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A high-numerical aperture objective focuses a laser to a steep intensity gradient. For a transparent particle with higher refractive index than its surroundings, refraction transfers photon momentum to the particle, giving a optical gradient force towards the intensity maximum. A forward scattering force pushes along the beam; a stable single-beam optical trap requires the restoring axial optical gradient force to exceed that push. In the small-particle limit, the same result follows from the induced-dipole potential energy, proportional to minus the local squared electric field. Close to the trap centre,An optical tweezers experiment can attach a bead to one end of a DNA molecule, hold the other end fixed, and obtain its force-extension relation from the bead displacement and calibrated trap stiffness. The measurement must distinguish the restoring force from Brownian motion and account for heating or optical damage.

Adopt the approximate mass-luminosity relation $L/L_{\odot }\simeq (M/M_{\odot }{)}^3$ and equal available fuel fractions. The main-sequence lifetime then scales as $M/L$, givingUsing an exponent $3.5$ instead gives about $5.4\times 10^7$ and $3.1\times 10^8$ years; these are model-dependent estimates. At $r=3r_s=6GM/c^2$ the Newtonian gravitational potential energy released from infinity is $GMm/r=m{c}^2/6$. Half remains as kinetic energy in a circular orbit, so the radiatable binding energy is $GMm/(2r)$:Using the whole potential energy loss would give $1/6$, which does not subtract the final kinetic energy. A fully relativistic calculation for a nonrotating black hole gives $1-\sqrt{8/9}\simeq 5.7\%$. When the surviving star exhausts hydrogen burning, it expands off the main sequence and can fill its Roche lobe. Binary mass transfer to the black hole can form an accretion disk and an X-ray binary; unstable binary mass transfer may instead produce a common envelope. This is conditional on the binary star remaining bound after the first remnant forms. The supplied black hole endpoint is an assumption, not a universal outcome for an isolated $8M_{\odot }$ star.

Let $C(x)$ be the particle number density, $U(x)$ the conservative potential energy and $F=-dU/dx$ the applied force. In the overdamped regime a viscous drag coefficient $\zeta$ gives drift velocity $v_d=F/\zeta$. Combining this drift with Fick's first law gives the number fluxAt thermal equilibrium, the Boltzmann distribution is $C_{\mathrm{eq}}\propto e^{-U/(k_{\mathrm{B}}T)}$, so $d{C}_{\mathrm{eq}}/dx=F{C}_{\mathrm{eq}}/(k_{\mathrm{B}}T)$. Setting its net flux to zero yieldsThis is the Einstein diffusion relation. The derivation assumes a dilute equilibrium suspension, uniform temperature, linear drift response and a well-defined long-time diffusion coefficient. With a nonzero conservative force, the full transport equation includes drift; the force-free equation in the next section describes spreading without that drift.

The one-dimensional time-independent Schrödinger equation isThe differential term is the kinetic energy operator acting on the wave function, $V(x)$ supplies the multiplicative potential energy, and their sum is the Hamiltonian operator. The eigenvalue $E$ is the stationary state's total energy. This form assumes a nonrelativistic particle, a time-independent scalar potential, and no additional internal degrees of freedom in the stated model.

The equation is $i\hslash {\partial }_t\psi =[-{\hslash }^2{\partial }_x^2/(2m)+V(x)]\psi$. The generator of time evolution is the sum of kinetic and potential energy operators. For a sharp stationary energy, insert $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$ to obtain $[-{\hslash }^2{\varphi }^{\prime \prime }/(2m)+V\varphi ]=E\varphi$. The probability in $dx$ is $|\psi {|}^{2}dx$, normalized to one; hence a normalized one-dimensional wavefunction has dimensions length${}^{-1/2}$. Its overall phase is unobservable, but relative phase affects interference.

For a generous illustrative roof area 100 m${}^2$, annual rain depth one metre and five-metre usable fall, maximum potential energy is $\rho A{h}_{\mathrm{rain}}gH\simeq 4.9$ MJ, only 1.36 kWh per year before conversion losses. Even 25 million roofs give only about 34 GWh annually, around $10^{-4}$ of a 300 TWh electricity system. Tiny intermittent flows and installation embodied energy make the practical result worse. Household roof-rain generation is therefore a very small carbon-saving resource under realistic domestic heights; the small gravitational drop and water flux explain why, without needing to assume an inefficient turbine.

For a stationary nonrelativistic state, $\boxed{[-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})]\psi =E\psi }$. The wavefunction $\psi$ is a probability amplitude, $m$ the particle mass and $E$ its energy eigenvalue; the potential energy $V$ is measured in the same energy units. In one dimension the Laplacian becomes $d^2/d{x}^2$.

Quantum tunnelling is transmission through a region where the particle's energy is below the classical potential energy. The stationary wavefunction decays within that region but need not vanish at its far boundary, so matching can leave a nonzero transmitted probability current.

This does not require a particle to borrow energy temporarily. Energy is conserved for a time-independent barrier; the classically forbidden region is described by an evanescent wave, with the same stationary energy as the incident and transmitted waves.

A pumped-storage plant moves water uphill using electrical power, then returns it through a turbine and generator when electricity is needed. The upper and lower reservoirs set the usable volume and head; pumps, penstocks, valves and reversible pump-turbines set the power and flow. The plant stores energy rather than generating net energy over a cycle.

For constant head, the gravitational energy is $E_g=\rho Vgh$. With the stated generation efficiency,At $3$ GW, $P=\eta \rho gQh$ givesHead variation, dead volume and hydraulic losses beyond the stipulated efficiency have been neglected.

To deliver $9$ kWh with the same head and efficiency needs $V\simeq 9.3{\mathrm{m}}^3$ of water, roughly a $2.1$ m cube, together with a $400$ m elevation difference and a lower reservoir. At a $10$ m head it would need about $370{\mathrm{m}}^3$. This contrasts with the small fuel volume: the useful energy density of lifted water depends on its external head, not its chemical composition.

Pumped storage offers long service life, fast dispatch, substantial power, and separate choices of storage volume and turbine rating. Its drawbacks include geographic constraints, construction cost and time, ecological impacts, evaporation and water requirements, plus round-trip losses. The quoted generation efficiency alone is not the round-trip efficiency because pumping also consumes more than the stored potential energy.

The rock cylinder has mass ${\rho }_r\pi r^2(2r)$ and rises through $r$. Equating its useful energy to the reservoir result cancels the common efficiency and gravity:Its height is about $262$ m and its lift about $131$ m, ignoring buoyancy as instructed. Stored energy scales as $r^4$, excavated volume roughly as $r^3$, and lined area as $r^2$. If excavation dominates cost, cost per unit stored energy decreases approximately as $r^{-1}$. Seals, increasing pressure, geology, leakage, supporting structure and power equipment can change that scaling; a volume law is not a complete cost model.

Diffusion and mobility. For a dilute isolated sphere in a Newtonian fluid, Stokes's law gives the drag magnitude $F_{\mathrm{drag}}=6\pi \eta Rv=\xi v$, where $\eta$ is the dynamic viscosity and $R$ is the hydrodynamic radius. The mechanical mobility is therefore $1/\xi$; the translational hydrodynamic mobility matrix is ${\xi }^{-1}I$.

To relate mobility to diffusion, put the particles in a weak external potential energy $U(x)$. Their drift speed is $-U^{\prime }(x)/\xi$. For number density $n(x)$, the sum of drift and Fick's first law fluxes isAt thermal equilibrium, the Boltzmann distribution gives $n\propto e^{-U/(k_{B}T)}$, and the net flux is zero. Substituting $n^{\prime }=-n{U}^{\prime }/(k_{B}T)$ into $J=0$ yields $D=k_{B}T/\xi$. Thus the Stokes–Einstein relation isHere $k_B$ is the Boltzmann constant and $T$ is absolute temperature. The assumptions include a no-slip approximately spherical particle, a continuum fluid, a small particle Reynolds number, and negligible hydrodynamic interactions with other particles or boundaries.

Gravitational sedimentation. The buoyant force must be included, since the mass of displaced water is comparable to the protein mass. Using $1\mathrm{Da}=1.66054\times 10^{-27}\mathrm{kg}$ givesso the buoyant mass in the stated spherical model isThe drag coefficient is $\xi =6\pi (10^{-3})(2.16\times 10^{-9})=4.0715\times 10^{-11}\mathrm{kg}{\mathrm{s}}^{-1}$. At terminal speed, $\xi v_g=m_{b}g$, soIgnoring buoyancy would give $1.47\times 10^{-11}\mathrm{m}{\mathrm{s}}^{-1}$ and would not use the supplied water density. This calculation treats the stated radius as both the hydrodynamic radius and the radius determining displaced volume; a real hydrated protein can require a separately measured partial specific volume.

Electrical drift. In the specified simple electrophoresis model, terminal force balance is $Ze\mathbf{E}=\xi \mathbf{v}$. Thereforewhere $\mu$ is the electrophoretic mobility and $e$ the elementary electric charge. For the given positive beta-lactoglobulin charge,in the direction of the electric field. This exceeds the gravitational sedimentation speed by about $8.6\times 10^6$.

Separation time. The two populations have equal drag coefficients because their radii are equal. Their difference in drift speed is therefore set by one elementary charge:Starting with coincident band centres, their separation is $\Delta vt$. HenceAt that time the original band has travelled $20\mathrm{mm}$ and the mutant $21\mathrm{mm}$. Their masses do not enter this ideal electrical force balance: equal-radius particles with different masses but the same charge would have equal electrophoretic mobility in this model.

Practical discrimination. A one-charge difference is enough for the stipulated separation, so the method is sensitive to mutations that change effective charge or friction. It cannot distinguish every mutation: substitutions that preserve charge and hydrodynamic size need not separate. Buffer pH changes protein ionization, and realistic electrophoresis also involves the ionic atmosphere, electro-osmosis, walls or gel interactions, and finite band widths.

For example, the Stokes–Einstein relation at $293.15\mathrm{K}$ gives $D=9.94\times 10^{-11}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. For an initially narrow band, diffusive broadening of electrophoretic bands givesat the calculated separation time. Thus the centres are about $3.15$ diffusion standard deviations apart, rather than forming infinitely sharp bands. The stated $1\mathrm{mm}$ centre-separation criterion is achievable in the model, but actual resolution also depends on the loading width and other dispersion mechanisms.

Equivalent centrifugal acceleration. For the original protein to attain its electrical drift speed by sedimentation, its buoyant centrifugal force must equal $20eE$:ConsequentlyThis is the acceleration at the protein's position in the rotating solution; no rotation rate can be assigned without a centrifuge radius. If the intended comparison is instead the relative separation speed of the two bands, use $\Delta m_b=600\mathrm{Da}$, since their displaced volumes are equal. Matching $|\Delta m_b|a_{\mathrm{eff}}=eE$ then gives $8.04\times 10^7\mathrm{m}{\mathrm{s}}^{-2}$. In that separation experiment the heavier original protein moves faster centrifugally, whereas the more highly charged mutant moves faster electrically.

Choose a coordinate $z$ measured from the sphere centre towards increasing distance from the block, so an atom in the sphere is at height $L+z$. Summing the atom–block potential energy with the sphere's number density givesFor $R/L\ll 1$, a uniformly valid Taylor series over the sphere isThe odd moments vanish by reflection symmetry. The sphere volume is $V_s=4\pi R^3/3$, and its second moment isConsequentlyThe first finite-radius correction is therefore $-4{\pi }^{2}B{n}^2{R}^5/(15L^5)$. There is no relative $R/L$ correction because the first moment about the centre is zero. The leading term is just the number of sphere atoms times the potential at its centre; the correction increases the attraction because the nearer atoms contribute more strongly than the farther atoms lose.

This is the far-field sphere-plane dispersion energy, with Hamaker constant $A_H={\pi }^{2}B{n}^2$. It must be distinguished from the near-contact sphere-plane dispersion force, whose small parameter is the surface gap divided by the radius. Here $L$ is explicitly the centre-to-plane distance.

Without stabilization, the potential energy is predominantly attractive outside the molecular core and drops into a deep primary minimum. With the charging treatment, the DLVO theory sum develops a positive barrier between distant particles and that primary minimum. The barrier must be large compared with $k_{B}T$ to inhibit aggregation on the experimental timescale. A shallow secondary minimum can remain at larger separation, depending on parameters.

The horizontal axis is the surface gap, so the molecular exclusion boundary is at its origin. Both interactions tend to zero at large separation. Raising salt concentration shortens the Debye–Hückel screening length and can reduce the barrier, allowing particles to stick again.

In a magnesium chloride chemical solution, the principal pair interactions are hard sphere exclusion, repulsion of overlapping electric double layers, and attractive Van der Waals forces. The positive surface attracts chloride counterions and repels magnesium coions; mobile ions screen the bare Coulomb's law interaction. The competition between screened electrostatic repulsion and Van der Waals force attraction is described by DLVO theory.

For well-separated identical weakly charged spheres, the approximate repulsive pair potential energy, at centre separation $R\ge 2a$, isIt is a screened Coulomb pair potential. An attractive Hamaker constant adds a negative short-range contribution, while overlap is forbidden. Very near contact, solvation and molecular surface structure can also matter. No polymer-induced interaction should be assumed before polymer is added in part (f).

For causal reachability on a spatial circle, put $P=2\pi R$ and use a representative $\bar{x}\in [-P/2,P/2)$ of the spatial circle. The lower-left panel of part c is a fundamental strip with its vertical edges identified. On the covering space the charged world line is still the solution of part b; on the quotient it isA jump between opposite strip edges is a coordinate seam, not a discontinuity of the physical world line. At its $n$th complete lap, $x=nP$, soFor $a=q{E}_x/m>0$, the speed increases on each return while remaining below $c$. Its kinetic energy increases byThere is no single-valued electrostatic potential energy $-q{E}_x\bar{x}$ on a circle: the nonzero electromotive force is $\oint \mathbf{E}\cdot d\mathbf{l}=E_{x}P$. A globally defined magnetic vector potential one-form $A_x=-E_{x}t$ instead makes the driving explicit. In an embedded circuit, Faraday's law realizes this electromotive force through changing magnetic flux. More generally an external apparatus maintaining the field supplies the increasing particle energy. In an isolated dynamical field-plus-particle system, backreaction would change the field. Thus this is not a cyclic machine extracting work from a conservative potential, and it does not violate conservation of energy.

Restoring units, the decaying Yukawa potential is $V(r)=C{e}^{-r/{\lambda }_m}/r$ with range ${\lambda }_m=\hslash /(mc)$. For $r_{\ast }=10^{-18}\mathrm{m}$,The geometric $1/r$ factor belongs in the potential ratio; the exponential alone is the additional suppression relative to massless exchange. Using $\hslash c=197.32698\mathrm{MeV}\mathrm{fm}$ givesThe following are ratios of potential magnitudes to the magnitude at $r_{\ast }$, with the same coupling held fixed:

For example, the pion's nuclear exponential factor is $e^{-0.68346}=0.505$, whereas the Higgs nuclear exponent is about $-632.83$. The huge atomic suppressions are most reliably reported through logarithms, not through floating-point exponentiation that silently underflows to zero. These are potential energy ratios, not force ratios or absolute comparisons of different interaction strengths.

Take positive $\theta$ toward positive $x$ from the downward vertical, as in the source diagram. The mass has position and velocityThus $T=m|\dot{\mathbf{r}}{|}^2/2$ and $V=mg(a_y-l\cos \theta )$. Expanding the squared speed givesThe prescribed support coordinates are functions of time, rather than additional dynamical variables. Terms depending only on time do not affect the Euler-Lagrange equation for $\theta$, but have been retained here to obtain the inertial-frame kinetic and potential energies directly.

Write the bead position in Cartesian coordinates as ${\mathbf{x}}_B=(a\cos {\theta }_B,a\sin {\theta }_B,b{\theta }_B)$. Differentiation gives $|{\dot{\mathbf{x}}}_B{|}^2=(a^2+b^2){\dot{\theta }}_B^2$, while the free particle has $|\dot{\mathbf{x}}{|}^2={\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. Thus the kinetic energy is the sum of these squared speeds weighted by half the respective mass. The squared spring length isA spring obeying Hooke's law with zero natural length has potential energy $k{\ell }^2/2$. Consequently $L=T-k{\ell }^2/2$, and deleting the additive constant $-k{a}^2/2$ givesThe spring length is the straight separation of its endpoints, not distance measured along the helix. The fixed frictionless constraint is already incorporated by using the generalized coordinate ${\theta }_B$; it supplies no extra potential energy.

Let $\theta$ and $\phi$ be the lower and upper azimuthal angles, measured from a common fixed direction. The squared spring length isBy Hooke's law, a spring with zero natural length has potential energy $k{d}^2/2$. ThereforeConstant gravitational energies can be omitted. It is useful to define the moments of inertia $I_1=m{a}^2$, $I_2=M{A}^2$ and the angular coupling $\kappa =kaA$. The part of the potential that affects motion is simply $-\kappa \cos (\theta -\phi )$. In particular, $h$ changes the constant spring energy but not the angular equations, because the spring has zero natural length.

Let ${\delta }_0={\theta }_0-{\phi }_0$ be $0$ or $\pi$, and write the angular displacements as ${\eta }_1,{\eta }_2$. To quadratic order the restoring potential energy is $\kappa \cos {\delta }_0({\eta }_1-{\eta }_2{)}^2/2$. The linearized normal mode equation isThe determinant gives ${\omega }^2[{\omega }^2{I}_1{I}_2-\kappa \cos {\delta }_0(I_1+I_2)]=0$. Thus the normal modes of two spring-coupled coaxial rotors areFor the aligned family, the relative mode oscillates with positive frequencyFor the antipodal family, ${\omega }_{\mathrm{rel}}^2=-{\Omega }^2$, so that displacement has solutions $e^{\pm \Omega t}$ and the equilibrium is unstable.

The common zero mode has ${\eta }_1={\eta }_2=C+Dt$. It describes a neutral displacement or uniform common rotation, not a restoring oscillation. Thus the aligned family is stable with respect to relative motion and has orbital stability modulo common rotations. A particular static point is not strictly Lyapunov stable against unrestricted small velocity perturbations: arbitrarily small nonzero $J$ eventually moves the common angle far from that point. With $J=0$, the common angle is fixed and the aligned relative equilibrium point is stable. This distinguishes the symmetry-induced drift from the exponential instability of the antipodal family.

Complete the square in the potential energy. With $x_F=F/(M{\omega }^2)$,The quantum harmonic oscillator is translated without changing its level spacing. If ${\phi }_n(x)$ is its normalized unforced energy eigenstate, thenEquivalently, $|n;F⟩=e^{-i{x}_{F}p/\hslash }|n⟩$. The ground state position is centered at $x_F$, and its position uncertainty remains $\sqrt{\hslash /(2M\omega )}$.