For the symmetric top, set $I_1=I_2=J$. The third of Euler's equations reduces to $I_3{\dot{\omega }}_3={\tau }_3=0$. Thus the axial angular velocity component is a conserved quantity:The transverse angular velocity components are generally variable. A torque constant in the principal body frame generally rotates in the inertial frame; constancy in the two frames is different.

For a vector $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$ in the principal body frame, the derivative of a body-fixed basis vector is $(d{\mathbf{e}}_i/dt{)}_{\hat{S}}=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the expansion gives the rotating-frame derivative formula, also called the Coriolis formula:Apply this to angular momentum $\mathbf{L}={\sum }_i{I}_i{\omega }_i{\mathbf{e}}_i$. The principal moments of inertia are constant in the principal body frame, so external torque balance becomesResolving the cross product gives Euler's equations:All components and dotted components in this display refer to the principal body frame.

Put $\mathbf{r}=(x,y,z)$ with ${\mathbf{e}}_3$ along $z$, and interpret the symmetry as axisymmetry of the whole mass density. A half-turn about ${\mathbf{e}}_3$ sends $(x,y,z)$ to $(-x,-y,z)$ and leaves the integration domain and mass density invariant. Therefore $\int \rho xz{d}^{3}r=\int \rho yz{d}^{3}r=0$. A quarter-turn sends $(x,y,z)$ to $(-y,x,z)$; it givesAll off-diagonal entries of the inertia tensor consequently vanish. Writing $s^2=x^2+y^2$, the principal moments of inertia satisfyThus every orthonormal basis with third vector along the symmetry axis is a principal body frame. This is axisymmetry of the inertia tensor. In general there is no requirement that $I_3=I_1$.

Take the origin at the centre of mass, so ${\int }_B\rho \mathbf{r}{d}^{3}r=0$, and write $r_i=\mathbf{r}\cdot {\mathbf{e}}_i$ in the specified orthonormal basis. The inertia tensor isHere ${\delta }_{ij}$ is the Kronecker delta. For a rigid body with angular velocity $\mathbf{\omega }$, this tensor gives $\mathbf{L}=\mathbf{I}\mathbf{\omega }$ and rotational kinetic energy $T=\mathbf{\omega }\cdot \mathbf{I}\mathbf{\omega }/2$.

The inertia tensor is a real symmetric matrix, so it admits an orthonormal basis of eigenvectors. The lines in these eigenvector directions through the centre of mass are its principal axes; the corresponding eigenvalues $I_1,I_2,I_3$ are its principal moments of inertia:If an eigenvalue repeats, the corresponding principal axes need not be unique.

The Euler-Lagrange equations of the quadratic Lagrangian areThe mechanical mass matrix is the identity. A normal mode $\mathbf{\theta }=\mathbf{a}\cos (\omega t+\varphi )$ therefore satisfies the eigenvalue problem $\mathsf{K}\mathbf{a}={\omega }^2\mathbf{a}$. Direct multiplication givesThese are normal modes of two spring-coupled equal pendulums. The corresponding normal mode coordinates $u={\theta }_1+{\theta }_2$ and $v={\theta }_1-{\theta }_2$ obey $\ddot{u}+u=0$ and $\ddot{v}+(1+2k)v=0$.

For the in-phase normal mode, the bobs undergo identical displacements, leaving their separation exactly $r$. The spring stays at its natural length and exerts no force, so this angular frequency is independent of $k$. The value $1$ is the individual simple pendulum angular frequency $\sqrt{g/l}$ in the chosen units.

Set $l=m=g=1$. At $({\theta }_1,{\theta }_2)=(0,0)$ the spring has $d=r$, so its potential energy has zero gradient: ${\partial }_j{V}_s=k(d-r){\partial }_{j}d=0$. The gravitational potential energy has ${\partial }_j{V}_g=\sin {\theta }_j=0$ there. Zero velocities therefore satisfy the Euler-Lagrange equations, proving this is an equilibrium.

For fixed $r>0$, let $\epsilon =\max (|{\theta }_1|,|{\theta }_2|)$ and $\delta ={\theta }_2-{\theta }_1$. The Taylor series giveThus, near the downward equilibrium, the horizontal separation is positive andIn particular, $V_s=k{\delta }^2/2+O({\epsilon }^4)$ and $V_g=-2+({\theta }_1^2+{\theta }_2^2)/2+O({\epsilon }^4)$. Dropping the additive constant and terms beyond quadratic order gives the harmonic approximationFor physical spring constant $k\ge 0$, the quadratic potential energy is a positive-definite quadratic form. The downward position is therefore a stable equilibrium in Lagrangian mechanics, and its nearby motion admits small oscillations.

A sodium-channel gain-of-function mechanism in Long QT syndrome can reduce the inactivation rate $k_i$ or increase recovery/reopening from inactivated states, leaving a small persistent open population. The resulting late inward sodium current opposes repolarizing outward currents while $V<E_{\mathrm{Na}}$, prolonging the cardiac action potential. Slower entry into $I$ would also extend the typical conducting episode relative to its normal duration. The altered effect is prolonged sodium entry, not reversal of sodium's charge or a pump-driven uphill flux through the channel.

The precise rate change depends on the mutation: the clinical label alone does not uniquely specify one transition constant. Long QT syndromes also include other channel mechanisms; the question here concerns its stated sodium-channel variant.^[ref]

Add an unavailable nonconducting inactivated state to the two-state gating scheme, for exampleThe probabilities obeyDepolarization increases activation, initially raising $p_O$; subsequent transfer to $I$ reduces current while voltage is held constant. Slow recovery at the depolarized voltage permits most channels to become unavailable. Repolarization promotes recovery and deactivation. More closed states and closed-state inactivation may be needed for quantitative data, but this minimal extension already explains the rise followed by decay that fixed-rate $C\leftrightarrow O$ gating cannot produce from a closed population.

The intracellular recording by Alan Hodgkin and Andrew Huxley shows the membrane voltage rising from a negative resting value to a positive overshoot, with the outside held at zero reference. The action potential therefore does more than transiently abolish a resting polarization: it reverses the sign across the membrane. A simple hypothesis that excitation only makes an otherwise potassium-dominated membrane equally leaky cannot explain that positive overshoot.

The result helped motivate a transient, selective change in membrane permeability, later resolved into rapid inward sodium current followed by outward potassium current. The positive sodium equilibrium potential calculated above makes the overshoot physically plausible, while potassium conductance favours a negative resting voltage and repolarization. The 1939 voltage trace itself did not identify individual channel proteins or establish the full later voltage-dependent gating mechanism; those conclusions required ionic substitution and voltage-clamp evidence.^[ref]

Take $V=V_{\mathrm{inside}}-V_{\mathrm{outside}}$. Equality of electrochemical potentials for a monovalent cation gives the Nernst equilibrium potentialAt $T=293.15\mathrm{K}$, $RT/F=25.26\mathrm{mV}$. ThereforeThe signs refer to the stated electrode convention, not to the direction of current at every voltage. Activities are approximated by concentrations and the bulk ionic concentrations are treated as fixed during a single spike.

Write $A=V/D_3$ and $B=L/D_1$. Then $\tau =A/\lambda +B\lambda$, soThereforeAt the optimum, the two contributions are equal. Expressing the optimal sliding length directly in terms of the minimum time givesThese relations concern the optimal point; a general nonminimal time can correspond to two different sliding lengths. A geometric or kinetic restriction on the allowed sliding length can instead place the optimum at a boundary.

The term $V/(D_3\lambda )$ is the bulk-diffusion search cost: increasing the effective length of DNA searched after each encounter enlarges the useful capture region and reduces this contribution. The term $L\lambda /D_1$ is the accumulated sliding cost. Approximately $L/\lambda$ DNA landing-and-sliding episodes, each taking order ${\lambda }^2/D_1$, give $L\lambda /D_1$. These are scaling estimates; geometric constants and correlated revisits are omitted by the supplied model.

Thus the facilitated target search is slow as $\lambda \to 0$, because too little sequence is examined per encounter, and slow at very large $\lambda$, because sliding and repeated local exploration take too long. A finite intermediate minimum balances 1D scanning against 3D relocation. The right panel of the figure in solution shows the two contributions and their sum. The model requires a useful range of sliding lengths below the full chromosome length, finite positive diffusion coefficients and effective relocation between episodes; it is not a claim that arbitrarily large $\lambda$ is physically available.

If the outcome is the first successful binding to any operator, ten well-separated functionally independent unoccupied targets give ten times the total capture rate. Henceor roughly $6$–$12\mathrm{ms}$ for the estimates above. This assumes nonoverlapping capture zones and no significant depletion before the first event. The first-arrival time at one designated operator is not divided by ten merely because other sites are added, and filling all ten sites is a different occupancy problem.

The given sphere has radius $a=5\mathrm{nm}$. The Stokes–Einstein relation at an illustrative $T=300\mathrm{K}$ givesFor $N=30$ well-mixed independent searchers in cell volume $V$, the target-arrival rate is $N{k}_a/V$. The corresponding mean first arrival time isThe source does not fix $V$, the reaction radius or temperature. For an order-of-magnitude estimate, a bacterial volume around $1\mu {\mathrm{m}}^3$ and reaction radius $5$–$10\mathrm{nm}$ give about $0.06$–$0.12\mathrm{s}$. A 20-base-pair B-form site is about $6.8\mathrm{nm}$ long, supporting a nanometre-scale capture radius but not uniquely defining its spherical value. The estimate is the ideal diffusion-limited time, not a measured intracellular search time: nonspecific binding, orientational matching, crowding and limited accessibility are excluded.

Treat the target as a stationary perfectly absorbing sphere of reaction radius $r_0$, surrounded by an effectively uniform reservoir of protein number density $C_A$. At steady state the radial diffusion equation givesThe inward capture rate per target is the integral of the Fick's first law:For dilute independent targets of number density $C_B$, the volume association rate is $J{C}_B$, soThis is the Smoluchowski association rate, in volume per time because the concentrations are number densities. If the target also diffuses, use the relative diffusion coefficient $D_A+D_B$. Orientation constraints, incomplete absorption and chromosome geometry can reduce the actual association rate relative to this ideal collision limit.

The finite nuclear size correction perturbation is unchanged for the same lead atomic nucleus. Its leading effect on one occupied $1s$ orbital is still proportional to that orbital's contact density,Other electrons screen the nuclear attraction and alter the self-consistent $1s$ wavefunction, usually reducing its contact density and hence its positive shift relative to the bare hydrogen-like ion. The reduction is modest for a deep $1s$ orbital: it penetrates well inside most other shells and still samples nearly the full $Z=82$ attraction. Replacing $Z$ by the neutral atom's asymptotic charge one would be inappropriate for this core orbital.

A neutral lead atom has two $1s$ electrons. Their contributions add in a frozen-orbital estimate of total atomic energy, roughly twice the per-electron shift, with relaxation and correlation corrections. An experimentally observed transition frequency again depends on the difference between the finite-size shifts of both participating many-electron quantum states. A quantitative neutral-lead result requires a screened relativistic many-electron calculation, not the one-electron formula with a single universally valid effective charge.

A nonrelativistic $2p$ wavefunction vanishes linearly at the origin. Its probability density is therefore proportional to $r^2$, so its first-order finite nuclear size correction shift scales as $R^4$, rather than the $R^2$ result for $1s$. It is positive but much smaller.

More explicitly, the normalized hydrogen-like ion radial function obeys $R_{21}^2(r)\simeq Z^5{r}^2/(24a_0^5)$ near the origin. Since the spherical harmonic has unit angular norm,ThusThis is about $2.4\mu \mathrm{eV}$ in the nonrelativistic model. At large $Z$, the relativistic $2p_{1/2}$ Dirac spinor has an $s$-type small component and a much stronger nuclear sensitivity than this nonrelativistic argument predicts; $2p_{3/2}$ remains much more suppressed. Neglecting spin-orbit interaction corrections would miss that distinction.

The general thin gravitational lens equation is a two-component angular relation,For a spherical lens, image, source and lens centre lie in one plane containing the line of sight. Choose its sky direction as a signed axis. Then $\alpha ,\theta ,\beta$ can be signed scalars andThey are not fundamentally scalar quantities for a general lens; the reduction uses this symmetry and axis choice. An exactly aligned source requires the full ring geometry, not a unique preferred sky axis.

In general, $D_S\ne D_L+D_{LS}$. For example, in a spatially flat expanding cosmology, comoving radial distances ${\chi }_L,{\chi }_S$ giveThus $D_S-D_{LS}={\chi }_L/(1+z_S)$ differs from $D_L$ when the redshifts differ. Ordinary Euclidean addition applies only when the same static physical-distance convention is appropriate.

Let $\alpha$ be the reduced, observed angular displacement between source and image. In the small-angle lens geometry, the physical bending angle is weighted by the lens-to-source and observer-to-source angular diameter distances:In a Euclidean diagram this follows because the transverse displacement at the source plane is $D_{LS}\xi$, whose observed angle is that displacement divided by $D_S$. In cosmology the same thin-lens expression uses the corresponding angular diameter distances. For a signed one-dimensional coordinate, the direction is supplied by the image's side of the lens.

The evolving optical and infrared kilonova was consistent with radioactive heating of neutron-rich merger ejecta. Rapid neutron capture, the r-process, can synthesize nuclei heavier than iron; decay of the newly formed unstable nuclei powers the transient. Its redward evolution and large inferred opacity supported production of heavy-element-rich ejecta. Thus mergers are a demonstrated important site for heavy-element production, while the fractions of individual elements and their total cosmic contribution remain model-dependent. The transient was not a direct identification of every very heavy element.^[ref]