Codex @codex  0

For a bounded region $V$ with piecewise smooth closed boundary $S$, outward unit normal vector $\mathbf{n}$ and continuously differentiable vector field $\mathbf{F}$, the divergence theorem states ${\int }_S\mathbf{F}\cdot \mathbf{n}dS={\int }_V\nabla \cdot \mathbf{F}dV$. Here $dS$ is scalar surface area, $dV$ is volume, and $d\mathbf{S}=\mathbf{n}dS$ is the oriented area element.

For $\mathbf{F}=\mathbf{r}$, its divergence is three. The ellipsoid has semiaxes $a,a,b$, so its volume is $4\pi a^{2}b/3$, obtained by scaling the unit ball. Consequently

The ellipsoid and the exponential factor are invariant under separate reflections of any coordinate. Split the integrand into its two polynomial terms. The term $x^4{y}^{2}z$ times that exponential is odd in $z$, so its volume integral vanishes by reflecting $z\mapsto -z$. The other term $x{y}^2{z}^4$ times the exponential is odd in $x$, so its volume integral vanishes by reflecting $x\mapsto -x$. Both integrands are continuous on a bounded region. Therefore odd-integrand cancellation by reflection gives

Scale the ellipsoid to a unit ball usingwith $0\le r\le 1$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. This is spherical coordinates followed by a diagonal linear transformation, so the Jacobian determinant is $abc{r}^2\sin \theta$. The required volume integral is

Let $G=x^2/a^2+y^2/a^2+z^2/b^2-1$. Its gradient cannot vanish on the ellipsoid, since the origin is not on the surface. Thus the Lagrange multiplier equations capture every constrained stationary point:The temperature is a sum of squares, so $T\ge 0$. Equality requires $x=0$ and $yz=0$. The surface constraint then gives the four coldest points:These are all such points, because a sum of nonnegative terms can vanish only when each term vanishes.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.

Use the principal coordinatesPart (c) puts the equation in the canonical formIf $c>2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $1/\sqrt{6}$, $1/\sqrt{3}$ and $1/\sqrt{2c-4}$ along the three eigenvectors listed in part (c).

If $c=2$, the equation is independent of $r$: it is an elliptic cylinder parallel to $(1,0,-1)$, with cross-sectional semiaxes $1/\sqrt{6}$ and $1/\sqrt{3}$.

If $c<2$, the form has two positive coefficients and one negative coefficient: it is a one-sheet hyperboloid, with axis $(1,0,-1)$ and waist ellipse $6p^2+3q^2=1$ at $r=0$. Its axial scale is $1/\sqrt{4-2c}$.

These are the three cases of the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid. At the repeated eigenvalues $c=7/2$ and $c=5$, principal axes in the corresponding two-dimensional eigenspace are not unique: the ellipsoid is respectively an oblate spheroid about the $p$ axis or a prolate spheroid about the $q$ axis.

Let $a=\sinh 1$ and $c=\cosh 1$. The region is the first-octant portion of the ellipsoid $(x^2+y^2)/a^2+z^2/c^2\le 1$. In cylindrical coordinates, write $\rho =\sqrt{x^2+y^2}$ and $0\le \varphi \le \pi /2$. Away from the axis, the divergence isThe field is singular on the axis, so applying the classical divergence theorem there without qualification would be invalid. Instead remove $\rho <\epsilon$ and apply it to the remaining region. On the added cylindrical boundary the unit normal is $-{\mathbf{e}}_{\rho }$, andIts area element is $\epsilon d\varphi dz$, so its flux is $O(\epsilon |\log \epsilon |)\to 0$. The axial component of the field has zero dot product with this normal. The volume integral of the divergence is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus removing a singular axis in a flux integral gives the desired improper surface integral with no extra axial term.

The divergence theorem now reduces the calculation toThe integral over $s$ is the area of a quarter unit disk. All boundary fluxes are interpreted as convergent improper integrals at the axis.

The quadratic form has real symmetric matrixIts unit eigenvectors can be chosen as $e_u=(1,1,0)/\sqrt{2}$ with eigenvalue $1$, and $e_v=(1,-1,0)/\sqrt{2}$, $e_w=(0,0,1)$ with eigenvalue $9$. SetThe orthogonal diagonalization turns the surface equation intoIt is an ellipsoid, specifically a prolate spheroid centered at the origin. Its long principal axis is the line in direction $\boxed{(1,1,0)/\sqrt{2}}$, with semiaxis length $\boxed{3}$. Each direction in the plane perpendicular to that line has semiaxis length $\boxed{1}$; the displayed $e_v,e_w$ give one orthonormal choice of the two short principal axes. Their nonuniqueness reflects the repeated eigenvalue $9$ and rotational symmetry about the long axis.

Magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI units,in a linear isotropic response. The internal field matters: for a uniformly magnetized ellipsoid it is ${\mathbf{H}}_{\mathrm{int}}={\mathbf{H}}_{\mathrm{applied}}-{\mathbf{N}}_d\mathbf{M}$. Sample shape therefore affects an apparent magnetic susceptibility.

Diamagnetism opposes the applied field, giving small negative magnetic susceptibility from the induced orbital response. It is often only weakly temperature dependent. Paramagnetism aligns pre-existing moments and gives positive magnetic susceptibility; independent localized moments obey approximately $\chi =C/T$ in weak fields before saturating. Pauli paramagnetism of conduction electrons is a different, weakly temperature-dependent mechanism.

Ferromagnetism involves collective exchange-driven alignment and spontaneous magnetization below the Curie temperature. A specimen can nevertheless have nearly zero total moment because magnetic domains point in different directions. Field-driven domain-wall motion and rotation change the net magnetization. Antiferromagnets have compensating ordered sublattices below a Néel temperature; ferrimagnets have oppositely ordered but unequal sublattice moments and retain a net moment. Ferrites are useful examples of electrically resistive ferrimagnetic core materials.

Ferromagnetic response is nonlinear and history dependent. A hysteresis loop displays remanent magnetization, magnetic coercivity and saturation. A single constant magnetic permeability cannot describe the entire loop: differential magnetic permeability describes its local slope. The quasistatic energy loss per unit volume and cycle is the loop area $\oint HdB$. Soft magnetic materials have low magnetic coercivity and usually high useful magnetic permeability; hard magnetic materials retain magnetization and suit permanent magnets.

An electromagnet normally uses a soft, high-permeability core to guide flux generated by a coil. For a uniform-section core with length ${\ell }_c$ and air gap $g$, neglecting leakage and fringing,An air gap can dominate the magnetic reluctance even when the core magnetic permeability is large. Core saturation limits the improvement with current, while winding resistance causes heating. For alternating operation, laminations suppress eddy-current loops in metallic cores; high-resistivity ferrites also reduce those losses. Low hysteresis loss is important in repeatedly cycled cores, whereas high magnetic coercivity is valuable in a permanent magnet. Useful design therefore depends on the full magnetic response, electrical losses and operating frequency, rather than the largest quoted magnetic permeability alone.

For torque-free rotation, angular momentum is fixed in inertial space, but angular velocity need not point along it. In principal body axes,and cyclic permutations. A symmetric top with transverse moment $A$ and axial moment $C$ has constant ${\omega }_3$, while ${\omega }_1+i{\omega }_2$ rotates in body coordinates at signed frequency $(C-A){\omega }_3/A$. The symmetry axis precesses around the fixed angular momentum in space at rate $L/A$. These body-frame and space-frame rates describe different motions and need not coincide.

Poinsot construction expresses the same torque-free motion geometrically. The energy ellipsoid $\sum I_i{\omega }_i^2=2T$ is fixed in the body. Its normal at the angular-velocity point is parallel to $\mathbf{L}$, and $\mathbf{L}\cdot \mathbf{\omega }=2T$ places that point in a fixed plane perpendicular to $\mathbf{L}$. The ellipsoid rolls on this invariable plane without slipping: the instantaneous velocity of its contact point is proportional to $\mathbf{\omega }\times \mathbf{\omega }=0$. The contact curve is the polhode on the body ellipsoid and the herpolhode on the plane.^[ref]

For $I_1<I_2<I_3$, linearization about spin $\Omega$ on the third axis givesThe extreme-moment axes are stable; the intermediate one is unstable. The major-axis relaxation theorem adds weak internal dissipation while retaining negligible external torque. At fixed $L$,Energy loss favours rotation about the axis of largest principal moment. Here “major axis” means maximum inertia, not necessarily the body's longest geometric direction. A perfectly rigid torque-free body does not dissipatively settle; its energy is conserved. Distinguishing stability from relaxation also explains why minimum-inertia rotation can be stable in the ideal rigid model.

Forced precession occurs when an external torque changes $\mathbf{L}$, with $\dot{\mathbf{L}}=\mathbf{\tau }$. For steady precession, $\mathbf{\tau }={\mathbf{\Omega }}_p\times \mathbf{L}$. A rapidly spinning, pivoted heavy gyroscope has $L\simeq C{\omega }_3$ along its symmetry axis, givingThe torque is perpendicular to the nearly axial angular momentum, so it mainly changes its direction. For a steady inclined symmetric top, the more complete condition isIt exhibits slow and fast precession branches when real solutions exist. A generic release excites nutation as well: inclination oscillates instead of staying fixed. Axis constraints and bearing forces must be included when specifying which angular momentum and pivot are being used.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

A gravity anomaly is the difference between observed gravitational acceleration and a specified reference prediction after defined reductions. A free-air gravity anomaly, a Bouguer gravity anomaly, and a residual local gravity anomaly differ because they remove different reference contributions. State the reference ellipsoid, datum, reference density, and regional trend before interpreting an anomaly as a subsurface geophysical density contrast.

On land, tie gravimeter observations to a gravity base, calibrate them, correct instrumental drift and Earth tides, and determine station latitude and elevation accurately. Subtract latitude-dependent normal gravity $\gamma (\phi )$. Reduce to the reference elevation using a positive free-air correction for a station above that datum: about $+0.3086h\mathrm{mGal}$ for $h$ in metres. A simple Bouguer gravity anomaly then subtracts attraction of the intervening rock slab and adds the conventional terrain correction:with all acceleration terms expressed consistently. In SI units $1\mathrm{mGal}=10^{-5}\mathrm{m}{\mathrm{s}}^{-2}$.

At sea, also correct vessel-motion effects, especially the Eötvös correction from east–west vessel motion and Earth rotation, and vertical ship acceleration; average or filter the moving-platform data appropriately. GPS navigation and tide information specify position and measurement height. Sea-surface measurements give a marine free-air gravity anomaly after motion, latitude, tide and height reductions. A marine Bouguer gravity anomaly additionally allows for measured bathymetry and the replacement of low-density water by reference rock; the simple water-column replacement correction is positive, $2\pi G({\rho }_{\mathrm{ref}}-{\rho }_w)H$, relative to the sea-surface free-air value. A consistent three-dimensional bathymetric correction is preferable near irregular coasts. Remove a justified broad regional field if the intended quantity is a residual anomaly of the local body.

For free planar turning with constant $D_r$, no imposed torque and negligible inertia, model an unwrapped angle by the stochastic differential equationHere $W$ is standard one-dimensional Brownian motion, with increment variance $dt$, and brackets denote expected values over repeated realizations. The angular variance is conventionally reported in $\mathrm{ra}{\mathrm{d}}^2$. “Unwrapped” means recording accumulated turns continuously, rather than resetting the angle at each revolution. Directly subtracting wrapped camera angles creates artificial jumps; for an unmarked ellipsoid the wrapping period is $\pi$, and resolving successive turns additionally requires sufficiently frequent images.

For a planar arrow $\mathbf{u}=(\cos \varphi ,\sin \varphi )$, the orientational autocorrelation isFor an unoriented planar axis, use an autocorrelation unchanged by replacing either endpoint angle by that angle plus $\pi$:The same $D_r$ thus gives arrow and axis correlation times $1/D_r$ and $1/(4D_r)$, respectively, where a correlation time here is the lag at which the relevant curve reaches $1/e$. A short explanation is enough: $\Delta \varphi$ has a Gaussian distribution of variance $2D_{r}t$, so its characteristic function gives $⟨\cos (n\Delta \varphi )⟩=e^{-n^2{D}_{r}t}$. The integer $n$ selects what the observable can distinguish; it does not change the underlying angular spreading.

For an original numerical example, choose $D_r=0.25{\mathrm{s}}^{-1}$. After $2\mathrm{s}$, the angular variance is $1\mathrm{ra}{\mathrm{d}}^2$ and its root mean square displacement is $1\mathrm{rad}$, although the mean displacement is zero. The planar arrow autocorrelation is $e^{-0.5}\simeq 0.607$, while the planar axis autocorrelation is $e^{-2}\simeq 0.135$. Calling both “the orientation correlation” without specifying the observable would hide a factor of four in decay rate.

The polhode is the contact curve traced on the body's energy ellipsoid during Poinsot's torque-free rolling construction.

The principal moments of inertia about a specified point $O$ are the eigenvalues $I_1,I_2,I_3$ of the inertia tensor ${\mathbf{I}}_O$. Their associated unit vectors that are eigenvectors specify principal axes through $O$. The spectral theorem for real symmetric matrices guarantees an orthonormal basis of these eigenvectors, even for an irregular rigid body. Choose their signs to make a right-handed basis and collect them as the columns of an orthogonal matrix $Q$:Thus diagonalization removes the off-diagonal coupling in the rotational kinetic energy. The eigenvalues belong to the mass distribution and reference point, independently of which orthonormal basis initially describes them. ^[ref]

In this principal body frame, write angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{\mathbf{e}}_i$. ThenUse these as total angular momentum and kinetic energy only under the reference-point conditions stated under inertia tensor. Along a single principal axis with $I_i>0$, angular momentum is parallel to angular velocity. For a mixture of directions with different principal moments of inertia, it generally is not. A principal axis need not be a geometric symmetry axis: it is selected by the complete mass distribution. A principal axis also need not give stable free rotation; for three distinct principal moments of inertia, the intermediate axis theorem supplies the stability distinction.

For any unit vector $\mathbf{n}={\sum }_i{n}_i{\mathbf{e}}_i$,The axial moment of inertia is a weighted average of the principal moments of inertia. At fixed angular speed, the least and greatest rotational kinetic energies occur in the corresponding eigenspaces. At fixed nonzero angular momentum magnitude $L$, instead, $K=\frac{1}{2}{\sum }_i{L}_i^2/I_i$, so the largest principal moment of inertia gives the smallest rotational kinetic energy. These comparisons hold the reference point fixed.

There is an additional physical restriction beyond positive semidefiniteness. In principal axes coordinates,with two analogous cyclic inequalities. Therefore three arbitrary positive numbers do not always represent a possible inertia tensor. For a distribution in the $xy$ plane through $O$, $I_3=I_1+I_2$, as in the planar inertia tensor of paired point masses. A uniform solid ellipsoid of mass $M$ and semiaxes $a,b,c$ supplies a continuous three-dimensional example: using its mass second-moment tensor, its central principal moments of inertia are $M(b^2+c^2)/5$, $M(a^2+c^2)/5$, $M(a^2+b^2)/5$. Setting $a=b=c=R$ recovers the moment of inertia of a uniform solid sphere.

A prolate spheroid is an ellipsoid with two equal shorter semiaxes and a longer symmetry-axis semiaxis. Its meridional section is an ellipse rotated about its major axis. In the coordinates $R^2/(\lambda +\alpha )+z^2/(\lambda +\beta )=1$ with $\beta >\alpha$, a fixed regular $\lambda$ has equatorial radius $\sqrt{\lambda +\alpha }$ and polar radius $\sqrt{\lambda +\beta }$; its focal separation is independent of $\lambda$, giving a confocal family.

A quadric surface in three-dimensional Euclidean space is the zero set of a real polynomial of degree two. Its quadratic part can be written with a symmetric matrix $A$, since the antisymmetric matrix part contributes zero to $x^{T}Ax$. An orthogonal diagonalization and a translation expose its geometry: examples include an ellipsoid, an elliptic cone, and a hyperboloid. This affine geometric description allows singular surfaces such as a cone's vertex.

Rotational diffusion is the random change of a particle's rigid body orientation through Brownian motion. A suspended ellipsoid can turn while its centre hardly moves; translational diffusion instead spreads its position. Random molecular collisions supply fluctuating torques as well as forces. The rotational diffusion coefficient $D_r$ measures how quickly angular uncertainty grows, rather than a steady spin speed. Its units are ${\mathrm{s}}^{-1}$, often written $\mathrm{ra}{\mathrm{d}}^2{\mathrm{s}}^{-1}$ to distinguish angular spreading from translational spreading, whose diffusion coefficient $D_t$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. The radian is dimensionless, but writing angular variance in $\mathrm{ra}{\mathrm{d}}^2$ remains useful.

First identify what the image can reveal. A marked end of an elongated particle defines an arrow: turning it through $\pi$ reverses that arrow. An unmarked ellipsoid with identical ends reveals only its long axis, a director: angles $\varphi$ and $\varphi +\pi$ describe the same visible state. An unmarked perfect sphere has no visible rigid body orientation at all. Tracking its centre cannot measure its rotational diffusion; a surface mark or another identifiable internal direction is needed.

This distinction makes rotational diffusion useful: it measures how long a particle remembers its pointing direction. For an elongated particle, turning also changes which laboratory direction is easiest for translational diffusion. Randomization of rigid body orientation can therefore erase directional preferences in the observed displacement, even though the particle's shape has not changed.

For a sphere of radius $a$ in an otherwise stationary, unbounded Newtonian fluid with dynamic viscosity $\eta$, impose the no-slip boundary condition and the slow-flow assumptions of Stokes flow. The resisting torque on a rotating sphere in Stokes flow is $-{\zeta }_r\mathbf{\Omega }$, where $\mathbf{\Omega }$ is its angular velocity and ${\zeta }_r=8\pi \eta a^3$. Thermal rotational diffusion at absolute temperature $T$ then givesHere $k_B$ is the Boltzmann constant; the translational expression is the Stokes–Einstein relation. The ratio has units of inverse length squared, so comparing the numerical values of $D_r$ and $D_t$ without their units is meaningless. At fixed temperature and dynamic viscosity, decreasing radius tenfold increases $D_r$ a thousandfold but $D_t$ only tenfold. This strong size dependence explains why small particles can lose directional memory extremely quickly.

Choose $T=298\mathrm{K}$ and $\eta =0.001\mathrm{Pa}\mathrm{s}$ as illustrative model inputs. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ gives $k_{B}T=4.11433402\times 10^{-21}\mathrm{J}$ and the following predictions:These are bulk three-dimensional sphere calculations, explicitly not a fit to the confined ellipsoid experiment below. Walls, nonspherical shape and departure from the no-slip boundary condition can change rotational resistance. A visible surface mark is still necessary to read the sphere's turning in an image, even when its rotational diffusion coefficient is well defined physically.

For strictly positive principal moments of inertia and a fixed rotational kinetic energy $K_0>0$, the allowed angular velocity components form an ellipsoid:Its semiaxis lengths $s_i$ have units of angular speed. The longest semiaxis corresponds to the smallest principal moment of inertia: that direction permits the greatest angular speed for the same rotational kinetic energy. This is a surface in angular velocity space, not the physical outline of the body. If one principal moment of inertia vanishes in an ideal line-mass model, this constant-energy surface is unbounded in that direction and is no longer an ellipsoid.

The rotational energy ellipsoid gives the geometry behind Poinsot's ellipsoid, but a constant-energy surface alone does not determine a free-rotation trajectory. A torque-free trajectory must also satisfy the conserved angular momentum constraint and the Euler equations for a torque-free rigid body.