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Use a two-sided top-hat line plume with full width $w(z)$, upward speed $U(z)$, and reduced gravity $b(z)$ uniform across the plume. Fluxes are measured per unit span along the line source: the volume flux is $V=wU$, the kinematic momentum flux is $M=w{U}^2$, and the buoyancy flux is $B=wUb$. If the inward edge speed is $eU$, where $e>0$ is the entrainment coefficient, the two exposed edges giveThese are the volume conservation, momentum conservation, and buoyancy flux balances for a Boussinesq approximation plume in an unstratified ambient. Entrained ambient fluid supplies neither vertical momentum nor reference buoyancy.

A pure plume has no persistent source length scale. Since a line-source $B$ has dimensions $L^3{T}^{-3}$, dimensional analysis gives constant $U\propto B^{1/3}$, $w\propto z$, and $V\propto z$. Substituting constant $U$ into $M=UV$ and the momentum balance determines the coefficients:Thus $\lambda =(2e{)}^{2/3}$ and $\mu =2e$ when width is full width and velocity and buoyancy have top-hat profiles. If width means half-width, $\mu =e$ instead. Other prescribed profile shapes change these numerical factors; the linear growth laws remain the same. The point-source $B$ in Question 1 has different dimensions, so its $z^{5/3}$ law must not be used here.

Let $C(x,z,t)$ be local contaminant concentration and let $c(z,t)=\int C(x,z,t)dx$ be horizontally integrated plume concentration. Its conserved amount per unit span is ${\int }_0^{\infty }cdz=V_c$. The dilute contaminant is a passive scalar; the maintained plume is unaffected by its impulsive release. A one-dimensional effective transport closure takes the integrated scalar flux to beThe constant advective speed and the eddy diffusivity scaling $K\sim Uw\propto B^{1/3}z$ follow from the line-plume scales. The entrainment coefficient alone does not determine the scalar dispersion coefficient: this Fickian closure for the integrated variable is an additional modelling assumption. Mass conservation $c_t+J_z=0$ then gives the displayed transport equation in the PDF. With this effective closure and advective speed chosen as $U$, $\alpha =(2e{)}^{-1/3}$; if $K=\kappa Uw$, then $\beta =\kappa (2e{)}^{2/3}$, with independent dimensionless mixing coefficient $\kappa$.

There is a second possible closure convention. If one instead applies local diffusive flux to horizontally averaged plume concentration $\overline{C}=c/w$, its integrated diffusive flux is $-Kw{\overline{C}}_z=-K(c_z-c/z)$. For $K=Dz$, the total scalar flux is then $(U+D)c-Dz{c}_z$. The same printed equation is obtained with $a=U+D$, rather than $a=U$. Thus the printed $\alpha$ and $\beta$ should be regarded as effective coefficients unless the averaging and closure conventions are specified; no equality between $\alpha$ and the velocity prefactor is universal.

Assume $a,D>0$, $z>0$, an initial impulse $V_c{\delta }_0$ at the origin, no further contaminant input, and zero endpoint scalar flux for $t>0$. For a similarity solution, putSubstituting into the advection-diffusion equation givesDecay at infinity makes the integrated constant zero. Hence $f^{\prime }/f=r/\eta -1$, and normalization with the gamma function gives $f(\eta )={\eta }^r{e}^{-\eta }/\Gamma (1+r)$. The resulting gamma impulse solution for linearly increasing diffusivity isIts integral over $z>0$ is $V_c$. Its flux is $J=(z/t)c$, which vanishes at both endpoints for $t>0$. The scale $Dt$ shrinks to zero as $t\downarrow 0$, so the normalized solution has weak convergence of probability measures to the required unit source impulse. The mass at the boundary is a full unit impulse on the half-line, not half of a whole-line impulse.

For the integrated quantity, ${\partial }_z\log c=r/z-1/(Dt)$, and consequentlyThis verifies the printed location for horizontally integrated plume concentration. It is a maximum because the derivative changes from positive to negative there. The normalized profile is a gamma distribution with shape $1+r$: its expected value is $(a+D)t$ and its variance is $(a+D)Dt$, so neither the mean position nor the spread should be mistaken for the modal position.

The PDF changes from “integral” to “averaged” concentration in its final request. A genuine horizontally averaged plume concentration divides by the growing width $w=\mu z$, and is thereforeIt is normalized by ${\int }_0^{\infty }w\overline{C}dz=V_c$, not by $\int \overline{C}dz$. When $a>D$, its interior maximum isWhen $a=D$ it decreases from a finite boundary supremum; when $0<a<D$ it is singular at the ideal point source and has no positive interior maximum. A finite source regularizes that singularity. Thus the final printed maximum is correct for the integrated variable $c$, or for an “average” using a fixed reference width, but is not generally the maximum of the local mean $c/w(z)$. Both quantities have been given rather than silently identifying them. If $D=0$, the smooth similarity formula is replaced by the advected impulse $V_c\delta (z-at)$; positive eddy diffusivity is essential to the gamma profile.

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

The dimensions of the injected power and ambient mass density are $[\dot{E}]=M{L}^2{T}^{-3}$ and $[{\rho }_1]=M{L}^{-3}$. Requiring ${\dot{E}}^a{\rho }_1^b{t}^c$ to have the dimensions of length givesThus dimensional analysis gives the adiabatic superbubble similarity solutionThe dimensionless coefficient $\alpha$ can depend on $\gamma$ and the specified inner structure, but dimensions alone cannot determine it. Constant power differs from a fixed-energy explosion because the energy available at time $t$ is proportional to $t$. This scaling assumes that ambient pressure, gravity, radiative losses and the finite injection radius introduce no relevant competing scale.

Put $m_0=M_0/{\rho }_0$ and $f_0=F_0/{\rho }_0>0$ under the literal mass convention. Their dimensions are $[m_0]=L^4{T}^{-2}$ and $[f_0]=L^4{T}^{-3}$. Dimensional analysis givesThe jet length compares source momentum with buoyancy. The region near the source behaves as a turbulent round jet; sufficiently far away, the forced plume behaves like a pure plume. With the constant entrainment coefficient retained explicitly, substantial turning occurs over a length of order ${\ell }_J=L_J/\sqrt{E}$, where $E=2\alpha \sqrt{\pi }$. Statements involving $s\ll L_J$ or $s\gg L_J$ normally hold this dimensionless coefficient fixed.

Near the point source, $m\sim m_0$ and the Batchelor entrainment hypothesis gives $q\sim E\sqrt{m_0}s$. ThereforeIntegrating the centreline tangent givesTo leading order the centreline is a straight turbulent round jet at its source inclination. For the horizontal source, the first curvature is the horizontal forced-plume trajectoryFor a vertical source, $\cos {\theta }_0=0$, the trajectory stays vertical rather than developing this cubic transverse displacement. If $F_0=0$, there is no finite buoyancy-induced jet length and the turbulent round jet remains straight.

Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

Let $f_0=F_0/(\pi {\rho }_0)$ as above. A self-similar starting plume has no source length or time scale; dimensional analysis therefore givesThe dimensionless constants depend on the entrainment coefficient and thermal closure. Thus $R/B$ is constant because the underlying Boussinesq point-source plume has $B=b(Z)=(6\alpha /5)Z$. The steadily increasing thermal volume scales as $t^{9/4}$, even though its reduced gravity decreases.

The specific angular momentum of a circular Keplerian orbit is $\sqrt{G{M}_{\ast }r}$. Therefore the disk angular momentum iswhereThe absence of an external or inner-boundary torque makes $J$, and hence $B$, constant.

From $\bar{\nu }=A{r}^{9/2}{\Sigma }^2$,The combination $A{B}^{2}t$ has dimension $L^{5/2}$. Dimensional analysis therefore gives

For the supplied similarity solution, put $\tau =t/t_0$ and $u=\sqrt{r/R(t)}$. Then $r=R{u}^2$ and $dr=2Rudu$. The total mass isSince $R^{1/2}=R_0^{1/2}{\tau }^{1/5}$,

In a cross-section normal to the cylinder axes, measure $y$ from the line of contact. The two circular boundaries have the parabolic approximationfor their separation. The contact point of the meniscus is at $y=w+o(w)$, so the leading cross-sectional area isThe area of the small meniscus cap is $O(w^4/a^2)$ and is lower order. At its upper end the gap has width $h(w)=w^2/a$. A tangent semicircle therefore has radius $w^2/(2a)$ and curvature

The Young–Laplace equation makes the liquid pressure, relative to the nearly uniform gas pressure,At a fixed $y$, lubrication theory gives a planar Poiseuille flow through a gap of width $h(y)$, with axial flux per unit $y$Integrating across the cusp,The continuity equation $A_t+Q_z=0$ now gives

Set $q=w^3$. Since $w^4{w}_z=\frac{1}{5}(q^{5/3}{)}_z$, this is the porous medium equationConservation of the fixed volumeand dimensional analysis give the self-similar solution $q=t^{-3/8}F(z/t^{3/8})$. One integration of the resulting ordinary differential equation yields the compactly supported Barenblatt solutionEquivalently,Its tip is $z_N=(C/B{)}^{1/2}{t}^{3/8}$. Usingin the volume constraint gives $C^2=8aV\sqrt{B}/\pi$. Therefore

For vertical cylinders at equilibrium, hydrostatic pressure gives $p-p_{\mathrm{atm}}=-\rho gz$. Balancing this with the capillary pressure $-2\gamma a/w^2$ gives the large-height profile

The Linearity of Stokes flow makes every velocity linear in ${\mathbf{G}}_1$. The only available isotropic polar vector built from the axial vector ${\mathbf{G}}_1$ and separation vector $\mathbf{R}$ is ${\mathbf{G}}_1\times \mathbf{R}$. Dimensional analysis then givesAn angular velocity is axial, so the two independent isotropic possibilities are ${\mathbf{G}}_1$ and $({\mathbf{G}}_1\cdot \mathbf{R})\mathbf{R}$. Hencefor dimensionless scalar functions ${\beta }_i,{\gamma }_i$.

In the midplane, the vacuum magnetic dipole field has scale $B_0\sim \mu /r^3$, while the result of part i gives $j\sim {\Omega }^2\mu /(c{r}^2)$. Ampère's law then gives the induced-field estimatewhere numerical factors are immaterial in this dimensional analysis. Thus $B_i\sim B_0$ atThis is the light cylinder radius: rigid corotation would have $u=\Omega R_e\sim c$. The nonrelativistic approximation and the assumed unmodified vacuum field therefore fail at the same scale, and a self-consistent pulsar magnetosphere must replace them.

Use the dimensional analysisand define the dimensionless sliding parameterThe dimensionless flux and steady conservation law areOn the right half-cap, let the snowline be $X=X_s$ and the nose be $X=X_n$. The zero-flux condition at the ice divide gives $Q=X$ in the accumulation region. Continuity of $Q$ gives $Q=2X_s-X$ in the ablation region, and zero nose flux gives

Introduce the increasing functionSince ${\Phi }_X=-Q$, the ablation profile satisfying $H(X_n)=0$ isAt the snowline $H=1$, soIn the accumulation region,and the divide thickness $H_0$ is fixed byThese two implicit formulas give a continuous thickness and flux at the snowline. Reflection across $X=0$ gives the full two-dimensional ice cap.

The arc length follows from the speed of the parametrization:Here $\sin (t/2)>0$ inside the given interval. ThereforeThe result scales as a length, while the preceding area scales as $a^2$, as required by dimensional analysis.

The dimensions of mass density, time and energy are $\mathsf{M}{\mathsf{L}}^{-3}$, $\mathsf{T}$ and $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$. Matching powers in the proposed dimensional analysis givesThusThis is the Sedov-Taylor blast wave scaling for a fixed released energy. Dimensional analysis cannot determine the dimensionless $C$; that requires the fluid equations and thermodynamic assumptions. The assumed list of controlling parameters excludes, for example, an independent ambient sound speed or a continuing energy supply.

In SI base dimensions,Equating current, time, mass and length powers respectively yieldsThusDimensional analysis cannot determine the dimensionless proportionality constant.

Use $[\eta ]=M{L}^{-1}{T}^{-1}$ and $[dp/dx]=M{L}^{-2}{T}^{-2}$. Equating mass, length and time exponents giveshenceDimensional analysis leaves $k$ undetermined. If positive $Q$ denotes flow along increasing $x$, the signed pressure derivative is negative and its minus sign belongs in the proportionality constant; alternatively use the pressure-drop magnitude.

The dimensions are $[\rho ]=M{L}^{-3}$, $[\sigma ]=M{T}^{-2}$ and $[\lambda ]=L$. Equating exponents in $L{T}^{-1}=M^{\alpha +\beta }{L}^{-3\alpha +\gamma }{T}^{-2\beta }$ yieldsThus the model has $v_{\mathrm{ph}}\propto \sqrt{\sigma /(\rho \lambda )}$. Its common dimensionless factor cancels in the ratio:They are nearly equal within this capillary-wave model. The assumed power law omits gravitational restoring forces and finite-depth effects; dimensional analysis cannot establish those omissions for an arbitrary real surface wave.

For a flexible uniform string with linear mass density $\rho$ and tension $T$, dimensional analysis gives $[T/\rho ]=L^2{T}^{-2}$ and hence $c\propto \sqrt{T/\rho }$. The small-slope force balance $\rho {\psi }_{tt}=T{\psi }_{xx}$ fixes the factor to one:Here $\rho$ has units kg/m, not the volume-density units used for a bulk fluid. The phase moves along the string at $c$, while a material point moves transversely with ${\psi }_t=A\omega \sin (kx-\omega t)$. Its maximum speed is $A\omega$, which is not the phase velocity. Small slope $Ak\ll 1$ makes $A\omega \ll c$ in this model.

The kinetic energy and extra elastic energy of a segment areThe second expression follows from extra length $\sqrt{1+{\psi }_x^2}dx-dx\simeq {\psi }_x^{2}dx/2$ against the pre-existing constant tension. Time averaging a harmonic wave gives $⟨{\mathcal{E}}_K⟩=\rho A^2{\omega }^2/4$ and $⟨{\mathcal{E}}_U⟩=T{A}^2{k}^2/4=\rho A^2{\omega }^2/4$. Therefore

At the driven endpoint the string pulls along its local tangent into $x>0$, with vertical component $T{\psi }_x$ for small slope. The point exerts the opposite force on the string, so the driven-string endpoint power isFor the outgoing wave $\psi =A\cos (kx-\omega t)$, at $x=0$ the derivatives are ${\psi }_x=Ak\sin \omega t$ and ${\psi }_t=-A\omega \sin \omega t$. Thus $P=T{A}^{2}k\omega {\sin }^2\omega t$ andThe energy flux of a transverse string wave follows by combining the energy density $\mathcal{E}=(\rho {\psi }_t^2+T{\psi }_x^2)/2$ and flux $P=-T{\psi }_x{\psi }_t$, which satisfy ${\partial }_t\mathcal{E}+{\partial }_{x}P=0$. For a pure right-going nondispersive wave, $P=c\mathcal{E}$ even instantaneously. energy is transported with the wave; a material string element merely oscillates about its unperturbed position.

Write the proposed Jeans mass as $M_J=K{v}^a{G}^b{P}^c$, with dimensionless $K$. The dimensions of speed, the gravitational constant and pressure areEquating the exponents of mass, length and time givesThus $a=4$, $b=-3/2$ and $c=-1/2$, so the pressure-confined Jeans mass scales asA larger speed of sound gives stronger support against gravitational collapse; stronger confinement at fixed speed of sound lowers the characteristic unstable mass. Dimensional analysis determines the scaling, not $K$, which requires a particular cloud model and instability criterion.

Use distinct symbols for the dimensions of mass, length, time, temperature and amount of substance. The molar gas constant has dimensions $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}{\mathsf{\Theta }}^{-1}{\mathsf{N}}^{-1}$, while the given $M$ is a molar mass with dimensions $\mathsf{M}{\mathsf{N}}^{-1}$. Matching dimensions in the proposed power law giveswith the amount-of-substance equation repeating $\alpha +\gamma =0$. ThereforeDimensional analysis fixes these exponents but not a dimensionless multiplier. For example, an open-tube mode number and the gas's heat-capacity ratio enter the actual resonance frequency without changing the dimensions.

Use dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The reduced Planck constant, gravitational constant and speed of light have dimensionsFor $l_p={\hslash }^{\alpha }{G}^{\beta }{c}^{\gamma }$ to have the dimensions of a length, dimensional analysis requiresThe first equation gives $\beta =\alpha$; the third gives $\gamma =-3\alpha$. Substitution in the second gives $2\alpha =1$. HenceWith $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, the Planck length isThe proposed monomial fixes the dimensionless prefactor to one; dimensional analysis on its own cannot select an arbitrary dimensionless multiplier.