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Let $k_S=\omega /c_S$ and $k_P=\omega /c_P$. Snell law for elastic waves fixes the reflected waves' common tangential wavenumber:Choose the reflected SV-wave and converted reflected P wave asThe polarizations are respectively perpendicular and parallel to their wavevectors. Phase matching makes all three exponential factors equal at $z=0$. The rigid boundary condition for an elastic wave ${\mathbf{u}}_I+{\mathbf{u}}_{RS}+{\mathbf{u}}_{RP}=0$ there givesSolving this two-by-two linear system yieldsThese fields give the Reflection of an SV-wave from a rigid plane.

The first $m$ letters of the blocked source contain exactly the first $mN$ letters of the original source, soIf the original information rate is $h$, thenThus fixed-length blocking multiplies information rate by $N$, exactly as recorded by information rate under fixed-length blocking.

Let $a$ denote the bubble size, $U$ its speed, and $\mathrm{Re}=Ua/\nu \gg 1$ the Reynolds number. Outside a thin boundary layer, inertia dominates and the flow is an irrotational flow with velocity scale $U$ and strain scale $U/a$. Its viscous dissipation has orderin three dimensions.

The boundary-layer thickness isFor a clean bubble the stress-free boundary condition prescribes tangential stress rather than tangential velocity. The outer flow already has tangential strain of order $U/a$, so cancelling that stress requires a velocity correction of onlynot a correction of order $U$. Its gradient remains $O(U/a)$, and its boundary-layer dissipation iswhich is asymptotically smaller.

In steady translation, the drag power $DU$ balances viscous dissipation. The leading estimate therefore follows from the known outer flow alone:in three dimensions. This is the dissipation estimate for high-Reynolds-number bubble drag.

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

Contracting the Riemann curvature tensor gives the nonzero components of the Ricci tensor:The Ricci scalar is the contraction with the inverse metric. ThereforeIt is independent of both coordinates, so the hyperbolic plane has constant negative scalar curvature.

Using the stated convention for the Riemann curvature tensor, the first independent component isThe other isusing the hyperbolic Pythagorean identity ${\coth }^{2}r-{\mathrm{csch}}^{2}r=1$. Antisymmetry in the final two indices supplies the components with those indices reversed. These are the curvature of the two-dimensional hyperbolic polar metric components.

For an affine parameter $\lambda$, the geodesic equation and the connection coefficients from part (a) givewhere dots denote derivatives with respect to $\lambda$.

On a line of constant $\varphi$, one has $\dot{\varphi }=0$. The equations reduce to $\ddot{r}=0$, so $r$ itself can be chosen as an affine parameter. Every such radial line is therefore a geodesic.

For a circle $r=r_0>0$, the first equation instead requiresBoth hyperbolic factors are positive when $r_0>0$, so $\dot{\varphi }=0$. This gives only a constant point, not a parametrized circle. Thus no circle of constant positive $r$ is a geodesic, in agreement with the coordinate geodesics of the hyperbolic polar metric.

For coordinates $x^{\alpha }$, the Christoffel symbols of the Levi-Civita connection arewhere $g^{\alpha \delta }$ is the inverse metric.

For the two-dimensional hyperbolic metric in polar coordinates,Only $g_{\varphi \varphi }$ has a nonzero derivative, namelyIt follows that the nonzero connection coefficients are

The condition $L\ll na$ implies $|\tau a|\ll 1$. The small-argument expansion of the hyperbolic tangent gives $\tanh x=x+O(x^3)$, soUsing $f=k_{B}T\tau$ yields the entropic Hooke law for a one-dimensional chainThus the effective spring constant is $k_{B}T/(n{a}^2)$.

At fixed tension, the exact extension isIncreasing the temperature decreases the positive argument of the hyperbolic tangent, so the chain contracts. In the small-extension regime this becomeswhich is inversely proportional to $T$.

If a physical tension $f$ acts on a state of extension $l_i$, its mechanical energy is $-f{l}_i$. The canonical ensemble therefore assigns that state the Boltzmann factorComparison with the statistical weight $e^{\tau l_i}$ givesEquivalently, $f$ is the thermodynamic conjugate variable to $L$ in the Gibbs free energy, just as pressure is conjugate to volume.

Substituting the fixed-tension partition function of a one-dimensional chain into the given free-energy definition givesDifferentiation with respect to the parameter conjugate to extension givesThe same result follows directly from the expected value under $p_i=e^{\tau l_i}/Z$:Consequently

Several microstates can have the same extension. If $g(l)$ denotes their state degeneracy, grouping equal terms in the state sum givesFor the one-dimensional freely jointed chain, a macrostate with $n_{+}$ positive links and $n_{-}=n-n_{+}$ negative links hasHenceThe binomial theorem now gives the fixed-tension partition function of a one-dimensional chain

The mean extension in an ensemble of $A$ molecules iswith constraints ${\sum }_i{a}_i=A$ and ${\sum }_i{a}_i{l}_i=AL$. For occupancies $\{a_i\}$, the number of assignments of the $A$ distinguishable ensemble members to the states is the multinomial coefficientThus maximizing the probability at fixed $A$ and $L$ is equivalent to maximizing the stated Lagrange multiplier expression.

Because $A\gg 1$, the Stirling formula gives $\log (a_i!)\simeq a_i\log a_i-a_i$. Differentiating with respect to each $a_i$ givesNormalization by ${\sum }_i{a}_i=A$ therefore yields the probability mass functionHere $Z$ is the fixed-tension partition function that normalizes the probabilities.

The extension of the one-dimensional freely jointed chain isAs the integer $n_{+}$ runs from $0$ to $n$, the possible extensions are thereforeFor fixed $n_{+}$ and $n_{-}=n-n_{+}$, choosing which $n_{+}$ of the $n$ links point in the positive direction determines the microstate. The state degeneracy is consequently the binomial coefficient