The dynamic modulus, or complex viscoelastic modulus, is the ratio of complex stress amplitude to complex strain amplitude in the periodic steady response:This defines the sign convention; for passive response the loss modulus is nonnegative at positive frequency. A Newtonian fluid obeys $\sigma =\eta \dot{ϵ}$, while a Hooke's law solid obeys $\sigma =G_0ϵ$. HenceThe first is purely dissipative and the second purely elastic. Here $ϵ$ denotes the scalar deformation used in the problem, with any shear-versus-extension geometric factors already included in its modulus and viscosity.

Use a rheometer, for example a cone-and-plate rheometer or a parallel-plate rheometer, to impose a small deformation and measure the required torque, or impose a torque and measure the deformation. The apparatus geometry converts these into shear stress and strain. First check that reducing the imposed amplitude does not change the normalized response; this identifies the linear regime and avoids yielding or structural rearrangement.

Three complementary protocols characterize the material's memory. In a stress relaxation experiment impose a small step ${ϵ}_{0}H(t)$ and hold the deformation fixed. The output is the relaxation modulus $G(t)=\sigma (t)/{ϵ}_0$, revealing how stored stress decays and whether an elastic plateau remains. In a creep compliance measurement impose a small stress step ${\sigma }_{0}H(t)$ and record $J(t)=ϵ(t)/{\sigma }_0$. Removing the load also distinguishes recoverable deformation from permanent flow.

In small-amplitude oscillatory shear, impose sinusoidal strain over a range of frequencies and measure both stress amplitude and phase. With $ϵ={ϵ}_0\cos \omega t$, fitThe outputs are the storage modulus $G^{\prime }$ and loss modulus $G^{\prime \prime }$, or equivalently amplitude ratio and loss tangent $\tan \delta =G^{\prime \prime }/G^{\prime }$. These measure elastic storage and dissipative response, respectively. The step experiments give time-domain information and the oscillatory experiment gives the corresponding frequency-domain response. All comparisons must use the same deformation convention and controlled temperature.

The polymer overlap concentration is reached when there is roughly one chain per coil volume. If $M$ is the chain molar mass, choose the explicit geometric convention $V_{\mathrm{coil}}=4\pi R_g^3/3$, where $R_g$ is the radius of gyration. ThenHere $c^{\ast }$ is mass per solution volume and $N_A$ is the Avogadro constant. A good solvent gives $R_g\propto N^{\nu }$ with $\nu \simeq 0.588$, approximated by $3/5$ in the Flory approximation. With a polystyrene repeat-unit molar mass of about $104.15\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the chain has approximately $N=960$ repeat units. This fixes the chain length but not the size amplitude: the problem supplies no statistical segment length or measured coil size.

For a numerical estimate, use the good-solvent size calibration for polystyrene,This empirical calibration is quoted for benzene or toluene near room temperature at molar masses above $2\times 10^5\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$; using it here is a modest extrapolation to $M=10^5\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. It gives $R_g\simeq 11.3\mathrm{nm}$. ThusThe unrounded calibration gives about $27\mathrm{g}{\mathrm{L}}^{-1}$. This is an estimate under a stated extra material input, not a unique number determined solely by molecular weight and the phrase good solvent. Choosing a different coil-volume convention changes its prefactor. The robust chain-length dependence is $c^{\ast }\propto M^{1-3\nu }\simeq M^{-4/5}$ for a flexible good-solvent coil.

At terminal speed, the downward excess weight is balanced by Stokes's law drag:The subtraction of the displaced water mass includes buoyancy. Hence the Stokes settling velocity isA density is needed but is not supplied. For the estimate, take a dense iron oxide represented by magnetite, with ${\rho }_p\simeq 5.2\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, and water with ${\rho }_w\simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. This givesUsing the diameter for the particle Reynolds number gives $\mathrm{Re}\simeq {\rho }_w{v}_t(2a)/\eta \simeq 1.8\times 10^{-5}$, which supports the creeping-flow assumption. The model assumes a dilute isolated sphere far from walls and no imposed flow. The precise numerical speed depends on the unspecified oxide phase and particle density; the estimate is of order $10\mu \mathrm{m}{\mathrm{s}}^{-1}$.

Assume steady fully developed laminar flow of a Newtonian fluid in a long circular tube, with no slip at a rigid wall. The Hagen-Poiseuille equation gives the volume flow ratewhere $R$ is the lumen radius, $\ell$ the tube length, $\eta$ the dynamic viscosity and $\Delta p$ the pressure drop. If $\Delta p$, $\ell$ and $\eta$ remain fixed, thenThus doubling the flow requiresThis estimate neglects pulsation, arterial compliance, entrance losses, the detailed shape of a stenosis and changes in the rest of the circulatory resistance. It treats blood as a homogeneous Newtonian fluid and assumes that enlarging the artery does not change its applied pressure drop. These assumptions supply the intended fourth-power scaling; an actual circulation need not keep that pressure drop fixed.

At any event on a massive observer's worldline, choose an orthonormal tetrad with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})={\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, so locally $d{s}^2={\eta }_{ab}d{X}^{(a)}d{X}^{(b)}$: the Minkowski metric. A local inertial frame can also remove the Christoffel symbols at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.

In general coordinates, $X^{\mu }(\tau )$ labels the trajectory, while the four-velocity and four-acceleration areCoordinate values $X^{\mu }$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the four-momentum is $P^{\mu }=m{U}^{\mu }$ and the four-force is $f^{\mu }=D{P}^{\mu }/d\tau =m{A}^{\mu }$. With the $+---$ metric signature,where $a$ is the proper acceleration. The middle identity follows by differentiating the constant velocity norm; in the instantaneous rest frame, $U=(c,0)$ and $A=(0,\mathbf{a})$, proving the last identity.

Measure $\alpha$ in the orthonormal tetrad of a static observer at $r>2\mu$. This specifies the local frame needed to define an angle. If $E_{\infty }$ is the conserved photon energy, gravitational redshift gives $E_{\mathrm{loc}}=E_{\infty }/\sqrt{1-2\mu /r}$. The tangential local momentum is $E_{\mathrm{loc}}\sin \alpha /c$, so the impact parameter isInside the photon sphere, an outward ray must have $b<3\sqrt{3}\mu$ to cross the potential barrier. The Schwarzschild photon escape cone therefore has critical half-angleThe upper bound is strict: a critical ray only approaches the circular orbit. At $r=3\mu$, the critical half-angle is $\pi /2$, and exactly tangential light is circular. As $r\downarrow 2\mu$, the cone shrinks to zero opening. No static observer exists exactly on the Schwarzschild event horizon, and no future-directed photon emitted there escapes to exterior infinity; the zero-angle result is an exterior limit.

A photon arriving from infinity crosses the barrier precisely when $c^2{k}^2/2$ exceeds the maximum in part (c):At infinity its spatial tangent has speed in affine-parameter units $ck$, while $h$ is its angular momentum per unit affine normalization. A straight asymptotic ray with impact parameter $b$ therefore has $|h|=bck$. ThusAt equality, the incoming ray approaches the unstable photon sphere asymptotically and is not captured. A larger impact parameter gives an outer turning point and scattering back to infinity.

For $h\ne 0$, the effective potential vanishes at $r=2\mu$, rises to a maximum, and then falls to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=h^2(3\mu -r)/r^4$, so the only circular null geodesic hasThe second derivative there is $V_{\mathrm{eff}}^{\prime \prime }=-h^2/(81{\mu }^4)<0$, proving instability by the effective potential stability criterion. These circular orbits make up the photon sphere. If the formula is continued inside the horizon, the potential is negative and diverges to $-\infty$ as $r\downarrow 0$. For purely radial light, $h=0$, the potential is identically zero and there is no circular orbit.

Set $E=0$ in part (a), substitute the two conserved quantities, and multiply the normalization by $f$:Consequently the null effective potential in Schwarzschild spacetime isOnly radii where the total radial energy is at least this effective potential are accessible.

Write $f(r)=1-2\mu /r$. In the equatorial plane, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}(f{c}^2{\dot{t}}^2-f^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)$. The cyclic coordinates $t$ and $\varphi$ give the conserved quantitiesThe squared tangent norm is constant along an affinely parametrized geodesic, since metric compatibility gives $d[g(\dot{X},\dot{X})]/d\tau =2g(D\dot{X}/d\tau ,\dot{X})=0$. HenceFor a massive particle parametrized by proper time, $E=c^2$; a photon has a null tangent, so $E=0$. This $E$ denotes the tangent norm, not the photon's physical energy. A future-directed exterior photon has $k>0$, proportional to its positive Killing energy. Its affine parameter can be rescaled, so there is no mass-shell normalization fixing $k$ to a particular positive value.

A nearby fundamental observer measures radial proper distance $d\ell =ad\chi$ and proper time $dt$. Thus the signed radial peculiar velocity isSubstitute $a^2\dot{\chi }=C$ and the massive normalization from part (b):Equivalently, for velocity $v_0$ at scale factor $a_0$,This is the exact law of redshifting of peculiar momentum. In the nonrelativistic limit, $v\propto a^{-1}$; an initially relativistic speed stays below $c$ and decreases as the universe expands.

Choose the null affine parameter so that the photon tangent is its four-momentum $P^{\mu }$. A fundamental observer has $U^{\mu }=(1,0,0,0)$ in cosmic-time coordinates, so the measured energy is $E=P_{\mu }{U}^{\mu }=c^2{P}^t$. The null normalization and conserved radial quantity in part (b) giveThus redshifting of peculiar momentum gives $E\propto a^{-1}$. Since $E=h\nu$, with $h$ the Planck constant, the received-to-emitted frequency ratio is ${\nu }_R/{\nu }_E=a(t_E)/a(t_R)$. By the definition of cosmological redshift,

For radial motion, use the geodesic LagrangianThe Euler-Lagrange equation for $t$ gives $c^2\ddot{t}+a(da/dt){\dot{\chi }}^2=0$. Since $\chi$ is a cyclic coordinate, its conjugate momentum is constant. HenceFor a massive particle, take $\tau$ as proper time; the four-velocity normalization is $c^2{\dot{t}}^2-a^2{\dot{\chi }}^2=c^2$. For a photon, use an affine parameter rather than proper time, and the corresponding tangent is a null vector. ThereforeThe normalization of the null affine parameter remains arbitrary.

A fundamental observer follows the coarse-grained cosmological matter, with fixed comoving coordinates. The cosmological principle asserts spatial homogeneity and cosmological isotropy on large scales for this family of observers. A moving observer generally sees a preferred direction, for example a Doppler dipole in the cosmic microwave background, even when the underlying cosmology is isotropic.

In an FLRW metric, these worldlines are orthogonal to homogeneous spatial slices. Synchronize their clocks to define cosmic time $t$, which is their proper time. Along a fixed spatial label, $d{s}^2=c^{2}d{t}^2$, giving $g_{tt}=c^2$. Orthogonality of the spatial coordinate directions to the observers' four-velocity gives $g_{ti}=0$. Thus, in the $+---$ metric signature,where the spatial matrix $g_{ij}$ is negative definite. Spatial homogeneity and cosmological isotropy further restrict it to $-a(t{)}^2{\gamma }_{ij}(\mathbf{x})$, with $\gamma$ a constant-curvature spatial metric.