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For the stipulated Strong-shock Rankine-Hugoniot conditions, ${\rho }_d=4{\rho }_u$ and $v_d=v_u/4$. The ideal gas law is $P_d={\rho }_d{k}_B{T}_d/\mu$, with $\mu$ a dimensional mean particle mass. The specific enthalpy of a nonrelativistic monatomic gas is $5k_{B}T/(2\mu )$. Dividing the energy flux equation in d by the common mass flux givesThereforeAs an independent check, conservation of momentum gives $P_d={\rho }_u{v}_u^2-4{\rho }_u(v_u/4{)}^2=3{\rho }_u{v}_u^2/4$, and division by ${\rho }_d=4{\rho }_u$ gives exactly the same temperature. The numerical coefficient relies on $\gamma =5/3$ and the absence of dynamically important radiative or magnetic stresses.

Use signature $(+---)$, coordinates $x^0=ct$ and four-velocity $u^{\mu }=d{x}^{\mu }/d\tau$ normalized by $u_{\mu }{u}^{\mu }=c^2$. The derivative is the metric-compatible Levi-Civita connection. Fix the curvature sign by $[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }=R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }$. Then the required contractions areWith $G^{\mu \nu }=R^{\mu \nu }-g^{\mu \nu }R/2$, the contracted Bianchi identity gives ${\nabla }_{\mu }{G}^{\mu \nu }=0$. The Einstein field equations $G^{\mu \nu }+\Lambda g^{\mu \nu }=(8\pi G/c^4)T^{\mu \nu }$, with constant $\Lambda$, therefore require covariant conservation of stress-energy. This is a consistency relation between geometry and the total matter/field source. It does not mean ${\partial }_{\mu }{T}^{\mu \nu }=0$ in arbitrary coordinates, nor guarantee a global conserved energy in an arbitrary curved spacetime. The geometric identity alone, without the field equations or matter dynamics, is not an equation of motion for a freely specified material source.

For relativistic dust, $\rho$ is proper rest-mass density, measured in the fluid rest frame, and $u^{\mu }$ is its timelike four-velocity with the $c^2$ normalization above. In a local Cartesian inertial frame $u^{\mu }=\gamma (c,\mathbf{v})$, soAccordingly $T^{0i}/c$ is momentum density and $c{T}^{0i}$ is energy flux in direction $i$. The equality of momentum-density and energy-flux components after the appropriate factors of $c$ reflects the symmetry of the stress-energy tensor. In the comoving frame the spatial stress is zero, as required for pressureless matter.

Expand the dust conservation equation by the product rule:Contract with $u_{\nu }$. metric compatibility and constant normalization give $u_{\nu }{\nabla }_{\mu }{u}^{\nu }={\nabla }_{\mu }(u_{\nu }{u}^{\nu })/2=0$, leaving $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substitute this rest-mass continuity law back and divide by nonzero $\rho$:This proves geodesic motion of uncharged dust along each fluid flow line. pressure forces or another exchange of momentum would change the argument; at zero density there is no dust worldline to constrain.

The electromagnetic field tensor is antisymmetric and has six independent components. In the stated local inertial coordinates choose $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, givingThe fields are those measured by that local observer; changing observer mixes them through the Lorentz transformation of $F$. With ${ϵ}^{0123}=+1$ and ${\tilde{F}}^{\mu \nu }={ϵ}^{\mu \nu \alpha \beta }{F}_{\alpha \beta }/2$, two independent quadratic contractions areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under an orientation reversal. Opposite choices for the orientation tensor change its written sign. A nonzero null electromagnetic field can have both contractions zero, so the two numbers do not determine whether the tensor itself vanishes.

For charged relativistic dust, insert $j^{\alpha }=(q/m)\rho u^{\alpha }$ into the force-density equation:Contraction with $u_{\nu }$ again gives the continuity law, because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$ by antisymmetry. After removing the continuity term,In a local inertial frame the spatial components reduce to $d(\gamma m\mathbf{v})/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and the temporal component gives $d(\gamma m{c}^2)/dt=q\mathbf{E}\cdot \mathbf{v}$. This is the relativistic Lorentz force, with gravity represented by the connection. The force is orthogonal to $u$ and preserves the rest-mass normalization. Matter alone now exchanges energy and momentum with the electromagnetic field; adding the field tensor's stress-energy restores zero divergence of the combined source.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

Orient the flow so $q>0$, and let subscripts 1 and 2 denote upstream and downstream. The mechanical energy flux is $\rho qB$, where $B=u^2/2+gh$. Inserting the relation for $q^2$ in i gives the hydraulic jump head loss,An admissible jump dissipates mechanical energy, so $\rho q(B_1-B_2)>0$. The denominator is positive, henceThe zero depth-change case is continuous flow. A depth-decreasing discontinuity would require an energy input at the jump; it is not the dissipative hydraulic jump selected by this model.

Specify the amplitude convention. For normal-stress/pressure amplitudes, the seismic reflection coefficient and seismic transmission coefficient areFor particle displacement or velocity resolved in one fixed normal coordinate, the reflected amplitude changes sign and the transmitted ratio is insteadThese are alternative conventions, not conflicting energy predictions. The reflected and transmitted energy flux fractions arewhich sum to one for the ideal lossless interface. Thus an amplitude transmission coefficient greater than one does not violate conservation of energy.

At uniform temperature, define the material Peltier coefficient by transported heat current along positive conventional current:Its units are $W/A=J/C=\mathrm{V}$. In one dimension, $q=\Pi j-\kappa dT/dx$, with electric current density $j$ and thermal conductivity $\kappa$. Here heat excludes the electrochemical potential contribution to total energy flux.^[ref]

Now orient positive conventional current $I$ from material A into material B, at a common junction temperature $T$. Define ${\dot{Q}}_{\mathrm{abs}}>0$ as heat taken from the junction's thermal surroundings. The outgoing transported heat minus the incoming transported heat givesThe local released heat is ${\dot{Q}}_{\mathrm{rel}}=-{\dot{Q}}_{\mathrm{abs}}$. Authors defining a junction coefficient by released heat therefore use the opposite sign. Exchanging A and B reverses the coefficient; reversing $I$ at fixed A and B reverses the junction's thermal response. An individual material coefficient and a junction difference are different quantities.

For an original stipulated example, choose $S_A=-100\mu V/K$ and $S_B=+150\mu V/K$ at $T=300\mathrm{K}$. The Kelvin thermoelectric relation gives ${\Pi }_A=-0.030\mathrm{V}$, ${\Pi }_B=+0.045\mathrm{V}$. A positive $2\mathrm{A}$ flowing A to B absorbs $(0.075\mathrm{V})(2\mathrm{A})=0.150\mathrm{W}$. A separate $0.010\Omega$ junction electrical resistance would dissipate $0.040\mathrm{W}$ for either direction. If all that dissipation heats the junction, its net absorbed power is $0.110\mathrm{W}$ forwards and $-0.190\mathrm{W}$ backwards. These numbers illustrate a chosen thermal accounting boundary, not measured material data or a universal allocation of contact heating.

The Poynting theorem is the local energy balance of classical electromagnetism. For vacuum field laws, with current density $\mathbf{J}$, the electromagnetic energy per unit volume isandThus a decrease in local electromagnetic energy can supply outward energy flux or work on electric charge. The term $\mathbf{J}\cdot \mathbf{E}$ is power transferred to matter per unit volume. Its sign matters: positive values remove field energy, while negative values describe matter supplying the field. The magnetic part of the Lorentz force supplies no work term.

For a fixed volume $V$, the divergence theorem gives the directly measurable balanceFor example, $2.0\mathrm{W}$ entering a region with $0.50\mathrm{W}$ leaving and $1.20\mathrm{W}$ transferred to matter means stored electromagnetic energy increases at $0.30\mathrm{W}$. In source-free vacuum only the boundary flux changes the stored energy; a nonzero local Poynting vector can still have zero net closed-surface flux.

A brief proof takes the dot product of the Ampère-Maxwell equation with $\mathbf{E}/{\mu }_0$ and uses Faraday's law with $\mathbf{B}/{\mu }_0$. The identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ turns the remaining terms into the divergence of the Poynting vector. ^[ref].

In vacuum the Poynting vectoris the local energy flux, measured in watts per square metre. Its surface integral $P={\int }_A\mathbf{S}\cdot \mathbf{n}dA$ gives signed power crossing a surface with unit normal vector $\mathbf{n}$. Positive flux follows the chosen normal; for a closed surface the usual convention is outward. Power depends on the normal component and area, rather than just the field magnitude.

The Poynting vector applies to steady fields as well as electromagnetic waves. In a driven resistive electrical circuit, surrounding fields can carry energy into the conductor, where electrical work becomes Joule heating. This turns a circuit-level power balance into a spatial description of energy delivery. ^[ref].

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

For radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

For a scalar complex wave amplitude $E$, the squared-amplitude scalar wave intensity convention is $I=E{E}^{\ast }$. It is nonnegative and unchanged by multiplying $E$ by a unit-modulus wave phase factor. This convention is useful in paraxial propagation and phase screen models. Conversion to physical energy flux also depends on the field variable and medium; an arbitrary superposition need not have its full flux determined by this scalar alone. For a random field, $⟨I⟩$ is the mean wave intensity, whereas $|⟨E⟩{|}^2$ is the squared coherent field; these generally differ.

Loss of spatial overlap between interacting finite beams because their energy flux directions differ. In an anisotropic crystal, the direction of the Poynting vector need not coincide with the wave vector. For beam radius $w$ and relative walk-off angle $\rho$, an overlap length is of order $w/|\tan \rho |$. A full conversion calculation also includes beam profiles, diffraction and phase mismatch.

Surface albedo is the fraction of incident solar energy flux reflected by a surface, averaged over the specified spectrum and illumination geometry. The locally absorbed fraction is $1-\alpha$. It is distinct from geometric albedo, which measures brightness in a particular viewing configuration.

The Townsend-Betchov cascade cartoon organizes inertial range vorticity into sheets by bi-axial strain, then uses sheet instability and roll-up to generate finer vortex tubes. Dissipative tube cores balance stretching with kinematic viscosity. Such differing mechanisms across scales challenge literal identical-step, memoryless cascade models without contradicting a mean energy flux.

The energy flux carried into a transmitted wave, integrated over the relevant cross section. The power transmission coefficient compares this with incident power. A displacement or field-amplitude transmission coefficient must be converted with the appropriate wave impedances; its squared modulus alone need not be the power fraction. This distinction applies both to electromagnetic waves and transverse string waves.

A wavefront connects points with the same wave phase at one specified time. It is a surface, rather than the trajectory of a particular piece of light. For a scalar monochromatic wave written as $u(\mathbf{r},t)=A(\mathbf{r})\cos \Phi (\mathbf{r},t)$, a wavefront is a connected level set of the phase function $\Phi$; different surfaces can share the same wave phase modulo $2\pi$. A plane wave has planar wavefronts. An outward wave from an ideal point source in a uniform medium has spheres as wavefronts, while an optical lens can turn nearly planar wavefronts into converging ones. The wavefront describes wave phase, not wave amplitude: drawing it does not say how bright the beam is. ^[ref]

The relation to light rays becomes precise in a stationary medium with isotropy, negligible loss and a real positive scalar refractive index $n(\mathbf{r},\omega )$. In the geometrical optics limit, spatial changes in the refractive index and wave amplitude are slow on the local wavelength scale. For a single smooth propagation branch, writeHere $\omega >0$ is the angular frequency, $c$ is the vacuum speed of light, $S$ has units of length, and the last relation is the optical eikonal equation. The local wavevector is $\mathbf{k}=\nabla \Phi =k_0\nabla S$. Its direction is normal to a smooth wavefront, since a gradient is normal to its level sets. A forward light ray has unit tangent vector $d\mathbf{r}/ds=\nabla S/n$, so it follows this normal under the stated assumptions. Along that light ray, $dS/ds=\nabla S\cdot (d\mathbf{r}/ds)=n$: the change in $S$ is the optical path length. This is the short geometric argument linking wavefronts, light rays and optical path length. ^[ref]

The diagram shows incident and transmitted branches separately, with the reflected branch omitted. Its stipulated refractive indices are $n_1=1$ and $n_2=3/2$, and its incident angle is $30^{\circ }$ to the interface normal. At a stationary interface, the frequency and tangential wavevector agree on the two sides: Snell's law gives $\sin {\theta }_2=1/3$, hence ${\theta }_2\approx 19.4712^{\circ }$. The wavefronts turn while their normal spacing decreases from ${\lambda }_0$ to $2{\lambda }_0/3$, where ${\lambda }_0$ is the vacuum wavelength. More explicitly, with $x$ normal to the interface and $y$ tangential, take $S_1=\sqrt{3}x/2+y/2$ for $x<0$ and $S_2=\sqrt{2}x+y/2$ for $x>0$. These functions agree at $x=0$; the drawn wavefronts are $S_j=m{\lambda }_0$ at $t=0$ with ${\varphi }_0=0$. This is an exact piecewise plane wave phase construction for the stipulated transparent media, not measured material data. ^[ref]

The normal-to-wavefront rule requires its assumptions. In a medium with anisotropy, the wavevector remains normal to a smooth wavefront, but the group velocity and energy flux need not follow it. ^[ref] Near a wave caustic, where multiple light rays meet, one smooth phase function may no longer describe the field; diffraction and the full wave amplitude matter. A wavefront is also not generally a surface of equal pulse arrival time. In a medium with optical dispersion, a monochromatic wave phase moves with the phase velocity, while a sufficiently narrow wave packet follows the group velocity. Even identical optical path lengths at one frequency can give different group delays. The usual equal-travel-time picture needs a common launch reference and a nondispersive ray model; an arbitrary broadband disturbance has no single monochromatic wavefront.^[ref]

For balanced anisotropic Alfvénic turbulence, suppose ${\tau }_A=l_{\parallel }/v_A\ll {\tau }_{\mathrm{nl}}=l_{\perp }/\delta z$ and resonant interactions preserve the parallel scale during the perpendicular cascade. Random encounter accumulation gives ${\tau }_{\mathrm{cas}}=v_A{l}_{\perp }^2/(\delta z^2{l}_{\parallel })$. Constant energy flux then gives $\delta z\propto l_{\perp }^{1/2}$ and the displayed perpendicular turbulent energy spectrum. Unlike isotropic phenomenology, the fixed parallel scale is independent of the decreasing perpendicular scale.

For $b_l{l}_{\parallel }/l_{\perp }\ll 1$, the weak wave cascade estimate gives ${\tau }_{\mathrm{cas}}\sim l_{\perp }^3/(v_A{d}_i{b}_l^2{l}_{\parallel })$. Constant energy flux $ϵ\sim v_A^2{b}_l^2/{\tau }_{\mathrm{cas}}$ yields the displayed amplitude. An additionally isotropic local cascade has $b_l\propto l^{1/2}$ and shell-integrated magnetic energy spectrum $E_B(k)\propto k^{-2}$. These statements are phenomenological consequences of the chosen locality and decorrelation assumptions.