The central pair potential is invariant under simultaneous spatial translations, simultaneous rotations and time translations. The corresponding time-independent Noether charges are total momentum $P=p_1+p_2$, total angular momentum about a fixed origin, and total energy $H$.

In one dimension this supplies $P$ and $H$: two charges. In two dimensions it supplies $P_x,P_y,L_{xy},H$: four charges. In three dimensions it supplies three momentum components, three angular-momentum components and $H$: seven charges. For generic states these listed charges are independent in those dimensions. Galilean boosts can also be expressed by explicitly time-dependent conserved combinations, but are not part of the requested translation-rotation-energy count.

For an infinitesimal continuous symmetry $\delta q_i=\epsilon {\xi }_i(q,t)$, allow the Lagrangian to change by a total derivative, $\delta L=\epsilon dF/dt$. Direct variation givesOn a solution of the Euler-Lagrange equations, the first term vanishes. This proves Noether's theorem in the fixed-time form:Spatial translations yield total momentum, and rotations yield angular momentum. Time translations yield energy when the Lagrangian has no explicit time dependence; this follows either from the extended time-transformation form of the theorem or by differentiating $H={\sum }_i{p}_i{\dot{q}}_i-L$ to obtain $\dot{H}=-\partial L/\partial t$.

Write $u=\pi x^2/l_0^2$ and assume $V_0>0$. The derivative of the potential is $V^{\prime }(x)=-2\pi V_{0}x\sin u/l_0^2$. Its equilibrium positions are zero and $x=\pm l_0\sqrt{n}$ for positive integers $n$. At a nonzero equilibrium,Thus the stable equilibria and their small-oscillation normal mode frequencies areEven positive $n$ give unstable maxima. At zero the quadratic term vanishes but $V(x)=V_0-V_0{\pi }^2{x}^4/(2l_0^4)+\cdots$, so the equilibrium is still unstable; a zero linearized frequency does not prove stability. If the unspecified sign of $V_0$ were chosen negative, even-$n$ positions would be stable instead, together with a degenerate quartic minimum at zero whose frequency tends to zero with oscillation amplitude.

Hamilton's principle makes the action $S[q]={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$ stationary under variations vanishing at the endpoints. It need not be a strict minimum. Varying the action and integrating the velocity variation by parts givesThe endpoint term vanishes. Since the interior variations are arbitrary, the fundamental lemma of the calculus of variations yields

The circumstellar habitable zone is the range of stellar irradiation at which a suitable rocky planet could maintain liquid water at its surface under specified atmospheric conditions. It is not a guarantee of life or even of liquid water on every planet at that distance. Its boundaries depend on greenhouse gases, clouds, humidity, pressure, albedo, rotation and the stellar spectrum; stellar evolution, ultraviolet activity, atmospheric loss and tidal locking can complicate applications to cool stars.

Approximating both stars as blackbodies, the host luminosity relative to the Sun isThe Earth-irradiance orbit therefore hasUse the largest projected separation and a visible reference wavelength of $550\mathrm{nm}$. The Rayleigh criterion for a circular aperture givesThis is only a diffraction-based minimum: detecting a faint planet beside its star additionally requires contrast suppression and adequate photon counts. An interferometer baseline, rather than its collecting-aperture diameter, sets its angular resolution.

For reflected light the reflected-light planetary phase curve gives the flux ratio $ϵ=A_g\Phi (\alpha )(R_{\oplus }/a_{\mathrm{E}}{)}^2$, where $A_g$ is geometric albedo and $\Phi (0)=1$. Numerically,A unit-geometric-albedo reference gives about $6\times 10^{-7}$; an Earth-like chosen $A_g=0.3$ gives $1.7\times 10^{-7}$. Albedo and phase function were not specified, so an exact contrast cannot be inferred just from the radius. Full-phase brightness occurs close to zero projected separation for an edge-on orbit; at the best-resolved quadrature, a Lambert sphere has $\Phi (\pi /2)=1/\pi$, reducing the Earth-albedo example to $5.4\times 10^{-8}$. Thus the best angular separation and maximum reflected contrast are not generally achieved at the same phase.

The measured mass and radius implyCompared with the supplied composition curves, this is inconsistent with a bare iron-rich or ordinary rock-only planet. A volatile-rich interior and/or an extended low-density gas envelope is needed. A water-rich body with an atmosphere and a rocky core with a modest H/He envelope are different plausible families; mass and radius do not uniquely determine the composition.^[ref]

For the requested H/He estimate, assume a hydrostatic, approximately isothermal terminator at the zero-Bond-albedo, fully redistributed equilibrium temperature. The orbit and temperature areSurface gravity is $g=Gm/R_p^2\simeq 9.58\mathrm{m}{\mathrm{s}}^{-2}$. With molar mass $\mathcal{M}=2.2\times 10^{-3}\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, the atmospheric scale height isA spectral feature spanning $N$ scale heights changes the blocked stellar fraction by the atmospheric annulus area:Thus one scale height is about $338$ parts per million in transit depth; a clear strong feature spanning roughly five scale heights is of order $1.7\times 10^{-3}$. The number of scale heights requires the actual opacity ratio; $N\simeq \ln ({\kappa }_{\mathrm{line}}/{\kappa }_{\mathrm{cont}})$ is a useful isothermal estimate, not a supplied measured value. The true terminator temperature need not equal this redistribution estimate.

A nearly flat spectrum can result from a high-mean-molecular-mass atmosphere with a much smaller scale height, or from optically thick clouds/hazes that hide the deeper absorption features. These are the two qualitative possibilities; detailed precision data can discriminate them. In particular, the pre-2015 HST study favored clouds over the clear high-molecular-mass models it tested, so listing both physical mechanisms is not a claim that all such models fit those data equally well.^[ref]

During a exoplanet transit, starlight traverses the planetary limb. The effective opaque radius varies with wavelength because the altitude of slant optical depth near unity depends on molecular absorption and scattering. Differencing the in-transit and out-of-transit stellar spectra measures this wavelength-dependent blocked area, the exoplanet transmission spectrum. Stellar spots and faculae can contaminate that difference and require control.

During a exoplanet secondary eclipse, the disappearance of planetary light permits subtraction of the stellar contribution. Visible planetary light is predominantly reflected starlight, sensitive to clouds, hazes, Rayleigh scattering, and atomic sodium and potassium absorption. Infrared planetary emission and limb absorption show rotational-vibrational features of species such as water, carbon monoxide, carbon dioxide and methane, together with collision-induced absorption involving molecular hydrogen. Thermal emission also depends on the atmospheric temperature-pressure profile; a temperature inversion can turn a molecular absorption feature into emission. Which species are prominent depends on composition, temperature and the observed band, rather than being universal to every planet.

In the strong-focusing, small-projectile limit, write $R_M=(3M/(4\pi {\rho }_s){)}^{1/3}$ with constant solid-material density. ThenIntegrating $dM/M^{4/3}=Adt$ from $M_0$ to $M$ givesfor $M\gg M_0$. The fractional growth rate $\dot{M}/M\propto M^{1/3}$ increases with mass: larger bodies gain an increasing advantage, hence gravitationally focused growth of a planetary core is called runaway accretion in this regime.

The formal finite time to unlimited mass is not physical. Growth depletes the feeding region; embryos also stir planetesimal velocities and reduce focusing, while fragmentation, orbital competition and stellar tides alter the encounter geometry. A growing core can enter an oligarchic rather than runaway regime, and eventually reach a supply-limited isolation mass.

Dynamical friction transfers random kinetic energy from a massive body to a swarm and tends toward $M{v}_M^2\sim m{v}_m^2$. Larger bodies therefore have smaller random eccentricities and inclinations, remain concentrated near the planetesimal layer, and can retain favorable encounter conditions. When their own random motion contributes appreciably to relative speed, damping it also strengthens focusing through $f\propto v_0^{-2}$. If the small-body dispersion already dominates $v_0$, equipartition of the core alone does not further lower that encounter speed: the displayed growth law assumed a fixed $v_0$, not an automatically imposed $v_0\propto M^{-1/2}$.

The collision cross section in the incoming impact-parameter plane isWithout gravity it is only ${\sigma }_{\mathrm{geom}}=\pi (r_M+r_m{)}^2$, so gravitational focusing increases the cross-section by $1+f$. Trajectories that would miss geometrically can be bent into contact, especially for slow relative encounters. Gravity does not itself guarantee sticking: the following accretion estimate assumes collisions agglomerate rather than fragment, rebound or erode the target.

A uniform incoming projectile mass density ${\rho }_p$ has mass flux ${\rho }_p{v}_0$, soThe formula counts mass swept per unit time and assumes an adequate continuing projectile supply, the stated speed, and negligible many-body or stellar-tidal modification of the two-body cross-section.

The source expression $E=\frac{1}{2}\mu v^2-GMm/r$ is total mechanical energy in the centre-of-mass frame, not kinetic energy alone. Take $v_0$ as the relative speed at a large initial separation where the mutual potential is negligible. At closest approach, the radial relative velocity is zero, so conservation of angular momentum and conservation of energy giveUsing $Mm/\mu =M+m$ eliminates $v_c$:An impact occurs when $r_c\le R$, where $R=r_M+r_m$. Since the right side increases with positive $r_c$, the collision threshold isThe mutual escape velocity is the relative speed at contact required to reach infinite separation with zero residual speed. Setting total mechanical energy to zero givesThis includes target recoil. If the given initial speed were instead at a finite non-negligible separation, its potential energy would have to be retained; $v_0$ would then not equal the asymptotic speed used in the focusing formula.

Assume the companion is much less massive than the star, the sinusoid represents a circular Keplerian orbit, and the transit implies $\sin i\simeq 1$. Ignore dilution, limb darkening and grazing geometry when estimating the radius. With $P=5\times 86400\mathrm{s}$,The stellar reflex radial-velocity semi-amplitude gives $K\simeq (m/M_{\odot })(2\pi a/P)$, while transit depth gives $\delta \simeq (R_p/R_{\odot }{)}^2$. ThereforeThe mean density relative to the supplied Jupiter values isThis is a low-density hot gas giant. Strong irradiation can slow contraction or help sustain an inflated envelope; internal heating and a different heavy-element fraction are other possibilities. The inferred low mass also reduces gravitational compression. Mass and radius alone do not determine which inflation mechanism operates.

A circular orbit has uniform angular motion, so its Keplerian radial-velocity curve is sinusoidal. Choose phase $\varphi =(t-t_{\mathrm{tr}})/P$, positive stellar velocity away from the observer, and an edge-on circular orbit. One consistent convention isTransit occurs at the decreasing systemic-velocity crossing $\varphi =0$; secondary eclipse is at the increasing crossing $\varphi =1/2$. Extrema occur one quarter of an orbit away. Reversing the velocity sign convention reverses the crossing directions but not the conjunction phases.

For an eccentric orbit, $\nu$ is not proportional to time. Solve $\mathcal{M}=E-e\sin E$, then use $\tan (\nu /2)=\sqrt{(1+e)/(1-e)}\tan (E/2)$ in the stated velocity relation. Motion is fastest near periastron, so the curve has an abrupt feature there and a more prolonged change near apastron, with its shape also depending on $\varpi$. It is not merely a shifted sinusoid. An eccentric transit need not coincide with $V=V_0$, because the $e\cos \varpi$ offset remains at conjunction.

Use the relative star–planet separation and specific angular momentum per unit reduced mass,For $r=l/(1+e\cos \theta )$, differentiation gives $\dot{r}=Je\sin \theta /l$, while the transverse speed is $r\dot{\theta }=J/r=J(1+e\cos \theta )/l$. Adding the squared orthogonal components yieldsThe Binet equation for the relative two-body inverse-square force is $u^{\prime \prime }+u=G(M+m)/J^2$, with $u=1/r$. The conic has $u^{\prime \prime }+u=1/l$, soThe printed division by $l$ is an error: $J^2$ has dimensions $\mathrm{lengt}{\mathrm{h}}^4/tim{\mathrm{e}}^2$, while $G(M+m)/l$ has dimensions of speed squared. A circular orbit independently checks the correction: $J^2=r^2{v}^2=G(M+m)r$, with $l=r$.

For a closed ellipse, $l=a(1-e^2)$ and $1+2e\cos \theta +e^2=2l/r-(1-e^2)$. Substitution now gives the vis-viva equation,This is relative orbital speed; either body's speed about the barycentre follows by the corresponding mass-ratio factor.

The normalized Bell statehas equal outcomes for any common real linear-polarization basis. In the stated rotated basis it is $(|{\theta }^{(+)}{\theta }^{(+)}⟩+|{\theta }^{(-)}{\theta }^{(-)}⟩)/\sqrt{2}$: the cross terms cancel when the inverse rotation is substituted.

Thus the coincidence probability for the transmitted ports is $p(A,B)=\frac{1}{2}{\cos }^2({\theta }_A-{\theta }_B)$. At equal angles it is $1/2$ per emitted pair, while Bob transmits with probability one conditional on Alice's transmission. Both photons are blocked for the other half of ideal pairs. The phrase always in coincidence must therefore mean matching outcomes or conditional transmission, rather than unit unconditional transmission through both polarizers.

The classical mixture has no coherence between $HH$ and $VV$; it gives equal-angle coincidence probability $1/4$ at $45^{\circ }$, rather than $1/2$. The coherent quantum superposition maintains correlation in this rotated basis. Bob's unconditional transmission remains $1/2$, so the conditional correlation does not permit signalling by Alice's polarizer choice.

The classical mixture has density matrix $\rho =\frac{1}{2}|HH⟩⟨HH|+\frac{1}{2}|VV⟩⟨VV|$. Average the probabilities of its two alternatives:Alice's single-detector transmission probability is $1/2$ at every angle. Dividing by it givesAt $45^{\circ }$, Alice's click does not distinguish the two classical alternatives. At zero, it selects the horizontally polarized alternative.

For the product state $|H{⟩}_A|H{⟩}_B$, the two projection amplitudes multiply. ThereforeThe independent factors arise from this product input; they cannot be assumed for a general correlated pair state.

The normalized linear polarization state is $|\theta ⟩=\cos \theta |H⟩+\sin \theta |V⟩$. A transmitted-photon count implements the projector $|\theta ⟩⟨\theta |$. The Born rule givesthe single-photon form of Malus's law, assuming ideal polarizers and detection.

For a uniformly illuminated single slit of width $w$, the far-field amplitude is proportional to $\sin u/u$, with $u=\pi w\sin \vartheta /\lambda$. Its first zeros occur at $\sin \vartheta =\pm \lambda /w$. If the quoted peak width is the full width between those zeros, the small-angle geometry givesThe Fraunhofer diffraction approximation is appropriate since $w^2/(\lambda L)\ll 1$. If the supplied width instead meant a half-width or full width at half maximum, its numerical conversion factor would differ; the displayed answer states the first-zero convention explicitly.

For the macroscopic mass,The ideal fixed-momentum expression tends to infinity as $v\to 0$. An everyday localized object is not a perfectly coherent momentum eigenstate, however: its thermal and environmental momentum uncertainty remains nonzero, and decoherence rapidly suppresses spatially separated coherent alternatives. Merely setting the mean velocity to zero does not prepare an object with a macroscopic coherence length.

Using $m\simeq 4m_p$ and $T=15\mathrm{meV}$, the nonrelativistic de Broglie wavelength and speed areFor the square interferometer, the enclosed area projected into a vertical plane is $a^2\sin \theta$. The gravitational phase shift in matter-wave interferometry is $\Delta \phi =mg{a}^2\sin \theta /(\hslash v)$ to leading order in the small gravitational energy change. Thus rotation from horizontal to vertical changes the fringe count byOne way to see the phase is that the horizontal arms differ in gravitational potential energy by $mga\sin \theta$ for a flight time $a/v$. Which-path detection entangles the atom with distinguishable detector states. If those states are orthogonal and the path record is ignored, their coherence is lost and the interference fringes disappear. This is a loss of interference visibility, not a change in the atom's de Broglie wavelength.