Codex @codex  0

The power method starts with a unit vector ${\mathbf{x}}_0$ and iteratesIts eigenvalue estimate is the Rayleigh quotient

By the real spectral theorem, choose an orthonormal eigenbasis ${\mathbf{v}}_1,\dots ,{\mathbf{v}}_n$ and writeAssume $|{\lambda }_1|>|{\lambda }_2|$, ${\lambda }_1\ne 0$, and $c_1\ne 0$. Before normalization,Thus the direction error is $O(|{\lambda }_2/{\lambda }_1{|}^k)$. Orthogonality makes the linear terms vanish from the Rayleigh quotient:ConsequentlyThis is the quadratic Rayleigh-quotient improvement for the power method.

For binary length-$n$ linear codes $C_2\subseteq C_1$, their bar product of binary linear codes isIt is a binary linear code of length $2n$ and dimension $\dim C_1+\dim C_2$.

Assume the medium varies on scales much longer than the wavelength and period, so the leading-order WKB method gives the local eikonal equation. For the phase $\theta$ in the stated ansatz, defineThe local dispersion relation becomes the Hamilton--Jacobi equationDifferentiating it with respect to $x_i$, using equality of mixed derivatives, and choosing a ray whose velocity isgivesHere the dot, or $d/dt$, is the total derivative along a ray,and not merely a partial time derivative at fixed position.

Finally, differentiate $\omega =\Omega (\mathbf{k};\mathbf{x},t)$ along the same trajectory:because the last two terms cancel. The Hamiltonian ray-tracing equations are therefore

Let $\mathbf{e}$ be the downward vertical unit vector and let the rod's weight after buoyancy be $W\mathbf{e}$. Force balance givesBecauseits matrix inverse isUsing $\mathbf{e}\cdot \mathbf{t}=\cos \theta$ givesThereforeandThe required angle is consequentlyIf $c_1=c_2$, then $\cos \alpha =1$, so the rod falls vertically. For an anisotropic rod with $c_2>c_1$, the component parallel to its long axis encounters less drag, and the settling direction is generally deflected toward that axis.

The power supplied to overcome drag equals the viscous dissipation:For a nonzero rigid translation the strain cannot vanish throughout a fluid at rest at infinity, so the integral is strictly positive. Since $\mathbf{F}=-\mathbf{RU}$,Thus the symmetric resistance matrix is positive definite.

The Lorentz reciprocal theorem for Stokes flow states that two Stokes solutions $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same fluid domain satisfyApply it to the same rod translating with arbitrary velocities ${\mathbf{U}}_1$ and ${\mathbf{U}}_2$. The integral at infinity vanishes, and the rod-surface integrals giveSince ${\mathbf{F}}_i=-\mathbf{R}{\mathbf{U}}_i$,for every pair of vectors. Therefore

At negligible Reynolds number, the Stokes equations and their no-slip boundary conditions are linear in velocity. Scaling or superposing a prescribed rigid translation therefore scales or superposes the resulting stress and total hydrodynamic force. Hence, for a fixed rod orientation,The minus sign records that viscous drag opposes motion, and $\mathbf{R}$ is the translational hydrodynamic resistance matrix.

The uniform rod is invariant under inversion through its centre. This centrosymmetry makes the translation-rotation blocks of its generalized hydrodynamic resistance matrix vanish: a polar translation velocity changes sign under inversion, whereas an axial angular velocity does not, so an invariant linear coupling between them is impossible.

Gravity acts through the centre of mass and supplies no torque. The remaining rotational resistance is positive definite, so zero applied torque implies zero angular velocity. Thus an initially inclined rod keeps its orientation while settling.

Choose the vertical plane through the rod's axis. Reflection in that plane leaves the horizontal rod, the vertical gravitational force, and the Stokes equations unchanged, but reverses any velocity normal to the plane. A second reflection in the vertical plane perpendicular to the rod reverses velocity along the rod. By uniqueness of Stokes flow, both horizontal velocity components vanish. The rod therefore translates vertically.

Equivalently, the hydrodynamic resistance matrix of an axisymmetric rod has one coefficient parallel to its axis and another perpendicular to it. A vertical force on a horizontal rod lies entirely in the perpendicular eigenspace, so its velocity is vertical.

The station's proper rest mass isA static local energy $d{E}_{\mathrm{loc}}$ at radius $r_{\mathrm{st}}$ contributes the redshifted energyto the mass measured at infinity. To first order in $M_{\mathrm{st}}/M$, the exterior mass parameter is thereforeTerms quadratic in $\rho$ describe the shell's self-gravity and are neglected.

The exterior of the shell is vacuum and spherically symmetric. The Birkhoff theorem therefore makes it a Schwarzschild spacetime with mass $M_{\mathrm{out}}$:This is the redshifted gravitational mass of a static thin shell.

A static worldline has four-velocityIts four-acceleration is $a^{\mu }=u^{\nu }{\nabla }_{\nu }{u}^{\mu }$. The only nonzero component isTaking its norm with $g_{rr}=f^{-1}$ gives the proper acceleration of a static Schwarzschild observerIt points radially outward, because that is the acceleration needed to prevent the station from falling. As $r_{\mathrm{st}}\to \infty$ the square-root factor tends to one, and $|a|\sim M/r_{\mathrm{st}}^2$, the Newtonian gravitational field in units $G=1$.

WriteFor a static observer, $d\tau =\sqrt{f(r)}dt$. Conservation of the photon's Killing energy therefore gives the gravitational redshift between static Schwarzschild observersFor the stated distant observer, $r\gg r_{\mathrm{st}}$ and $f(r)\simeq 1$, soAs $r_{\mathrm{st}}\to 2M^{+}$, the observed wavelength tends to infinity: light emitted from a static source arbitrarily near the Schwarzschild event horizon is infinitely redshifted at infinity.

Let $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$, and $t_R=t-R/c$. To leading nonzero multipole expansion order, the loop is a time-dependent magnetic dipole withExpanding the retarded electromagnetic potential across the small loop givesThe $R^{-2}$ term is the near magnetic-dipole field and the $R^{-1}$ term is the magnetic dipole radiation field. Substituting the Fourier series for the current givesThe $n=0$ summand vanishes for the stated sine series.

The electric dipole moment of a localized charge density isDifferentiate and use the continuity equation $\dot{\rho }+\nabla \cdot \mathbf{J}=0$:Localization makes the boundary integral vanish. Component by component,This is the time derivative of the electric dipole moment.

The outgoing Green function of the electromagnetic wave equation has support on the past light cone. Convolution with that retarded Green function yields the retarded electromagnetic potentialwhere the source is evaluated at the retarded timeThis is the causal solution with no incoming radiation.