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During ejection the mitral valve is closed, so $Q_v=0$. Adding the arterial and ventricular balances cancels aortic flow:where the last approximation assumes $p_v\ll p_a$.

Model the rapid contraction as a switch to approximately constant systolic elastance $E_S$, with $C_s=1/E_S$ the systolic ventricular compliance. A sufficiently small aortic-valve resistance makes $p_{LV}\simeq p_a$ after a short pressure-equalization transient. The systolic pressure must be high enough to open the valve, $E_S(V^{+}-V_0)>p_a^{+}$. During that transient neglect systemic runoff. Conservation of combined arterial and stressed ventricular volume givessoThis is an approximation across a rapid transient, not a discontinuity at finite valve resistance. The subsequent common pressure obeys $R_c(C_a+C_s){\dot{p}}_a\simeq -p_a$, givingThe source's $R_C$ in this expression denotes the same systemic resistance $R_c$.

At the end of firing, the remaining ventricular volume is $V_{\mathrm{end}}\simeq V_0+C_s{p}_a(\Delta t_F)$. The stroke volume is $V^{+}-V_{\mathrm{end}}$, henceThe assumptions are a closed mitral valve throughout ejection, negligible venous pressure, rapid valve equalization relative to systemic runoff, and nearly constant systolic elastance during the interval measured by $\Delta t_F$. These give the ventricular ejection pressure and stroke-volume approximation.

During isovolumetric contraction, ventricular pressure lies between the venous and arterial pressures, $p_v<p_{LV}<p_a$. Both valves are closed, so $Q_v=Q_a=0$ and ventricular volume remains at its end-filling value $V^{+}$. ConsequentlyHere $V^{+}$ is end-diastolic ventricular volume and $V_0$ is the zero-pressure/unstressed volume.

The arterial and venous reservoirs still exchange systemic blood:If contraction is short compared with both $R_c{C}_a$ and $R_c{C}_v$, the fractional changes are small andwhere the superscript values refer to the beginning of contraction. Thus pressure can rise rapidly in the left ventricle through increasing elastance while the other pressures are approximately fixed. This stage ends when $p_{LV}$ reaches $p_a$ and the aortic valve opens.

Let $p_a,p_v,p_{LV}$ be pressures in the arterial reservoir, venous reservoir and left ventricle. Constant vascular compliances $C_a,C_v$ relate volume changes to pressure changes. The ventricular elastance relation iswhere $V_0$ is the unstressed ventricular volume and $E_{LV}(t)>0$ is the time-dependent inverse compliance. Define the positive part $[x{]}_{+}=\max (x,0)$. The systemic, aortic-valve and mitral-valve flows areEach $R$ is a positive hydraulic resistance. The aortic valve permits outflow only when ventricular pressure exceeds arterial pressure, and the mitral valve permits filling only when venous pressure exceeds ventricular pressure.

Mass conservation, with blood treated as incompressible, givesSubstitution and multiplication of the first two equations by $R_c$ give exactlyAdding the three original balances gives zero total volume change. This is a three-compartment ventricular circulation model.

In the Hodgkin-Huxley model, $V$ is the membrane potential, the voltage inside the axon relative to outside. The space clamp assumption makes it spatially uniform. The gating variables lie between zero and one: $m$ describes sodium activation, $h$ describes availability against sodium inactivation, and $n$ describes potassium activation. Thus large $h$ means channels remain available, not that inactivation is strong.

The left side $C_m\dot{V}$ is capacitive electric current per unit membrane area. The imposed $I$ is inward-positive applied electric current. The other terms are outward-positive ionic electric currents: sodium with electrical conductance $g_{Na}{m}^{3}h$, potassium with conductance $g_K{n}^4$, and a passive leak with conductance $g_L$. Each current is its conductance times the difference between $V$ and the corresponding reversal potential $V_{Na}$, $V_K$ or $V_L$. The coefficients $g_{Na},g_K$ are maximal conductances; the gate factors represent independent permissive gates in an ion channel.

For each gate, $k_{\infty }(V)$ is the equilibrium opening/availability fraction at fixed voltage and ${\tau }_k(V)>0$ its exponential relaxation time. The equation follows from two-state ion-channel gating: if $\dot{k}={\alpha }_k(V)(1-k)-{\beta }_k(V)k$, thenVoltage feedback couples these otherwise elementary gating equations to the membrane potential and produces action potentials.

The coordinate derivatives of the smooth two-parameter map commute. With a torsion-free Levi-Civita connection,because ${\Gamma }^a{}_{bc}$ is symmetric in its lower indices. Each member of the family is a geodesic, so $DV/Ds=0$. ConsequentlyThe commutator of covariant derivatives is the action of the Riemann curvature tensor. With the source's index and sign convention this givesThis is geodesic deviation: free fall relative acceleration measures curvature, even where an individual observer can eliminate the connection at one point.

For Earth's exterior Newtonian field, $\varphi =-GM/r$ and acceleration is $a_i=-{\partial }_i\varphi$. Nearby freely falling particles have relative acceleration $\delta a_i=-{\varphi }_{,ij}{Y}^j$. At the observation point align $z$ radially upward and $x,y$ horizontally. Since $GM/R_E^2=g$,The weak-field geodesic-deviation equation therefore identifiesIn these units time is measured in seconds as in the stated Newtonian comparison; if the zeroth coordinate is $ct$ instead, the curvature components acquire the corresponding $c^{-2}$ factor. A small freely falling cloud is squeezed in the two horizontal directions and stretched radially. These are tidal effects, not the uniform acceleration removed by a freely falling frame.

Their characteristic inverse-square time is $g/R_E$, so $\boxed{t_{\mathrm{tide}}=\sqrt{R_E/g}\simeq 8.1\times 10^2\mathrm{s}\simeq 13.4\min }$. One concrete thought experiment is an ideal satellite moving in a circular orbit just above a spherical Earth, ignoring atmosphere and terrain. Its orbital angular frequency is $\sqrt{GM/R_E^3}=\sqrt{g/R_E}$ and its period is $2\pi t_{\mathrm{tide}}\simeq 84$ minutes. Equivalently, initially nearby free fall trajectories develop order-one tidal changes on this characteristic scale. This is the surface tidal time scale of a spherical gravitating body.

Under a smooth coordinate change $x\mapsto x^{\prime }$, a contravariant vector obeysThe scalar-contraction hypothesis gives $U_a^{\prime }(\partial x^{\prime a}/\partial x^b)T^b=U_b{T}^b$ for every choice of $T^b$. Equality of all coefficients and multiplication by the inverse Jacobian matrix give $\boxed{U_a^{\prime }=(\partial x^b/\partial x^{\prime a})U_b}$, the covector law.

Let $V^a=d{x}^a/ds$ be a timelike geodesic tangent. Metric compatibility and the geodesic equation giveA Killing vector therefore gives a conserved contraction. Conversely, if that contraction is conserved along every timelike geodesic, then at any point its symmetric tensor $S_{ab}={\nabla }_{(a}{T}_{b)}$ satisfies $S_{ab}{V}^a{V}^b=0$ for every unit timelike initial tangent. Homogeneity extends this to the entire timelike cone, an open set of tangent vectors. A quadratic polynomial vanishing on an open set has every coefficient zero, so $\boxed{{\nabla }_{(a}{T}_{b)}=0}$. Considering a single selected geodesic would not suffice. This is the Converse geodesic characterization of a Killing vector.

For the given exterior transformation, differentiating at fixed $r$ givesThe transformation also gives $UV=(2m-r)e^{r/(2m)}$. This determines $r$ smoothly for $r>0$ throughout $UV<2m$. The flow of the displayed vector field rescales $U\mapsto e^{-\lambda /(4m)}U$, $V\mapsto e^{\lambda /(4m)}V$. It leaves $UV$, $r$, $dUdV$ and the angular metric unchanged. Thus it preserves the full metric everywhere in that region and extends as a global Killing field, not merely by a formal exterior coordinate substitution.

Its squared norm isIt is timelike for $UV<0$ ($r>2m$), spacelike for $0<UV<2m$ ($0<r<2m$), and a nonzero null vector on either horizon $UV=0$ away from $U=V=0$. At $U=V=0$ it is the zero vector, on the whole bifurcation surface of angular points. Zero must be distinguished from a nonzero null vector. This is the Schwarzschild time-translation Killing field in Kruskal coordinates.

Use metric signature $(+---)$ and the curvature sign fixed by the stated Newtonian limit. In flat spacetime the source-free Maxwell equations are ${\partial }_a{F}^{ab}=0$ and ${\partial }_{[a}{F}_{bc]}=0$. The stress-energy tensor in mixed form is $T^a{}_b=F^{ac}{F}_{cb}+{\delta }_b^a{F}_{cd}{F}^{cd}/4$. Its divergence isContracting the cyclic Maxwell equation with $F^{ac}$ gives $F^{ac}{\partial }_a{F}_{cb}=-{\partial }_b(F_{cd}{F}^{cd})/4$. The first term also vanishes, proving $\boxed{{\partial }_a{T}^{ab}=0}$. The component $b=0$ expresses local electromagnetic energy conservation, and the spatial components express momentum conservation. Integrating converts them into the usual energy and momentum flux balances.

In curved spacetime the tensor equations areThey agree with the flat equations in coordinates of a local inertial frame at a point, where the connection vanishes, and covariance then extends them to arbitrary coordinates. Antisymmetry makes the homogeneous equation equivalent to $dF=0$. Metric compatibility permits the same contraction proof, now giving $\boxed{{\nabla }_a{T}^{ab}=0}$. In general,which need not vanish. Covariant local balance remains valid; ordinary coordinate components do not provide a universal conserved global energy-momentum vector in a curved geometry. Additional spacetime symmetries can furnish conserved currents. A locally inertial frame removes the connection at a point, not the curvature or the global obstruction.

The alternating array with fixed component ${ϵ}_{0123}=1$ is appropriate in oriented inertial special-relativistic coordinates, whose proper Lorentz transformations have determinant one. It is not itself a covariant tensor under general coordinate changes. For an oriented Lorentzian metric, replace it by the metric metric volume tensor ${\epsilon }_{abcd}=\sqrt{-\det g}[abcd]$. The genuine dual isThis is the Hodge star; the determinant factor, orientation and raising conventions are essential. Even in flat spacetime arbitrary curvilinear coordinates require the metric volume tensor, so the distinction is about coordinate covariance rather than curvature alone.

The isolated equation for $R_{00}$ singles out a coordinate component and cannot be a generally covariant law. Replacing it by a proportionality $R_{ab}\propto T_{ab}$ still fails: the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}={\nabla }_{b}R/2$, whereas stress-energy conservation requires ${\nabla }^a{T}_{ab}=0$. Such a law would force $R$ constant and hence impose an unphysical restriction on a varying relativistic dust density. Missing units or a coupling constant in the proposed proportionality would not repair this structural problem.

The Einstein tensor $G_{ab}=R_{ab}-R{g}_{ab}/2$ is divergence-free. Thereforeis consistent with covariant stress-energy conservation in the source's curvature convention. Its trace gives $R=8\pi GT$ and hence $R_{ab}=-8\pi G(T_{ab}-T{g}_{ab}/2)$. For slowly moving relativistic dust with $V^a{V}_a=1$, $T_{00}\simeq \rho$, $T\simeq \rho$, $g_{00}\simeq 1$, so $R_{00}\simeq -4\pi G\rho$, exactly matching $-\Delta \varphi$ and Poisson's equation.

Finally, electromagnetic stress-energy is trace-free: $T^a{}_a=F^{ac}{F}_{ca}+F_{cd}{F}^{cd}=0$ by antisymmetry. The traced field equation then gives $\boxed{R=0}$ for a purely electromagnetic source. This does not say $R_{ab}=0$ or that spacetime is flat; its trace-free Ricci curvature can still be nonzero.

The no-cloning theorem states that no fixed deterministic quantum operation can perfectly duplicate every unknown pure state. For a unitary copier with a fixed blank, suppose it cloned two normalized vectors $|a⟩,|b⟩$. Preservation of inner products would give $⟨a|b⟩=⟨a|b{⟩}^2$, impossible for a nonorthogonal distinct pair with overlap strictly between zero and one in modulus. Including an environment does not help: its final overlap has modulus at most one, so one would require $|⟨a|b⟩|\le |⟨a|b⟩{|}^2$. Thus a universal perfect copier is impossible. Orthogonal known alternatives do not contradict this restriction.

Label the three qubits as input $S$, Alice's entangled qubit $A$ and Bob's qubit $B$. Initially their joint state isThe controlled-NOT gate from $S$ to $A$ changes this toApply $H|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H|1⟩=(|0⟩-|1⟩)/\sqrt{2}$ to the first qubit and group by Alice's two computational-basis outcomes. The result isIn expanded form the last two Bob factors are ${\alpha }_0|0⟩-{\alpha }_1|1⟩$ and ${\alpha }_0|1⟩-{\alpha }_1|0⟩$. Their Alice labels must be $10$ and $11$; the repeated $00$ labels in the source are printing errors.

Alice measures her two qubits and sends the two classical bits $(a,b)$ to Bob. Each outcome has probability $1/4$ and Bob's conditional state is $X^b{Z}^a|\psi ⟩$. He applies $\boxed{Z^a{X}^b}$ to recover the input, up to physically irrelevant global phase. Explicitly the corrections for $00,01,10,11$ are $I,X,Z,ZX$. This is quantum teleportation: Alice's input is consumed by measurement, so no second surviving copy is created. Before her classical message, Bob's reduced density operator is $I/2$, independent of the input, so the protocol also supplies no faster-than-light communication.

Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.

Use the normalized Rayleigh quotient $R(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its real directional derivative in an arbitrary complex direction $\eta$ isIt vanishes for every direction precisely when $(H-R)\psi =0$, since testing $\eta =(H-R)\psi$ gives its squared norm. Conversely an eigenvector makes every derivative zero. Stationarity over all states is stronger than stationarity in a restricted trial family.

Normalize the Gaussian as ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, with $\alpha >0$. Its moments are $⟨X^2⟩=1/(2\alpha )$, $⟨X^4⟩=3/(4{\alpha }^2)$ and $⟨P^2⟩={\hslash }^2\alpha /2$. The printed potential energy consequently givesPut $a=\alpha \hslash /(m\omega )$. Then $\mathcal{E}/(\hslash \omega )=a/4+1/(4a)+15/(256a^2)$, whose second derivative is $1/(2a^3)+45/(128a^4)>0$. It diverges at both positive endpoints, so its sole stationary point is the unique global minimum. Its equation isThe variational principle therefore yields the optimized upper boundThe printed minimizer is inconsistent with the printed quartic coefficient. At $a=3/2$ the derivative is $5/48>0$, so this is not the stationary Gaussian for that potential energy. It is nevertheless a legitimate trial vector and gives the simple, weaker bound $\boxed{E_0\le 109\hslash \omega /192}$. If the quartic coefficient were instead $5m^2{\omega }^3/(16\hslash )$, four times the printed value, then $a=3/2$ would indeed minimize the family and its energy would be $31\hslash \omega /48$. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is Gaussian trial for a quadratic-plus-quartic oscillator.

The orthogonal coordinate change preserves the kinetic energy and takes the particle exchange operator to $(x_{+},x_{-})\mapsto (x_{+},-x_{-})$. Thus the spatial wave function is even in $x_{-}$ for bosons and odd for fermions.

The total potential energy and Hamiltonian separate asAssume $1+2\kappa >0$ so the relative oscillator is confining. Its spectrum is $E_{n_{+},n_{-}}=\hslash \omega (n_{+}+1/2)+\hslash \Omega (n_{-}+1/2)$, and its parity in the two-particle relative coordinate is $(-1{)}^{n_{-}}$.

The lowest boson state uses $n_{+}=n_{-}=0$, whereas the lowest fermion state uses $n_{+}=0$, $n_{-}=1$. ThereforeFor an oscillator of frequency $\Omega$, $⟨x_{-}^2{⟩}_n=(n+1/2)\hslash /(m\Omega )$. Since $(x_1-x_2{)}^2=2x_{-}^2$,The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied oscillator wave functions correspond to kinetic energy $P^2/(2m)$; the extra factor $1/2$ printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional kinetic energy dictated by particle mass $m$.