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For a stationary rotating apparatus driven by a source with optical coherence and proper frequency ${\nu }_0$, define ${\lambda }_0=c/{\nu }_0$. The source wave phase advances by $2\pi {\nu }_0\Delta \tau$ between the emission times of the two contributions that meet at one readout spacetime event. Define the signed phase lag $\Delta \varphi$ of the $+$ contribution and the corresponding displacement $N$ in units of one interference fringe byThe sign of a photographed spatial displacement also depends on the imposed fringe orientation; this definition fixes the theoretical sign through the delayed beam. In an equal-amplitude output, a representative light intensity is $I(x)=I_{\max }[1+\cos (Kx-\Delta \varphi )]/2$, where $K$ sets the chosen spatial interference fringe spacing $2\pi /K$. Moving $\Delta \varphi$ by $2\pi$ shifts the pattern by one fringe. A fixed output exactly at a bright or dark fringe has zero first-order intensity slope, so a known phase bias or a spatial fringe readout is useful for determining a small signed delay.

Choose a square of side $0.30\mathrm{m}$, a rectangle $0.40\mathrm{m}$ by $0.20\mathrm{m}$, ${\lambda }_0=633\mathrm{nm}$ and $\Omega =+10\mathrm{rad}{\mathrm{s}}^{-1}$, with angular velocity parallel to the positive normal vectors. Both perimeters are $1.20\mathrm{m}$, but the areas are $0.090{\mathrm{m}}^2$ and $0.080{\mathrm{m}}^2$. Using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the leading predictions areThese are original model choices and predictions, not measured points. The square has $9/8$ times the response despite the equal path lengths. Reversing from $+\Omega$ to $-\Omega$ gives $N(-\Omega )-N(+\Omega )=-2N(+\Omega )$: the displacement magnitude between opposite rotations is twice the displacement magnitude from rest. Confusing these two comparisons introduces a factor-of-two error.

In 1913 Georges Sagnac mounted a lamp, collimator, beam splitter, four mirrors and photographic plate on a horizontal rotating platform. The mirrors defined a closed polygon; the two beams traversed it in opposite senses. A slight beam splitter adjustment produced visible interference fringes. A motor brought the platform to steady rotation, and switching the lamp exposed the plate. Comparing opposite senses of rotation revealed a reversing displacement. Changing the beam splitter adjustment supplied a further check against optical deformation. The report compared measured and calculated displacements. Georges Sagnac interpreted this as evidence for a luminiferous aether; the relativistic derivation below reproduces the delay without that hypothesis.^[ref]

The readout measures a relative wave phase, not two stopwatch readings separated by an optical transit time. Reversal isolates a contribution odd in angular velocity: if a signed interference fringe position is $q(\Omega )=q_0+k\Omega$ to leading order, then $[q(\Omega )-q(-\Omega )]/2=k\Omega$ cancels the fixed offset $q_0$. Changes of alignment or temperature correlated with reversal can survive this subtraction. Thus the experimental inference concerns a reproducible rotating-loop phase difference, with controls for disturbances; it is not by itself a unique test of an underlying propagation hypothesis.

A secular tidal torque is the torque remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned tidal deformation lets the companion's gravity change the spin angular momentum; the orbit receives the opposite torque. Tidal dissipation converts mechanical energy into heat, while total angular momentum is conserved in an isolated system. Thus the spin-orbit coupling exchanges angular momentum rather than destroying it. ^[ref].

For intuition near synchronous rotation on a circular orbit, hold mean motion $n$ and moment of inertia $I$ fixed and adopt the illustrative linear dissipative lawHere $\omega$ is spin angular velocity, $\theta$ is a marked surface direction in an inertial frame, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss areThe orbital contribution to this power is $-n⟨\tau ⟩$, and the spin contribution is $\omega ⟨\tau ⟩$; their sum becomes heat. Holding $n$ fixed is a local approximation to the coupled evolution. Actual secular tidal torque depends on the deformation response and orbital eccentricity, so this simple law does not establish which spin-orbit resonance an eccentric body captures into.

For a spherical pendulum with nonzero conserved vertical angular momentum $h$, eliminate the azimuthal angular velocity using $h=m{l}^2{\sin }^2\theta \dot{\varphi }$. The polar effective potential is $V_{\mathrm{eff}}=h^2/(2m{l}^2{\sin }^2\theta )-mgl\cos \theta$. A conical pendulum at $0<{\theta }_0<\pi /2$ is a minimum satisfying $h^2=m^{2}g{l}^3{\sin }^4{\theta }_0/\cos {\theta }_0$. Linearization at fixed $h$ yields polar angular frequency ${\omega }_{\theta }^2=V_{\mathrm{eff}}^{\prime \prime }({\theta }_0)/(m{l}^2)={\Omega }^2(1+3{\cos }^2{\theta }_0)$, where ${\Omega }^2=g/(l\cos {\theta }_0)$ is the conical revolution frequency squared. Holding $\dot{\varphi }$ fixed during the perturbation would describe a different, externally driven constraint. The polar perturbation is a nutation; the exactly planar $h=0$ limit requires separate treatment.

In solid-body rotation the angular velocity is independent of distance from the rotation axis, so tangential speed is proportional to cylindrical radius. A galactic region with this galaxy rotation curve has radial epicyclic frequency $\kappa =2\Omega$.

A prograde spin-orbit resonance relates mean spin angular velocity to mean motion by a ratio of small positive integers, $⟨\omega ⟩/n=p/q$: the body makes $p$ turns in $q$ orbital periods. Synchronous rotation is $1:1$; Mercury's $3:2$ spin-orbit resonance is nonsynchronous. The relevant combination of spin and orbital angles must stay bounded around a preferred orientation, not merely have a near-integer frequency ratio. For an elongated body on an eccentric orbit, the companion's gravitational torque on the permanent shape can sustain the resonance while tidal dissipation permits capture. ^[ref].

The spin-rotation coupling contribution to a rotating-frame Hamiltonian couples angular velocity to angular momentum. When discussing spin alone, identify which part of the total angular momentum enters the effective model.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

Use the rotating frame and positive gyromagnetic ratio convention of free induction decay. Let $\Delta ={\omega }_0-\omega$ be each component's fixed detuning, and normalize its magnetization by the equilibrium value. An instantaneous resonant pulse along $+u$ rotates through $-\pi /2$ about $u$, preparing $\mathbf{m}(0^{+})=(0,1,0)$. A second pulse along $+v$ rotates through $-\pi$ about $v$. These ideal pulses require a drive much stronger than the relevant detunings and negligible relaxation during either pulse. The corresponding rotation matrices, acting on column vectors, areHere $F_{\Delta }(t)$ is free Larmor precession in the rotating frame, before including transverse relaxation. Define a calibrated complex signal $s=-i(m_u+i{m}_v)$ so the prepared component has $s(0^{+})=1$. The pulse $P$ sends $m_u+i{m}_v$ to its negative complex conjugate, hence sends $s$ to $s^{\ast }$. Direct multiplication givesThe phase accumulated before the pulse is reflected; the angular velocity during subsequent free evolution keeps its sign. At $2\tau$ every fixed detuning has zero residual phase. Equivalently, $P{F}_{\Delta }(t)P=F_{\Delta }(-t)$, so $F_{\Delta }(\tau )P{F}_{\Delta }(\tau )=P$. This is not the identity on arbitrary three-dimensional magnetization: it returns the prepared $+v$ component because $P(0,1,0)=(0,1,0)$. The two transverse pulse axes specified here also fix the receiver sign of the echo.

For an original ensemble example, let the static detunings have a Gaussian distribution with mean zero and standard deviation $\sigma$, independent of a common homogeneous $T_2$. Averaging $s_{\Delta }$ uses the characteristic function of that Gaussian distribution:Choose $\sigma /(2\pi )=25\mathrm{Hz}$, $\tau =40\mathrm{ms}$ and $T_2=160\mathrm{ms}$. Immediately before the refocusing pulse the normalized signal is only $2.08\times 10^{-9}$, yet $S(80\mathrm{ms})=e^{-1/2}=0.60653$. A one-pulse free induction decay at 80 ms would instead be about $3.11\times 10^{-35}$. These chosen numbers describe a calculation, not a measured trace. Increasing $\sigma$ narrows the echo without changing $S(2\tau )$ in this ideal model. Because this Gaussian distribution produces a Gaussian factor rather than an exponential decay, its inhomogeneous decay is not characterized by a unique exponential $T_2^{\ast }$.

The exact phase-refocusing time is $2\tau$. A slowly falling transverse relaxation envelope moves the maximum of this particular echo shape slightly earlier: differentiating $\log S$ after the pulse gives $t_{\max }=2\tau -1/({\sigma }^2{T}_2)$, or $79.747\mathrm{ms}$ here. Measuring consistently at $2\tau$ avoids confusing this small envelope effect with a failure of refocusing. A series of such measurements gives $\log S(2\tau )=-2\tau /T_2$ only when the assumed homogeneous exponential decay, pulse quality and negligible additional attenuation are appropriate.

An elongated nonaxisymmetric stellar structure in a galaxy, often approximated as a rotating perturbation of an axisymmetric or spherical gravitational potential. Its pattern angular velocity need not equal an individual star's orbital angular velocity. The Galactic bar is the particular example in the Milky Way.

For a forced sphere and a distant equal force-free, torque-free sphere, the leading incident Stokeslet strain is proportional to $(\mathbf{U}\cdot \mathbf{n})(I-3\mathbf{nn})$. Its reflected stresslet is axisymmetric about the line of centers and has zero vorticity on that line. The nominal fifth-order self-spin coefficient therefore vanishes. Finite-size incident-strain corrections and higher force-free scattering multipoles first supply a generic seventh-order angular velocity. Purely longitudinal forcing has zero spin exactly by axial symmetry. The leading translation correction is fourth order for a generic orientation, but sixth order for purely transverse forcing. These cancellations matter when using the method of reflections for Stokes flow to infer orders from far-field decay alone.

Let a symmetric top have principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Apply a torque $({\tau }_1,{\tau }_2,0)$ constant in a principal body frame. Then the axial angular velocity $c={\omega }_3$ is a conserved quantity. Define $z={\omega }_1+i{\omega }_2$ and $\nu =(I_{\parallel }-I_{\perp })c/I_{\perp }$. The first two of Euler's equations combine into the equation in the title. This derivation uses constancy in the principal body frame, which generally does not mean constant torque in an inertial frame.

Multiplying by the integrating factor $e^{-i\nu t}$ proves, for $\nu \ne 0$,Equivalently $z$ rotates about the constant point $z_{\ast }=i({\tau }_1+i{\tau }_2)/(\nu I_{\perp })$ in the complex transverse plane. When $\nu =0$, the solution is instead $z(t)=z(0)+({\tau }_1+i{\tau }_2)t/I_{\perp }$. The angular velocity is $(\mathrm{Re}z,\mathrm{Im}z,c)$.

Consider a symmetric top with fixed vertex $Q$, axial unit vector $\mathbf{e}$, principal moments of inertia $A,A,C$, and centre of mass at $h\mathbf{e}$. Let a force $\kappa \ell (\mathbf{k}-\mathbf{e})$ act at $\ell \mathbf{e}$, where $\mathbf{k}$ is the upward vertical unit vector. Together with gravity, it produces torque $(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ and potential energy $mgh\mathbf{k}\cdot \mathbf{e}+\kappa {\ell }^2(1-\mathbf{k}\cdot \mathbf{e})$.

The axial angular velocity of a symmetric top $n$, vertical angular momentum $\mathbf{k}\cdot \mathbf{L}$ and total mechanical energy are conserved. For steady precession, $\dot{\mathbf{e}}=\Omega \mathbf{k}\times \mathbf{e}$ and $\mathbf{k}\cdot \mathbf{e}=\cos \alpha$. Substituting $\mathbf{e}\times \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\times \mathbf{k}$ into $A\mathbf{e}\times \ddot{\mathbf{e}}+Cn\dot{\mathbf{e}}=(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ gives the condition in the title for nonvertical inclination. If the total angular velocity itself is $\Omega \mathbf{k}$, then $n=\Omega \cos \alpha$, and the condition becomes $\kappa {\ell }^2=mgh+(A-C){\Omega }^2\cos \alpha$.

For the Schwarzschild metric with $\mu =GM/c^2$, let $k=(1-2\mu /r)\dot{t}$ and $h=r^2\dot{\varphi }$ on an equatorial timelike geodesic. The normalization of the four-velocity gives ${\dot{r}}^2+V(r)=c^2(k^2-1)$, where $V(r)=(1-2\mu /r)h^2/r^2-2\mu c^2/r$. A circular geodesic obeys $V^{\prime }(R)=0$, giving the displayed constants and requiring $R>3\mu$. Its coordinate angular velocity is ${\Omega }^2=GM/R^3$, and its proper clock rate is $d\tau /dt=\sqrt{1-3\mu /R}$. Differentiating $V$ twice at the circular solution gives stability for $R>6\mu$.

For constant circular speed $v_0$, angular velocity is $v_0/r$ and radial epicyclic frequency is $\sqrt{2}{v}_0/r$. Constant effective random speed and surface density then give $Q=r_Q/r$. The local axisymmetric fluid model is stable inside $r_Q$ and unstable outside it. This statement is conditional on the thin-disk local approximation, rather than a proof of global or collisionless stability.

For an axially symmetric rigid body with $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$, no external torque means that angular momentum $\mathbf{L}$ is fixed in an inertial frame. The axial angular velocity ${\omega }_3$ is constant by Euler equations for a torque-free rigid body, and the symmetry axis obeysIts angle to $\mathbf{L}$ is constant and it precesses in space at $L/I_{\perp }$. The transverse angular-velocity components instead rotate in body coordinates at signed rate $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. These are different frame descriptions and must not be substituted for one another.

For a smooth strictly convex rigid body with a smooth unique contact point, a configuration consists of its orientation $R\in SO(3)$ and horizontal position $c_{\parallel }\in {\mathbb{R}}^2$; its height is fixed by contact. Translation makes this a principal bundle over $SO(3)$ with group ${\mathbb{R}}^2$. If $r(R)$ points from its reference point to contact and $\Omega$ is its spatial angular velocity, rolling without slipping requires $\dot{c}+\Omega \times r=0$. The displayed translation-valued principal connection vanishes on precisely these velocities. It is translation-invariant and equals the identity on vertical translation vectors. A prescribed orientation curve consequently has a unique horizontal lift from a given initial position. No-slip alone leaves spin free and does not specify a connection over the plane with orientation as its fiber; an additional no-twist constraint is needed for that familiar spherical model.

Use vorticity to distinguish the local spin of a fluid element from the motion of its center along a pathline. Circular streamlines can occur both with and without vorticity: the two velocity fields below even have the same circulation around the unit circle.

For a continuously differentiable velocity field $\mathbf{u}=(u_x,u_y,u_z)$, vorticity is the curl vectorEach partial derivative divides velocity by distance, so vorticity has units ${\mathrm{s}}^{-1}$. In a planar velocity field independent of $z$, only ${\omega }_z={\partial }_x{u}_y-{\partial }_y{u}_x$ survives. Positive ${\omega }_z$ points along $+{\mathbf{e}}_z$: the right-hand rule associates it with counterclockwise spin viewed from positive $z$ toward the plane. ^[ref]

The local rotational angular velocity is half the vorticity, ${\mathbf{\Omega }}_{\mathrm{spin}}=\mathbf{\omega }/2$. To see the factor, split the velocity gradient $L_{ij}={\partial }_j{u}_i$ into its symmetric rate-of-strain tensor $E=(L+L^T)/2$ and antisymmetric spin tensor $W=(L-L^T)/2$. A nearby displacement $\delta \mathbf{x}$ has relative velocity $L\delta \mathbf{x}$; its rotational part satisfies $W\delta \mathbf{x}=(\mathbf{\omega }/2)\times \delta \mathbf{x}$. This is a proof outline of the local spin interpretation. The rate-of-strain tensor can also change the angle of a particular material line, so that line's angular velocity need not equal $\mathbf{\omega }/2$.

For solid-body rotation, $\mathbf{u}=(-\Omega y,\Omega x,0)$ gives $\mathbf{\omega }=(0,0,2\Omega )$ and ${\mathbf{\Omega }}_{\mathrm{spin}}=(0,0,\Omega )$. Taking $\Omega =1{\mathrm{s}}^{-1}$, the vorticity is $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$, although the angular velocity is only $1{\mathbf{e}}_z{\mathrm{s}}^{-1}$. For the potential line vortex, $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the planar curl is $r^{-1}{\partial }_r(r{u}_{\theta })=r^{-1}{\partial }_{r}K=0$ wherever $r>0$. Its circular pathlines therefore coexist with zero local spin.

This distinction makes vorticity useful for recognizing shear flow as well as swirling motion; taking the curl of the Euler equations also removes a pressure gradient and leads to the vorticity equation. A historical landmark is Hermann von Helmholtz's 1858 paper on fluid vortex motion, an early study devoted to vortex dynamics. ^[ref]

Use a linear shear flow to see why straight pathlines can still carry local spin. Faster layers pull past slower ones, and a small symmetric paddle can turn clockwise while its center travels straight ahead.

For the planar simple shear flow $\mathbf{u}=(\gamma y,0,0)$ with constant signed shear rate $\gamma$, direct differentiation givesWith $\gamma =1{\mathrm{s}}^{-1}$, these are $-1{\mathrm{s}}^{-1}$ and $-0.5{\mathrm{s}}^{-1}$: both signs indicate clockwise orientation under the right-hand rule. Yet a particle initially at $(x_0,y_0)$ follows the straight pathline $x(t)=x_0+\gamma y_{0}t$, $y(t)=y_0$. The curl measures differences of velocity between neighboring particles, rather than bending of one particle's pathline.

The velocity gradient has $L_{xy}=\gamma$ as its only nonzero entry. Its spin tensor supplies clockwise rotation, while its rate-of-strain tensor supplies simultaneous deformation. For a material line making angle $\alpha$ to the horizontal, the equation $\dot{\delta \mathbf{x}}=L\delta \mathbf{x}$ gives $\dot{\alpha }=-\gamma {\sin }^2\alpha$. Initially horizontal and vertical material lines therefore have angular velocities $0$ and $-\gamma$; their mean is $-\gamma /2$. At $t=0.5\mathrm{s}$ with $\gamma =1{\mathrm{s}}^{-1}$, an initially vertical displacement becomes $(0.5,1)$ times its original length and has angle $\arctan 2\approx 63.435^{\circ }$, a clockwise change of $26.565^{\circ }$. The plotted cross deforms; an actual paddle's response also depends on its shape and interaction with the fluid. The exact local spin statement concerns the spin tensor, rather than every tracer's orientation.