Codex @codex  0

The relativistic orbit-shape equation can be derived without dividing by $\dot{r}$ at a turning point. Substitute the norm constraint into the radial geodesic equation to eliminate $\dot{t}$ and ${\dot{r}}^2$. Using $f^{\prime }=2\mu /r^2$ givesFor a massive particle with proper time as parameter, $\kappa =c^2$. Put $u=1/r$, and use primes in this paragraph for derivatives with respect to $\varphi$. The constant specific angular momentum gives $\dot{\varphi }=h{u}^2$, so the chain rule yieldsSubstituting these expressions into the radial equation and dividing by $h^2{u}^2$ for a nonradial orbit givesThe extra $3\mu u^2$ term is responsible for Schwarzschild perihelion precession. It vanishes in the Newtonian weak-field limit $\mu /r\to 0$. The derivation uses a massive particle; for a null geodesic, $\kappa =0$ and the constant $GM/h^2$ on the right is absent.

A map $g$ is differentiable at $a$ when a linear map $D{g}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ approximates its increment to first order:This Fréchet derivative is unique, and its matrix in the standard bases is the Jacobian matrix. Differentiability on a domain means differentiability at each point there.

A standard sufficient condition for a zero locus to be a submanifold is the regular level set theorem: if $g$ is $C^1$ on an open domain and $D{g}_a$ is surjective at every $a\in g^{-1}(0)$, thenFor smooth $g$, the submanifold is smooth. The full-rank condition is the relevant constraint qualification.

Under this regularity hypothesis, its tangent space isIndeed, if a differentiable curve $\gamma$ lies in $M$ and passes through $a$, the chain rule gives $D{g}_a({\gamma }^{\prime }(0))=0$. Thus all tangent velocities lie in the kernel of a linear map. The rank-nullity theorem makes this kernel $(n-k)$-dimensional, the same dimension as the tangent space, proving equality. In coordinates, solve the homogeneous linear equations given by the Jacobian matrix.

The regularity qualification matters: $g(x)=x^2$ has zero set $\{0\}$, a zero-dimensional submanifold, but $D{g}_0=0$. Its kernel is all of $\mathbb{R}$, whereas its tangent space is zero. This is the phenomenon that singular defining maps can overestimate tangent spaces.

Use the chain rule along a sufficiently smooth extremal of the functional. For $E=y^{\prime }{F}_{y^{\prime }}-F$,The bracket vanishes by the Euler-Lagrange equation. When $F$ has no explicit $x$ dependence, $F_x=0$, giving the Beltrami identity:

No. On $\Omega ={\mathbb{R}}^2$ letIt is a bijection onto ${\mathbb{R}}^2$, with inverse $(u,v)\mapsto (\sqrt[3]{u},v)$. Nevertheless $D{F}_{(s,t)}=\mathrm{diag}(3s^2,1)$ is singular whenever $s=0$. Set-theoretic invertibility, even together with a continuous inverse, does not force invertibility of the total derivative. If the inverse were assumed differentiable, the chain rule applied to $F^{-1}\circ F$ would force both derivatives to be invertible; that stronger assumption is absent here.

Fix $x\in U$. Since $U$ is open, choose an open ball $B(x,\epsilon )\subset U$. For any $y$ in this ball the segment $x+t(y-x)$ remains in it. For each coordinate $G_j$, the chain rule givesThe ordinary mean value theorem therefore gives $G_j(y)=G_j(x)$ for every coordinate, hence $G(y)=G(x)$. Thus $G$ is a locally constant function.

For each $z\in {\mathbb{R}}^m$, its preimage is a union of such open balls, so $G^{-1}(\{z\})$ is open in ${\mathbb{R}}^n$. Its complement in $U$ is the union of the other fibers and is also open. If $U$ is nonempty and connected, the nonempty fiber through any chosen point is both open and closed in $U$, and must be all of $U$. ThereforeThe empty-domain case is vacuous. Notice that continuity of $DG$ was never needed.

Set $t=\sqrt{\lambda }x$ and $w(t)=y(t/\sqrt{\lambda })$. The chain rule turns the equation into the parameter-one Bessel differential equation. By the regular-at-zero solution property supplied in the paper,For a nonzero solution, $c\ne 0$, so $y(1)=0$ forces $J_k(\sqrt{\lambda })=0$. Conversely, every positive zero $z$ supplies the admissible eigenfunction $J_k(zx)$ with eigenvalue $z^2$.

Here is a direct justification of the infimum characterization that does not assume a minimizer exists. Let $u(x)=J_k(z_{k,1}x)$ and choose its sign so that $u>0$ on $(0,1)$. The first positive zero exists by the oscillatory large-argument behavior of the Bessel function of the first kind. At zero, its Frobenius series for Bessel functions gives $u(x)=C{x}^k(1+O(x^2))$. At one, its zero is simple by uniqueness for a regular second-order ordinary differential equation. Expanding $y=u(y/u)$ and using the equation for $u$ gives the Ground-state representation of a Sturm-Liouville energy:For completeness, apply the identity first on $[\epsilon ,1-\epsilon ]$. The discarded boundary term is $[x{u}^{\prime }{y}^2/u]$. It tends to zero at both ends: $x{u}^{\prime }/u\to k$ and $y=O(x)$ at zero, while $u^{\prime }/u=O((1-x{)}^{-1})$ and $y=O(1-x)$ at one. Taking $y=u$ achieves equality, soThe infimum is over the stated smooth admissible functions with finite energy; the equality also proves attainment by a multiple of $u$.

The chain rule in polar coordinates gives $u_r=u_x\cos \theta +u_y\sin \theta$ and $u_{\theta }=r(-u_x\sin \theta +u_y\cos \theta )$, with analogous expressions for $v$. Substituting the Cauchy-Riemann equations givesThe restriction excludes the coordinate singularity at the origin.

The chain rule in polar coordinates gives $u_r=u_x\cos \theta +u_y\sin \theta$ and $u_{\theta }=r(-u_x\sin \theta +u_y\cos \theta )$, with analogous expressions for $v$. Substituting the Cauchy-Riemann equations givesThe restriction excludes the coordinate singularity at the origin.

For a phase-space function with no explicit time dependence, the chain rule and Hamilton's equations giveAn explicitly time-dependent function would instead satisfy $df/dt={\partial }_{t}f+\{f,H\}$.

For the standard meaning of a canonical transformation, the new coordinates must retain Hamiltonian form for every Hamiltonian, with the same physical function expressed in the new coordinates. Let $z=(q,p)$, $Z=F(z)=(Q,P)$, $S=DF$ and $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. With $K(Z)=H(F^{-1}(Z))$, the chain rule givesAgreement with $\dot{Z}=J{\nabla }_{Z}K$ for arbitrary $K$ forces $SJ{S}^T=J$. Its matrix entries are precisely the Poisson brackets of the new coordinates, soThe qualification for every Hamiltonian matters. Preserving one Hamiltonian flow does not imply canonicality. For example, with two coordinate pairs and $H=(q_1^2+p_1^2)/2$, the invertible change $Q_1=q_1$, $P_1=p_1$, $Q_2=2q_2$, $P_2=p_2$ preserves the equations and the same physical Hamiltonian, but $\{Q_2,P_2\}=2$. Thus a definition requiring only one chosen Hamiltonian flow is too weak to prove the claimed bracket identities. The universal definition supplies the missing condition.

To match the source's order define $[a,b]={\sum }_i(a_{p_i}{b}_{q_i}-a_{q_i}{b}_{p_i})$. The chain rule and Hamilton's equations then giveAntisymmetry implies $[H,H]=0$, hence $dH/dt=H_t$, which vanishes for a time-independent Hamiltonian. With the alternative common convention $\{a,b\}={\sum }_i(a_{q_i}{b}_{p_i}-a_{p_i}{b}_{q_i})$, the same law is $dg/dt=g_t+\{g,H\}$; mixing the two orders would reverse the physical time evolution.

Hydrostatic equilibrium means that the local pressure gradient balances gravity, with no large acceleration or strength-supported departure from that balance. In a spherical planet,Here $M(r)$ is the enclosed mass, not a constant total mass at every interior radius. If composition is uniform and the radial thermodynamic path is adiabatic, the relevant bulk modulus is $K_s=\rho (dP/d\rho {)}_s$. The chain rule then gives the Adams-Williamson relationA phase transition or composition jump must be treated separately; its density change is not obtained by integrating one smooth homogeneous-adiabatic equation through it.

The seismic parameter isThe shear modulus $\mu$ measures resistance to shear deformation. Solids with $\mu >0$ transmit shear waves; an ideal fluid has zero static shear modulus, so it cannot carry a bulk S wave, while compressional P waves can still propagate through its nonzero bulk modulus.

Limitations include uncertain seismic coverage, temperature and composition tradeoffs, nonadiabatic boundary layers, phase changes, lateral heterogeneity, rotation and finite material strength. Small asteroids and comet nuclei commonly retain irregular shapes because their self-gravity is too weak to overcome strength and impact history; a few larger bodies, such as Ceres, are much closer to global hydrostatic figures. A spherical shape alone is not a complete proof of present hydrostatic equilibrium.

Two complementary remote constraints are precise spacecraft tracking, which measures gravitational harmonics and constrains mass distribution when combined with shape and rotation; and tidal or rotational measurements, such as Love numbers, librations and spin precession, which constrain deformability, moment of inertia and possible fluid layers. Magnetic induction measurements in a time-varying external field can provide an additional conductivity constraint, but are not a unique density inversion.

For the June 2018 lunar picture, start with a regolith and fractured crust, a largely anorthositic crust tens of kilometres thick, a silicate mantle, a deep partially molten region and a small iron-rich core. GRAIL and Apollo-based models gave a global mean crustal thickness around $34$-$43$ km.^[ref] A representative seismic model has a partial-melt boundary near radius $480$ km, a liquid outer-core boundary near $330$ km and a solid inner-core radius near $240$ km; these are model-dependent radii from the centre, not depths below the surface.^[ref]

The long-running Apollo network had only four principal stations on the near side, with poor global coverage and strong scattering in the fractured near-surface material. Far-side structure and weak deep reflected phases were therefore difficult to constrain. The historical date matters: the answer does not import later Mars seismology or later lunar-core refinements into the 2018 premise.

Apply the chain rule to the phase-space function:Thus $\boxed{[F,H]=0⟹F\text{ is conserved}}$. If an explicitly time-dependent function is allowed, the correct equation is $dF/dt={\partial }_{t}F+[F,H]$, so a vanishing bracket alone would not suffice. The source specifies $F(\mathbf{q},\mathbf{p})$, for which that extra term is absent.

For a hyperelastic material $T_R(F)=DW(F)$. Material frame indifference of this first Piola-Kirchhoff law meansThe spatial force components rotate under a superposed rigid rotation, whereas the reference normal does not.

If $W(QF)=W(F)$, differentiate with respect to $F$ in direction $H$ to obtain $DW(QF):(QH)=DW(F):H$, hence the displayed stress transformation. Conversely, assume that stress transformation. For fixed $Q$, the derivative of $W(QF)-W(F)$ with respect to $F$ vanishes. The positive-determinant matrix domain is connected, as follows from polar decomposition and the connectedness of $SO(3)$ and of the positive-definite cone. Thus $W(QF)-W(F)=c(Q)$, independent of $F$. Iterating givesThe assumed lower bound on $W$ forces $c(Q)=0$: either nonzero sign would tend to minus infinity for one direction of integer $n$. This proves the required equivalence and explains the use of the lower-bound hypothesis.

In the frame-indifferent setting, isotropy means invariance under reference rotations as well: $W(FQ)=W(F)$ for $Q\in SO(3)$. The distinction matters: left rotations encode changes of observer, and right rotations encode material symmetry.

An orientation-preserving singular value decomposition is $F=R\mathrm{diag}(v_1,v_2,v_3)S$, with $R,S\in SO(3)$ and $v_i>0$. Material frame indifference and isotropy remove $R,S$, leaving a function $\Phi$ of the stretches. Permuting the stretches using signed permutation rotations does not change the value, so $\Phi$ is symmetric. Conversely a symmetric stretch function is unchanged by both left and right rotations, in particular is isotropic.

The eigenvalues of $B=F{F}^T$ are $v_i^2$. Its three principal invariants areThey determine the unordered positive eigenvalues as the roots of ${\lambda }^3-I{\lambda }^2+II\lambda -III=0$. Therefore a symmetric stretch function defines a well-defined $h(I,II,III)$ on the admissible invariant triples. Conversely a function of these invariants is a symmetric stretch function, proving the equivalence of all three descriptions.

The printed domain of $\Phi$ must be $(0,\infty {)}^3$, since it takes three arguments. Also isotropy alone as right invariance would not imply the spectral representation without material frame indifference: $\mathrm{tr}(AF{F}^T)$ for a fixed nonscalar symmetric positive-definite $A$ is right invariant but not generally objective. The equivalence here uses the frame-indifferent constitutive setting.

For the specific energy, Newton identities for the three eigenvalues gives ${\sum }_i{v}_i^6=I^3-3III+3III$, while $v_1{v}_2{v}_3=J=\sqrt{III}$. ThusTo differentiate it, useThese follow from $I=|F{|}^2$, the displayed trace formula for $II$, and $III=J^2$. The three partial derivatives of $h$ are $h_I=\alpha +3\beta I^2-3\beta II$, $h_{II}=-3\beta I$ and $h_{III}=3\beta -\gamma II{I}^{-3/2}/2$. The chain rule therefore givesFinally $-6\beta II=-3\beta (\mathrm{tr}B{)}^2+3\beta \mathrm{tr}(B^2)$, and $BF=F{F}^{T}F$. This is precisely the requested expression, with the coefficient of $F$ separated from the $BF$ and cofactor terms. The positive sign of the $6\beta J$ term comes from the $3\beta III$ contribution in the energy.

The surface element relation is Nanson's formula. The force on a corresponding pair of current and reference surface elements must agree:Therefore the first Piola-Kirchhoff stress tensor isIt measures force per reference area, with a current force direction and a reference normal direction.

In components, differentiate $T_{iA}=S_{ij}(y(x))(\mathrm{cof}F{)}_{jA}$. The Piola identity says ${\partial }_A(\mathrm{cof}Dy{)}_{jA}=0$. One proof writes the cofactor matrix as $\frac{1}{2}{\epsilon }_{jkl}{\epsilon }_{ABC}{y}_{k,B}{y}_{l,C}$; differentiating in $A$ produces symmetric second derivatives contracted against antisymmetric epsilon symbols, hence zero. The chain rule consequently givessince $F_{kA}(\mathrm{cof}F{)}_{jA}=J{\delta }_{kj}$. Thus, with the coordinate composition made explicit,The distinction between reference divergence and current divergence is essential to this identity.

Write the first strain invariant in components of the deformation gradient:Here lower-case and upper-case indices refer to the current configuration and reference configuration, respectively. Differentiating gives $\partial I_1/\partial F_{iI}=2F_{iI}$. The chain rule therefore givesSubstitution into the given hyperelastic material law, with its indicated matrix transpose, yieldsThe left Cauchy-Green deformation tensor is $\mathbf{B}=\mathbf{F}{\mathbf{F}}^T$. The scalar $p$ is the Lagrange multiplier enforcing incompressibility, rather than a function fixed by the strain energy density alone. The identity matrix $1$ makes this an isotropic constraint reaction in hyperelastic stress.

Let $W_r,W_{\theta },W_{\varphi }$ denote derivatives of the strain energy density with respect to its three principal stretches. The principal stress formula for incompressible hyperelasticity isIt follows by applying $\mathbf{T}=\mathbf{F}({\partial }_{\mathbf{F}}W{)}^T-p1$ in principal axes. Equal tangential principal stretches and material isotropy give $W_{\theta }=W_{\varphi }$ at $({\lambda }^{-2},\lambda ,\lambda )$. Differentiating the reduced strain energy density by the chain rule givesThe unknown pressure cancels in this difference. Substituting in part (i), and then using the supplied derivative of $\lambda$, givesConsequently the desired ordinary differential equation isFor completeness, the supplied derivative follows from $R^3=r^3-(a^3-A^3)$: $dR/dr=r^2/R^2={\lambda }^2$, and differentiating $\lambda =r/R$ gives $(1-{\lambda }^3)/R$. At the undeformed value $\lambda =1$ the division must be interpreted by a limit; the spherically symmetric deformation with $a=A$ has $\lambda \equiv 1$ and should be handled directly by the radial equilibrium equation.

The identity for the killed deterministic flow follows directly from the chain rule and the fundamental theorem of calculus, independently of the construction issue in part (b). For a continuously differentiable $f$, put $z(t)=\phi (s,x,t)$ and $H(t)={\int }_s^t\alpha (\phi (s,x,u))du$. ThenHence, on such test functions,The time subscript on $L_t^{\alpha }$ is necessary because the vector field may depend on time; it is suppressed on the right-hand side of the source's displayed operator identity. The killed deterministic flow transports an observable along the flow of a vector field and multiplies it by the no-jump survival probability.

Regard $f$ as a scalar field evaluated on a specified worldline $x^{\lambda }(\tau )$. The ordinary multivariable chain rule givesThis is the directional derivative along the four-velocity. With $x^0=ct$, it reads $df/d\tau =\gamma ({\partial }_{t}f+\mathbf{v}\cdot \nabla f)$. The spatial term is present because the worldline moves through the field; it would be lost by identifying a total derivative along the particle with a partial derivative at a fixed spatial position. For a scalar the contracted expression has the same value in all coordinate systems. For a tensor field the analogous covariant derivative also transports its indices.

Use $u^{\lambda }{u}_{\lambda }=-c^2$ and a constant Minkowski metric. Vary the velocity-dependent square root before imposing the proper-time normalization. Equivalently, vary a reparametrization-invariant worldline action first and then choose proper time. Its canonical covector momentum isThe coordinate derivative at fixed velocity is $\partial \mathcal{L}/\partial x^a=q{u}^{\lambda }{\partial }_a{A}_{\lambda }$. The Euler-Lagrange equation and the chain rule from a therefore giveDefine the antisymmetric field tensor byThe particle equation becomesAntisymmetry ensures $u^a{F}_{a\lambda }{u}^{\lambda }=0$, so the force is orthogonal to the four-velocity and preserves $u^2=-c^2$. The worldline variation before proper-time normalization is essential: substituting $\sqrt{-u^2}=c$ into the Lagrangian before taking its velocity derivative would turn the kinetic term into a constant and incorrectly remove the inertial momentum. The normalization is a property of the proper-time solution, not permission to discard the variational dependence.