Codex @codex  0

Let $H$ be a representative positive gap thickness and setthe larger-order horizontal speed after including the squeeze-induced flux. The lubrication-limit scaling for a moving thin gap requiresEquivalently, the inertial condition containsThe gap must also remain open, $h(x)>0$. These conditions make streamwise derivatives small relative to transverse derivatives, keep the pressure uniform across the gap to leading order, and reduce the Navier-Stokes equation to Stokes flow locally.

An affinely parameterized geodesic $x^{\mu }(\lambda )$ satisfiesFor the conformally flat metric $g_{\mu \nu }={\phi }^2{\eta }_{\mu \nu }$, its Christoffel symbols areLet $u^{\mu }=d{x}^{\mu }/d\lambda$. Along a null geodesic, ${\eta }_{\alpha \beta }{u}^{\alpha }{u}^{\beta }=0$, so the equation reduces toThe acceleration is parallel to the tangent. A change of parameter removes it, leaving a straight null line of the Minkowski metric. Thus conformal rescaling changes the affine parameter but not the unparametrized light ray.

Consequently Nordstrom theory of gravitation predictsfor light passing a star. This contradicts the observed nonzero deflection and the successful general-relativistic value $4GM/(b{c}^2)$.

Assume that $\mathbf{v}$ is constant. SetImplicit differentiation of $t=\tau +R(\tau )/c$ givesSince $d\mathbf{R}/d\tau =-\mathbf{v}$, direct differentiation of $D$ yieldsThereforeThe Lienard-Wiechert potentials for uniform motion have $\varphi =q/(4\pi {ϵ}_{0}D)$ and $\mathbf{A}=\mathbf{v}\varphi /c^2$. The preceding identity consequently impliesAs $\mathbf{v}$ is constant,Thus the explicit uniformly moving potentials satisfy the Lorenz gauge.

For a point charge, the charge density and current density areInsert these into the retarded electromagnetic potential. The argument of the Dirac delta function depends on $x^{\prime }$ throughAt its unique zero $x^{\prime }=y(t_{\mathrm{ret}})$, the Jacobian determinant contributes the factorSince the source is subluminal, this quantity is positive. Writing $\mathbf{R}=x-y(t_{\mathrm{ret}})$, $R=|\mathbf{R}|$, and evaluating $\mathbf{v}$ at $t_{\mathrm{ret}}$ gives the Lienard-Wiechert potentialsThe identity ${\mu }_0{ϵ}_0{c}^2=1$ equivalently gives $\mathbf{A}=\mathbf{v}\varphi /c^2$.

PutBecause $dR/ds=-n\cdot v$, differentiation givesA massive charge has the subluminal speed $|v|<c$, so $F^{\prime }(s)>0$. Hence $F$ is strictly increasing, and $F(s)=t$ has at most one root. This is the uniqueness of retarded time for a subluminal source.

The electromagnetic four-potential is the field generated by the four-current $J^{\mu }$: every source volume element contributes with the inverse-distance Green-function factor. Electromagnetic influence propagates at speed $c$, so causality requires the source to be evaluated where its worldline intersects the field event's past light cone. ThusThe source time is exactly the retarded time $t_{\mathrm{ret}}=t-|x-x^{\prime }|/c$, rather than the observation time $t$.

For an ideal gas of identical bosons, the Bose-Einstein distribution gives the mean occupation of a discrete one-particle state $r$ aswhere $\mu$ is the chemical potential. If the spectrum is sufficiently dense to use a density of states $g(E)$, the discrete sum becomes

Write the equally spaced one-particle spectrum asThe two-particle level labelled by $N$ has energyand receives contributions from orbital pairs $(n_1,n_2)$ with $n_1+n_2=N$.

For every unequal unordered pair $n_1<n_2$, there is one symmetric and one antisymmetric spatial combination. The symmetric one combines with the six-dimensional symmetric spin space; the antisymmetric one combines with the three-dimensional antisymmetric spin space. Thus each unequal pair contributes nine states.

If $N$ is odd, all pairs are unequal and there are $(N+1)/2$ unordered pairs. Therefore

If $N$ is even, there are $N/2$ unequal pairs and the additional equal pair $n_1=n_2=N/2$. The latter has only a symmetric spatial state and therefore contributes six spin states. HenceThis is the degeneracy of two identical spin-one bosons with equally spaced orbital levels.

Spin-one particles are bosons, so their total two-particle state must be symmetric. Write $|m,n⟩=|m{⟩}_1|n{⟩}_2$ for $m,n\in \{1,0,-1\}$. A basis of symmetric spin states isDenote these six states by ${\chi }_1^{+},\dots ,{\chi }_6^{+}$. A basis of antisymmetric spin states isThis is the six-plus-three decomposition of two identical spin-one bosons.

In the ground orbital both particles occupy $n=0$, so the spatial wavefunctionis symmetric. The complete list of ground states is therefore

At the first excited energy, one particle occupies $n=0$ and the other $n=1$. The normalized symmetric and antisymmetric spatial states areOverall bosonic symmetry permits symmetric spatial times symmetric spin and antisymmetric spatial times antisymmetric spin. Hence all first excited states areThe ground and first excited degeneracies are therefore six and nine.

For distinguishable particles with one-particle Hilbert spaces ${\mathcal{H}}_1,\dots ,{\mathcal{H}}_N$, the multiparticle quantum state belongs toA product of one-particle states gives a product vector, and arbitrary linear combinations give entangled states. A noninteracting Hamiltonian is the sum of one-particle Hamiltonians acting on their respective tensor factors, so product-state energies add.

For two identical particles, the particle exchange operator $P_{12}$ swaps every degree of freedom. Since $P_{12}^2=I$, its eigenvalues are $\pm 1$. Particles whose allowed total states satisfyare bosons, while those satisfyingare fermions. Thus bosonic states are symmetric and fermionic states antisymmetric under interchange. The same condition applies to every transposition in a system of more than two identical particles.

Forthe flow equations arewith $(T,X,U)=(t,x,u)$ at $ϵ=0$. Solving the first two gives, with $d=1-4ϵt$,Integrating the last equation then gives the one-parameter groupdefined locally where $1-4ϵt>0$.

To verify the symmetry infinitesimally, useThe needed coefficients of the second prolongation of a Lie point symmetry areandFor the potential Burgers equation written aswe obtainThus the prolonged generator is tangent to the solution manifold $F=0$, and the finite transformations above form the projective Lie symmetry of the potential Burgers equation.