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Let the equilateral triangle have side length $a$. Its altitude is $H=\sqrt{3}a/2$, and its incircle has radius $r=H/3=a/(2\sqrt{3})$: the centre lies on a median, one third of the altitude above the base. The area of the triangle is $aH/2=\sqrt{3}{a}^2/4$, while the circle has area $\pi r^2=\pi a^2/12$. ThereforeThe side length cancels, as required for a dimensionless ratio between similar configurations.

A cross-section perpendicular to the rotation axis is a disk of radius $\sin x$. Disc integration therefore givesThe result is in cubic coordinate units; the full rotation is already accounted for by the disk area.

In cylindrical coordinates, $x=r\cos \theta$ and $y=r\sin \theta$, so $x^2+y^2=r^2$. Thus the circular elliptic paraboloid has equationAt height $h\ge 0$, the rim radius is $\sqrt{ah}$.

Use the convention $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$, with $\theta$ measured from the positive vertical axis. The squared radius isSince the horizontal radial distance is $1/\sqrt{1+t^2}>0$ for finite $t$,The horizontal angle advances with $t$. Thus the spherical coordinates areAn unwrapped continuous azimuth is simply $\varphi =t$; a principal-angle convention introduces harmless jumps by $2\pi$. The polar angle stays in $(0,\pi )$ and approaches the two poles only as $t\to \pm \infty$.

The polar area element is $rdrd\varphi$. Integrating first in radius and using symmetry givesThis is the specialization $a=1$, $c=3/2$ of the polar cap between a cardioid and a circle area formula.

The radial interval is nonempty exactly when $1+\cos \varphi \ge 3/2$, or $\cos \varphi \ge 1/2$. Choose the representative angular range $-\pi /3\le \varphi \le \pi /3$. The region of integration is thereforeIt is the right-hand cap inside the cardioid and outside the circle of radius $3/2$, symmetric about the horizontal axis. Its boundary intersections have coordinates $(3/4,\pm 3\sqrt{3}/4)$, and the cap reaches $(2,0)$. The radial segment on the horizontal axis runs from $3/2$ to $2$.

Use cylindrical coordinates $x=\rho \cos \varphi$, $y=\rho \sin \varphi$. The velocity field becomesFor each fixed height, particles follow concentric circular streamlines around the vertical axis. At $z>0$ the motion is anticlockwise viewed from above, while at $z<0$ it is clockwise. Within a horizontal plane the angular velocity is $\alpha z$, independent of radius, and the speed is $\alpha |z|\rho$. Both the plane $z=0$ and the vertical axis are stationary. The rotation rate varies with height, so the entire liquid is not one rigidly rotating body.

Use completing the square to obtainThe locus is therefore a circle with centre $(1,0)$ and radius $2$. It meets the horizontal axis at $(-1,0)$ and $(3,0)$ and the vertical axis at $(0,\pm \sqrt{3})$; its highest and lowest points are $(1,\pm 2)$.

The first curve is a circle of radius $1$ centred at $(2,1)$. A line through the origin with equation $kx-y=0$ intersects it precisely when the perpendicular distance from that centre is at most the radius. The distance from a point to a line givesHence the full parameter range isThe two endpoint lines are tangent, at $(2,0)$ and $(6/5,8/5)$ respectively; interior values give two real intersections. Both endpoints are therefore included.

For real $\lambda >0$, squaring the two nonnegative moduli gives $|z-i{|}^2={\lambda }^2|z+i{|}^2$. In Cartesian coordinates this isAssume $\lambda \ne 1$ and put $h=(1+{\lambda }^2)/(1-{\lambda }^2)$. Dividing by $1-{\lambda }^2$ yields $x^2+y^2-2hy+1=0$. Since $z{z}^{\ast }=x^2+y^2$ and $b{z}^{\ast }+b^{\ast }z=2\mathrm{Re}(b{z}^{\ast })$, the requested complex equation hasIndeed these choices make the mixed term $-2hy$. Completing the square givesThe Circles of Apollonius therefore haveThe radius is positive for the stated range. No spurious point at $z=-i$ has been introduced: at that point the undivided equation would require $4=0$. Positive $\lambda <1$ gives a circle above the real axis, while $\lambda >1$ gives one below it.

The complex modulus $|z-(-1-i)|$ is the Euclidean distance from $z$ to $-1-i$. Thus the locus isa circle with centre $(-1,-1)$ and radius $8$. Its extreme horizontal coordinates are $-9$ and $7$, and its extreme vertical coordinates are also $-9$ and $7$.

At each height, the removed circular sector has the same angle and $4/5$ of the original radius. Its area and hence its volume are therefore multiplied by $(4/5{)}^2=16/25$. Equating the removed volume to that of the reformed solid sphere givesConsequently

At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore

Take the dumbbell axis to be the $z$ axis, with its midpoint at $z=0$. The right-hand sphere has centre $z=\alpha a$ and radius $R=\beta a$, while its junction with the cylinder is at $z=\gamma a$. Put $d=(\alpha -\gamma )a$, so $0<d<R$. The Pythagorean theorem in a meridional cross-section gives the handle radius $H$ from $H^2+d^2=R^2$. Thus

Use cylindrical coordinates $(r,\varphi ,z)$ about the dumbbell axis. Their Jacobian determinant contributes the volume factor $r$, so the handle's mass is the volume integralInserting the radius from part (a) yields

Completing the square directly with the complex modulus givesFor $|b{|}^2>c$, this is the circle with centre $-b$ and radius $\sqrt{|b{|}^2-c}$. Equality gives just the point $-b$, and $|b{|}^2<c$ gives the empty locus. In particular, the nondegenerate answer is

There is a single point exactly when the radius is zero:Then $|z+b{|}^2=0$ forces this unique value of the complex number $z$.

An accelerating point charge has a transverse radiation field in addition to its near field. At nonrelativistic speed and large distance $R$,The acceleration is evaluated at retarded time. Radiation fields decay as $1/R$, so their outward energy flux integrated over a sphere is independent of radius; the near electric field instead decays as $1/R^2$. The radiation vanishes along the acceleration and is strongest perpendicular to it.^[ref]

Using the Poynting vector givesThe integral uses $\int {\sin }^2\vartheta d\Omega =8\pi /3$. This is the Larmor formula. A harmonically accelerated charge with peak acceleration $a_0$ has mean power $q^2{a}_0^2/(12\pi {ϵ}_0{c}^3)$; external work supplies this radiated energy in steady operation. Uniform inertial motion does not radiate.

At relativistic speed the angular pattern becomes directionally concentrated and retarded-source factors matter. The total instantaneous radiated power generalizes to $q^2{\gamma }^6[a^2-|\mathbf{\beta }\times \mathbf{a}{|}^2]/(6\pi {ϵ}_0{c}^3)$, with $\mathbf{\beta }=\mathbf{v}/c$. The displayed simple ${\sin }^2\vartheta$ pattern is restricted to the nonrelativistic limit.

A positive-parity $K=0$ nuclear rotational band in an even-even nucleus has $J=0,2,4,\dots$ and excitation energiesThe four nonzero levels give $E_J/[J(J+1)]=15.17,14.85,14.48,14.01\mathrm{keV}$. These approximately constant coefficients, together with the even-spin positive-parity sequence, support a rotational interpretation. The roughly eight-percent decline means a constant-inertia rigid rotor is only an approximation: the effective moment of inertia increases along the sequence.

Using the lowest excited level givesUsing the highest level instead gives $2.48\times 10^{-54}\mathrm{kg}{\mathrm{m}}^2$, showing the scale of the departure from a constant inertia. This is a comparison of nuclear rotational inertia relative to a rigid sphere, not evidence for an exactly spherical rotating intrinsic state.

For a conventional size estimate, explicitly take a uniform sharp radius $R=r_0{A}^{1/3}$ with $r_0\simeq 1.2\mathrm{fm}$ and mass $M\simeq A{m}_u$. These are standard scale approximations, not additional measured properties of this isotope supplied by the paper. They give $R\simeq 6.70\mathrm{fm}$ andThus the fitted rotational inertia is substantially smaller than the rigid-body scale. Nuclear pairing provides a physical reason collective nuclear inertia need not equal the classical rigid-body value. The comparison's numerical radius dependence is explicit; it is not an independently determined radius.

The deuteron is the bound proton–neutron system. Its ground state has $J^P=1^{+}$ and isospin $T=0$, with binding energy approximately $2.225\mathrm{MeV}$ and no excited bound state. The small binding compared with characteristic nuclear energy scales produces an extended tail. With reduced mass $\mu \simeq m_N/2$, the asymptotic radial wavefunction decays as $e^{-\kappa r}$, whereThe decay length is not a sharp deuteron radius or the force range. It illustrates weak binding outside the region of strong attraction.

The observed total spin is consistent with aligned nucleon spins, $S=1$, and a predominantly $L=0$ spatial state, ${}^3{S}_1$. A central attraction identical in both spin–isospin channels would not distinguish this channel from a spin-singlet proton–neutron channel. In fact the triplet channel binds while the corresponding singlet channel does not. A realistic nuclear force must therefore distinguish these spin–isospin channels, for example through a term $V_{\sigma }(r){\mathbf{\sigma }}_p\cdot {\mathbf{\sigma }}_n$, whose eigenvalues are $+1$ and $-3$ in triplet and singlet states. The absence of a bound dineutron and diproton reinforces the need to distinguish channels; proton–proton Coulomb repulsion is an additional effect, not a complete explanation of the singlet nuclear interaction.

The clearest evidence against a purely central force is the nonzero electric quadrupole moment. A pure isotropic S-wave charge distribution would give zero. The deuteron quadrupole evidence for a tensor force instead points to a nuclear tensor force,It depends on how the spins point relative to the line joining the particles. It preserves total $J$ but couples orbital angular momenta differing by two, producingBoth terms have $J^P=1^{+}$, and their interference permits a quadrupole moment. The 1939 theoretical analysis explicitly connected the measured quadrupole moment with such directional dependence.^[ref]

The nuclear magnetic moment provides a second check. A pure aligned-spin S wave would give approximately ${\mu }_p+{\mu }_n\simeq 0.880{\mu }_N$, close to but larger than the observed deuteron value approximately $0.857{\mu }_N$. The discrepancy is consistent with non-S-wave structure and interaction-current corrections; it is not by itself a model-independent measurement of a D-state probability. The deuteron D-state admixture depends on the adopted Hamiltonian and electromagnetic current. The magnetic moments and mass differences used for these scales are tabulated physical constants.^[ref]

A useful schematic interaction therefore contains central, spin-dependent and tensor terms, and possibly spin–orbit terms in other channels. Long-range one-pion exchange naturally supplies a tensor component, with its finite range set by the pion mass. The deuteron establishes an attractive triplet channel and a noncentral tensor interaction; a spin-independent central potential is insufficient.