Codex @codex  0

Mean energy imparted by ionizing radiation per unit mass; its unit is the gray.

For a spherical drop of constant mass density $\rho$, radius $a>0$ and downward velocity $v>0$, suppose its volume grows at rate $c\pi a^{2}v$ by collecting stationary mist and its upward quadratic drag is $k\rho \pi a^2{v}^2$, with $c,k>0$. The drop mass is $4\pi \rho a^3/3$. Its incoming mass accretion rate and the variable-mass system momentum law give the displayed equations. The accreted matter must be accelerated from rest, producing the additional $c$ term alongside drag.

Growth of an astronomical object's mass by capture of surrounding matter. During planetary scattering, a collision can lead to accretion if material is retained by the planet; not every collision retains all material. An accretion disk is one particular geometry supplying material.

For constant supply, negligible seed mass and fixed emitting radius, $M=\dot{M}t$ and $L_{\mathrm{acc}}\propto t$. Thermal emission with $L=4\pi r_0^2{\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$ therefore gives the quarter-power law. It describes an effective surface temperature. A homologous ideal-gas interior at the same fixed radius instead has virial temperature proportional to mass and hence to time; its absolute temperature requires a structural model.

Choose $D=7.5\mathrm{kpc}$, $\alpha =0.110$ arcseconds, $P=18$ years, $e=0.70$ and $i=55$ degrees. The small-angle conversion gives $a=825\mathrm{AU}$. Taking one model year as $365.25$ days gives $\mu \simeq 2.30\times 10^{26}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ and $K\simeq 1566km/s$.

Dividing by the nominal solar mass parameter, $(GM{)}_{\odot }^{\mathrm{N}}=1.3271244\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, gives $1.73\times 10^6$ nominal solar masses. The nominal parameter is an exact conversion convention, rather than an assertion that the Sun's measured mass is exact.^[ref]

The chosen periapsis radius is $r_p=a(1-e)=247.5\mathrm{AU}$. Its corresponding Schwarzschild radius, $r_s=2\mu /c^2$, is about $0.0342\mathrm{AU}$. Thus this illustrative orbit remains roughly $7200$ Schwarzschild radii from the centre at closest approach. Inferring a very concentrated mass from such an orbit is different from directly resolving its event horizon. Every number here follows from the chosen orbit, not from a fit to the actual Galactic-centre measurements.

Adding a load $M$ to a spring-supported table shifts its static downward position by $\Delta =Mg/k$. If the load is deposited with negligible impulse and moves with the table, the moving mass is $m+M$. Starting at the former equilibrium with zero velocity, an underdamped response isIts transient envelope decays over $1/\beta$. Critical damping $b=2\sqrt{k(m+M)}$ gives $x=\Delta [1-(1+{\omega }_{0}t)e^{-{\omega }_{0}t}]$. A nonzero impact impulse changes the initial velocity and therefore the transient.

Two uniform discs of mass m and radius a stick at first edge contact. One initially translates at speed v and the other rests, with neither initially spinning. If the incoming velocity makes angle $\theta$ to the line of centres, the composite axial moment of inertia about its centre of mass is $3m{a}^2$. Conserved angular-momentum magnitude is $mav|\sin \theta |$, so the angular velocity magnitude is $v|\sin \theta |/(3a)$. The centre-of-mass speed is v/2. Splitting kinetic energy into translation and rotation gives $K_f=m{v}^2(1/4+{\sin }^2\theta /6)$. The difference from $m{v}^2/2$ is dissipated internally; it cannot be removed from angular-momentum conservation.

For a Hermitian adjoint scalar field with kinetic term $\mathrm{tr}(D_{\mu }\Phi (D^{\mu }\Phi {)}^{†})$, use $D_{\mu }\Phi ={\partial }_{\mu }\Phi +ig[A_{\mu },\Phi ]$ and generators normalized by $\mathrm{tr}(t^a{t}^b)={\delta }^{ab}/2$. Expanding at a constant vacuum expectation value ${\Phi }_0$ and matching ${\mathcal{L}}_{\mathrm{mass}}=A_{\mu }^a(m^2{)}^{ab}{A}^{b\mu }/2$ gives the displayed expression. A real combination of generators $X=x_a{t}^a$ has quadratic form $2g^2\mathrm{tr}([X,{\Phi }_0{]}^{†}[X,{\Phi }_0])\ge 0$. Thus the massless directions are exactly the generators commuting with the vacuum.

For ${\Phi }_0=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$, the two Hermitian generators associated with each matrix unit pair $E_{ij},E_{ji}$ have squared mass $g^2({\lambda }_i-{\lambda }_j{)}^2$. The surviving group is the centralizer of a Hermitian matrix in SU(n).

The translational Helmholtz free energy cost of restricting a dilute ideal gas from a volume $V$ to an area $A$ at the same temperature. For identical particle number and mass,Here $\Lambda$ is the thermal de Broglie wavelength. This contribution is positive for $V/A\gg \Lambda$; a favourable binding energy must be added separately to predict net adsorption.

The fate of an Asymptotic giant branch star depends on whether its envelope is removed before its core reaches the next ignition or instability threshold. The approximate envelope-loss time is $t_{\mathrm{env}}=M_{\mathrm{env}}/|\dot{M}|$, while the time to add a required core mass $\Delta M_c$ is $t_{\mathrm{grow}}=\Delta M_c/{\dot{M}}_c$. Shell stellar nuclear fusion grows the core; third dredge-up can reduce the net growth. If $t_{\mathrm{env}}\ll t_{\mathrm{grow}}$, shell fuel disappears first and a white dwarf remnant is favored. The opposite ordering permits further core growth but does not by itself specify whether carbon burning is stable, explosive, or followed by further evolution. Both rates change with time, so the estimates guide rather than replace a stellar model.

A physicist whose work connects special relativity, general relativity, Brownian motion and the photoelectric effect. He worked at the Institute for Advanced Study from 1933 until his death in 1955.^[ref]

In 1905, Albert Einstein used the photoelectric effect to explore quantized light, related Brownian motion to molecular agitation, formulated special relativity, and connected changes in mass and energy. These problems link visible measurements to otherwise inaccessible structure: emitted electron energies, suspended-particle displacements and comparisons of clocks.^[ref]

The Millikan photoelectric experiment follows one of those predictions into the laboratory. Its stopping-voltage measurements test a quantitative relation; the experiment's apparatus, surface preparation and electrical corrections matter to that inference. The Einstein field equations and general relativity lead to a different set of tests involving gravity and spacetime.

The Nobel Prize in Physics in 1921 recognized his theoretical work, particularly the law of the photoelectric effect; the award was presented in 1922.^[ref]

Amount of substance measures a count of specified entities, such as chemical atoms, molecules, ions, or electrons, using the mole as its SI base unit. It differs from mass: equal amounts of substance of different chemical substances contain equal entity counts but generally have different masses. If $N$ entities are present, $n=N/N_{\mathrm{A}}$, where $N_{\mathrm{A}}$ is the Avogadro constant. The specified entity matters: one mole of water molecules contains two moles of hydrogen chemical atoms. ^[ref]

The Anderson positron track experiment identified a light positively charged particle, now called the Positron, by combining magnetic bending, loss of kinetic energy in lead, and particle range. Each observation answered a different question: bending constrained momentum, the lead supplied a direction of travel, and particle range tested the proposed mass.

Carl David Anderson's The Positive Electron (1933), pp. 491–493, describes the 1932 discovery in a vertical cloud chamber with a $15,000\mathrm{gauss}$ magnetic field. A $6\mathrm{mm}$ lead plate separated two photographed track segments. The effective viewing depth was $1\mathrm{cm}$ along the magnetic field, with $14\mathrm{cm}$ transverse diameter: this restricted out-of-plane motion. The photographic readout supplied curvature on both sides. Comparisons with electron tracks under the same conditions constrained ionization; specific ionization was not directly measured for these events. He tested accidental alignment of unrelated tracks and rejected a reversed electron interpretation requiring an increase of kinetic energy in lead. Assuming unit electric charge, the mass bound was about twenty electron masses, rather than an exact mass determination. The paper credits Robert Andrews Millikan with the design and Seth Henry Neddermeyer with assistance.^[ref]

Preparation used rapid piston expansion and a water–alcohol mixture in the cloud chamber; cooling permitted droplets to form along ionization trails. A camera photographed these droplets. Initially, expansions occurred at random, so many photographs contained no tracks. Later counter triggering improved the useful yield.^[ref]

In The Production and Properties of Positrons, his 1936 lecture, Carl David Anderson describes the chamber inside an electromagnet and explains why the lead plate was decisive: after losing kinetic energy, a particle bent more sharply. He describes a proton hypothesis predicting only $5\mathrm{mm}$ of particle range, against an observed path longer than $50\mathrm{mm}$. These are historical values from the lecture, not the stipulated example below. The point was the incompatibility of particle range with the momentum inferred from curvature, rather than positive electric charge alone.^[ref]

Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

A Keplerian disk with surface density $\Sigma \sim r^{-p}$ at infinity has finite outer mass for $p>2$ and finite outer angular momentum for $p>5/2$, since the relevant integrals are $\int \Sigma rdr$ and $\int \Sigma r^{3/2}dr$. The borderline $p=5/2$ has logarithmically divergent angular momentum even though its mass converges.

Take nonrelativistic quantum mechanics, a short-range interaction, a fixed heavy target, and a common projectile mass $m$. An incoming plane wave in atomic state $a$ has unit coefficient. In an open scattering channel $i$, the target finishes in a normalized atomic state ${\chi }_i$, and the projectile has positive velocity magnitude $v_i=\hslash k_i/m$. At large distance,Here $\Omega$ denotes direction, $d\Omega$ is a solid angle element, and ${\epsilon }_i$ is an internal atomic energy. Different ${\chi }_i$ are orthogonal. The scattering amplitude $f_{ia}$ has units of length. A closed scattering channel has no propagating outgoing probability current at this total energy.

The incoming probability current density is $j_{\mathrm{in}}=v_a$. The outgoing radial probability current density in channel $i$ is $j_{r,i}\sim v_i|f_{ia}{|}^2/r^2$. Multiplying by the area $r^{2}d\Omega$ and dividing by $j_{\mathrm{in}}$ gives the differential scattering cross-section:The $r^{-2}$ dilution cancels the detector's area growth. The velocity ratio remains: a squared wave coefficient alone is not generally a flux fraction. A differential scattering cross-section has units of area per solid angle, rather than being a normalized directional probability density. For selected scattering events in acceptance $D$, an ideal detector with equal efficiency in all retained channels instead givesThe denominator specifies the event selection; it need not equal one square unit of area. Absolute rates require incident flux, target number and detector efficiency in addition to these relative weights. The next finite-mode model normalizes outgoing probability current to one incoming event and predicts event fractions directly.

A real cosmological deformation adds $3m^{2}e/{\kappa }^2+2me{\bar{\psi }}_{\mu }{\gamma }^{\mu \nu }{\psi }_{\nu }$ and changes the gravitino variation to ${\mathcal{D}}_{\mu }ϵ/\kappa$. The modified commutator identity is ${\gamma }^{\mu \nu \rho }{\mathcal{D}}_{\nu }{\mathcal{D}}_{\rho }ϵ=(G^{\mu \nu }+\Lambda g^{\mu \nu }){\gamma }_{\nu }ϵ/2$. It fixes the negative cosmological constant in terms of the gravitino mass parameter. A positive cosmological constant is not obtained by this real unbroken minimal deformation.

Apparent weight measures support rather than the gravitational force itself. In an ideal scale model, it is the magnitude $N$ of the normal force exerted on the body. A scale calibrated to display mass divides that support force by its calibration value of $g$.

Take upward as positive in a constant downward gravitational acceleration of magnitude $g$. If a body stays in contact with a horizontal scale and accelerates upward at $a$, Newton's second law gives $N-mg=ma$, hence $N=m(g+a)$. At rest, $N=mg$. In common free fall, $a=-g$, so $N=0$ despite the nonzero gravitational weight $mg$. If the formula would require $N<0$, ordinary supporting contact cannot realize it: the body loses contact instead.

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.