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Consider two equal balls of mass $m$, four light rods each of length $L$, a fixed upper pivot and a light sliding lower collar. Each upper and lower rod makes angle $\theta$ with the spindle. A spring of spring constant $k$ is uncompressed at $\theta =0$ and is compressed by collar motion. Its compression is $\delta =2L(1-\cos \theta )$. Collar balance gives $2T_l\cos \theta =k\delta$, while vertical ball balance gives $(T_u-T_l)\cos \theta =mg$. Radial Newton's second law gives $(T_u+T_l)\sin \theta =m{\omega }^{2}L\sin \theta$. Eliminating the rod tensions proves the displayed relation for nonzero $\theta$.

Attach a spring of spring constant $k$ to the axle of a round body of mass $m$, radius $R$ and axial moment of inertia $I$. Assume rolling without slipping, a horizontal drag force $-b\dot{x}$ applied through the axle, and no additional resisting torque. The translational force equation is $m\ddot{x}=-kx-b\dot{x}+f$, where $f$ is signed static friction. The rotational equation and rolling constraint give $f=-I\ddot{x}/R^2$, yielding the displayed equation. Equivalently, the kinetic energy is $(m+I/R^2){\dot{x}}^2/2$. This added effective inertia lowers the undamped angular frequency to $\sqrt{k/(m+I/R^2)}$. A distributed viscous torque would add a separate damping term and cannot be inferred from the drag force alone.

A pendulum bob and a horizontally moving support of equal mass $m$, with support spring spring constant $k$, have the quadratic potential energy $k{x}_1^2/2+mg(x_2-x_1{)}^2/(2l)$. With $\kappa =k/m$, $q=g/l$, the normal mode eigenvectors can be normalized to $(1,r_{\pm })$, where $r_{\pm }=(\kappa /2\mp \sqrt{{\kappa }^2/4+q^2})/q$. The lower mode is in phase and the upper mode in opposite phase. The positive eigenvalues of the symmetric stiffness-per-mass matrix give the displayed squared angular frequencies.

The parameter multiplying a scalar quadratic mass term. A negative value describes an instability of the expansion point, rather than a negative value of the physical mass of a stable excitation. Nonnegative quartic terms do not stabilize a negative quadratic term along an exactly flat quartic ray.

For real scalar fields with quartic potential energy density $V_4=\lambda ({\varphi }_1^4+{\varphi }_2^4)+2\mu {\varphi }_1^2{\varphi }_2^2$, nonnegativity in every direction is equivalent to the displayed conditions. Necessity follows on an axis and on ${\varphi }_1={\varphi }_2$. Sufficiency follows by writing $V_4=\lambda ({\varphi }_1^2-{\varphi }_2^2{)}^2+2(\lambda +\mu ){\varphi }_1^2{\varphi }_2^2$. With a positive quadratic mass term, equality cases still give a stable isolated classical vacuum at the origin. The quartic alone is strictly positive away from the origin precisely when $\lambda >0$ and $\mu >-\lambda$. Rotation symmetry additionally requires $\mu =\lambda$.

In static coordinates on anti-de Sitter spacetime, write $f=1+r^2/L^2$, conserved energy per unit rest mass $E=f\dot{t}$, and conserved specific angular momentum $h=r^2\dot{\varphi }$. Timelike normalization gives ${\dot{r}}^2=E^2-V_{\mathrm{eff}}$ with $V_{\mathrm{eff}}=f(1+h^2/r^2)$. A circular geodesic at $r_0>0$ obeys $V_{\mathrm{eff}}^{\prime }=0$, so $h^2=r_0^4/L^2$ and $E=f(r_0)$. Thus $\dot{t}=1$ and $\dot{\varphi }=\pm 1/L$. Both its static-coordinate period and its own proper time period are $2\pi L$, independent of radius. Stability follows from $V_{\mathrm{eff}}^{\prime \prime }(r_0)=8/L^2>0$. A static clock at the orbital radius measures the different period $2\pi L\sqrt{f(r_0)}$, which does depend on radius.

The sum of specific enthalpy and macroscopic kinetic energy per unit mass. In a steady adiabatic flow without shaft work, its change is determined by the change in external potential energy. At essentially constant external potential energy, it is conserved through a stationary nonradiating shock wave even though entropy increases.

The product $\mu =GM$ of the gravitational constant and a central mass sets the strength of the exterior Newtonian gravitational field. Its dimensions are length cubed divided by time squared. In the exact relative two-body problem, the parameter is $G(M_1+M_2)$; using $GM$ for one source assumes the test particle approximation.

The combination is useful because orbital and falling-body observations constrain $\mu$ directly. A gravitational circular orbit of radius $a$ has oscillation period $T=2\pi \sqrt{a^3/\mu }$, so measuring $a$ and $T$ gives $\mu =4{\pi }^2{a}^3/T^2$. That measurement alone does not separately determine $G$ and $M$. Similarly, radial free fall from rest has a natural time scale $\sqrt{r_0^3/\mu }$.

Other interactions require a different model. Inverse-square radiation pressure can reduce the effective attraction for a dust Kepler orbit; calling that effective coefficient a gravitational parameter must not hide the added force.

For a steady Keplerian disk with a zero-torque inner edge $R_{\mathrm{in}}$, the flux summed over both faces is

A star cluster is a gravitationally associated population of stars formed in a compact region.

Star formation is the gravitational collapse and fragmentation of cold interstellar gas into protostars and eventually stars. Cooling, turbulence, magnetic fields, rotation, and feedback determine its efficiency and characteristic masses.

Starobinsky inflation is an inflationary model whose Einstein-frame scalar potential can be writtenThe exponentially flat large-field plateau supports slow-roll inflation, while the potential is quadratic near its minimum.

An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

For $T=-[\cos \varphi {]}_{-\infty }^{+\infty }/2$, complete the static energy into squares $({\varphi }_x\mp m\sin \varphi {)}^2$ and ${\sin }^2\varphi {\alpha }_x^2$. The boundary term is $\pm rmT$. A $T=1$ saturating kink is $\varphi =2\arctan e^{m(x-x_0)}$, with constant $\alpha$ and mass $rm$. This is a model-specific Bogomolny bound.

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]

For a Keplerian disk with evolution ${\partial }_t\Sigma =(2\pi r{)}^{-1}{\partial }_r\mathcal{F}+S$ and $\mathcal{F}=6\pi r^{1/2}{\partial }_r(r^{1/2}\bar{\nu }\Sigma )$, take zero viscous torque at $r_1>0$ and zero flux outside the source region at $r_2>r_1$. The steady state gives ${\mathcal{F}}^{\prime }=-2\pi rS$, and its outer condition gives the displayed flux. Integrating once more from the zero-torque inner boundary condition yieldsThe equation determines the viscosity-density product, so a thermodynamic viscosity closure can be imposed afterward. For nonnegative integrable $S$, it is nonnegative and the inner accretion rate is exactly the total supplied mass per unit time.

The Stefan number compares sensible heat $c_p\Delta T$ with latent heat $L$ per unit mass. Some solidification literature uses its reciprocal as the Stefan parameter, so the defining formula must accompany the convention.

Time-dependent stellar phenomena such as starspots and chromospheric emission that can change brightness and spectral-line shapes, complicating radial-velocity planet searches.

The stellar age is the elapsed time since a star formed. A constant number formation rate on a finite interval gives a uniform distribution of present ages if every formed object remains counted, including stellar remnants. An age distribution inferred only from surviving main sequence stars generally differs because their lifetimes depend on mass.