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Choose $x$ tangential to the solid boundary and locally along the outer inviscid streamlines, and let $y$ be the inward normal coordinate. Write $u$ and $v$ for the corresponding tangential and normal velocity field components, $U(x)$ for the outer tangential velocity, and $\nu$ for the kinematic viscosity.

For a layer of streamwise scale $L$ and thickness $\delta \ll L$, incompressible flow givesso $u=O(U)$ and $v=O(U\delta /L)$. The normal Navier-Stokes equation then gives $p_y=0$ to leading order, while the outer Euler equations giveIn the tangential Navier-Stokes equation, streamwise viscous diffusion is smaller than normal diffusion by $O((\delta /L{)}^2)$. Retaining the leading inertial and normal-diffusion terms produces the Prandtl boundary-layer equationIt is supplemented by $u=v=0$ at a stationary wall and $u\to U(x)$ on matching to the outer flow.

In coordinates $x^{\mu }$, write the type-$(1,1)$ tensor asFor a covector $\eta ={\eta }_{\mu }d{x}^{\mu }$ and vector $V=V^{\nu }{\partial }_{\nu }$,Since this must equal $\eta (V)={\eta }_{\nu }{V}^{\nu }$ for every $\eta$ and $V$, the components areThus $T$ is the identity tensor.

The tensor component transformation law givesThe components are therefore the same Kronecker delta in every coordinate system.

In the radiation zone, both fields are transverse to $\hat{\mathbf{x}}$ and obeyThe outward Poynting vector isso its time average isAn absorbing shell receives the incident momentum rather than reversing it. Hence radiation pressure on a perfectly absorbing surface givesAt $r=R$,and thereforeIf $\vartheta$ is the angle between $\hat{\mathbf{x}}$ and ${\mathbf{p}}_0$, this isIts spherical average is ${\mu }_0{\omega }^4|{\mathbf{p}}_0{|}^2/(48{\pi }^2{R}^2{c}^2)$; the vector force integrates to zero by the inversion symmetry of the electric dipole radiation pattern.

Let $\mathbf{\sigma }$ be the stated Maxwell stress tensor. Taking its divergence and usinggivesThe Maxwell equations in vacuum with sources areSubstitution, together with antisymmetry of the cross product, yieldsTherefore the required vector is the electromagnetic momentum densityandThis is local conservation of electromagnetic momentum. The tensor component ${\sigma }_{ij}$ is the flux of $i$-momentum in the $j$ direction in the convention of the question, $\mathbf{g}$ is field momentum per unit volume, and $\rho \mathbf{E}+\mathbf{J}\times \mathbf{B}$ is the Lorentz force density transferred to matter.

The first law of thermodynamics isso, regarding $S$ as a function of $(T,V)$,Equality of mixed partial derivatives gives the thermodynamic derivation of blackbody radiation pressureWriting $a={\pi }^2/(15{\hslash }^3{c}^3)$, we have $E=aV{T}^4$, and henceEquivalently, $d(P/T)/dT=a{T}^2$, soThe condition $P=o(T)$ as $T\to 0$ forces $C=0$. Therefore the radiation pressure is

Expanding $(e^x-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nx}$ and using the Gamma integral givesThus the energy of blackbody radiation is

In a box of volume $V$, the electromagnetic mode density in a box, including the two transverse polarizations, isEach photon has energy $\hslash \omega =\hslash ck$, and the given mean occupation number is $(e^{\hslash ck/T}-1{)}^{-1}$. Omitting the temperature-independent zero-point energy, the total energy is thereforeWith the change of variables $x=\hslash ck/T$,

For a real short-range spherically symmetric potential, the stationary wavefunction must solve the Schrödinger equation, be regular at the origin whenever the potential is finite there, and satisfy the scattering boundary conditionHere $e^{ikz}$ is the incident plane wave, the second term is an outgoing spherical wave, and its coefficient $f(\theta )$ is the scattering amplitude.

Comparing the stated asymptotic expression with the partial-wave expansion of a scattering amplitude giveswhere $P_l$ is a Legendre polynomial. The outgoing-to-incoming coefficient in channel $l$ isPartial-wave unitarity requiresor equivalentlyThe real scattering phase shift ${\delta }_l$ is defined modulo $\pi$ by

The transformationsform a one-parameter group with infinitesimal generatorUnder this scaling, ${\partial }_X$ has weight $-1$ and ${\partial }_T$ has weight $1$, so $u_{XT}$ has weight zero, as does $\sin u$. Thus $u_{XT}=\sin u$ is invariant and $g^s$ is a Scaling symmetry of the sine-Gordon equation.

The productis invariant under the group. A group-invariant solution therefore has the form $u(X,T)=F(z)$. Sincethe Sine-Gordon equation reduces to

Now put $w=e^{iF}$. ThenSubstitution and multiplication by $iw$ giveHenceThis is the Painlevé III equation with parameters $(\alpha ,\beta ,\gamma ,\delta )=(1/2,-1/2,0,0)$.

Write $\xi =V_1$, $\eta =\varphi$, and $u_j=d^{j}u/d{x}^j$. On the jet space of a scalar ordinary differential equation, definewhere the total derivative operator isThe order-$N$ prolongation of a vector field is

In the coordinates $(x,u)$, the given action of the special orthogonal group isDifferentiating at $s=0$ gives the infinitesimal generator of a Lie point symmetryso $\xi =u$ and $\eta =-x$. The recursive formula givesandTherefore the second prolongation of the rotation generator for plane graphs is

For ${\Delta }_0=u_{xx}$,which vanishes when ${\Delta }_0=0$. Forwe obtainwhich likewise vanishes on the equation. This is the infinitesimal invariance criterion for both equations.

Geometrically, $u_{xx}=0$ describes straight lines. The signed curvature of a plane graph isso the second equation says $\kappa =1$ and describes consistently oriented arcs of unit circles. Rotation preserves straight lines, circles, and signed curvature, explaining both invariances.