A spatially uniform external gravitational acceleration moves a body's centre of mass without stretching it. Tidal forces arise from the variation of that acceleration across the body. Let a distant mass $M$ lie at distance $D$ along unit vector $\mathbf{n}$, and let $|\mathbf{r}|\ll D$ denote position relative to the body's centre. Subtract the centre's acceleration. To first order,The tidal tensor is $T_{ij}=GM(3n_i{n}_j-{\delta }_{ij})/D^3$. It stretches along the line to the source with eigenvalue $2GM/D^3$ and compresses in the two perpendicular directions with eigenvalues $-GM/D^3$. Its trace vanishes outside the source, consistently with Laplace's equation for the gravitational potential.

After removal of the constant and uniform-acceleration terms, the tidal potential isAn ideal equilibrium ocean on a sphere of radius $R$ follows an equipotential of the body's gravity plus this perturbation. For surface gravity $g$, the leading displacement isup to a constant fixing total water volume. The quadrupolar form produces bulges on both the near and far sides; the far-side bulge follows from differential acceleration, without postulating a repulsive gravitational force.

The Moon and Sun both raise terrestrial tides, with relative strength set mainly by $M/D^3$, rather than by the total gravitational pull $M/D^2$. Their contributions reinforce near new and full Moon to produce spring tides, and partly cancel near quadrature to produce neap tides. Earth's rotation relative to the tidal pattern gives a strong semidiurnal component in the simplest geometry. Real tide heights and timing depend on ocean depth, coastlines, basin resonances, Coriolis acceleration, friction and the declination of the forcing bodies. Tidal dissipation produces phase lags and torques, exchanging rotational and orbital angular momentum and gradually altering the spin and orbit. The equilibrium bulge is therefore a useful starting model, not an exact local tide forecast.

Take a fixed sphere of radius $R$ in an unbounded incompressible inviscid fluid, with uniform velocity $U{\mathbf{e}}_z$ at infinity. Assume irrotational flow, so $\mathbf{v}=\nabla \varphi$ and incompressibility gives ${\nabla }^2\varphi =0$. Axisymmetry and the dipolar angular dependence of the far field suggestMatching the far field gives $A=U$. The no-penetration condition $v_r(R)=0$ then gives $B=U{R}^3/2$. ThusThe surface has stagnation points at its front and rear poles, while the equatorial speed is $3U/2$. Fluid slips tangentially past the sphere; imposing a no-slip condition would contradict this inviscid potential-flow model.

With gravity omitted or the hydrostatic pressure subtracted, steady Bernoulli givesAt the surface,The front–rear symmetry makes the integral of pressure force along the flow vanish: terms proportional to $\cos \theta$ integrate to zero over the sphere. This is D'Alembert's paradox, not a prediction that real spheres have no drag. Viscosity creates boundary layers; separation and a wake break the fore–aft symmetry. Potential flow remains valuable for the outer velocity field when viscous regions are thin, but it cannot supply their drag by itself.

For a slender elastic beam undergoing small bending, cross-sections remain approximately plane and perpendicular to the centreline. At distance $z$ from the neutral axis, longitudinal strain is ${ϵ}_{xx}=-z\kappa$, where $\kappa$ is curvature. Linear elasticity gives ${\sigma }_{xx}=E{ϵ}_{xx}$, so the bending moment has magnitudeHere $I$ is the second moment of cross-sectional area, not a mass moment of inertia. For width $b$ and depth $h$ in the bending direction, $I=b{h}^3/12$. The cubic depth dependence explains why a deep beam resists bending more effectively than the same material spread sideways.

For a uniform beam with transverse displacement $y(x)$, $\kappa \simeq y^{\prime \prime }$ and bending energy is $\frac{1}{2}\int EI(y^{\prime \prime }{)}^{2}dx$. Balancing transverse load $q(x)$ yields the Euler-Bernoulli beam equation $EI{y}^{\prime \prime \prime \prime }=q$ under a consistent sign convention. Supports specify values of deflection, slope, moment or shear; the equation alone does not fix the response.

Now apply a compressive end load $P>0$. A small deflection shortens the axial projection by $\frac{1}{2}\int (y^{\prime }{)}^{2}dx$, so the total potential energy relative to the straight configuration isIts stationary equation isFor pinned ends, $y(0)=y(\ell )=0$ and the end moments vanish, so $y^{\prime \prime }(0)=y^{\prime \prime }(\ell )=0$. The candidate shapes $y=A\sin (n\pi x/\ell )$ require $P=EI(n\pi /\ell {)}^2$. The first loss of stability is consequentlyAt smaller $P$, the quadratic energy is positive; beyond this threshold the straight configuration no longer minimizes it. End restraint changes the threshold: clamped–clamped gives $4{\pi }^{2}EI/{\ell }^2$ and clamped–free gives ${\pi }^{2}EI/(4{\ell }^2)$, equivalently an effective-length correction. A column buckles in the direction of smaller flexural rigidity. Euler buckling is a geometric instability within elastic response; yielding or crushing can occur first for a short thick strut. Large deflections, shear deformation and imperfections require refinements of the ideal slender-beam model.

Let the rotating frame have origin acceleration ${\mathbf{A}}_0$ and angular velocity $\mathbf{\Omega }(t)$ relative to an inertial frame. For a vector referred to the moving axes, the transport rule isApply it twice to position. With rotating-frame velocity $\mathbf{v}$ and acceleration ${\mathbf{a}}_R$,Thus Newton's equation expressed in the rotating frame isThe added terms are translational, Coriolis, Euler and centrifugal fictitious forces. They reflect the frame's acceleration, rather than additional interactions with nearby objects.

The centrifugal force points outward from the rotation axis and, for constant $\mathbf{\Omega }$, has potential $-m|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. The Coriolis force is perpendicular to velocity, so does no work on the particle measured in that rotating frame; it changes direction rather than speed directly. The Euler force is present when the frame's rotation rate or direction changes.

On Earth, gravity and centrifugal acceleration combine into effective gravity. For slow horizontal motion at latitude $\lambda$, the vertical component of Earth's rotation gives Coriolis parameter $f=2\Omega \sin \lambda$: an initially northward motion deflects eastward in the northern hemisphere. This supplies a useful local explanation of rotating flows and Foucault precession. A bucket of rotating liquid illustrates the centrifugal potential: its steady free surface is an equipotential of $gz-{\Omega }^2{r}^2/2$, giving $z={\Omega }^2{r}^2/(2g)+\mathrm{constant}$. These equations require a declared frame; adding a centrifugal force while retaining inertial-frame acceleration would count the same kinematic effect twice.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

In a homogeneous Ohmic conductor, $\mathbf{J}=\sigma \mathbf{E}$. Combining this with Maxwell's equations givesFor a good conductor, $\sigma \gg \omega ϵ$, so conduction dominates displacement current. With time dependence $e^{-i\omega t}$ and depth $z>0$, take $E\propto e^{ikz}$. ThenThe field is consequently proportional to $e^{-z/\delta }{e}^{iz/\delta }$: it both attenuates and changes phase with depth. The skin depth is the distance for amplitude to fall by $e^{-1}$; power density falls as $e^{-2z/\delta }$, so its attenuation length is $\delta /2$.

Physically, changing fields induce currents whose magnetic field opposes penetration. Equivalently, after neglecting displacement current, magnetic diffusion obeys ${\partial }_t\mathbf{B}=(\mu \sigma {)}^{-1}{\nabla }^2\mathbf{B}$. Faster changes have less time to diffuse into the metal. Increasing frequency, conductivity or magnetic permeability reduces $\delta$, provided the relevant material parameters remain applicable.

For a thick conductor the tangential surface electrical impedance is $Z_s=(1-i)/(\sigma \delta )$ with the stated time convention and consistent field orientation. Its real part $1/(\sigma \delta )$ represents dissipative resistance per square. Current crowding near the surface raises the AC resistance of a wire when $\delta$ is small compared with its radius; thin strands and laminated magnetic cores reduce suitable eddy-current losses. When $\sigma /\gg \omega ϵ$, the full complex propagation constant must replace the good-conductor approximation. In a magnetic metal the magnetic permeability can itself be nonlinear and frequency dependent.

A dielectric responds to an electric field mainly by displacing or orienting bound charges, rather than by sustaining a steady conduction current. The polarization density $\mathbf{P}$ is electric dipole moment per unit volume. It creates bound chargeSeparating these charges from free charge motivatesFor an isotropic linear dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, soThe electric susceptibility ${\chi }_e$ describes material polarization and is not the magnetic susceptibility. More generally it is a tensor and can depend on frequency, field strength and temperature.

Electronic polarization displaces an electron cloud relative to its nucleus; ionic polarization displaces positive and negative ions; orientational polarization partially aligns permanent molecular dipoles against thermal disorder. For independent permanent moments of magnitude $p$, weak-field orientational response gives ${\chi }_e\simeq n{p}^2/(3{ϵ}_0{k}_{B}T)$ when local-field corrections and interactions are negligible. This explains a common inverse-temperature contribution without asserting that every dielectric follows that law.

At a static interface, $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$ and the tangential electric field is continuous. Bound surface charge is already included through $\mathbf{P}$. For a homogeneous filled capacitor, $C=ϵA/d$, and a reversible linear dielectric stores energy density $u=\mathbf{E}\cdot \mathbf{D}/2$. At fixed voltage, insertion raises the stored field energy but the voltage source also supplies energy; force arguments must include that source. At fixed charge the capacitor energy instead decreases on insertion.

At frequencies too high for a polarization mechanism to follow the field, its contribution weakens. A complex permittivity represents dissipative dielectric loss as well as phase lag; the sign of its imaginary part depends on the time convention. Strong fields can cause nonlinear response or dielectric breakdown, beyond the reversible linear model.

In a homogeneous linear medium, write $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$. Without free charge or current, Maxwell's equations becomeTaking the curl of Faraday's law, using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, givesTheir plane-wave solutions have ${\omega }^2=k^2/(\mu ϵ)$ and speed $v=1/\sqrt{\mu ϵ}$; in vacuum this is $c=1/\sqrt{{\mu }_0{ϵ}_0}$. For a wave $\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, the divergence equations give $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$, while Faraday's law gives ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$. Thus the fields and propagation direction form an orthogonal triad, with $B_0=E_0/v$.

The two independent transverse electric components allow linear, circular and elliptical polarization of an electromagnetic wave. Their relative amplitudes and phase determine the polarization state. In a lossless nondispersive medium the energy density and energy flux areA travelling plane wave has equal electric and magnetic energy contributions and time-averaged intensity $⟨S⟩=E_0^2/(2Z)$, where $Z=\sqrt{\mu /ϵ}$ is its wave impedance and $E_0$ denotes peak electric-field amplitude. Reflection and refraction follow by imposing the electromagnetic boundary conditions. Frequency-dependent constitutive response introduces dispersion; conductivity introduces attenuation, so the simple real constant-speed result has a specified domain of validity.

Initially and finally the block and lake are at the same temperature. After the falling block has come to rest, its lost gravitational potential energy has dissipated into internal energy of the lake and surroundings. In the idealization that all $mgh$ is ultimately thermalized in the lake at $T_0=283.15\mathrm{K}$,The block returns to its original thermal state, so its net entropy change is zero. Mechanical potential energy carries no thermal entropy that must be subtracted. ThusUse of $g=9.80\mathrm{m}{\mathrm{s}}^{-2}$ gives the same three-significant-figure answer. Splashing and any temporary heating do not change this final-state balance if their mechanical energy is eventually dissipated in the same reservoir.

The units supplied for 130 indicate specific heat capacity, so the block's total heat capacity is $C=mc=0.5(130)=65\mathrm{J}{\mathrm{K}}^{-1}$. Treat it as constant, and the lake as a thermal reservoir at $T_0=283.15\mathrm{K}$. The initial block temperature is $T_i=373.15\mathrm{K}$.

Compute the block's entropy change using a reversible comparison path, even though the actual cooling is irreversible:The lake receives $Q=C(T_i-T_0)=5850\mathrm{J}$ at constant temperature, soThereforeThis is entropy generated by cooling a finite body in a reservoir. The positive total is consistent with the Second law of thermodynamics, although the cooling block's own entropy decreases.

Replacing the fixed surface temperature by Newton's law of cooling changes the boundary condition but not the radial derivative in steady state. At the surface,HenceHere $\gamma$ is a surface heat transfer coefficient in $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, unrelated to the surface tension symbol used in C10. An independent energy-balance check uses a length $\ell$: the electrical input $I^2\ell /(\sigma \pi a^2)$ must equal the surface loss $2\pi a\ell \gamma [T(a)-T_{\mathrm{air}}]$, which gives the same answer.

The complete temperature profile isThermal conductivity controls the internal gradient, but cancels from the surface-temperature rise because all generated power must eventually cross the prescribed surface. This assumes positive constant $\gamma$, uniform electrical conductivity and negligible end or additional radiative losses; temperature-dependent material properties could make the heating and boundary problem nonlinear.

Origin of diffusion. In a material near local thermal equilibrium, microscopic carriers transfer energy from hotter regions to colder regions. On distances long compared with their mean free path and times long compared with relaxation times, the local constitutive law is Fourier's law,Combining this temperature-gradient-driven flux with local conservation of energy produces a diffusion equation. This description assumes diffusive thermal conduction; it is not generally valid for ballistic transport, and flow would add advective transport.

Energy balance with heating. Choose any fixed volume $V$, let $c$ be its uniform volumetric heat capacity and $H$ its heat generation rate per unit volume. The rate of increase of internal energy equals heat generation minus outward heat flux:Since $V$ is arbitrary, for constant thermal conductivityThe ratio $\kappa /c$ is the thermal diffusivity in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Here $c$ is in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, $\kappa$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H$ in $\mathrm{W}{\mathrm{m}}^{-3}$, so both right-hand terms have units $\mathrm{K}{\mathrm{s}}^{-1}$. If thermal conductivity varies with position or temperature, the divergence $\nabla \cdot (\kappa \nabla T)$ must be retained rather than replacing it by $\kappa {\nabla }^{2}T$.

In a long wire away from its ends, the steady temperature is independent of axial position and azimuth. The heat equation reduces toMultiplying by $\rho$ and integrating givesA finite temperature and nonsingular flux at the axis require $C_1=0$. Applying $T(a)=T_0$ determines $C_2=T_0+H{a}^2/(4\kappa )$. ThereforeThis is the steady temperature of a uniformly heated cylindrical wire. Its hottest point is at the centre, with $T(0)-T_0=I^2/(4{\pi }^2{a}^2\kappa \sigma )$. The outward heat flux is $J_Q(\rho )=-\kappa T^{\prime }(\rho )=H\rho /2$, so the power crossing a cylindrical surface per unit length, $2\pi \rho J_Q(\rho )=H\pi {\rho }^2$, equals the heating generated inside it.

Distinguish electrical current density ${\mathbf{J}}_e$ from heat flux ${\mathbf{J}}_Q$. Uniform current across the circular section givesby Ohm's law. The local Joule heating rate is electrical work per unit volume and time:For a segment of length $\ell$, integration gives $H\pi a^2\ell =I^2\ell /(\sigma \pi a^2)=I^{2}R$, confirming the usual resistor power. The model takes electrical conductivity constant and ignores temperature-induced current redistribution.

Over the plotted range, approximate the surface tension by $\gamma (T)={\gamma }_0-\beta T$ with constant positive $\beta =-{\gamma }^{\prime }$, and take $C_A=70\mathrm{mJ}{\mathrm{K}}^{-1}=0.070\mathrm{J}{\mathrm{K}}^{-1}$. A reversible adiabat hasIntegrating givesThe arbitrary fixed reference temperature $T_{\ast }$ makes the logarithm dimensionless; changing it only shifts the integration constant. This is the dimensionally precise version of writing $C_A\ln T+\beta A$ as constant. Each term has units of entropy. Between two states,If the graph had substantial curvature, the entropy differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.

Reading the original PDF graph gives approximatelyFor example, the points near $330\mathrm{K}$ and $450\mathrm{K}$ are about $22$ and $6.5\mathrm{mN}{\mathrm{m}}^{-1}$, respectively, giving this slope. The graph supports approximate values rather than exact fitted constants. The area increment is $\Delta A=10^{-4}{\mathrm{m}}^2$.

At fixed temperature, the surface tension is constant during this stretch. The work done on the film is positive:For a reversible isothermal process, the heat supplied is $Q=T\Delta S=-T{\gamma }^{\prime }\Delta A$. ThusThe film absorbs heat while its temperature stays constant. The first law gives the useful consistency check $\Delta U=Q+W_{\mathrm{on}}\simeq 6.5\mu \mathrm{J}$, agreeing with $(\gamma -T{\gamma }^{\prime })\Delta A$.

Treat the film at fixed material content, with area $A$ the coordinate conjugate to the surface tension in the stated work term. The Helmholtz free energy is a Legendre transform of the internal energy:Its natural independent variables are therefore $T,A$, soEquality of mixed second partial derivatives gives the Maxwell relation for the thermodynamics of a liquid film:The area convention matters: the supplied fundamental relation already specifies the coefficient of $dA$. There is no additional factor of two unless one changes from total interfacial area to a projected film area and changes the work term consistently.

Substitute the entropy differential into the fundamental relation:ThereforeThe coefficient of $dA$ is the change of internal energy per added area at fixed temperature, whereas $\gamma$ is the change of Helmholtz free energy per added area. Their difference $-T{\gamma }^{\prime }$ is the heat absorbed per added area in a reversible isothermal stretch. Both $\gamma$ and $T{\gamma }^{\prime }$ have units $\mathrm{J}{\mathrm{m}}^{-2}=\mathrm{N}{\mathrm{m}}^{-1}$.

The definition of heat capacity at constant area and the Maxwell relation give the entropy differentialFor a reversible adiabatic process, $dS=0$. Hencefor positive heat capacity $C_A$. Reversible adiabatic stretching cools the film. Reversibility, or negligible entropy production, is essential to this inference. Adiabatic isolation alone only sets the heat transfer to zero; an irreversible stretch can generate entropy, so its temperature change is not fixed by this isentropic derivative.

For the specified increment,Using the adiabat, or its first-order Taylor expansion,This is a very small cooling, about $61\mu \mathrm{K}$. There is no heat exchange, and the film is stretched against positive surface tension, so the internal energy increases, withThe cooling does not contradict this increase. In the linear model, $\gamma -T{\gamma }^{\prime }={\gamma }_0\simeq 0.065\mathrm{J}{\mathrm{m}}^{-2}$, soCreation of surface raises the internal energy more than the cooling lowers its temperature-dependent part. The change in surface tension during this tiny temperature fall changes the work only at a negligible higher order.