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About a fixed origin, the total momentum and angular momentum areA time-independent translation ${\mathbf{x}}_i^{\prime }={\mathbf{x}}_i-\mathbf{b}$ leaves every velocity unchanged. ThereforeThe cross product $\mathbf{b}\times \mathbf{P}$ is perpendicular to $\mathbf{P}$, giving the invariantThe other requested transformation follows from the vector triple-product identity:The actual PDF asks about $\mathbf{L}\times \mathbf{P}$; the TeX's occurrence of $\mathbf{F}$ is a transcription error.

If $\mathbf{P}=0$, the translation term vanishes and angular momentum is independent of origin. If $\mathbf{P}\ne 0$, chooseThen $\mathbf{b}\times \mathbf{P}=\mathbf{L}-(\mathbf{L}\cdot \mathbf{P})\mathbf{P}/|\mathbf{P}{|}^2$, and soThus the new angular momentum is parallel to the total momentum. The freedom along $\mathbf{P}$ gives the central axis of a momentum system. If the invariant dot product is zero, the new angular momentum is the zero vector. All expressions assume a translated fixed origin, not a moving reference frame.

With no electric force, the Lorentz force equation isIts scalar product with velocity proves conservation of energy for the particle's kinetic energy:If $\mathbf{B}=\mathcal{B}(\mathbf{x},t)\hat{\mathbf{b}}$ with a fixed unit direction $\hat{\mathbf{b}}$, then $m\ddot{\mathbf{x}}\cdot \hat{\mathbf{b}}=0$, so the parallel component of velocity is constant even when the field magnitude varies.

The angular momentum about the origin does not generally remain constant, becauseFor example, at an instant with $\mathbf{x}=(r_0,0,0)$, $\dot{\mathbf{x}}=(v_0,0,0)$ and $\mathbf{B}=(0,0,B)$, its derivative is $(0,0,-eB{r}_0{v}_0)$, nonzero when those factors are nonzero. A force perpendicular to velocity can preserve kinetic energy while exerting a nonzero torque.

For the constant axial field, Cartesian equations are $m\ddot{x}=eB\dot{y}$, $m\ddot{y}=-eB\dot{x}$, and $\ddot{z}=0$. Use the signed projected areal rate $\dot{A}=(x\dot{y}-y\dot{x})/2=r^2\dot{\theta }/2$. ThenThusThe second equation is the magnetic axial angular-momentum invariant, replacing ordinary conservation of $L_z=m{r}^2\dot{\theta }$. In the symmetric gauge, the added term is the magnetic contribution to canonical angular momentum; it is the axial component that is conserved, not the full mechanical angular-momentum vector.

Let $v^2={\dot{x}}^2+{\dot{y}}^2$ be the constant squared transverse speed. If the motion has a nonzero parallel speed, $v^2=2K/m-{\dot{z}}^2$, where $K$ is total kinetic energy. Solving the invariant and using $v^2={\dot{r}}^2+r^2{\dot{\theta }}^2$ givesThe TeX transcription omits the angular equation, which is present in the PDF and restored here. The PDF's positive radial root describes the outward branch only. An inward branch needs the minus sign, with sign changes at radial turning points; polar coordinates themselves are singular at $r=0$. These qualifications are needed for a global description of a cyclotron trajectory whose radius about an arbitrary origin can both increase and decrease.

The initial time-dependent-field discussion treats the stipulated magnetic force as a prescribed particle model. If one additionally imposes full Maxwell equations with electric field zero throughout a region, Faraday law requires the magnetic field there to be time independent. The energy calculation is nevertheless valid for the stated force model; it does not assert existence of a Maxwell-consistent time-dependent field with zero electric field everywhere.

Write $H=\dot{R}/R$. A fluid element in the homologous spherical flow obeys $\dot{\mathbf{r}}=H\mathbf{r}$, so $\mathbf{r}/R$ remains constant. Also $\nabla \cdot {\mathbf{u}}_0=3H$. Along every element, the continuity equation therefore gives $D{\rho }_0/Dt=-3H{\rho }_0$. Every element starts with the same mass density, and hence has the same mass density at subsequent times:This proves preservation of uniformity, rather than assuming it when conserving the total mass.

By the shell theorem, only the mass interior to radius $r$ contributes to its gravitational acceleration. That mass is $4\pi {\rho }_0{r}^3/3$. The same result follows directly from the Poisson equation for Newtonian gravity: spherical symmetry gives $(r^2{\Phi }_r{)}_r=4\pi G{\rho }_0{r}^2$, and regularity at the centre sets the integration constant to zero. Hence ${\Phi }_r=4\pi G{\rho }_{0}r/3$ andThe material derivative of the background velocity is $(\dot{H}+H^2)\mathbf{r}=(\ddot{R}/R)\mathbf{r}$. There is no pressure force, and the momentum equation consequently reduces toMultiplication by $\dot{R}$ and integration, using collapse from rest, givesChoose the negative square root for contraction. Put $C^2=2K/R_0$ and $R=R_0{\cos }^2(\varphi /2)$. For $0<\varphi <\pi$, the energy integral then gives $\dot{R}=-C\tan (\varphi /2)$, while $dR/d\varphi =-R_0\sin \varphi /2$. Their ratio isIntegrating from the initial turning point proves the cycloid parametrization of pressureless homologous spherical collapse:It covers the entire contracting branch, with complete collapse at $\varphi =\pi$ and $t_c=\sqrt{3\pi /(32G{\rho }_{00})}$. At $R=0$ the uniform dust configuration is singular, so the smooth calculation is restricted to $t<t_c$.

Each Fourier mode among the comoving perturbations of a homologously collapsing cloud has its phase fixed on each fluid element. In fact, for $\mathbf{k}=\mathbf{q}/R$,Its physical wavelength contracts in proportion to $R$, while its wavelength in the comoving coordinate $\mathbf{r}/R$ stays fixed. This is the physical reason to use a time-dependent wavenumber; holding the physical wavenumber fixed would not follow the contracting pattern. Boundary effects are being excluded in this bulk Fourier mode calculation.

Let ${\rho }^{\prime }={\rho }_1(t)e^{i\mathbf{k}\cdot \mathbf{r}}$ and let the perturbed velocity and Newtonian gravitational potential have amplitudes ${\mathbf{u}}_1(t)$ and ${\Phi }_1(t)$. Linearizing the continuity equation, the uniform background eliminates ${\mathbf{u}}_1\cdot \nabla {\rho }_0$. The advected phase eliminates its own time derivative, leavingFor the density contrast $\delta ={\rho }_1/{\rho }_0$, use ${\dot{\rho }}_0=-3H{\rho }_0$ to obtain $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$. The linear momentum equation and Poisson equation for Newtonian gravity areDifferentiating the density contrast equation must include $\dot{\mathbf{k}}=-H\mathbf{k}$:Thus the compressive Fourier mode with $k\ne 0$ satisfiesThe pressureless approximation removes the usual acoustic restoring term of Jeans instability. A transverse velocity perturbation instead obeys ${\dot{\mathbf{u}}}_{1\perp }+H{\mathbf{u}}_{1\perp }=0$ and does not drive this density contrast.

Since $d/dt=(C/R)d/d\varphi$, the density contrast equation becomesFor a direct verification, put $s=\tan (\varphi /2)$. As $d/d\varphi =(1+s^2)d/(2ds)$, the first-derivative terms cancel, reducing the equation toThe function ${\delta }_g=s(1+s^2)/2$ has second derivative $3s$ and satisfies this equation. Transforming back gives the required growing branch:To establish what the other solution does, use reduction of order, not just the growth of this example. The coefficient of ${\delta }_{\varphi }$ is $d\log R/d\varphi$, so an independent solution is proportional to ${\delta }_g\int d\varphi /(R{\delta }_g^2)$. Choose the integration endpoint at complete collapse to select the decaying branch:For example, evaluating the integral with $a=\tan (s/2)$ gives integrand $4/[a^2(1+a^2{)}^2]$ and antiderivative $-4/a-6\arctan a-2a/(1+a^2)$. Its endpoint value gives the displayed expression. Its Wronskian with ${\delta }_g$ is $-(1+\cos \varphi {)}^{-1}$, so the two solutions are independent on $0<\varphi <\pi$.

For $ϵ=\pi -\varphi \to 0$, $R\sim R_0{ϵ}^2/4$, ${\delta }_g\sim 4{ϵ}^{-3}$ and the integral in ${\delta }_d$ is ${ϵ}^5/40+O({ϵ}^7)$. Consequently the two density modes of a collapsing pressureless sphere haveEquivalently, with $\tau =t_c-t$, the local powers are ${\tau }^{-1}$ and ${\tau }^{2/3}$. There is one independent growing branch: every solution is $A{\delta }_g+B{\delta }_d$, and only the branch with $A=0$ decays at collapse. The uniqueness is up to normalization and addition of the decaying solution; it does not mean that only one literal function can grow. A fixed nonzero initial perturbation will eventually violate $|\delta |\ll 1$, so the divergent linear result is not a nonlinear solution through collapse.

Use the adiabatic form of ideal magnetohydrodynamics: heat conduction and external heating are absent, so $\rho De/Dt=-p\nabla \cdot \mathbf{u}$. Infinite electrical conductivity and zero viscosity alone would not remove the thermal source terms on the cover page. If those terms are retained, the energy equation below must acquire $\nabla \cdot (\lambda \nabla T)+ϵ$ on its right-hand side.

Expanding the continuity equation gives its conservative form ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Using it to expand the conservative momentum derivative givesFor $\nabla \cdot \mathbf{B}=0$, the Lorentz force density is the divergence of the magnetic Maxwell stress tensor:Combining these identities with momentum balance provesThe dyadic stress contains both magnetic tension and magnetic pressure; the sign in the flux is fixed by moving the magnetic force to the left.

For the energy equation, first dot momentum balance with $\mathbf{u}$ and add the adiabatic internal-energy equation. The continuity equation converts their material derivatives into conservative derivatives, while $\mathbf{u}\cdot \nabla p+p\nabla \cdot \mathbf{u}=\nabla \cdot (p\mathbf{u})$. The result isThe ideal magnetohydrodynamic induction equation gives the magnetic counterpart. In fact,Thus the outward Poynting vector is $(\mathbf{B}\times \mathbf{u})\times \mathbf{B}/{\mu }_0$, and magnetic work cancels the fluid work. Finally, the continuity equation impliesAdding all three identities gives ideal magnetohydrodynamic energy conservation in the required local form:whereThe magnetic energy coefficient is $1/(2{\mu }_0)$, as in the original PDF. For a self-consistent, time-dependent gravitational field this local balance should not be confused with a global self-gravitating energy integral: the latter uses gravitational field energy, or the familiar one-half factor in $\int \rho \Phi$ after the field is eliminated.

Work in the shock frame of a thin planar discontinuity, with normal speed $u>0$. A smooth gravitational potential has no finite jump through a vanishingly thin front; its advected contribution cancels using the common mass flux. Integrating the conservation laws gives constant normal mass, momentum and energy fluxes, together with the normal-field and tangential-electric-field conditions of the ideal magnetohydrodynamic shock conditions.

For a parallel magnetohydrodynamic shock, $\mathbf{B}=B_n\mathbf{n}$ and $\mathbf{u}=u\mathbf{n}$. The magnetic solenoidal constraint makes $B_n$ the same on both sides. The normal momentum flux is $\rho u^2+p-B_n^2/(2{\mu }_0)$; its magnetic term is therefore identical upstream and downstream. The Poynting vector is zero because $\mathbf{u}\times \mathbf{B}=0$. Hence the nontrivial jumps are exactly the Rankine-Hugoniot conditions for a perfect gas:with specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$. Physically, motion along the straight field neither bends it nor compresses its transverse flux. The field consequently provides no changed stress or energy flux in this purely parallel geometry. This conclusion does not apply to a shock that develops a transverse magnetic field.

For a perpendicular magnetohydrodynamic shock, the magnetic field is tangential and the velocity normal. Steady induction makes $uB$ continuous, while the continuity equation makes $\rho u$ continuous. Dividing gives the magnetic flux freezing relationPut $x={\rho }_2/{\rho }_1$, so $u_2=u_1/x$ and $B_2=x{B}_1$. Constant normal momentum flux now reads $\rho u^2+p+B^2/(2{\mu }_0)$. Substitution givesThe magnetic correction is negative for compression: part of the incoming ram pressure builds magnetic pressure instead of thermal pressure.

Here the Poynting vector has normal component $u{B}^2/{\mu }_0$. Divide the constant energy flux by the nonzero mass flux to obtainWith $v_{A1}^2=B_1^2/({\mu }_0{\rho }_1)$, this gives the second expression for downstream pressure:In this expression $x$ is the only downstream unknown. Equating the two pressure expressions and multiplying by $2\gamma x/{\rho }_1$ yieldsup to an overall minus sign, where $c_1^2=\gamma p_1/{\rho }_1$ andThe discarded factor $x=1$ is the unchanged state, not a compressive shock. This derives the polynomial for the compression ratio of a perpendicular magnetohydrodynamic shock without omitting its trivial branch.

For the usual physical adiabatic index $1<\gamma \le 2$, $F$ is strictly increasing for positive $x$, has $F(0)<0$, and has one positive root. Sincethat root is greater than one precisely when $u_1^2>c_1^2+v_{A1}^2$. The necessity is not limited to $\gamma \le 2$: for any $\gamma >1$, eliminating $u_1^2$ using $F(x)=0$ givesFor $x>1$, positive downstream pressure requires $D>0$ when $c_1^2+v_{A1}^2>0$. Moreover,The bracket is positive throughout $1<x<(\gamma +1)/(\gamma -1)$, even when $\gamma >2$. The exactly cold unmagnetized limit gives the familiar strong-shock upper compression ratio directly. Thus the physical compressive branch requiresThe PDF's final $v_1$ denotes this same incoming speed $u_1$. The right side is the squared speed of a perpendicular fast magnetosonic wave: the upstream flow must be super-fast-magnetosonic so that this compressive signal cannot run upstream against the incoming flow. At equality the shock amplitude vanishes into a characteristic disturbance.

Use $S(z)=\sigma +k_{x}U(z)$, $k^2=k_x^2+k_y^2>0$ and primes for $z$ derivatives. Time dependence is $e^{i\sigma t}$, so exponential growth means $\mathrm{Im}\sigma <0$. The constant mass density and incompressible velocity giveAll magnetic factors below use the source convention ${\mu }_0=1$.

Linearizing the ideal magnetohydrodynamic momentum equation, the advective perturbation is $iS{\mathbf{u}}_1+U^{\prime }w{\mathbf{e}}_x$. The magnetic force is $(\nabla \times \mathbf{b})\times B{\mathbf{e}}_x$, whose $x$ component vanishes. Thus the three component equations areIn particular the vertical equation has the pressure-gradient and magnetic signs printed in the PDF.

The incompressible ideal magnetohydrodynamic induction equation can be written $D\mathbf{B}/Dt=(\mathbf{B}\cdot \nabla )\mathbf{u}$. Linearizing it givesFor $S\ne 0$, these areThe $U^{\prime }{b}_z$ term is essential: a perturbed vertical magnetic component is stretched by the background shear flow. Differentiating these amplitudes and using incompressibility also verifies $i{k}_x{b}_x+i{k}_y{b}_y+b_z^{\prime }=0$.

For completeness, substitution into all three momentum equations giveswhere $\zeta =i{k}_{x}v-i{k}_{y}u$ is the vertical vorticity amplitude. Multiplying the $y$ equation by $k_x$ and subtracting $k_y$ times the $x$ equation eliminates pressure:Therefore, away from $\rho S^2=k_x^2{B}^2$,This is the nonresonant reduction relevant to the instability calculation. Division at the Alfvén wave resonance would be invalid. For a concrete counterexample to an unrestricted reading, take constant $U$, $k_y=0$, $S=\pm k_{x}B/\sqrt{\rho }$, $v=a(z)$ and $b_y=k_{x}Ba(z)/S$, with all other amplitudes zero. Every smooth $a(z)$ solves the original linear system, but $\zeta =i{k}_{x}a(z)$ need not vanish. These transverse resonant waves do not drive $w$. A genuinely growing mode in a real background has complex $S$ and cannot satisfy either $S=0$ or $S^2=k_x^2{B}^2/\rho$ on the real $z$ axis. This explains exactly when the requested reductions may be used.

On that nonresonant branch, multiply the horizontal equations by $k_x$ and $k_y$ and add. Incompressibility gives $k_{x}u+k_{y}v=i{w}^{\prime }$, soCombining incompressibility with vorticity, multiply its first equation by $k_x$ and subtract $k_y$ times the definition of $\zeta$. This givesConsequentlySubstitute this and the derivative of the displayed pressure relation into vertical momentum. Multiplication by $k^2/i$ givesExpanding the second derivative once gives exactly the requested vertical-velocity equation:

To take the discontinuous-layer limit without multiplying distributions, introduce $f=w/S$. It is proportional to the normal displacement: the material kinematic condition is $w=iS\eta$, so $f=i\eta$. Since $S^{\prime }=k_x{U}^{\prime }$, $S{w}^{\prime }-S^{\prime }w=S^2{f}^{\prime }$. The same equation becomes the displacement equation for a parallel magnetic shear flowThe physical interface condition is continuity of $f$, because both fluids move the same interface. For constant $U_j$ on each side, away from resonance the coefficient is constant and nonzero, so $f^{\prime \prime }-k^{2}f=0$. Requiring decay at infinity and matching $f(0)$ yieldsThe exponent is $z$, as in the actual PDF; the converted TeX's $x$ there is a transcription error.

Integrate the conservative displacement equation across a layer of thickness $2\epsilon$. Its undifferentiated term has an integral tending to zero, while the derivative givesEquivalently, the perturbed magnetohydrodynamic total pressure is $p_1+B{b}_x=(\rho S^2-k_x^2{B}^2)f^{\prime }/(i{k}^2)$ and is continuous. With $f_{+}^{\prime }=-kA$ and $f_{-}^{\prime }=kA$, a nontrivial amplitude requiresThis is also the rigorous thin-layer integration of the original vertical equation, after combining the derivative terms before taking the limit.

Put $\overline{U}=(U_1+U_2)/2$ and $\Delta U=U_2-U_1$. Completing the square gives the dispersion relationFor $(\Delta U{)}^2<4B^2/\rho$ both interface frequencies are real, so there is no exponentially growing interface mode. Above that threshold, the growing sign has growth rate $|k_x|\sqrt{(\Delta U{)}^2/4-B^2/\rho }$. This is magnetic stabilization of an equal-density vortex sheet: magnetic tension from bending the field overcomes the kinetic shear driving Kelvin-Helmholtz instability. Equality is marginal and the strict stable condition avoids its double-root degeneracy. The case $k_x=0$ has no streamwise shear drive and has the neutral interface frequency $\sigma =0$; the nonzero-frequency elimination should be interpreted by the original equations or a limiting displacement formulation in that case. The horizontal $k_y$ affects the exponential decay length through $k$, but not the stated threshold.

Approximate the gap by steady Couette flow, neglecting edge effects and additional losses from membrane deformation. Its velocity gradient is $U/h$, so the shear stress is $\eta U/h$. The force on the gliding bacterium and the rate of viscous dissipation areThe same power follows by integrating $\eta (U/h{)}^2$ over gap volume $Ah$. In SI units, $A=10^{-10}{\mathrm{m}}^2$, $h=10^{-7}\mathrm{m}$, and $U=10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$. ThusAn energy needs a duration or distance: $E=10^{-18}\Delta t\mathrm{J}$ for $\Delta t$ in seconds, equivalently $E=F_{d}s$ for travel distance $s$. Moving one micrometre takes one second and dissipates $10^{-18}\mathrm{J}$. This is the mechanical cost of the sheared film. A numerical biological efficiency would additionally require the motor's input power and the losses associated with its peristaltic actuation.

Set $v=\dot{x}$, $\zeta =6\pi \mu a$, and $\tau =m/\zeta$. To keep dimensions and the thermal amplitude explicit, write the force covariance as $⟨F(t)F(s)⟩=A\delta (t-s)$. The printed unit coefficient is a choice of noise normalization, not a general dimensional thermal force. Assume the initial velocity is independent of the future white noise. The Underdamped Langevin dynamics has integrating-factor solutionTaking its expectation and using the delta covariance in the double integral givesHere the noise contribution is $A{m}^{-2}{\int }_0^t{e}^{-2(t-s)/\tau }ds$. The equipartition theorem requires the long-time value $⟨m{v}^2/2⟩=k_{B}T/2$. Consequently the fluctuation-dissipation relation for a Langevin particle isWith an initially thermal velocity, $⟨v_0⟩=0$ and $⟨v_0^2⟩=k_{B}T/m$, so the velocity variance is stationary. If one insists on the printed coefficient $A=1$ in dimensional units, the model has temperature fixed by $k_{B}T=1/(2\zeta )$ in those units; it cannot describe arbitrary temperature while retaining that coefficient.

Take $x(0)=0$. Integrating the velocity solution givesThe initial/noise cross term vanishes under the independence assumption. Evaluating the remaining integral yields the general integrated Ornstein-Uhlenbeck displacementUse the thermal initial velocity and the fluctuation-dissipation relation for a Langevin particle. With diffusion coefficient $D=k_{B}T/\zeta$, this simplifies toFor $t\ll \tau$, expansion gives $⟨x^2⟩=(k_{B}T/m)t^2+O(t^3)$: displacement is initially $v_{0}t$, so the particle is ballistic. For $t\gg \tau$, $⟨x^2⟩=2Dt-2D\tau +o(1)$, hence the leading behavior is one-dimensional diffusion with $D=k_{B}T/(6\pi \mu a)$.

The ballistic conclusion depends on the initial state. For a particle prepared exactly at rest, $v_0=0$, the first expression gives $⟨x^2⟩=A{t}^3/(3m^2)+O(t^4)$ at very short times. A thermal initial velocity, or a nonzero initial velocity variance, is required for the stated $t^2$ mean-square law. It should not be replaced silently by a zero-velocity initial condition.

Model the polymerization Brownian ratchet by a rigid tip and a Brownian load. Let $a$ be the elementary load advance, ${\zeta }_L$ its drag coefficient, and $D=k_{B}T/{\zeta }_L$ its diffusion coefficient. For a spherical load, ${\zeta }_L=6\pi \mu R_L$, but no load radius is specified, so the rate is naturally expressed in terms of $D$. Assume addition is effectively instantaneous once the gap $h$ reaches $a$, and removal is negligible. The gap reflects at zero, first arrival at $a$ causes insertion, and insertion resets it to zero.

With no opposing force, the mean first-passage time satisfies $D{T}^{\prime \prime }=-1$, $T^{\prime }(0)=0$, $T(a)=0$. Integration gives $T(h)=(a^2-h^2)/(2D)$. One advance per passage givesThe advance-event rate is $2D/a^2$, while $V_0$ is a length per time.

A constant resisting force $F\ge 0$ produces drift $-u_F=-DF/(k_{B}T)$. The diffusion-limited Brownian ratchet with constant load now satisfiesAn integrating factor gives $T^{\prime }(h)=[1-e^{u_{F}h/D}]/u_F$. Integrating again from $h$ to $a$ gives $T(0)=D(e^z-1-z)/u_F^2$, where $z=Fa/(k_{B}T)$. ThereforeAt weak load the ratio is $1-z/3+O(z^2)$; at large load $V\sim (D/a)z^2{e}^{-z}$. Replacing the whole force dependence by $e^{-z}$ would discard the first-passage kinetics of this regime.