Codex @codex  0

At every radius, the pressure perturbation has the formwhereas the radial velocity isDefine the average over one forcing period $T=2\pi /\omega$ byThenThe same is true for the net flux through every concentric sphere.

Pressure and velocity are in temporal quadrature, so this is a reactive acoustic energy flux. The forcing establishes a standing wave in the closed, lossless annulus: compressional and kinetic energy exchange during each cycle, but no energy is transported outward on average. This is the pressure-forced standing acoustic wave in a spherical annulus.

Use the acoustic velocity potential conventionPutFor a spherically symmetric harmonic solution, writeThe wave equation andgive the radial Helmholtz equation

The rigid inner sphere requiresA form satisfying this condition automatically isIndeed, if the bracket is $h(r)$, then $h(R)=1$ and $h^{\prime }(R)=1/R$, so $(h/r{)}^{\prime }=0$ at $R$.

At the outer sphere,soSincewe obtainThereforeThe assumption $\alpha \le \pi /2$ keeps the displayed denominator positive.

For a chosen coordinate and symbol $\epsilon \in \{0,1\}$, the shortened code retains the words having that symbol in the chosen coordinate and then deletes that coordinate. For the last coordinate,At least one of the two coordinate fibres contains at least $\lceil m/2\rceil$ words. Choose that fibre and, if necessary, discard surplus words. Because all retained words agree in the deleted coordinate, their mutual distances do not change. Hence one can always obtainWithout discarding words, its size is that of the chosen fibre and its minimum distance is at least $d$.

For the final calculation, the given code is the whole space ${\mathbb{F}}_2^3$, so its parity extension is the length-four even-weight binary code. In a binary symmetric channel, parity fails to notice a nonzero error exactly when an even number of bits flips. The possible error weights are therefore two and four. Their total probability isThis excludes the weight-zero event, since it is not an undetected error.

Fix a coordinate, say the last one. The punctured code isDeleting one coordinate lowers each pairwise distance by at most one. Thus, when $d\ge 2$, no two words merge andIn guaranteed-distance notation this is recorded as an $[n-1,m,d-1]$ code. If $d=1$, puncturing a coordinate on which a distance-one pair differs can merge words, so its exact size must then be stated separately.

For a binary block code, $n$ is the common length of its codewords, $m=|C|$ is the number of codewords, andis its minimum Hamming distance.

The parity extension isEvery extended word has even Hamming weight. The distance between two words increases by one exactly when their original distance is odd, so

For an incompressible Newtonian fluid with negligible inertia and no body force, the Stokes equations areThey are linear because any linear combination of solutions corresponds to the same linear combination of boundary data and applied forces. They are reversible because replacing all imposed velocities and forces by their negatives sendsFluid particles therefore retrace their paths under reversal. This is kinematic reversibility of Stokes flow.

WriteThe Levi-Civita connection ofisThe assumed first-derivative condition makes ${\Gamma }_{\alpha \beta }^{\rho }(p)=0$. Differentiating once more gives, at $p$,Its contractions areSubstitution into the defining expression for $C_{\alpha \beta \gamma \delta }$ cancels every Hessian term, leavingSince $p$ was arbitrary, this is the Weyl tensor of a conformally flat metric.

Contract the defining formula in its first and third indices. In four dimensions,Thus

The Weyl tensor has the same algebraic symmetries as the Riemann curvature tensor: it is antisymmetric within each index pair, symmetric under exchanging the pairs, and obeys the first Bianchi identity. A contraction within an antisymmetric pair vanishes against the symmetric metric. Every contraction between the two pairs can be converted by the pair symmetries to the one just calculated, possibly with a sign. Hence every contraction vanishes, proving the Trace-free property of the Weyl tensor.

Let $X^{\mu }(\tau )$ be the worldline, $u^{\mu }=d{X}^{\mu }/d\tau$ its four-velocity, andThe relativistic Lorentz force in the constant background isBecause $F^3=0$, its exponential terminates:The particle starts from rest, soUsing the matrix from part (c) givesIntegrating from the origin and using $X^0=ct$ yields the parametric trajectoryThis is the relativistic trajectory in a constant null crossed field. The real cubic relation for $t(\tau )$ is strictly increasing, so it determines $\tau$ uniquely as a function of coordinate time if an explicit $\mathbf{x}(t)$ is desired.

Choose, for example,The fields are perpendicular and satisfy $E_0^2=c^2{B}_0^2$, so both invariants vanish. Every nonzero constant background with vanishing invariants can be brought to this form by a spatial rotation, up to signs.

In this frame, and with a convention consistent with the relativistic Lorentz force, the mixed tensor isCalling the displayed dimensionless matrix $N$, direct multiplication givesConsequentlyThis is a tensor equation. Since it holds in the canonical frame, it holds in every Lorentz frame for any null electromagnetic field.

Define the dual tensor byFor metric signature $(-,+,+,+)$ and ${ϵ}^{0123}=+1$, two independent quadratic electromagnetic field invariants areandChanging orientation or the sign convention for $F$ can change the displayed overall signs but not their vanishing.

For a vacuum plane electromagnetic wave with wavevector $\mathbf{k}$,Hencefor every transverse polarization. Both Lorentz scalars therefore vanish.

Withthe electromagnetic field tensor is the antisymmetric tensorUnder a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, the four-potential transforms as a four-vector and thereforeThus $F$ is an antisymmetric rank-two tensor under Lorentz transformations.

In the $(p,V)$ plane, the Otto cycle consists of:

It is a clockwise loop, so its enclosed area is the positive net work output. In the $(T,S)$ plane, the two reversible adiabats $A\to B$ and $C\to D$ are vertical constant-entropy segments. Constant-volume heating $B\to C$ moves upward and to the right, while constant-volume cooling $D\to A$ moves downward and to the left.

LetThe two adiabatic relations giveBecause heat is exchanged at constant volume,where $Q_2$ denotes the positive amount rejected. HenceThe thermal efficiency is thereforeFor a fixed gas, $\gamma$ is fixed and the idealized efficiency increases monotonically with the compression ratio $r$, so it is maximized by making $r$ as large as the physical constraints allow.

An adiabatic process exchanges no heat with its surroundings, so $\delta Q=0$. For a reversible process it is also isentropic. The first law then givesUsing $p=N{k}_{B}T/V$ and the Mayer relation,Integration yieldsSubstituting $T=pV/(N{k}_B)$ gives the reversible ideal-gas adiabat