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Put $M=m_1+m_2$ and let $\mathbf{R}=(m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2)/M$ be the center of mass. Applying Newton's second law separately to the two point masses givesThus the center of mass has constant velocity, $\mathbf{R}(t)=\mathbf{R}(0)+t\dot{\mathbf{R}}(0)$. For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, subtraction givesHere $\mu$ is the reduced mass; it converts the relative motion into a one-body equation of motion.

For the given linear force, the relative position obeys simple harmonic motion, $\ddot{\mathbf{r}}+{\Omega }^2\mathbf{r}=0$, where ${\Omega }^2=k/\mu$. The initial rest condition givesFor $d>0$, collision first occurs when the relative position vanishes, namely when $\Omega t=\pi /2$. ThereforeThe separation $d$ cancels because the frequency of this simple harmonic motion is independent of its amplitude.

The likelihood-ratio statistic isThe null hypothesis $\pi =0$ is at the boundary of the permitted mixture weights. Assuming the remaining parameters are identifiable and interior, including $\tau >0$, an efficient normalized score for $\pi$ is asymptotically $Z\sim N(0,1)$ under the null hypothesis. The locally unconstrained quadratic log-likelihood maximizer is proportional to $Z$, while the constrained maximizer truncates it at zero. Thus $G^2\Rightarrow (\max (0,Z){)}^2$, giving the one-sided boundary likelihood-ratio limitHere ${\delta }_0$ is a point mass at zero. For a positive observed statistic the upper-tail probability is half the ordinary ${\chi }_1^2$ upper tail. The 5% critical value is the 90th percentile of ${\chi }_1^2$, about 2.7055, displayed as 2.71. Hence the reported likelihoods give a very marginal rejection of $\pi =0$ at 5%, with $p\simeq 0.04955$.

Using the ordinary 95th-percentile cutoff 3.84 would be inappropriate for this one-sided boundary test. The conclusion is close enough to the threshold that likelihood rounding matters: since each log-likelihood is printed to two decimals, the unrounded statistic can differ from 2.72 by nearly 0.02. The intended calculation rejects using the reported values, but an exact decision from the fitted data requires unrounded likelihoods. If the nuisance dispersion is also on its boundary or the mixture is not identifiable, the stated half-and-half null distribution needs reconsideration.

Here $I_{ij}$ denotes the unreduced second mass moment tensor, distinct from its trace-free mass quadrupole moment. In coordinates referred to the chosen origin,This expression uses $c=1$. In SI units the leading nonrelativistic mass density is $T^{00}/c^2$. For slowly moving point masses, $T^{00}\simeq {\sum }_A{m}_{(A)}{\delta }^{(3)}(\mathbf{y}-{\mathbf{y}}_{(A)}(t))$, soKinetic corrections to the energy density are higher order in the velocity. The second mass moment tensor is not the mechanical moment of inertia tensor, which instead has components ${\delta }_{ij}{I}_{kk}-I_{ij}$.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

Take the oscillation to be $z(t)=L\cos \omega t$; a different initial phase does not affect a period average. In the mass-density normalization of the quadrupole formula, the second mass moment tensor of the point mass has only $I_{33}=m{z}^2$ nonzero. Its trace-free mass quadrupole moment is thereforeThe constant part does not radiate. Since $d^3({\cos }^2\omega t)/d{t}^3=4{\omega }^3\sin 2\omega t$, the tensor contraction isUsing $⟨{\sin }^2(2\omega t)⟩=1/2$ in the quadrupole formula gives the averaged positive radiated powerThe mass quadrupole moment oscillates at twice the source's angular frequency, and the source energy loss has the opposite sign. The expression uses the leading slow-motion weak-field approximation, $\omega L\ll c$. With SI stress-energy tensor components, the mass-density integrand is $T_{00}/c^2$; the displayed $c^{-5}$ quadrupole formula uses that mass normalization. This is the point-mass contribution requested by the model: an apparatus maintaining an accelerating mass would also contribute its own mass quadrupole moment.

Differentiate the cumulative distribution function for $y>0$, using ${\Phi }^{\prime }(z)=(2\pi {)}^{-1/2}{e}^{-z^2/2}$ and $d\sqrt{2y}/dy=1/\sqrt{2y}$. The probability density function isIts value at zero can be assigned arbitrarily without changing any probability. Although this probability density function diverges as $y\downarrow 0$, $F_Y(0)=0$ and there is no point mass at zero. Equivalently, $Y$ has a gamma distribution with shape $1/2$ and rate one.

Assuming $g$ is measurable, applying it separately to independent and identically distributed random variables preserves those properties. Thus $W_i=g(X_i)$ have common mean ${\mu }_g$ and variance ${\sigma }_g^2$. For ${\sigma }_g^2>0$, the central limit theorem givesHence, for large $n$, the approximate probability distribution isThe second argument is the variance. If ${\sigma }_g^2=0$, this sample mean equals ${\mu }_g$ almost surely, so the appropriate description is a point mass at ${\mu }_g$.

For any closed surface enclosing a region $V$, Gauss's law for gravity states that the outward flux of the gravitational field isHere $G$ is the gravitational constant and $M_V$ is the enclosed mass; the negative sign expresses attractive gravity.

Use the gravitational potential sign conventions specified by this paper: $\mathbf{g}=\nabla \phi$. Thus $\phi$ is the negative of the usual Newtonian gravitational potential $\Phi$ defined by $\mathbf{g}=-\nabla \Phi$. For example, setting the value at infinity to zero gives $\phi =GM/r$ for a point mass. The divergence theorem now givesSince $V$ is arbitrary, continuity of the integrands implies the local Poisson's equationConversely, integrating this Poisson's equation over any $V$ and applying the divergence theorem recovers Gauss's law for gravity. At a discontinuity of the mass density, the local equation is interpreted piecewise, or globally in the sense of distributions.

On a sphere of radius $r$, the radial gravitational field has constant magnitude. Thus Gauss's law for gravity becomeswhere $M(r)$ is the enclosed mass. In the hollow core $M(r)=0$. In the material shell, put $u=r/r_0$ and use the spherical coordinates volume element:Outside the shell the enclosed mass is the totalConsequently,At $r=0$ the displayed value refers to the radial derivative of the constant core value. The expressions match at both shell boundaries. There is no surface sheet of mass, so a jump in the volume mass density does not produce a jump in the gravitational field. The core has zero gravitational field, and the exterior gravitational field equals that of a central point mass $M_{\ast }$, as in the shell theorem.

For a fixed attachment ratio $\delta >0$, making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, $\alpha /\delta =a/d\to 0$, its moment of inertia about its own center is negligible compared with $M{d}^2$. ThusThis is a simple pendulum with a point mass at the disc's center and effective length $d$; it is length $\ell$ only if $d=\ell$. Even before the small-angle approximation, its equation tends to $\ddot{\theta }=-(g/d)\sin \theta$.

If $\delta$ is allowed to tend to zero together with the other parameters, the bare conditions $\alpha \ll 1$ and $\mu \gg 1$ do not alone justify that reduction. The needed dominance conditions are $\mu \delta \gg 1$, $\mu {\delta }^2\gg 1$, and $\alpha /\delta \ll 1$, so that the rod contributes negligible gravitational torque and moment of inertia, and the disc behaves as a point mass. They follow automatically in the combined limit with fixed $\delta >0$.

Use the ideal transverse-motion model of the question: constant tension, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$, since $\rho$ is linear mass density. Newton's second law givesDividing by $\Delta x$ and taking its limit proves the wave equation on a string on each side of the mass:The string on the right pulls the point mass with vertical component $T{y}_x(0^{+},t)$, while the string on the left contributes $-T{y}_x(0^{-},t)$. HenceThe fixed supports and attachment impose the boundary conditionsTogether with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical acceleration. Requiring the slope to be continuous would incorrectly remove its dynamics.

gravitational lensing measures the gravitational influence of matter without requiring that matter to shine. In general relativity a point mass produces small-angle deflectionFor a thin lens, with angular-diameter distances $D_d,D_s,D_{ds}$,Image multiplicity, distortion and magnification therefore constrain projected gravitating mass, with source and lens geometry measured separately.

Strong lensing gives multiple images, arcs and Einstein rings. Their positions and, where available, time delays constrain a central mass distribution. Weak lensing measures correlated small image distortions over a larger field. Using the critical surface density for gravitational lensing and reduced gravitational shear, the convergence and shear satisfyBecause intrinsic galaxy shapes are unknown, weak shear is inferred statistically using many sources. Point-spread-function anisotropy, shear calibration, photometric redshifts, intrinsic alignments and foreground contamination must be controlled.

An inferred lensing mass can be compared with the independently estimated stellar and hot-gas mass. The bullet Cluster dark-matter separation provides a particularly informative separation of components: the collisional X-ray-emitting gas is displaced from the galaxies, while lensing mass peaks approximately follow the less-collisional galaxy component. The primary analysis combined wide-field and Hubble imaging rather than inferring lensing mass from X-ray hydrostatic equilibrium.^[ref] This favors a dominant relatively collisionless unseen component. Translating a merger offset into a particle self-interaction cross section also requires a merger model and projection uncertainties; lensing does not directly reveal a WIMP's mass or quantum numbers.

Microlensing tests a different possibility: dark matter in compact objects. A compact foreground lens can create an unresolved, time-dependent magnification of a background star. Event frequency and duration constrain the abundance and mass range of compact lenses only after survey cadence, detection efficiency, blending, source sizes and halo geometry are included. Achromaticity and the expected light-curve shape help distinguish simple lensing from intrinsic variability. A null survey result does not exclude every compact-object mass at every location.

Weak-lensing correlations and small-scale strong-lens structure also test the matter power spectrum and halo substructure, potentially distinguishing cold from warm free-streaming behavior. Baryonic feedback can change the same small-scale mass distribution. There are intrinsic inversion ambiguities: the mass-sheet transformationleaves reduced gravitational shear unchanged. Source-redshift information, absolute magnification or other data can restrict such degeneracies. Lensing establishes where gravitating matter is and tests its distribution and interactions; identification of its particle species requires complementary evidence.

Let $\theta$ measure the rim point mass's displacement from vertically below the fixed axis. The disk has moment of inertia $I_D=\frac{1}{2}(2M)R^2=M{R}^2$, and the rim mass contributes another $M{R}^2$. Its varying gravitational potential energy is $-MgR\cos \theta$. ThusThe disk's centre of mass remains fixed, so its own gravitational contribution is a constant and has been omitted. The total fixed-axis moment of inertia is $2M{R}^2$.

The uniform cylindrical torso contributes $I_c=M{r}^2/2=50(0.20{)}^2/2=1.00\mathrm{kg}{\mathrm{m}}^2$. Adding both arm point masses about the same vertical axis givesWith initial angular frequency ${\omega }_i=2\pi \mathrm{rad}{\mathrm{s}}^{-1}$, conservation of angular momentum gives $I_i{\omega }_i=I_f{\omega }_f$. ThereforeThis corresponds to $3.47788$ rotations per second; it is important not to confuse that frequency with the requested angular frequency.

Using rotational kinetic energy $K=I{\omega }^2/2$ at the two steady endpoint configurations,At fixed angular momentum $J$, $K=J^2/(2I)$ increases when the moment of inertia decreases. The skater supplies work by pulling the arms inward; chemical energy in the muscles is converted into additional rotational kinetic energy, with possible extra dissipation in a real skater. This does not require an external axial torque.

For the ideal rotating point masses, an inward radial force supplies the required centripetal acceleration. During contraction both that inward force and the radial displacement point inward, so their scalar product gives positive work. At the initial and final steady configurations radial motion has ceased; any transient radial kinetic energy must also be included in a full intermediate-time balance. The quoted difference compares the rotational endpoint energies only.

Take $\theta$ from the downward vertical. Under the conventional light-rod idealization, only the endpoint point mass contributes appreciably. Its speed is $l\dot{\theta }$, and its gravitational potential energy is $-mgl\cos \theta$. ThereforeAn additive constant in the potential energy has no effect on the Euler-Lagrange equation.

Rod-mass qualification: “thin” and “uniform” do not imply negligible mass. If the uniform rod has a finite unspecified mass $M_r$, its pivot moment of inertia is $M_r{l}^2/3$ and its centre of mass lies $l/2$ from the pivot. For a pendulum with a uniform rod and an endpoint mass, the general answer isThis formula covers either convention without inventing a rod mass. A result involving only the specified $m,l,g$ uses $M_r=0$ unless an additional relation for $M_r$ is stated.

Let $\mathbf{r}$ be the orbiting point mass's position relative to the attracting body, and let its velocity be $\mathbf{v}=\dot{\mathbf{r}}$. Its angular momentum isConservation of angular momentum keeps both the magnitude and direction of $\mathbf{L}$ fixed. Since $\mathbf{L}$ is perpendicular to $\mathbf{r}$, a nonradial orbit lies in a fixed plane. During a short interval $dt$, the radius vector sweeps a thin triangle of areaThus the areal velocity is constant:Equal time intervals give equal swept areas: this is Kepler's second law. In plane polar coordinates, the same statement is $L=m{r}^2\dot{\varphi }$ and $\dot{A}=r^2\dot{\varphi }/2$. The attracting body is at the relevant focus of the ellipse, not its geometric centre. The area argument needs conserved angular momentum, not specifically an ellipse; a central force provides this conservation because its torque $\mathbf{r}\times \mathbf{F}$ vanishes.

Treat the asteroid as a point mass with negligible influence on the planet. Outside a spherical planet, Newton's law of universal gravitation gives the specific gravitational potential $-GM/r$. The conserved specific energy and specific angular momentum areAt the periapsis $r_p$, the radial velocity vanishes, so conservation of energy givesThe right-hand side increases with $r_p>0$. The grazing threshold $r_p=R$ is therefore the grazing impact parameter in gravitational focusing relation. For $d>R$ its equivalent speed isThe threshold is customarily called the minimum avoiding speed, with tangency counted as its limiting boundary. If any contact with the surface counts as a hit, it is an infimum, not an attained minimum. For $d\le R$, no finite incoming speed avoids contact: gravitational focusing bends the path inward, and even its infinite-speed straight-line limit has closest approach $d\le R$. The finite radius of the asteroid has been neglected.

Take $+y$ upwards in the source diagram. Write $w$ for $B$'s incoming speed, so the initial velocities areA contact-direction assumption is needed for a unique answer. The diagram specifies the incoming velocities but does not specify the contact normal or an outgoing angle. First adopt the usual intended model: the impulse is horizontal, so neither particle changes its $y$ component of velocity. This is the smooth oblique collision model with a horizontal line of centres at impact. Bringing $A$ to rest in the $x$ direction then brings it entirely to rest.

Let $b$ be $B$'s outgoing $x$ component of velocity. Conservation of momentum and conservation of kinetic energy givePut $q=(\sqrt{3}/2)w$. The second equation becomes $3u^2+q^2=(q-3u{)}^2$, so $q=u$. ThereforeThe outgoing kinetic energies areThe outgoing speed of $B$ is $u\sqrt{13/3}$, directed above the negative $x$ axis at angle ${\tan }^{-1}[1/(2\sqrt{3})]\simeq 16.1^{\circ }$. The initial kinetic energy is also $(3/2+2/3)m{u}^2=13m{u}^2/6$, as required for an elastic collision.

For the perfectly inelastic collision with these same incoming velocities, both particles share a common outgoing velocity. Conservation of momentum givesIts direction is again $16.1^{\circ }$ above the negative $x$ axis. The separate translational kinetic energies of the constituents and their total areThe loss of translational kinetic energy isThis becomes internal energy, for example deformation and heating; total energy is conserved. In the elastic collision no kinetic energy is lost and the particles separate with different velocities. In the perfectly inelastic collision relative motion disappears and the common velocity is the original centre of mass velocity.

The limitation of the printed data can be made explicit. Without the horizontal-impulse assumption, write the final velocity of $A$ as $(0,h)$. Then conservation of momentum determinesand conservation of kinetic energy gives onlyThere is a family of possibilities. For a concrete second solution, $h=u/\sqrt{3}$ givesThese velocities conserve both components of momentum and conserve kinetic energy, and they satisfy the specified vanishing $x$ component for $A$, while requiring a different incoming speed. They can arise from a smooth contact normal proportional to $(1,1/\sqrt{3})$. Thus the horizontal-impulse result is conditional, rather than uniquely implied by the printed statement. For any member of this family the corresponding sticking result remains well determined once $w$ is chosen:The last expression is the lost relative kinetic energy in a perfectly inelastic collision of these two point masses.

Axis theorems. The parallel axis theorem stateswhere $I_G$ is the moment of inertia about an axis through the centre of mass, $I_O$ is about a parallel axis, $m$ is the body's total mass, and $d$ is the perpendicular separation of the axes. Expanding $\int [(x-d{)}^2+y^2]dm$ proves the result because $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane:All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let $\theta$ be the displacement from downward vertical, and let ${\mathbf{e}}_r$ point along the rod from pivot to head, with ${\mathbf{e}}_{\theta }$ in the direction of increasing $\theta$. DefineThe two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf{R}$ but no torque about its axis. The total restoring torque and rotational equation of motion areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force:The negative radial component points towards the pivot. These follow from the centre of mass acceleration $-d{\dot{\theta }}^2{\mathbf{e}}_r+d\ddot{\theta }{\mathbf{e}}_{\theta }$ and the total weight $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. They specify the external forces even when the oscillation is not small.

With the small-angle approximation, $\sin \theta \simeq \theta$, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ is the complete elliptic integral of the first kind; $K(0)=\pi /2$ recovers $T_0$.

For $m=M/10$ and $r=l/20$, compare with the simple pendulum value $T_s=2\pi \sqrt{l/g}$:For small oscillations, the hammer's oscillation period is about $0.794\%$ shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.