mathbb{R}}^2$ represented by the rectangular matrixTo compute its kernel, solve the homogeneous linear system $x+2y=0$, $y+z=0$. Taking $y=t$ gives $x=-2t$, $z=-t$, soThus $v$ represents a null direction of a linear map: decreasing the first input by two, increasing the second by one, and decreasing the third by one changes neither output. Solving the homogeneous linear system is the concrete computation behind a kernel; the column space instead decides which outputs are attainable.^[ref]

Now prescribe the nonzero output $b=(3,2{)}^T$. The linear system $Au=b$ becomes $x+2y=3$, $y+z=2$. Again take $y=t$:Direct substitution gives $A{u}_0=(3,2{)}^T$, and $A(u_0+tv)=A{u}_0+tAv=b$. Conversely, the two linear equations force exactly this form, so no solutions have been omitted. This fiber is an affine subspace of dimension one in the three-dimensional input vector space. It is parallel to the kernel, but does not pass through the origin. For instance, $u_0=(3,0,2{)}^T$ and $u_1=(1,1,1{)}^T$ both give $b$, while their difference $u_1-u_0=v$ gives zero. Adding $u_0$ and $u_1$ gives output $2b$, which is another way to see that this nonzero-output fiber is not a vector subspace.

The first and third columns of $A$ are $(1,0{)}^T$ and $(0,1{)}^T$, a basis of the output vector space. Hence $A$ has matrix rank two and reaches every output. The rank-nullity theorem predicts nullity $3-2=1$ before we solve any linear system. It provides a useful completeness check: one nonzero kernel vector already forms a basis here, so a second independent null direction of a linear map would signal a calculation error. In general, matrix rank $r$ for an $m\times n$ matrix leaves $n-r$ independent kernel directions, rather than $m-r$.^[ref]

A null direction of a linear map $T$ is represented by a nonzero vector $v$ in its kernel. Every scalar multiple $tv$ is an undetectable displacement: $T(x+tv)=T(x)$. The vector subspace $\mathrm{span}\{v\}$ describes the direction independently of the chosen nonzero representative. More generally, a basis of the kernel supplies all independent null directions of a linear map; their number is the nullity. The zero vector belongs to the kernel but supplies no direction.

A Hohmann transfer uses two tangential impulses and one half of an elliptic Kepler orbit to move between two coplanar circular orbits.

A zero-energy inverse-square orbit is parabolic and has eccentricity one.

For unit mass with inverse-square potential $-Q/r$, $Q>0$, and positive conserved angular momentum $h$, a bounded orbit has $-Q^2/(2h^2)\le E<0$. Its periapsis distance and apoapsis distance are roots of $|E|r^2-Qr+h^2/2=0$, so their sum is $Q/|E|$ and product $h^2/(2|E|)$. Speed is largest at the periapsis and smallest at the apoapsis, because $v^2=2(E+Q/r)$ decreases with $r$. Radial velocity is zero there, so the two speeds are $h/r_{\min }$ and $h/r_{\max }$. Their sum is $h(r_{\min }+r_{\max })/(r_{\min }{r}_{\max })=2Q/h$, independent of $E$, including the circular limit.

In axes directed along the periapsis and transverse to it, a Kepler orbit with gravitational parameter ${\mu }_g$ and specific angular momentum $h$ has the displayed velocity. The velocity tip traces a circle of radius ${\mu }_g/h$ centred at $(0,e{\mu }_g/h)$. To derive it, differentiate $r=h^2/[{\mu }_g(1+e\cos f)]$, use $r^2\dot{f}=h$, and resolve $\dot{r}\hat{\mathbf{r}}+r\dot{f}\hat{\mathbf{\theta }}$. A rotation of the spatial axes rotates the entire hodograph by the same angle.

For two kappa-filtrations at $\kappa ={\omega }_1$ exhausting sets of size ${\aleph }_1$, any bijection between the sets carries corresponding stages onto one another on a club set of indices. Bound both forward and inverse images at each countable stage by a function $g:{\omega }_1\to {\omega }_1$. The limit ordinals closed under all earlier $g$-values form a club; continuity gives equality at those ordinals.

For independently left-truncated and right-censored survival data, let $E_i$ and $Y_i$ be entry and exit times on the same clock. At each distinct event time $t_j$, the risk set is $R_j=\{i:E_i<t_j\le Y_i\}$, with size $r_j$, and $d_j$ events occur. The Kaplan–Meier estimator isBefore entry a subject contributes neither an event nor time at risk. Thus adding observed entrants can increase successive risk sets; pretending that all subjects were observed from time zero distorts the estimated hazard function. The estimator identifies the survival function over ages covered by observation; normalization from birth requires observation arbitrarily close to time zero or additional knowledge of earlier survival.

For the Kaplan–Meier estimator, the estimated variance isThe formula follows by summing approximate variances of the logarithms of the conditional survival factors and applying the delta method to their product. Its square root is the estimated standard error. The same expression uses the actual delayed-entry risk sets for the Kaplan–Meier estimator with delayed entry. When $0<\hat{S}(t)<1$ and every contributing $r_j>d_j$, an approximate log-scale confidence interval is $\exp \{\log \hat{S}(t)\pm z_{1-\alpha /2}\widehat{\mathrm{se}}(\hat{S}(t))/\hat{S}(t)\}$, with the upper endpoint capped at one. This interval can be inaccurate with very small risk sets.