Interpret the stated free energy as per lattice site of volume $v_0$. Flory–Huggins solution theory counts polymer chains, giving the entropy term $(\phi /N)\ln \phi$, and solvent molecules, giving $(1-\phi )\ln (1-\phi )$. The interaction term penalizes unlike contacts when $\chi >0$; a simple contact model has $\chi =z[{ϵ}_{ps}-({ϵ}_{pp}+{ϵ}_{ss})/2]/(k_{B}T)$. Connectivity reduces polymer translational entropy relative to independent monomers. This is an incompressible mean-field model with composition-independent $\chi$ for the calculations below.^[ref]

Expanding the dilute osmotic pressure identifies the effective second virial coefficient. A good solvent has $\chi <1/2$ and positive excluded volume, promoting swollen chains. A poor solvent has $\chi >1/2$, favouring contraction and potentially phase separation. The theta-solvent condition is $\chi =1/2$ in this long-chain mean-field description: the second virial term vanishes, but higher terms remain. For finite $N$, this condition is not the exact critical point, which instead has ${\chi }_c=(1+N^{-1/2}{)}^2/2$.

Let $g=f_{\mathrm{mix}}/(k_{B}T)$. The osmotic pressure in Flory-Huggins solution theory isThus $\Pi =(k_{B}T/v_0)[\phi /N+(1/2-\chi ){\phi }^2+{\phi }^3/3+\cdots ]$. For mass concentration $c={\rho }_p\phi$ and chain molar mass $M_n=N{\rho }_p{v}_0{N}_A$,osmometry uses a membrane permeable to solvent but retaining every polymer chain, at known uniform temperature. Extrapolate $\Pi /c$ to zero concentration to remove interactions; the intercept gives the number-average molar mass. Aggregation, leakage of small chains, ions with a Donnan contribution, or a poor-solvent two-phase sample invalidate the simple molecular-weight interpretation.

The measured ratios give an exactly linear reduced-pressure fit:The reduced osmotic-pressure intercept is $185\mathrm{Pa}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$. The data determine $M_n/T=0.04494\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$; the measurement temperature is needed to convert this into an absolute molar mass. For an explicitly assumed room temperature $T=300\mathrm{K}$, $M_n\simeq 13.5\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}=1.35\times 10^4\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The boxed expression is the general answer for an arbitrary experimental temperature.

Use the Monge representation $\mathbf{r}(x,y)=(x,y,h(x,y))$ for a single-valued surface without overhangs. Its area element is $dA=\sqrt{1+|\nabla h{|}^2}dxdy$; at small slopes, $dA\simeq [1+|\nabla h{|}^2/2]dxdy$ and twice the mean curvature is $2\mathcal{H}\simeq {\nabla }^{2}h$, up to orientation sign. For zero spontaneous curvature, the quadratic excess Helfrich energy isThe fluid membrane shear response has no equilibrium static in-plane shear modulus: constituents can rearrange to relax shear. Bending and isotropic tension still cost energy; viscous shear stresses can matter during motion. A Gaussian curvature term is omitted under the stated approximation; for fixed closed-surface topology its integral is constant, but boundaries and topology changes require care. The small-slope Monge expression describes weak deformations, not an entire fully wrapped sphere.^[ref]

For spherical-particle wrapping by a fluid membrane, integrate the spherical area element from polar angle $0$ to $\alpha$:Its footprint radius is $r_{\parallel }=r_0\sin \alpha$, soFor a spherical surface $2\mathcal{H}=2/r_0$, making the bending cost $\kappa (2/r_0{)}^{2}A/2=4\pi \kappa \xi$. adhesion contributes $-2\pi w{r}_0^2\xi$, and tension contributes $\sigma \Delta A$. ThereforeAll three contributions are energies; $\bar{w}$ and $\bar{\sigma }$ are dimensionless. This is the specified bound-patch model: deformation of the free membrane and the neck is omitted.

The allowed interval $0\le \xi \le 2$ corresponds to no adhesion, partial wrapping and full wrapping. At $\xi =2$, the cap covers a whole sphere and its footprint shrinks to zero. This is a closed neck in the geometric limit. It does not by itself prove membrane scission: creating a detached enclosure changes topology and involves neck physics absent from this energy.

For positive membrane tension, $E^{\prime \prime }/(\pi \kappa )=2\bar{\sigma }>0$. Projecting the unconstrained minimum $(\bar{w}-2)/\bar{\sigma }$ onto the allowed interval givesThe tension-dependent wrapping phase diagram boundaries are $\bar{w}=2$ and $\bar{w}=2+2\bar{\sigma }$. At zero membrane tension, the energy is linear: the minimum jumps from $0$ to $2$ as $\bar{w}$ crosses $2$, while all indentation values are degenerate at $\bar{w}=2$. Negative membrane tension is outside the stable-flat-membrane interpretation used here. Increasing membrane tension makes the excess area required for wrapping more expensive, broadening the partially wrapped region.

TODO: user-deferred biological application. The final request for biological relevance is intentionally left unsolved; all preceding generic geometry and membrane mechanics are included.

Write $s$ for the number of grafted chains per unit area and $\Sigma =s{a}^2$ for its dimensionless form; this separates the physical grafting density from a dimensionless coverage. A free dilute chain has size $R_0\sim a{N}^{\nu }$, with $\nu =1/2$ in a theta solvent and approximately $3/5$ in a good solvent. An isolated end-grafted chain loses translations and configurations that cross the substrate, but retains the same leading size exponent if the surface is nonadsorbing. When the lateral graft spacing $D\sim s^{-1/2}$ exceeds $R_0$, the layer consists of polymer mushrooms. When coils overlap, crowding stretches the chains away from the surface into a polymer brush.

The polymer mushroom-to-brush crossover isFor the stipulated statistical-segment interpretation of $N$ and $a$, the scaling estimates areThus ${\Sigma }_{\ast }\simeq 2.50\times 10^{-3}$ and $7.54\times 10^{-4}$ respectively. The numerical constants are scaling conventions: using a radius of gyration rather than an end-to-end size changes order-one amplitudes, not the powers of $N$.

In the Alexander-de Gennes brush model, take blobs of lateral size $D$. Within a blob, $D\sim a{g}^{\nu }$; beyond it, the $N/g$ blobs are aligned away from the wall. ThusWith coarse-grained segment volume $a^3$, conservation gives the polymer-brush volume fraction through $\varphi H\sim N{a}^{3}s$, henceThe theta-solvent brush height includes three-segment repulsion even though the effective two-segment interaction vanishes. These are scaling estimates for a roughly uniform brush, not an assertion that every chain end lies at the same height in an actual brush.

The many-segment-blob requirement $g\gg 1$ gives $D\gg a$, or $\Sigma \ll 1$. Together with overlap, the regime is $N^{-2\nu }\ll \Sigma \ll 1$. At the dense-brush packing limit, $\varphi \sim 1$, $g\sim 1$ and $H\sim Na$. Here $s\sim a^{-2}=2.04\mathrm{n}{\mathrm{m}}^{-2}$ and $H\sim 280\mathrm{nm}$. If one formally continues a dense incompressible layer above this boundary, conservation would instead giveThis is the requested formal dense-layer expression, but for $\Sigma >1$ it gives $H>Na$, which violates finite extensibility of a polymer chain. There is no physically admissible continuation above this limit in the stated segment-volume model. A Kuhn length segment is not necessarily a cube of material: with an independently specified material volume $v_0$ per counted segment, the dense-layer law is $H=N{v}_{0}s$, and the packing bound is $s\le a/v_0$. The data do not determine $v_0$ separately. The simple model therefore establishes its own breakdown, rather than a universal microscopic maximum grafting density.

grafting-to uses preformed chains whose lengths and dispersities can be characterized before attachment. As coverage rises, each incoming coil must penetrate a crowded brush and lose conformational entropy to bring its end to the surface; the growing insertion barrier limits achievable coverage. grafting-from supplies small monomers to surface-bound growing chains and avoids insertion of a whole coil. Choose grafting-from for high grafting density. Initiator coverage and reaction control still matter, and inaccessible active sites, termination and unequal growth can broaden the resulting chain-length distribution. Neither route removes the elastic and packing costs of the final brush.

A generic application is steric stabilization of synthetic colloids in a good solvent. Bringing coated particles together compresses and interpenetrates their brushes, increasing osmotic pressure and conformational free-energy cost. This repulsion can oppose aggregation. The chains must remain attached and sufficiently cover the surface; poor-solvent collapse, sparse coverage or interparticle bridging can defeat stabilization.

Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

Use signature $(+---)$, coordinates $x^0=ct$ and four-velocity $u^{\mu }=d{x}^{\mu }/d\tau$ normalized by $u_{\mu }{u}^{\mu }=c^2$. The derivative is the metric-compatible Levi-Civita connection. Fix the curvature sign by $[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }=R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }$. Then the required contractions areWith $G^{\mu \nu }=R^{\mu \nu }-g^{\mu \nu }R/2$, the contracted Bianchi identity gives ${\nabla }_{\mu }{G}^{\mu \nu }=0$. The Einstein field equations $G^{\mu \nu }+\Lambda g^{\mu \nu }=(8\pi G/c^4)T^{\mu \nu }$, with constant $\Lambda$, therefore require covariant conservation of stress-energy. This is a consistency relation between geometry and the total matter/field source. It does not mean ${\partial }_{\mu }{T}^{\mu \nu }=0$ in arbitrary coordinates, nor guarantee a global conserved energy in an arbitrary curved spacetime. The geometric identity alone, without the field equations or matter dynamics, is not an equation of motion for a freely specified material source.

For relativistic dust, $\rho$ is proper rest-mass density, measured in the fluid rest frame, and $u^{\mu }$ is its timelike four-velocity with the $c^2$ normalization above. In a local Cartesian inertial frame $u^{\mu }=\gamma (c,\mathbf{v})$, soAccordingly $T^{0i}/c$ is momentum density and $c{T}^{0i}$ is energy flux in direction $i$. The equality of momentum-density and energy-flux components after the appropriate factors of $c$ reflects the symmetry of the stress-energy tensor. In the comoving frame the spatial stress is zero, as required for pressureless matter.

Expand the dust conservation equation by the product rule:Contract with $u_{\nu }$. metric compatibility and constant normalization give $u_{\nu }{\nabla }_{\mu }{u}^{\nu }={\nabla }_{\mu }(u_{\nu }{u}^{\nu })/2=0$, leaving $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substitute this rest-mass continuity law back and divide by nonzero $\rho$:This proves geodesic motion of uncharged dust along each fluid flow line. pressure forces or another exchange of momentum would change the argument; at zero density there is no dust worldline to constrain.

The electromagnetic field tensor is antisymmetric and has six independent components. In the stated local inertial coordinates choose $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, givingThe fields are those measured by that local observer; changing observer mixes them through the Lorentz transformation of $F$. With ${ϵ}^{0123}=+1$ and ${\tilde{F}}^{\mu \nu }={ϵ}^{\mu \nu \alpha \beta }{F}_{\alpha \beta }/2$, two independent quadratic contractions areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under an orientation reversal. Opposite choices for the orientation tensor change its written sign. A nonzero null electromagnetic field can have both contractions zero, so the two numbers do not determine whether the tensor itself vanishes.

For charged relativistic dust, insert $j^{\alpha }=(q/m)\rho u^{\alpha }$ into the force-density equation:Contraction with $u_{\nu }$ again gives the continuity law, because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$ by antisymmetry. After removing the continuity term,In a local inertial frame the spatial components reduce to $d(\gamma m\mathbf{v})/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and the temporal component gives $d(\gamma m{c}^2)/dt=q\mathbf{E}\cdot \mathbf{v}$. This is the relativistic Lorentz force, with gravity represented by the connection. The force is orthogonal to $u$ and preserves the rest-mass normalization. Matter alone now exchanges energy and momentum with the electromagnetic field; adding the field tensor's stress-energy restores zero divergence of the combined source.