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For constant injected power $\dot{E}$ and ambient mass density ${\rho }_1$, dimensional analysis gives $R=\alpha (\dot{E}/{\rho }_1{)}^{1/5}{t}^{3/5}$. In a similarity solution, set $\eta =r/R$, $\rho ={\rho }_{1}f(\eta )$, $p={\rho }_1{\dot{R}}^{2}g(\eta )$ and $u=\dot{R}h(\eta )$. The strong-shock values on the inner side of $\eta =1$ are $f=(\gamma +1)/(\gamma -1)$ and $g=h=2/(\gamma +1)$.

Take a reservoir with constant horizontal cross-sectional area $A$, one small orifice of area $a$, no inflow, and constant stipulated discharge coefficient $C_d>0$. Let the initial head be $H>0$. For constant mass density $\rho$, the stored mass above the outlet level is $\rho Ah$ and outgoing mass flow rate is $\rho Q$. Mass conservation cancels $\rho$ and gives the initial value problemThe downward speed of the free surface is $Q/A$, smaller than the ideal jet speed by the factor $C_{d}a/A$. Thus a small opening relative to the reservoir supports the negligible-surface-velocity assumption.

For $h>0$, separation of variables or the chain rule gives $d\sqrt{h}/dt=-k/2$. Integrating yieldsThe model reaches zero head in finite time even though volumetric flow rate tends to zero. Extending the squared expression beyond $T$ would make head grow again and violate the original ordinary differential equation. The zero-head continuation enforces no further outflow. For an elevated side opening, zero head means the surface has reached that opening, leaving fluid below it; $T$ is not the time to remove every drop from such a tank.

For an original synthetic example, stipulateThe initial volume above outlet level is $AH=0.0250{\mathrm{m}}^3=25.0\mathrm{L}$. The initial ideal jet speed is $2.215\mathrm{m}{\mathrm{s}}^{-1}$, the initial volumetric flow rate is $Q_0=0.1373\mathrm{L}{\mathrm{s}}^{-1}$, and the predicted zero-head time is $T=364.1\mathrm{s}$. These are consequences of stipulated parameters, not experimental measurements. At $t=T/2=182.1\mathrm{s}$, head is $H/4=0.0625\mathrm{m}$, stored volume is $6.25\mathrm{L}$, and volumetric flow rate is $Q_0/2=0.06866\mathrm{L}{\mathrm{s}}^{-1}$. Half the initial volume has already left at $t=T(1-1/\sqrt{2})=106.7\mathrm{s}$.

An independent mass conservation check integrates the outgoing volumetric flow rate:Multiply both sides by mass density to obtain discharged mass. A dimensional analysis check gives $[k]={\mathrm{m}}^{1/2}{\mathrm{s}}^{-1}$, $[Q]={\mathrm{m}}^3{\mathrm{s}}^{-1}$ and $[T]=\mathrm{s}$. The triangular area under $Q(t)$ explains why the initial-rate estimate $AH/Q_0$ is only half the actual model time. Doubling $A$ doubles $T$; doubling $a$ or $C_d$ halves $T$, within the same assumptions.

This finite-time result is an idealized prediction. The constant $C_d$ and quasi-steady approximation need not survive very small head or the initial startup. Appreciable viscosity in a narrow outlet tube requires a different pressure balance, as in Poiseuille flow; appreciable surface tension can prevent zero head altogether.

Let an electron start with negligible kinetic energy and move through an increase $U>0$ in electric potential. Its electric charge is $q=-e$, where $e>0$ is the elementary charge, so the decrease in potential energy is $eU$. In a field-free region after acceleration,This is an electron formula, not a universal conversion from voltage to wavelength. It assumes negligible initial kinetic energy, negligible losses, and $eU\ll m_e{c}^2$, where $c$ is the speed of light. It describes the de Broglie wavelength outside the crystal; the potential energy inside a material can change the local kinetic momentum.

For $U=54\mathrm{V}$, take $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, and $m_e\simeq 9.1093837\times 10^{-31}\mathrm{kg}$. The rounded electron mass is consistent with the 2022 recommended constants.^[ref]The electronvolt is an energy unit: $54\mathrm{eV}$ must be converted to joules before combining it with $m_e$ in SI units. Dimensional analysis checks both steps:The resulting velocity is about $4.3584\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, or $0.01454c$, so the nonrelativistic assumption is consistent. For a larger accelerating voltage, the relativistic energy-momentum relation gives $pc=\sqrt{K(K+2m_e{c}^2)}$, with $K=eU$, while $\lambda =h/p$ still holds. At $54\mathrm{V}$ the relativistic wavelength is shorter by about $0.00264\%$.

Explicitly, the exact special relativity formula and its nonrelativistic quantum mechanics limit areHere $K=eU$ is the gained kinetic energy; the total energy is $E=m_e{c}^2+K$. At $80\mathrm{kV}$, the nonrelativistic quantum mechanics expression gives $4.336\mathrm{pm}$, about $3.84\%$ larger than the exact $4.176\mathrm{pm}$. This difference matters when using electron diffraction ring positions to infer lattice plane spacings.

The gravitational constant $G$ sets the strength of Newton's law of universal gravitation. Two point masses $M$ and $m$ a center-to-center distance $r$ apart attract with force magnitudeFor fixed masses and separation, doubling $G$ would double the force. For bodies with spherical symmetry, the same exterior formula follows from the spherical shell theorem; for arbitrary extended bodies, add the attractions of their mass density elements instead. The gravitational constant connects a laboratory attraction to the Newtonian gravitational field controlling a Kepler orbit.

The gravitational constant $G$ and local gravitational acceleration $g$ are different quantities. Applying Newton's second law to a test particle gives $g=F/m=GM/r^2$ for one spherical source. The source's mass and the observer's position change $g$; $G$ remains the same parameter of the law. A local falling-body experiment therefore measures a field, not $G$ alone. On rotating Earth, an instrument fixed to the surface also responds to centrifugal acceleration, while surrounding mass density and altitude change the gravitational contribution.

Dimensional analysis explains the units. Denote the base physical dimensions of length, mass and time by $\mathsf{L},\mathsf{M},\mathsf{T}$. Since force has dimension $\mathsf{ML}{\mathsf{T}}^{-2}$,By contrast, $[g]=\mathsf{L}{\mathsf{T}}^{-2}$ and the standard gravitational parameter $\mu =GM$ has dimension ${\mathsf{L}}^3{\mathsf{T}}^{-2}$. A small numerical value of a dimensional constant depends on the units chosen: it is the resulting force for specified physical masses and separation that conveys the laboratory scale.

NIST's current complete listing givesfor the 2022 CODATA adjustment, the latest adjustment listed when checked on 14 October 2026. Parentheses indicate a standard uncertainty $u(G)=0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, approximately $22$ parts per million relative to $G$. The adjustment year is 2022; NIST's May 2024 data update is a different date.^{[ref] [ref]}

For an original scale calculation, stipulate two $1\mathrm{kg}$ point masses separated by $1\mathrm{m}$ and insert this recommended $G$: $F=6.67430\times 10^{-11}\mathrm{N}$. The corresponding acceleration of either mass is $6.67430\times 10^{-11}\mathrm{m}{\mathrm{s}}^{-2}$ at that instant. Reducing the separation to $0.10\mathrm{m}$ multiplies both by $100$. These are ideal model predictions, not observations; their tiny scale explains why stray forces matter.

Being a constant of a physical law does not make the numerical value exact. The SI fixes the numerical values of seven defining constants, including the speed of light and Planck constant; $G$ is determined experimentally. Exact unit definitions still require physical realizations with measurement uncertainty: calibrating a length or a mass does not become error-free.^[ref]

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

A characteristic proper timescale for strong bound-state formation, estimated by dimensional analysis from the QCD scale. A scale of a few tenths of a GeV corresponds to a few $10^{-24}\mathrm{s}$. It is an order-of-magnitude guide rather than a universal sharply defined formation time.

Under single-length dynamical scaling of binary-fluid coarsening, mass density $\rho$, dynamic viscosity $\eta$ and surface tension $\sigma$ determine the displayed unique length and time scales by dimensional analysis. Writing $L=L_{0}f(t/t_0)$ reduces the force-density estimate to $\alpha f^{\prime \prime }+\beta f^{\prime 2}/f=\gamma f^{\prime }/f^2+\delta /f^2$. Resistive and driving terms need opposite signed geometric coefficients. Viscous hydrodynamic coarsening has $f\sim u$ and inertial hydrodynamic coarsening has $f\sim u^{2/3}$. These follow respectively from density-independent and viscosity-independent powers in $L\propto {\eta }^{2-3p}{\rho }^{p-1}{\sigma }^{2p-1}{t}^p$. They require thin interfaces, statistically self-similar bicontinuous domains, dominant advection and no additional relevant macroscopic scales. A degeneracy in the signed inertial coefficients can invalidate the pure-power force balance without invalidating the dimensional calculation.

For source kinematic momentum flux $m_0$ and buoyancy flux $f_0>0$, dimensional analysis gives $L_J=m_0^{3/4}/f_0^{1/2}$. Under constant Batchelor entrainment, the actual turning length also contains an entrainment coefficient factor of order ${\alpha }^{-1/2}$.

In the locally isotropic equilibrium range, the small-scale velocity increment statistics depend on the separation, mean viscous dissipation and kinematic viscosity, independently of large-scale forcing details. Dimensional analysis leaves the Kolmogorov microscales as normalization scales and gives the displayed second-order longitudinal structure function. Universality of $F$ is part of the hypothesis, not something proved by dimensional analysis.

In the inertial range, velocity increment statistics also become independent of kinematic viscosity. Combined with the Kolmogorov first similarity hypothesis, this gives $S_p(r)={\beta }_p(ϵr{)}^{p/3}$ by dimensional analysis. Signed integer moments and absolute noninteger moments must be distinguished; the signed third-order coefficient is negative in a forward energy cascade.

Length measures spatial extent or separation. It is a base physical dimension in dimensional analysis, distinguished from the choice of units used to express a particular value.

Efficiency has several denominators. Specify a steady electric current $I>0$, positive terminal electric potential difference $V_f$, and a simplified narrow emission spectrum with photon energy $E_{\gamma }$. Assume at most one radiative photon per injected electron and electron hole pair. Let ${\eta }_{\mathrm{inj}}$ be the fraction of the electrical pair-supply rate $I/q$ reaching the active layer, ${\eta }_{\mathrm{rad}}$ the fraction of those pairs recombining radiatively, and ${\eta }_{\mathrm{ext}}$ the fraction of internally emitted photons that escape. Then counting the successive events givesDefine internal quantum efficiency here conditionally on pairs reaching the active layer: ${\eta }_{\mathrm{IQE}}={\eta }_{\mathrm{rad}}$. The internal photon yield per electrical pair is instead ${\eta }_{\mathrm{inj}}{\eta }_{\mathrm{IQE}}$. Stating this convention matters when another definition folds injection into internal quantum efficiency. The external quantum efficiency and wall-plug efficiency areThese are photon-count and energy-flow identities for the declared model. For a broad spectrum, replace $E_{\gamma }$ by the mean escaping photon energy, weighted by photon number, when computing optical power.

A stipulated example chooses $I=20\mathrm{mA}$, $V_f=3.20\mathrm{V}$, $\lambda =450\mathrm{nm}$, ${\eta }_{\mathrm{inj}}=0.90$, ${\eta }_{\mathrm{IQE}}=0.80$ and ${\eta }_{\mathrm{ext}}=0.50$. For every 100 electrical pairs, 90 reach the active layer, 72 emit internally, and 36 photons escape. Thus ${\eta }_{\mathrm{EQE}}=0.36$, not $0.80$. Using $q=1.602176634\times 10^{-19}\mathrm{C}$ givesThe dimensional analysis is transparent: $A/C={\mathrm{s}}^{-1}$, $J/s=\mathrm{W}$, and $\mathrm{CV}=\mathrm{J}$. The optical and electrical efficiencies differ because $E_{\gamma }/q=2.75520\mathrm{V}$ differs from the chosen $V_f$.

The factors suggest different improvements. Better confinement can reduce injection leakage; suppressing nonradiative recombination raises the radiative fraction; improving light extraction efficiency helps already-emitted photons escape. A photon trapped by total internal reflection has already passed the radiative step and can still be lost through absorption. Improving extraction therefore need not change the internal radiative fraction.

Shuji Nakamura's lecture cautions that why indium gallium nitride emits efficiently despite many defects was not completely settled in that account.^[ref]

As an independent consistency check on the 1994 reported values, treating its broad spectrum as monochromatic at the $450\mathrm{nm}$ peak givesThis agrees at the quoted precision with the reported $2.7\%$. Its optical/electrical power ratio is instead $1.5\mathrm{mW}/(20\mathrm{mA}\times 3.6\mathrm{V})=2.0833\%$. The first estimate uses a peak in place of a photon-number-weighted mean, so agreement is a consistency check rather than an exact reanalysis of the measured spectrum.

Newton's law of cooling models an object's temperature $y$ using a positive constant rate $k$ and ambient temperature $T_a$. The displayed ordinary differential equation assumes one temperature represents the object, the environment remains at $T_a$, and the proportional rate stays fixed over the modeled range. It also describes warming when $y<T_a$. These are modeling assumptions: ^[ref] explain the proportional law through a tangent line approximation and caution about large temperature differences.

Dimensional analysis requires $k$ to have units of inverse time. The product $k(t-t_0)$ in the exponential function solution is therefore dimensionless. The equilibrium point $y=T_a$ attracts both warmer and colder starting states: the excess temperature undergoes exponential decay. See initial value problem for computed cooling times, direction field for the family of trajectories, and separable differential equation for a derivation.

Once swept-up matter dominates the ejecta and radiative cooling is negligible, the Sedov-Taylor blast wave has its shock radius fixed by dimensional analysis:Here $\xi$ is a dimensionless constant of order unity fixed by the adiabatic index. Treating the ambient matter as hydrogen gives ${\rho }_0\simeq 10^6{m}_H\simeq 1.67\times 10^{-21}\mathrm{kg}{\mathrm{m}}^{-3}$, while $t\simeq 3.16\times 10^{10}\mathrm{s}$. ThusFor a monatomic gas, the usual similarity normalization $\xi \simeq 1.15$ instead gives about $5.3\mathrm{pc}$. An estimate of a few parsecs is therefore appropriate. Negligible ejecta mass does not by itself guarantee this energy-conserving stage: substantial radiation losses or a strongly nonuniform ambient mass density would invalidate the assumed similarity law.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.

A one-dimensional probability density has units $L^{-1}$, while $D$ has units $L^2{T}^{-1}$. With no other length scale, dimensional analysis requiresNormalization gives ${\int }_{-\infty }^{\infty }F(\xi )d\xi =1$ and unbiased diffusion gives an even $F$. Dimensional arguments alone do not specify $F$.

For ordinary Fickian diffusion, impose ${\partial }_{t}n=D{\partial }_x^{2}n$ with $n(x,0)=\delta (x)$. Substitution of the similarity form yieldsA decaying symmetric solution obeys $F^{\prime }=-\xi F/2$, so $F=C{e}^{-{\xi }^2/4}$. Normalization fixes $C=1/\sqrt{4\pi }$:This heat kernel integrates to one, has variance $2Dt$, and tends to the initial Dirac delta distribution as $t\downarrow 0$.

Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

Choose the stream function convention $u=-{\psi }_y$, $v={\psi }_x$, so that this is an incoming stagnation point flow rather than the reversed flow. The velocity field isAt the plate $v(x,0)=0$, satisfying the no-penetration boundary condition. The nonzero tangential velocity $u(x,0)=Ax$ is permitted for an inviscid flow; the no-slip boundary condition is not imposed on the Euler equations for an inviscid fluid.

The steady convective acceleration is $(A^{2}x,A^{2}y)$. Thus the Euler equations for an inviscid fluid requireIntegrating givesThe printed formula omits $\rho$ from its quadratic term. The source uses physical pressure, not pressure divided by mass density, as is clear from its $\rho gy$ term and its Navier-Stokes equation. The omitted factor is also required by dimensional analysis. For a concrete substitution check, the printed expression gives $-p_x/\rho =A^{2}x/\rho$ rather than the required $A^{2}x$ when $\rho \ne 1$ in the chosen numerical units.

mass and energy signs. The enclosed mass is $M(r)={\int }_0^{r}4\pi s^2\rho (s)ds$, so $dM/dr=+4\pi r^2\rho$. The negative sign printed in that equation cannot describe positive enclosed mass. Use the physical sign and define gravitational potential energy by

Boundary-pressure virial theorem. Multiply hydrostatic equilibrium by $4\pi r^3$ and integrate. The left side is $[4\pi r^{3}p{]}_0^{r_0}-3{\int }_0^{r_0}4\pi r^{2}pdr$, while the right side is $\Omega$. The central boundary contribution vanishes when $r^{3}p\to 0$ and the outer pressure is $p(r_0)=p_0$. ThusA finite external pressure is compatible with a gravitationally self-contained cloud, but is not the zero-surface-force notion of isolation. External ram stress or mass flow would require additional terms.

radiation budget of monatomic cloud collapse. For an ideal gas with fixed $\gamma$, the specific internal energy is $u=p/[\rho (\gamma -1)]$. Therefore the pressure integral is $3(\gamma -1)U$. Cold monatomic hydrogen has $\gamma =5/3$ when only translational degrees of freedom matter. With negligible final surface pressure, $2U+\Omega =0$, so $E_f=U+\Omega =\Omega /2$. If the initial cloud is sufficiently extended and cold that $E_i\simeq 0$, the emitted energy isBeing initially at rest alone does not set its binding energy to zero. If the initial total energy is appreciable, retain $E_{\mathrm{rad}}=E_i-\Omega /2$; a different equation of state also changes the half-energy result.

Inside-out accretion. The combination of $a_T$ and $G$ with dimensions of mass per time iswhere dimensional analysis leaves the dimensionless coefficient $C$ undetermined. An isothermal gas keeps $a_T$ constant, giving constant supply in the self-similar idealization. A front at radius $a_{T}t$ and an isothermal mass scale $M(r)\sim a_T^{2}r/G$ give the same scaling. The front's swept mass is not automatically the accreted central mass; the coefficient requires a dynamical similarity solution.^[ref]

constant-radius accretion energy budget. If all available accretion power is radiated at the fixed radius, $L_{\mathrm{acc}}=GM\dot{M}/r_0$. Neglecting the seed mass,This equals $-\Omega /2$ precisely whenThe positive sign in the final printed energy condition is inconsistent with the earlier use of negative gravitational potential energy. Equivalently its positive right side must denote the binding-energy magnitude $|\Omega |$. A general structural coefficient $\Omega =-qG{M}^2/r_0$ leaves a mismatch unless $q=1$ under these other idealizations. Radiation trapped in the object, outflows or release over a different radius also change the budget.

Which temperature? With $M=M_s+\dot{M}t$ and fixed $r_0$, $L_{\mathrm{acc}}\propto M$. If this power emerges thermally from area $4\pi r_0^2$, the accretion-powered effective-temperature scaling iswhen the seed mass is negligible. This is the radiating surface's effective temperature, inferred from the preceding accretion-luminosity calculation. A homologous ideal-gas interior instead has virial temperature $T_c\sim \mu m_{H}GM/(k_B{r}_0)\propto t$. Thus the surface-effective and interior-virial scalings are $t^{1/4}$ and $t$, respectively, under their stated assumptions. Inferring a detailed interior temperature profile would additionally require a structural and transport model.

A typical net ice accumulation curve decreases downstream: snow gain at high elevation, zero at the equilibrium line, and negative values in the ice ablation zone. In a steady glacier with zero volume flux at its head, $Q(x)={\int }_0^{x}a(X)dX$. If its terminus also has zero volume flux, a natural glacier length $L$ is the positive root of ${\int }_0^{L}a(X)dX=0$, with the integral positive between the endpoints. Such a root requires a suitable given accumulation profile; it is not guaranteed by dimensional analysis alone.

Let $a_0$ be a representative accumulation rate and $K=A(\rho g\sin \alpha {)}^n$. ChooseUsing $x=L\hat{x}$, $H=H_{\ast }\hat{H}$, $Q=Q_{\ast }\hat{Q}$, $t=t_{\ast }\hat{t}$, $a=a_0\hat{a}$, both terms in mass conservation have scale $a_0$. The dimensionless flux isDropping hats gives the required partial differential equation. The parameter measures the surface-thickness slope correction relative to the bed slope. For the linear example $a=1-2x$, accumulation changes sign at $x=1/2$ and its integrated balance vanishes at $x=1$.