Codex @codex  0

A measured mercury column represents pressure through its actual mass density and local gravitational acceleration. An equivalent reference column therefore obeys ${\rho }_0{g}_s{h}_s=\rho (T)gh$. If the indicated scale length expands as $1+{\alpha }_s\Delta T$ and mercury volume as $1+\beta \Delta T$, the first-order thermal factor multiplying the scale reading is $1+({\alpha }_s-\beta )\Delta T$. These factors describe a pressure-unit reduction, not a physical change in local gravity.^[ref]

A directed stream of molecules used to prepare samples for spectroscopy.

A molecular beam also includes a stream of chemical atoms, then called an atomic beam. An oven supplies particles; aligned slits restrict their directions, and a vacuum limits subsequent collisions. Narrowing the directions does not by itself make every particle's speed equal. This gives a useful separation of tasks: prepare a path, select or measure speed, apply an interaction, and read the resulting particle counts or deflection.

The advantage is unusually tangible. A magnetic field can separate spin components in the Stern-Gerlach experiment; even gravity can sort a beam by flight time. The effusive beam speed distribution explains which particles leave the source, while gravity deflection of an atomic beam explains how that distribution becomes a pattern on a detector.

Neutral buoyancy means a fully immersed body's upward buoyancy equals its downward force from gravity. For a body of fixed volume in a constant-mass density fluid under a uniform gravitational field, this occurs when its mean mass density equals the surrounding mass density. A $0.10\mathrm{m}$ cube of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ in the stipulated water has mass $1.000\mathrm{kg}$, so both forces are $9.81\mathrm{N}$.

Ideally it can remain at rest at any depth for which it stays fully immersed and clear of boundaries. Raising or lowering it changes the absolute pressure, but neither its displaced volume nor the difference between the vertical pressure forces. There is therefore no vertical restoring force toward a preferred depth in this model. Neutral buoyancy is a vertical force balance, not a claim about resistance to tilting or about a compressible body whose volume changes with pressure.

In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

Awarded in equal shares to Russell Alan Hulse and Joseph Hooton Taylor Jr. for discovering a new type of pulsar that opened new possibilities for studying gravity.^[ref]

The rubidium experiment used laser cooling, a magneto-optical trap and magnetic trapping of neutral atoms with a rotating bias field. Radio frequency transitions removed energetic atoms; elastic collisions restored thermal equilibrium. Before absorption imaging, the trap was decompressed, then nearly released with a gradient supporting against gravity. Varying the final evaporation cut revealed a narrow anisotropic core within an isotropic thermal cloud. The broad component supplied a simultaneous comparison through the same optics. Resolution and size calibration limited inference; interactions broadened the core relative to an ideal gas.^[ref]

The oil drop experiment infers an oil drop's electric charge from its motion under gravity, buoyancy, drag and a calibrated electric field. Repeated changes of a single drop's electric charge reveal discrete steps; see the model, experimental procedure and limitations in elementary charge.

Optical molasses surrounds slow chemical atoms with opposing red-detuned laser beams, usually three pairs along perpendicular axes. In the elementary Doppler cooling picture, each pair damps one component of velocity. The result resembles viscous drag, even though the atoms are moving through a vacuum rather than a liquid.

Uniform balanced optical molasses supplies velocity damping, without a restoring force towards a chosen position. An atom at rest has zero mean force wherever it sits within the uniform illuminated region. Without noise or external forces, $m\dot{v}=-\alpha v$ givesThe atom approaches a position determined by its initial condition, not the beam intersection. Recoil-driven diffusion and gravity eventually remove atoms from finite beams. A long residence time in optical molasses is therefore compatible with the absence of position trapping. An optical trap provides spatial confinement by a different force mechanism; a magneto-optical trap adds a position-dependent magnetic field and suitable beam electromagnetic polarization to create a restoring force.

A point mass constrained to the fixed graph $y=y(x)$ has velocity $(\dot{x},y^{\prime }\dot{x})$. Under uniform downward gravity, its Lagrangian is $L=m(1+y^{\prime 2}){\dot{x}}^2/2-mgy$, so its canonical momentum is $p_x=m(1+y^{\prime 2})\dot{x}$, not the horizontal mechanical momentum. The Euler-Lagrange equation gives the equation in the title, and the conserved energy is $m(1+y^{\prime 2}){\dot{x}}^2/2+mgy$. A straight segment with constant slope $s$ has constant horizontal acceleration $-gs/(1+s^2)$; zero curvature does not imply zero acceleration. A cart held to a track satisfies this mechanical constraint only while the assumed contact or retaining mechanism remains effective.

Use the magnetic energy $E_B=(2{\mu }_0{)}^{-1}{\int }_V|\mathbf{B}{|}^{2}dV$. Vary the Euler potentials while keeping their boundary values fixed. The field variation isLet $\delta \mathbf{A}=\delta \alpha \nabla \beta -\delta \beta \nabla \alpha$. Integration by parts givesbecause $\delta \mathbf{A}=0$ on the boundary. ThereforeThe two interior variations are independent. At a smooth minimum their coefficients must vanish, so the curl is perpendicular to both potential gradients and hence parallel to $\mathbf{B}$ wherever $\mathbf{B}\ne 0$:This is a force-free magnetic field: the Lorentz force density vanishes. Taking the divergence also gives $\mathbf{B}\cdot \nabla \lambda =0$, so the proportionality factor is constant along a field line but need not be the same on different lines.

The fixed-label variation preserves the encoded field-line connection rather than allowing arbitrary reconnection during minimization. The argument proves a necessary condition for a smooth minimum, not that every force-free stationary field is a stable energy minimum. It also does not include gas pressure or gravity in the minimized functional.

The equation describes small plane-wave perturbations of a uniform, stationary plasma in ideal magnetohydrodynamics, with propagation parallel to the background magnetic field. The nonzero field defines $\hat{\mathbf{b}}={\mathbf{v}}_A/v_A$; $c_s$ is the sound speed for the assumed thermodynamic closure. Neglect viscosity, electrical resistivity, gravity and microscopic dispersive effects. A uniform background flow would instead introduce its Doppler-shifted frequency.

Decompose $\mathbf{u}={\mathbf{u}}_{\perp }+u_{\parallel }\hat{\mathbf{b}}$. The displayed matrix equation then becomesThus there are two independent transverse Alfvén wave polarizations with ${\omega }^2=k^2{v}_A^2$, restored by magnetic tension, and a longitudinal acoustic mode with ${\omega }^2=k^2{c}_s^2$, restored by pressure. The transverse modes have no first-order density compression; a strictly longitudinal displacement along the field does not bend it. These are the parallel-propagating ideal magnetohydrodynamic wave branches. In the general oblique problem, fast and slow magnetosonic modes mix compression and magnetic response; their parallel limits coincide with the faster and slower of these speeds. If $c_s=v_A$, all three velocity polarizations share the same frequency magnitude. The velocity equation alone does not enumerate nonpropagating entropy perturbations.

The Rayleigh-Taylor instability occurs when a denser fluid is supported against an effective acceleration by a lighter fluid. Let density ${\rho }_h$ occupy the upper half-space, density ${\rho }_l<{\rho }_h$ the lower half-space and gravity point downward. A displaced heavy-fluid finger lowers gravitational potential energy as it sinks; neighboring light fluid rises. For incompressible inviscid half-spaces without surface tension, a disturbance $\eta \propto e^{\sigma t+ikx}$ hasThe density ratio in this expression is the Atwood number. With the density ordering reversed, ${\sigma }^2<0$ and the interface supports stable gravity-wave oscillations in this ideal model.

Surface tension suppresses sufficiently short wavelengths, and viscosity changes the growth dynamics. A magnetic field parallel to the interface can resist perturbations that bend its lines; interchange directions that do not bend the field need separate consideration. In astrophysics, effective deceleration can make dense ejecta or a swept-up shell supported by a lower-density hot interior unstable. The effective acceleration and density ordering must be evaluated in that moving interface's frame; simply seeing a dense shell does not fix the stability sign.

For a steady inviscid barotropic fluid in a conservative gravitational potential, define $h(\rho )=\int dp/\rho$. Dotting the Euler equation with the velocity givesThe Bernoulli function is therefore constant along a streamline. It balances kinetic energy, pressure work or enthalpy, and gravitational potential energy per unit mass. Different streamlines can have different constants; for an irrotational flow, the identity $\nabla H=\mathbf{u}\times (\nabla \times \mathbf{u})$ makes the constant common throughout a connected region.

For an incompressible fluid, $h=p/\rho$ up to a constant, giving $u^2/2+p/\rho +\varphi =C$. For a constant-entropy perfect-gas streamline, $h=\gamma p/[(\gamma -1)\rho ]$. Together with mass conservation, this explains conversion of enthalpy into kinetic energy in a nozzle and the need to include gravity in an escaping wind. Viscous losses, heating and entropy production require their corresponding energy balance. Across an adiabatic shock the total stagnation enthalpy can still be conserved, but the upstream isentropic pressure-density law cannot simply be continued through the entropy jump.

Follow a fluid element with position $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives the material derivativeApply Newton's second law to a material element of volume $dV$ and mass $\rho dV$. The net pressure force is $-\nabla pdV$, and gravity gives $-\rho \nabla \varphi dV$. Dividing by the element mass, neglecting viscosity and using the chain rule componentwise gives the Euler equation

For a static spherical isothermal gas, $p=c_s^2\rho$ and $dp/dr=-GM(r)\rho /r^2$. Trial density $\rho =A/r^2$ gives enclosed mass $M(r)=4\pi Ar$, soThis is the singular isothermal sphere. Its central density diverges, although the enclosed mass tends to zero linearly and its total mass inside a finite radius is finite. It is not the finite-central-density regular isothermal solution.

Truncate at $r_0$ with boundary pressure $p_0$. Then $M=4\pi A{r}_0=2c_s^2{r}_0/G$ and $p_0=c_s^2\rho (r_0)$, givingThe external medium is being treated as a prescribed pressure boundary; its own gravitational field is excluded from the cloud self-energy calculation.

For a monatomic perfect gas, the equilibrium is isothermal but the internal-energy relation still uses its adiabatic index $\gamma =5/3$:The gravitational self-energy follows by assembling shells:Both integrals remain finite at the singular center. ConsequentlyThe pressure-confined singular isothermal gas sphere therefore satisfies the virial theorem including the work of the confining surface pressure. One can see its origin directly by integrating $\mathbf{r}\cdot \nabla p$: integration by parts gives $\oint p\mathbf{r}\cdot d\mathbf{S}-3\int pdV$, whose surface term is $4\pi r_0^3{p}_0$. An isolated vacuum-boundary version would omit that term and could not truncate this same positive-temperature isothermal profile at finite radius. The equality expresses equilibrium force and energy balance, rather than proving dynamical stability.

For a barotropic fluid, define $h(\rho )=\int dp/\rho$. Dotting the steady Euler equation with $\mathbf{u}$ givesThus the Bernoulli function $H=u^2/2+h+\psi$ is constant along each streamline. Different streamlines can have different constants; steady flow alone does not imply one spatially uniform $H$.

In a quasi-one-dimensional flow through area $A(z)$ with negligible gravity, the axial momentum equation and $dp=c_s^{2}d\rho$ giveMeanwhile the mass flow rate $\dot{M}=\rho uA$ is constant, so $d\ln \rho =-d\ln u-d\ln A$. Substitution proves the nozzle area-velocity relationA subsonic flow accelerates where $A$ decreases, whereas a supersonic flow accelerates where $A$ increases. A smooth accelerating transonic branch in a de Laval nozzle therefore crosses $u=c_s$ at a minimum-area throat. Suitable reservoir and downstream pressures are also required: a converging-diverging shape by itself does not exclude a wholly subsonic solution or a downstream shock.

For the stipulated constant-$K$ polytropic flow, $c_s^2=\gamma K{\rho }^{\gamma -1}$. At the throat $u_m=c_{s,m}$, soThe specific enthalpy is $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)=c_s^2/(\gamma -1)$. Evaluating $H$ at the throat and substituting $\rho =\dot{M}/(Au)$ elsewhere givesOn the outward supersonic branch, $A\sim C{r}^2$ lets the gas expand and its specific enthalpy tend to zero for $\gamma >1$. ConsequentlyFor $\gamma =5/3$, $u_{\infty }=2c_{s,m}$. Also $T\propto {\rho }^{\gamma -1}\propto r^{-2(\gamma -1)}$. These are the adiabatic nozzle terminal speed and expansion laws for a lossless escaping branch; a subsonic branch could instead approach finite pressure with $u\propto A^{-1}$. A pressure-shaped astrophysical jet is being approximated by a nozzle here: the assumed geometry does not by itself explain the confinement or account for magnetic forces or relativistic motion in an actual active galactic nucleus.

In an adiabatic outflow without additional heating, the reservoir's specific enthalpy supplies both escaping kinetic energy and the work against gravity. For a monatomic ideal gas, $\gamma =5/3$ and $h=c_s^2/(\gamma -1)=3c_s^2/2$, where $c_s$ is the reservoir's adiabatic sound speed. The Bernoulli equation, measured from a reservoir initially at rest at radius $r$, givesThe maximum terminal speed uses all available specific enthalpy, so $h_{\infty }\to 0$ andA freely escaping adiabatic flow needs a nonnegative radicand; equality is marginal escape. If the radicand is negative, this reservoir cannot supply an unheated escaping outflow. The initial rest condition is a stagnation-reservoir approximation: an exactly stationary finite-density surface cannot simultaneously carry a nonzero steady mass flow rate. Further heating, radiation forces or an imposed confining pressure would change the terminal-energy budget.

The Rayleigh-Taylor instability extracts gravitational potential energy when a heavy fluid is supported above a lighter one. A small downward excursion of heavy material and upward excursion of light material can grow instead of experiencing restoring buoyancy. The Kelvin-Helmholtz instability instead extracts kinetic energy from a velocity difference across an interface or shear layer. The interface displacement and its induced flow reinforce one another by tapping the background shear; a reversed density ordering is not required.

A useful common model has two inviscid, incompressible, semi-infinite fluids separated at $z=0$, with lower density/velocity ${\rho }_1,U_1$, upper density/velocity ${\rho }_2,U_2$, downward gravity $g$, and surface tension $\sigma$. For a disturbance proportional to $e^{i(kx-\omega t)}$, with $k>0$, the two-fluid interfacial gravity-shear dispersion relation isA negative right side gives exponential growth. With no shear or surface tension and ${\rho }_2>{\rho }_1$, the growth rate of the Rayleigh-Taylor instability is $s=\sqrt{\mathcal{A}gk}$, where $\mathcal{A}=({\rho }_2-{\rho }_1)/({\rho }_2+{\rho }_1)$ is the Atwood number. With equal densities and no restoring force, the growth rate of the Kelvin-Helmholtz instability is $s=k|U_1-U_2|/2$. Heavy-over-light stratification drives Rayleigh-Taylor growth; velocity shear drives Kelvin-Helmholtz growth. A heavy lower fluid supplies gravitational stabilization, and surface tension stabilizes sufficiently short waves in this ideal interface model.

Finite shear-layer thickness, viscosity, compressibility and magnetic tension can change the growth rates and unstable wavelength range. In particular, unlimited growth with increasing $k$ is a limitation of the zero-thickness, zero-surface-tension idealization. Density-interface overturning and jet-boundary mixing are appropriate physical applications of the two mechanisms, but the nonlinear mixed state is not determined by the linear growth rate alone.

Write $\mathbf{\omega }=\nabla \times \mathbf{u}$ and use the vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$. Taking the curl of the momentum equation eliminates all pure gradients, including gravity and the $\nabla (\nabla \cdot \mathbf{u})$ viscous contribution. Before invoking barotropy, the pressure term givesThis is baroclinic vorticity generation. For a barotropic fluid, $p=p(\rho )$, so $\nabla p$ is parallel to $\nabla \rho$ and this source vanishes. Constant $\nu$ permits the curl and Laplacian to commute. The result isNo incompressibility assumption was used. Since $\nabla \cdot \mathbf{\omega }=0$, an equivalent form isThe terms represent vortex stretching/tilting, compression/dilatation and viscous diffusion. In particular, removing the baroclinic source does not make the vorticity of a three-dimensional parcel constant. A spatially varying entropy-dependent $K$ in $p=K{\rho }^{\gamma }$ need not define a single barotropic law; the common-pressure-law assumption matters.

Use the idealized fully developed, laminar cylindrical-flow model implied by the fixed cross-section and zero velocity at its sides. Take the jet length to be $R$, as for a central source reaching the cloud boundary; if the source is off-centre, the actual path length must replace $R$. Let $s$ denote distance from the jet axis and write the axial velocity as $u(s)$. Then $(\mathbf{u}\cdot \nabla )\mathbf{u}=0$, the pressure gradient is $dp/dz=-p_0/R$, and the axial Navier-Stokes equation reduces toRegularity at $s=0$ excludes a logarithmic term. The imposed no-slip boundary condition $u(r_0)=0$ gives the Hagen-Poiseuille flow profileThe kinetic energy crossing each area element per unit time is $(\rho u^2/2)udA$, so the velocity must be cubed before integration. ThusConsequentlyThe mean velocity is $\bar{u}=U_{\max }/2$. Replacing the profile by $\bar{u}$ in $\frac{1}{2}\rho \pi r_0^2{\bar{u}}^3$ underestimates the flux by two; the kinetic energy correction factor is two for this profile. The radius dependence follows from $u\propto r_0^2$ and cross-sectional area $\propto r_0^2$.

The specified $P$ is outgoing bulk kinetic-energy flux, not the pressure-work rate $p_{0}Q$, which scales as $r_0^4$. In a fully developed section the profile is independent of $z$ and pressure work is dissipated as heat; entrance acceleration and energy supply establish the profile. A turbulent or freely entraining astrophysical jet need not have this no-slip parabolic structure. The model assumes a long enough, laminar confined region and neglects entrance losses and gravity.

For an inverted free surface, gravity promotes a displacement instead of restoring it. The surface tension term still resists curvature. Hence the gravity-capillary wave relation becomes, with $q=|k|$,For $0<q<k_c$, where $k_c=\sqrt{\rho g/\sigma }$, ${\omega }^2<0$. Writing the displacement as $\eta \propto e^{st+ikx}$ gives the Rayleigh-Taylor instability growth rateShorter wavelengths, $q>k_c$, remain oscillatory because surface tension dominates. The marginal scale is determined by the capillary length,A sufficiently small aperture excludes the long unstable modes: this is finite-aperture stabilization of an inverted liquid surface. The estimate requested is therefore a container opening of order one centimetre, with $1.7\mathrm{cm}$ the marginal full wavelength rather than a geometry-independent maximum diameter.

The exact largest opening cannot be fixed from the stated infinite-surface dispersion relation alone. It requires the aperture shape and contact-line boundary conditions. For example, a one-dimensional span of width $L$ whose contact line is pinned at both ends has a half-sine lowest mode with $q_{\min }=\pi /L$. In that simple model,Different geometries change the numerical factor. This is a small-disturbance stability estimate with negligible air inertia and no leakage or contact-line failure; a guarantee that a real inverted container retains its water needs those additional boundary details.