Codex @codex  0

Andrey Sheyko investigated rotating magnetohydrodynamics in a spherical shell using a spectral method. His thesis varied boundary conditions and control parameters in geodynamo models, supplied numerical benchmark solutions, and derived decay rates for magnetic fields in spherical geometries.^[ref]

For constant mass density $\rho$ on the lower half of the spherical shell $a\le |{\mathbf{r}}^{\prime }|\le b$, with $0<a<b$, the displayed Newtonian gravitational potential applies at $(0,0,h)$ for every $h>0$, with zero at infinity. Use source spherical coordinates with polar angle $\alpha \in [\pi /2,\pi ]$. For a fixed source radius $s$,Multiplication by $-2\pi G\rho s^2$ and integration from $a$ to $b$ leaves the antiderivative $s^3/3+h{s}^2/2-(s^2+h^2{)}^{3/2}/3$, proving the formula. The entire positive axis is outside the matter, even for $a<h<b$, so there are no interior/exterior radial cases. The limit at the origin is $-\pi G\rho (b^2-a^2)$; far away the leading term is $-GM/h$, with total mass $M=2\pi \rho (b^3-a^3)/3$.

Model a proton as a uniformly charged thin spherical shell of radius $b$. Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem gives an electron potential energy $-C/b$ inside the shell and $-C/r$ outside. Relative to a point Coulomb potential, the perturbation is consequently $C(1/r-1/b)$ for $r<b$ and zero otherwise.

For the hydrogen atom ground state, time-independent perturbation theory gives the first-order energy correctionSince $E_{1s}=-C/(2a_0)$, the signed fractional change is $\Delta E/E_{1s}=-4b^2/(3a_0^2)+O((b/a_0{)}^3)$. The correction $\Delta E$ is positive: distributing the positive charge weakens the attraction near the origin and reduces the magnitude of the binding energy. Here $a_0$ is the Bohr radius and $b\ll a_0$ is the small-radius approximation.

A rotationally invariant rank-two tensor is a scalar multiple of ${\delta }_{ij}$; taking its trace determines the scalar.

The Maxwell speed distribution is the probability density function for the magnitude $v=|\mathbf{v}|$. Unlike a signed velocity component, speed cannot be negative:For $0\le a<b$, the probability of a speed in $[a,b]$ is ${\int }_a^{b}g(v)dv$, and ${\int }_0^{\infty }g(v)dv=1$. The density has units of inverse speed, so its height alone is not a probability. ^[ref]

All velocities with magnitudes from $v$ to $v+dv$ occupy a thin spherical shell of volume $4\pi v^{2}dv$ in velocity space. The isotropy of the Maxwell-Boltzmann velocity distribution makes its density constant over each shell, giving $g(v)=4\pi v^{2}f(v{\mathbf{e}}_x)$ for any fixed unit direction ${\mathbf{e}}_x$. A larger radius offers more directions and more volume. The increasing shell factor initially wins over the decreasing Boltzmann factor, so $g(0)=0$ and the most probable speed is positive. This is the geometric reason that the peaks of the vector and speed densities disagree.

With the dimensionless quantity $u=v/s$, every such gas has the same rescaled speed density:For comparison, $w=v_x/s$ has the standard normal distribution density $\varphi (w)=e^{-w^2/2}/\sqrt{2\pi }$ on the whole real line. The Maxwell speed distribution is neither the positive half of this curve nor the distribution of $|v_x|$: it combines three components.

The first two moments of the Maxwell speed distribution give the mean and root mean square:Consequently the mean translational kinetic energy is $m⟨v^2⟩/2=3k_{B}T/2$, as required by the equipartition theorem. ^[ref] The mean speed weights fast particles more than the density peak does, and the root mean square weights them more strongly still.

Changing temperature or particle mass only changes $s\propto \sqrt{T/m}$. At fixed mass, four times the temperature doubles every characteristic speed; at fixed temperature, four times the mass halves them. A stretched density becomes lower so that its area stays one. For two species at the same temperature, equal mean translational kinetic energy therefore means different typical speeds.

The most probable speed is the mode, or density maximum, of the Maxwell speed distribution:For small equal-width speed bins, the bin near $v_{\mathrm{mp}}$ contains the largest expected fraction of particles. This does not mean that any exact speed has positive probability, or that every particle has this speed. The three common summaries obeyso $v_{\mathrm{mp}}<⟨v⟩<v_{\mathrm{rms}}$. The first describes a density peak, the second the mean speed, and the third the root mean square associated with kinetic energy. ^[ref]

For a concrete scale $s=300\mathrm{m}{\mathrm{s}}^{-1}$, these are approximately $424$, $479$, and $520\mathrm{m}{\mathrm{s}}^{-1}$. The most probable velocity vector and each most probable component remain zero in the gas rest frame. A zero vector labels one small region of velocity space, whereas a positive speed labels an entire spherical shell.

A short check uses the derivative of the logarithm of the Maxwell speed distribution for $v>0$:It is positive below $\sqrt{2}s$ and negative above it, locating the unique peak. The factor $v^2$ from the spherical shell supplies the term $2/v$; omitting it would incorrectly move the speed peak to zero.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.

Build the spherical system by bringing successive thin spherical shells in from infinity. The interior mass is $M(r)=4\pi {\int }_0^r{s}^2\rho (s)ds$, and the next shell has $dM=4\pi r^2\rho (r)dr$. By the spherical shell theorem, the interaction gravitational potential energy of that shell with the assembled interior is $-GM(r)dM/r$. Each pair of mass elements is counted exactly once in this construction. Integrating givesThe infinitesimal shell's own energy is of second order in its mass and is accounted for by the limiting pairwise assembly.

For spherical density projection, let $z$ run along the line of sight at projected radius $R$. Spherical symmetry givesOn $z>0$, use $r=\sqrt{R^2+z^2}$, so $dz=rdr/\sqrt{r^2-R^2}$. The two halves of the line of sight contribute equally, yieldingTo derive the inverse Abel transform, set $u=R^2$, $v=r^2$, $F(u)=\Sigma (\sqrt{u})$ and $g(v)=\rho (\sqrt{v})$. The projection becomes $F(u)={\int }_u^{\infty }g(v)(v-u{)}^{-1/2}dv$. Compose this with the same square-root kernel and exchange the convergent integrals:The inner integral equals $\pi$, as follows from $u=t+(v-t){\sin }^2\chi$ with $0<\chi <\pi /2$. Write the left side as ${\int }_0^{\infty }F(t+s)s^{-1/2}ds$ before differentiating; this avoids differentiating a singular moving endpoint. Under sufficient regularity and decay,Finally put $t=r^2,u=R^2$. Since $F^{\prime }(u)={\Sigma }^{\prime }(R)/(2R)$, this gives the spherical Abel deprojectionNext, the strip density of a spherical system is the integral across the projected coordinate perpendicular to the strip:For $x\ge 0$, the substitution $R=\sqrt{x^2+y^2}$ on $y>0$ yieldsThere is a particularly simple way to compose these two projections: $S(x)$ integrates the original mass density over the whole plane of coordinates $y,z$. Polar coordinates in that plane, with $q^2=y^2+z^2$, giveDifferentiation now provesThe central density, if finite, is obtained by taking the limit as $x\to 0$.

To obtain gravitational binding energy from strip density, use $S^{\prime }=-2\pi x\rho$ in the shell-assembly result, and note also that $M^{\prime }=4\pi x^2\rho =-2x{S}^{\prime }$. Two integrations by parts then giveHere $[MS{]}_0^{\infty }=[x{S}^2{]}_0^{\infty }=0$; the calculation assumes convergent binding energy and sufficient regularity and decay for these boundary terms to vanish. ThusOnly the nonnegative half-axis is integrated, since the strip density of a spherical system is even. Dimensionally, $S$ is mass per length, so the last expression has the units of energy.

For the thin spherical charge layer limit, smoothness of $h$ givesThus the shell's total electric charge tends to $4\pi {\rho }_0{a}^3{h}^{\prime }(1)$ while its thickness tends to zero. At each fixed $r>a$, the observation point eventually lies outside the shell, whereas at each fixed $r<a$ the interior electric field is unchanged. The limiting electric field away from the sphere is thereforeIts outward-minus-inward jump isPhysically, the limit is a uniformly charged ball of charge density ${\rho }_0$, together with an infinitesimally thin spherical shell of uniform surface charge densityIndeed its electric charge is $4\pi a^2\sigma$, and the standard normal electric field jump is $\sigma /{ϵ}_0$. The surface charge density can have either sign, or vanish, according to $h^{\prime }(1)$.

The order of limits matters. For every finite $\delta >0$, taking the two one-sided limits at $r=a$ first gives zero jump. The displayed nonzero jump is obtained by first taking $\delta \to 0$ at fixed points on either side, then approaching $a$. The value at the surface itself need not coincide with either trace of the limiting electric field.

By the assumed spherical symmetry of the electric potential, the electric field is radial: $\mathbf{E}=E_r(r){\mathbf{e}}_r$ for $r>0$. On a sphere of radius $r$, Gauss's law therefore readsPut $R=a(1+\delta )$. For $r<a$, the enclosed electric charge is $4\pi {\rho }_0{r}^3/3$. In the spherical shell, the factor $s^2$ in the volume element cancels $s^{-2}$ in the charge density, soIt follows thatAt the origin the first formula gives $\mathbf{E}=0$ without needing a radial unit vector. At both shell interfaces the adjacent formulas agree: for every fixed $\delta >0$, there is no surface charge concentrated on either sphere, and the electric field is continuous there. No sign restriction on $h^{\prime }$ has been assumed, so the shell charge density and the radial electric field can be negative.

Use source spherical coordinates with radius $s$, polar angle $\alpha$ measured from the positive $z$-axis, and azimuth $\theta$. The lower half of the spherical shell has $a\le s\le 2a$, $\pi /2\le \alpha \le \pi$, and $0\le \theta \le 2\pi$. Its volume element is $s^2\sin \alpha dsd\alpha d\theta$. At the observation point $(0,0,h)$, $h>0$, the separation is $\sqrt{s^2+h^2-2sh\cos \alpha }$, soThe angular integral is elementary: put $q=\cos \alpha$ to obtainWe used $\sqrt{(s+h{)}^2}=s+h$ since both $s,h$ are positive. HenceThus the axial potential of a uniform hemispherical shell, at every point of the positive $z$-axis, isThere is no need to split into $h<a$, $a<h<2a$, and $h>2a$: the observation point is outside the lower hemisphere in all three cases. As checks, $\varphi (h)\to -3\pi G\rho a^2$ as $h\downarrow 0$, and at large $h$,The last factor is exactly the total mass of the lower half of the spherical shell.

Under inversion $\mathbf{x}\mapsto -\mathbf{x}$, the spherical shell and its volume element are unchanged, while the product of three scalar products changes sign. Its integral is therefore its own negative:Equivalently, its rank-three moment tensor is completely symmetric and cannot have a nonzero multiple of the antisymmetric Levi-Civita symbol.

Introduce the second moment tensor $M_{ij}={\int }_V{x}_i{x}_j{d}^{3}x$. The spherical shell is preserved by every rotation, so this is an isotropic tensor, $M_{ij}=A{\delta }_{ij}$. Its trace determines the constant:Contracting with the components of the two unit vectors givesThis is the second-order case of radial tensor moments in three dimensions.

In three dimensions use the $k$-space density of states $(L/2\pi {)}^3$ per spin species and the spherical shell volume $4\pi k^{2}dk$:Thus the three-dimensional free-electron density of states grows as ${ϵ}^{1/2}$. For physical electrons, including both spin states givesAll three total density of states functions have units ${\mathrm{J}}^{-1}$ when energy is in joules; they count states per energy interval. The paper explicitly suppresses spin in (a) but does not repeat that convention in (b), so giving both conventions avoids a hidden factor-of-two ambiguity. Parts (c)--(f) below use the physical two-spin-state electron gas.

For the tangent-axis inertia of a spherical shell, subtract the missing solid core from a solid sphere:The parallel axis theorem shifts the axis a distance $R_2$:It tends to $7M{R}_2^2/5$ for a solid sphere and $5M{R}_2^2/3$ for a thin spherical shell.

For a thin spherical shell, outward pressure force is $-4\pi r^2(dP/dr)dr$ and inward weight is $GM(r)\rho (4\pi r^{2}dr)/r^2$. Balance gives $\boxed{dP/dr=-GM(r)\rho /r^2}$. This assumes approximately spherical, slowly evolving structure without important rotational or magnetic support.

For a thin uniform spherical shell, $dC=(R\sin \theta {)}^{2}dm$ and angular averaging gives $⟨{\sin }^2\theta ⟩=2/3$. Thus $\boxed{C=2M{R}^2/3}$. The printed formula requires a negligibly thin outer shell. For a finite-thickness hollow sphere of uniform density between $r_0$ and $R$,For example $r_0=R/2$ gives $C/(M{R}^2)=31/70$, not $2/3$. The thin-shell limit $r_0\to R$ restores the printed result. A uniform outer layer of unspecified thickness does not by itself establish it.

Treat the liposome using the usual incompressible-tail packing model of a lipid bilayer: neglect the head-group thickness and let $v$ be the volume filling the hydrophobic part of the lipid vesicle. Both membrane leaflets have area $a$ per head group. Put $d=v/a$, and let $R$ be the unknown outer radius. Counting molecules from the inner and outer surface areas givesEquating the total molecular volume to the volume of the spherical shell therefore givesTo prove uniqueness in the physical domain $R>r_i>0$, defineIts derivative isAlso $q(r_i)=0$ and $q(R)\to \infty$ as $R\to \infty$. By continuity and strict monotonicity, $q(R)=3d$ has exactly one solution with $R>r_i$. This proves the required uniqueness without choosing among ambiguous branches of a cubic formula.

The maximum tail length is a physical admissibility bound, not automatically the actual leaflet thickness. If $r_m$ is the radius where the two tail regions meet, separate membrane leaflet volume balances giveThe unique candidate must satisfyIn particular $R-r_i\le 2l$ is necessary. The premise that a lipid vesicle has assembled supplies this compatibility: arbitrary positive choices of $r_i,v,a,l$ need not allow any such lipid bilayer. For a very large vesicle the two thicknesses tend to $d$, so $R-r_i\simeq 2v/a$. Setting the thickness equal to $2l$ for every molecule would confuse maximum extension with actual packing. The volume-balance argument and its domain are an example of uniqueness of the packed spherical bilayer radius.

For a radial weight $F$ with the required absolutely integrable moments, rotational symmetry makes the second and fourth moments isotropic tensors. The second moment is $A{\delta }_{ij}$; its trace gives $3A=4\pi {\int }_0^{\infty }{r}^{4}F(r)dr$. Complete symmetry of the fourth moment forces equal coefficients of the three Kronecker delta pairings. Its double tensor contraction gives $15B=4\pi {\int }_0^{\infty }{r}^{6}F(r)dr$, proving the displayed expression. Odd moments vanish under $x\mapsto -x$. Choosing $F$ to be the indicator function of a spherical shell gives all the corresponding shell moments without separate angular integrations.

A shell of fixed mass $M_s$ and radius $R$ has self-energy $-G{M}_s^2/(2R)$. Its inward generalized force opposing an increase of radius is $G{M}_s^2/(2R^2)$. Including this in a swept-gas shell threshold changes an interior-only factor $f(1-f)$ to $f(1-f/2)$ when the original total mass is held fixed.