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The minimal polynomial of a linear transformation $T$ is the unique monic polynomial of least degree among the nonzero polynomials $p\in F[x]$ satisfying $p(T)=0$:Such a polynomial exists because the endomorphism space is finite-dimensional, so sufficiently many powers $I,T,T^2,\dots$ are linearly dependent. Monic normalization is possible over the field $F$. Two monic polynomials annihilating $T$ of the same least degree have a difference of smaller degree annihilating $T$, so their difference is zero. For the zero vector space the convention is $m_T=1$.

The Doppler factors for the two discrete blobs are ${\mathcal{D}}_{\pm }=[\gamma (1\mp \beta \cos \theta ){]}^{-1}$. Their ratio is $(1+x)/(1-x)=77/33=7/3$. Lorentz-invariant specific intensity means $I_{\nu }/{\nu }^3$ is unchanged; with $\nu =\mathcal{D}{\nu }^{\prime }$ and the same rest-frame power-law spectrum,The brightness ratio at the common observed frequency therefore givesThe stated identical intrinsic specific intensities must refer to comparable intrinsic states. If the blobs fade with their proper ages, equal observer times sample different ages and need that correction. A steady continuous jet has a different boosting exponent, $2+\alpha$; the discrete ejecta and the specified specific intensity transformation justify $3+\alpha$ here.

For optically thin synchrotron radiation from an isotropic electron distribution $dN/dE\propto E^{-p}$, the characteristic emitted frequency scales as $B{E}^2$ and the total single-particle power as $B^2{E}^2$. Thus $j_{\nu }\propto E^{-p}{E}^2/(d\nu /dE)\propto {\nu }^{-(p-1)/2}$; integration over the full single-particle spectrum preserves this exponent. HenceThe test-particle strong nonrelativistic diffusive shock acceleration prediction is $p=(r+2)/(r-1)=2$ for compression ratio $r=4$, giving $\alpha =0.5$. The inferred spectrum is steeper. If interpreted literally within that uncooled formula, $p=2.50$ would correspond to $r\simeq 3$, but cooling, escape, and other acceleration conditions can also steepen it.

The observed Galactic cosmic ray differential energy spectrum below the knee is approximately $E^{-2.7}$, steeper still; its formal synchrotron radiation equivalent would be $\alpha \simeq 0.85$. Thus $p\simeq 2.50$ lies between the simplest source prediction and the propagated cosmic ray slope. The comparison is qualitative: most measured cosmic rays are nuclei, whereas the radio slope here probes electrons, and propagation and radiative losses alter the spectra differently.

Relativistic beaming concentrates the emission into a forward cone about the instantaneous velocity. The aberration of light maps a rest-frame ray at ${\theta }^{\prime }=\pi /2$ to $\cos \theta =\beta$, so the characteristic cone half-angle is $\theta \simeq {\gamma }^{-1}$. Thus its conventional full width is $2/\gamma =2\times 10^{-6}\mathrm{rad}$.

The isotropic-emission assumption has no beam nulls. For example, its received bolometric angular power is proportional to $D^4$, where the Doppler factor is $D=[\gamma (1-\beta \cos \theta ){]}^{-1}>0$. In particular it is nonzero at $\theta =1/\gamma$, disproving a literal null there. A true dipole pattern for an accelerated electron is not isotropic in its rest frame and needs a specified polarization and viewing plane. The width just quoted is a characteristic beaming angle, not a null-to-null width under the stated premise.

For a circular orbit seen in its orbital plane, let the velocity direction sweep past the observer through $-1/\gamma <\psi <1/\gamma$. The emission interval is $\Delta t_{\mathrm{em}}\simeq 2/(\gamma {\omega }_g)=1.14\times 10^{-2}\mathrm{s}$. The observed interval must also include light-travel compression:Its ratio to the gyromotion period is $\Delta t_{\mathrm{obs}}/T_g\simeq 2/(3\pi {\gamma }^3)=2.1\times 10^{-19}$. Since the boundary of a smooth beam is conventional, only the scale ${\omega }_g^{-1}{\gamma }^{-3}$ is invariant under reasonable pulse-width definitions. For helical motion the viewing geometry and pitch angle are additional inputs; the numerical estimate assumes transverse motion and a sightline intercepted by the cone.^[ref]

Set the ground-state energy to zero. Bose-Einstein occupation of an excited level is $[e^{(ϵ-\mu )/(k_{B}T)}-1{]}^{-1}$, with $\mu \le 0$. As particles are added, $\mu$ approaches zero and the excited-state population saturates. Separating the ground state from the density-of-states integral therefore givesThe excited-state integral must actually converge: $g\to 0$ alone is not a sufficient condition in complete generality. A power law $g\propto {ϵ}^s$ with $s>0$ suffices at low energy, as in the stated three-dimensional box.

Let $\mathcal{C}=V{m}^{3/2}/(\sqrt{2}{\pi }^2{\hslash }^3)$ denote the density-of-states prefactor, not a heat capacity. Substitution $x=ϵ/(k_{B}T)$ givesThe condensate has zero energy on this reference, so a further factor of $ϵ$ in the integral givesRelative to the classical value, the ratio is $\zeta (5/2)/[\eta \zeta (3/2)]\simeq 0.5134/\eta <0.5134$. Macroscopic occupation of the zero-energy state and the Bose distribution both lower the average energy.

At fixed volume the condensed gas has $E\propto T^{5/2}$. Unbiased random loss keeps average $E/N$ constant, henceRemoving a condensate atom removes essentially no energy and by itself leaves the excited cloud's temperature unchanged. Removing an excited atom removes its energy; during re-equilibration, some zero-energy condensate atoms join the excited population and share the remaining energy, lowering temperature. Averaging over unbiased loss gives the stated constant energy per particle, even though each individual type of loss need not preserve it separately.

Using $m\simeq 87u$ and the supplied volume,There is only one occupied internal spin state, so no factor $2F+1$ belongs in the density of states. During loss, $\eta \propto N/T^{3/2}\propto N^{2/5}$, hence $\eta /{\eta }_i=T/T_i$. Condensation disappears at ${\eta }_f=1$:As a consistency check, $N_f=N_i{\eta }_i^{-5/2}\simeq 8.15\times 10^4$ agrees with $N_c(T_f)$.

Both this process and evaporative cooling of atoms require particle loss and re-equilibration. Ordinary evaporation preferentially removes energetic atoms, reducing $E/N$, and can cool an uncondensed classical gas. Here loss is energy independent, $E/N$ stays constant, and cooling relies on the condensed gas's unusual equation of state; random loss of a classical ideal gas would leave its temperature unchanged. Exact noninteraction would prevent collisional rethermalization: the imposed equilibrium assumption can be understood as weak thermalizing interactions whose equation-of-state corrections are neglected.

Here $k$ is a mass absorption coefficient, because $k\rho$ has units of inverse length. The Beer-Lambert law is $dF/dz=-k\rho F$, giving $F(z)=F_0{e}^{-k\rho z}$. A horizontal slab of area $A$ and thickness $dz$ absorbs power $A[F(z)-F(z+dz)]$. Dividing by its volume and taking the limit gives

Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

Yes, for genuine precession of a nonspherical disc. Let its unit normal be $\mathbf{n}=(\sin \alpha \cos \omega t,\sin \alpha \sin \omega t,\cos \alpha )$. Its trace-free mass quadrupole moment has the axisymmetric formThe $xz,yz$ entries vary at $\omega$, and $xx-yy,xy$ at $2\omega$. Their nonzero second and third time derivatives give an oscillating far-field metric perturbation and positive average radiated power. For $\alpha =\pi /2$ the $\omega$ terms vanish but the $2\omega$ terms remain. The precise condition is $\omega \ne 0$ and $\sin \alpha \ne 0$: the literal inequality $\alpha \ne 0$ also allows $\alpha =\pi$, where the normal is fixed and no such gravitational radiation occurs. For a uniform thin disc of mass $M_d$ and radius $b$, $Q_0=-M_d{b}^2/4$, confirming that it is not zero.

The annual, zonal-mean absorbed sunlight falls from the tropics towards the poles much more strongly than outgoing thermal radiation. Thus Earth's top-of-atmosphere radiation budget has a tropical surplus and a polar deficit, with changes of sign around $35^{\circ }$ latitude in a basic sketch. The curves below are illustrative, rather than measured data; their area-weighted global means agree.

In a statistically stationary climate, local conservation of energy requires the surplus to be exported. If $R(\varphi )$ is the net downward radiative heat flux and $\mathcal{H}(\varphi )$ the northward total power carried across the latitude circle, thenHere $a$ is Earth's radius. The terrestrial atmosphere and ocean circulation carry this energy poleward by moving warm and cold fluid, including atmospheric transport of latent heat. Without this transport the assumed steady radiation budget could not persist: the tropics would warm and polar regions cool until the budget changed. The tropical atmospheric contribution includes the Hadley cell: rising warm air near the equator, poleward flow aloft, subtropical descent, and an equatorward lower return.

For a parcel moving without an axial torque, conservation of angular momentum applies in an inertial frame. The specific axial angular momentum at latitude $\varphi$ is“Initially at rest” means $u=0$ relative to the rotating surface at the equator, so $M=\Omega a^2$. Neglecting atmospheric height compared with $a$ givesWith $\Omega =7.292\times 10^{-5}{\mathrm{s}}^{-1}$ and $a=6.371\times 10^6\mathrm{m}$, the eastward velocity at $30^{\circ }\mathrm{N}$ isThis is the ideal angular-momentum-conserving upper branch of the Hadley cell. Real jet streams exchange angular momentum through eddies and friction, so this estimate is not an observational claim about the actual wind speed.

The upper-state density is clamped near threshold once the laser operates, so excess pumping mainly increases photon density and output power. For approximately pump-independent absorption efficiency, $r=P_p/P_{p,\mathrm{th}}$, and both $P_{\mathrm{out}}$ and $f_{\mathrm{ro}}^2$ should rise approximately linearly with $P_p-P_{p,\mathrm{th}}$.

The left ordinate is the square of the frequency, in ${\mathrm{kHz}}^2$, rather than frequency in kHz. Reading the original triangles gives a threshold near $12$–$13\mathrm{mW}$ and a slope about $1.0{\mathrm{kHz}}^2/\mathrm{mW}$. A representative straight-line fit to visually read marker centers gives $P_{p,\mathrm{th}}\simeq 12.2\mathrm{mW}$ and $A=d{f}_{\mathrm{ro}}^2/d{P}_p\simeq 1.02\times 10^6{\mathrm{Hz}}^2/\mathrm{mW}$. ThereforeThe displayed precision is limited by graph reading and scatter; a lifetime around $0.15\mu \mathrm{s}$ is the meaningful result. The output squares have a compatible threshold and an approximately constant slope above it.

The round-trip time isIf mirror transmission is the only appreciable passive loss, energy decays by $R_1{R}_2=e^{-t_{\mathrm{rt}}/{\tau }_c}$ per round trip. For $R_1=1$,a power reflectance of roughly 65%. The exact exponential is preferable to replacing $1-R_2$ by $t_{\mathrm{rt}}/{\tau }_c$, since the inferred round-trip loss is not tiny. If distributed fibre loss is appreciable, the lifetime instead fixes the total round-trip retention; without a separate loss measurement, the bare mirror reflectance is not uniquely determined and is at least this retention factor.

Q-switching stores excitation in a pumped laser gain medium while preventing oscillation, then rapidly changes the optical cavity from high loss, or low quality factor, to low loss, or high quality factor. The initially suppressed stimulated field can then grow rapidly from a seed and deplete a large population inversion in a short pulse. The resulting peak power is much higher than for continuous operation at the same average pumping rate. This is energy storage and release, rather than the phase locking of many longitudinal modes used in mode locking.

An active implementation uses a Pockels cell and a polarizer. An applied voltage makes the cell change the intracavity polarization, causing the polarizer to introduce sufficient loss to prevent lasing while pumping builds the inversion. Rapidly changing the voltage removes that loss and opens the high-Q state. The switching time should be short relative to the pulse build-up and depletion times. A saturable absorber provides an alternative passive Q switch: its absorption decreases as the intracavity intensity rises.

During the pulse, the photon density reaches its maximum when $\dot{n}=0$ with $n>0$, so $N^{\ast }=N_{\mathrm{th}}^{\ast }$. Evaluating i at this inversion givesThe ideal instantaneous output power is $P(t)=V_c\hslash {\omega }_{L}n(t)/{\tau }_c$. ThereforeFor $r\gg 1$, almost all the extractable energy is stored briefly as intracavity photons, and $P_{\max }\simeq E_{\max }/{\tau }_c$. The leading rise while the inversion remains near its initial value is $n\propto e^{(r-1)t/{\tau }_c}$, with short growth time ${\tau }_c/(r-1)$. Depletion then takes the inversion below threshold. At late times $N^{\ast }\simeq N_f^{\ast }$ andwhich approaches a tail with decay time ${\tau }_c$ at large $r$. The pulse is consequently asymmetric, with a rapid rise and a slower approximately exponential trailing edge. The negligible-seed build-up delay depends on the seed strength. The pumping and spontaneous population decay have been neglected over the short pulse; parasitic cavity loss multiplies useful power by the output fraction described in ii.

Put $s=z-z_0$. In the source's complex beam parameter convention,Substitution separates the transverse wave amplitude and phase:Thus the Gaussian beam is circularly symmetric, has a Gaussian function transverse light intensity, and narrows to a beam waist at $z_0$ before spreading by diffraction. Its wavefront is planar at the beam waist, converging before it and diverging after it. The quadratic transverse phase identifies the signed radius of curvature:Define spot size as the wave amplitude $1/e$ radius, equivalently the light intensity $1/e^2$ radius. ThenThe Rayleigh range is where $w=\sqrt{2}{w}_0$; the far-field beam divergence half-angle is $\lambda /(\pi w_0)$. The factor $1/q$ also produces a longitudinal phase shift relative to a plane wave, changing by $\pi$ across the beam waist; its sign here follows the given $e^{i(kz-\omega t)}$ convention. The on-axis light intensity decreases as $1/(s^2+z_R^2)$ while transverse integration leaves the power constant. These are scalar paraxial approximation results.

For a symmetric confocal optical resonator, the mirror separation equals either mirror's curvature, $R_2=L_2$, so its symmetric fundamental Gaussian beam hasThe same optical mode matching calculation applies, now with $f_{\min }=\sqrt{z_A{L}_2/2}$. Keeping the example's $L_2=10\mathrm{m}$ gives $z_B=5\mathrm{m}$ and $f_{\min }=0.707107\mathrm{m}$, compared with $0.930605\mathrm{m}$ for the nonconfocal target. With $f=1.5\mathrm{m}$, the two branches becomeAgain only the positive-$v$ branch fits the stated downstream geometry. The confocal target's smaller Rayleigh range means a smaller fundamental beam waist at the same wavelength and a less stringent minimum $f$.

A confocal optical resonator also has transverse normal mode degeneracies: in the ideal paraxial spectrum, increasing transverse order by two shifts the resonance by one longitudinal spacing. Such a cavity can transmit combinations of higher transverse normal modes at a fundamental resonance, so transmitted power alone is a weaker diagnostic of pure fundamental optical mode matching. The degeneracy does not remove the complex beam parameter matching requirement if the objective is specifically the symmetric fundamental Gaussian beam.

The conventional relaxation-loss peak near $\omega \tau =1$ applies to fixed strain amplitude, not to fixed stress amplitude. This distinction matters because the question explicitly describes oscillating stress.

The full complex modulus of the series Maxwell element isFor imposed strain $\epsilon ={\epsilon }_0\cos \omega t$, the energy dissipated per volume per cycle isSince $x/(1+x^2)$ has its maximum at $x=1$, fixed-strain loss per cycle peaks at ${\omega }^2={\tau }^{-2}$. At that scale the imposed cycle time is comparable with the relaxation time, and substantial viscous flow occurs while stress remains appreciable.

For the literally stated comparison with fixed stress amplitude, use the complianceThe dashpot strain rate is ${\dot{\epsilon }}_d=\sigma /\eta$. Its dissipation integrated over a period givesThis decreases monotonically with frequency; there is no finite-frequency maximum at fixed stress amplitude in the ideal Maxwell model. Its mean dissipated power is ${\sigma }_0^2/(2\eta )$, independent of frequency. The formal low-frequency divergence in energy per cycle accompanies very large strain excursions and cannot be extrapolated beyond the linear model's range. The graph identifies both results rather than assigning the fixed-strain peak to the wrong loading condition.

For a sinusoidal drive of variable positive angular frequency, the average absorbed power is proportional toIt peaks at $\omega =\Omega$. The absorbed-power bandwidth of a viscously damped oscillator is the separation of its half-power frequencies: solving $|{\Omega }^2-{\omega }^2|=2\gamma \omega$ givesThe usual quality factor is $\boxed{Q=\Omega /(2\gamma )=\Omega /\Delta \omega }$. Large $Q$ means a narrow response band relative to the resonant frequency. The post-drive motion has amplitude envelope $e^{-\gamma t}$, connecting a narrow bandwidth to a long ringdown. For weak damping, the cycle-averaged energy decays as $e^{-2\gamma t}$ and $Q$ is also $2\pi$ times stored energy divided by the energy lost per cycle. The exact instantaneous energy has an oscillatory factor, so this cycle interpretation uses the weak-damping approximation.

Dividing Newton's equation by the mass gives the stated model: $\ddot{z}$ is the acceleration, $\gamma \dot{z}$ represents viscous damping, $(k/m)z$ is the spring's restoring acceleration, and the right side is the applied driving force per unit mass. The complex notation is a convenience; the physical displacement is the real part. Put ${\omega }_0^2=k/m$. The homogeneous characteristic roots areFor $m,k>0$ and damping $\gamma \ge 0$, oscillations occur when ${\gamma }^2<4{\omega }_0^2$, the underdamped case of the damped harmonic oscillator.

For a steady response $z=Z{e}^{i\omega t}$, substitution givesTaking the modulus and phase givesThe phase is between zero and $\pi$ for positive damping and frequency; an unqualified arctangent loses the correct quadrant above resonance.

With the real drive $F_0\cos \omega t$ and real response $r\cos (\omega t-\varphi )$, the average driving power equals the dissipated viscous power:For $\gamma >0$, write this asIt is largest when the squared term vanishes, givingThis power peak is at ${\omega }_0$ even when the displacement-amplitude peak is shifted. The frequency bandwidth of viscous oscillator power absorption follows from the half-power condition $|{\omega }_0^2-{\omega }^2|=\gamma \omega$:In fact the bandwidth identity is exact in this model. Sharp resonance additionally gives ${\omega }_{1,2}\simeq {\omega }_0\mp \gamma /2$. With zero damping at exact resonance, there is no bounded steady harmonic response to which these finite-power formulas apply.

In the steady periodic response, the cycle average of the mechanical energy derivative vanishes. The applied average power therefore equals the power dissipated by the viscous force:Using solution givesThe same result follows from averaging the instantaneous input power $F(t)\dot{z}(t)$, using the response phase relative to the drive.

For positive $m,k,\gamma$ and $\omega >0$, rewrite this asThe last factor is at most one, with equality precisely when $k-m{\omega }^2=0$. Thus the power peak occurs at ${\omega }_0=\sqrt{k/m}$ andThis power-resonance angular frequency is exactly ${\omega }_0$; it need not equal the angular frequency maximizing displacement amplitude. Positive damping is essential: at $\gamma =0$ and exact resonance there is no bounded steady periodic response to nonzero forcing, so the divergent expression is not a finite-power solution.

The instantaneous energy flux towards positive $x$ is the rate at which the string on the left does work on the string on the right. To first order its transverse force is $-T{y}_x$, soFor a right-travelling sinusoidal wave $y=A\cos (kx-\omega t)$, $y_x=-Ak\sin (kx-\omega t)$ and $y_t=A\omega \sin (kx-\omega t)$. ThusThe printed constant formula is therefore the time-average power, not the instantaneous power. The latter oscillates between zero and twice its average. A left-travelling wave carries negative signed flux through the same positively oriented section; its transported power magnitude has the same average formula with positive $k$.

Place the massless junction at $x=0$, with common tension $T$ on both sides. Using a common time factor $e^{-i\omega t}$, write the left-hand wave as $A_i{e}^{i{k}_{1}x}+A_r{e}^{-i{k}_{1}x}$ and the right-hand wave as $A_t{e}^{i{k}_{2}x}$. A stationary junction does not change the frequency, while $k_j=\omega \sqrt{{\rho }_j/T}$.

displacement continuity requires $A_i+A_r=A_t$. Transverse force balance on a massless junction requires equal tension-weighted slopes, giving $k_1(A_i-A_r)=k_2{A}_t$. Hence the signed displacement coefficients areHere ${\rho }_2=4{\rho }_1$, so $k_2=2k_1$ and $\boxed{A_r=-A_i/3,A_t=2A_i/3}$. The negative reflected coefficient signifies inversion. If “reflection and transmission coefficients” instead denotes power fractions, their values are $\mathcal{R}=1/9$ and $\mathcal{T}=8/9$, as derived separately below; power and displacement coefficients must not be conflated.

The reflected wave remains on the incident string, so its wavenumber magnitude is still $k_1$. Its outward power magnitude isThus $|P_r|+P_t=(1/9+8/9)P_i=P_i$, verifying energy conservation. With signed positive-$x$ fluxes, $P_r=-P_i/9$ and the same statement is $P_i+P_r=P_t$. The minus sign of the reflected displacement is a phase reversal, whereas the minus sign of its flux records travel in the opposite direction; these are different uses of a sign.