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A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.

The separation is $|x_1-x_2|=\sqrt{2}|y_{-}|$. The normalized $y_{+}$ factor integrates out. In the fermionic ground state, the relative-coordinate density is $a^{-1}{\sin }^2(\pi y_{-}/a)$, givingFor the bosonic ground state it is $a^{-1}{\cos }^2(\pi y_{-}/(2a))$, givingBoth integrals follow by replacing squared trigonometric functions by their constant and cosine parts and integrating once by parts. Henceand the difference is $\boxed{2\sqrt{2}a/{\pi }^2}$. The odd relative state vanishes at coincidence, which increases its expected separation.

Under an additional thermalization assumption, an opaque star emits approximately black-body radiation characterized by the temperature near the layer from which thermal photons escape. Planck's law gives a Rayleigh-Jeans law low-frequency tail $B_{\nu }\propto T{\nu }^2$ and a Wien approximation high-frequency tail $B_{\nu }\propto {\nu }^3{e}^{-h\nu /(k_{B}T)}$.

Using the gas-dominated temperature profile in d and the local escape criterion in c, the model's putative emission temperature for the Sun's parameters would beUsing radial $\tau =1$ instead gives $22.8\mathrm{K}$. The frequency form of Wien's displacement law for the maximum of $B_{\nu }$ is $h{\nu }_{\max }/(k_{B}T)=2.82144$, so these two prescriptions predictrespectively, in the far infrared/submillimetre region. These are conditional estimates for this model, not observed solar peaks. A realistic solar effective temperature near $5.8\times 10^3\mathrm{K}$ gives ${\nu }_{\max }\simeq 3.4\times 10^{14}\mathrm{Hz}$ in $B_{\nu }$. The maximum of $B_{\lambda }$ occurs near $0.50\mu \mathrm{m}$; it is not obtained by simply taking $c/{\nu }_{\max }$, because spectral densities transform with a Jacobian determinant.

The model is unsuitable for the solar photosphere: fully ionized hydrogen at $16$–$23\mathrm{K}$ is inconsistent with thermal ionization, and the assumed opacity and near-surface temperature gradient do not resemble the Sun's. Even its central temperature is substantially below the actual solar value, about $1.6\times 10^7\mathrm{K}$.

There is a further logical limitation: the question retains only Thomson scattering. In its coherent, cold-electron limit, this redistributes photon directions without generating a Planck photon distribution. For a concrete counterexample, monochromatic injected radiation remains monochromatic after any number of such elastic scatterings; large optical depth cannot turn it into black-body radiation. True absorption and emission, or a specified pre-existing thermal radiation field with appropriate energy exchange, are missing. Thus an approximately blackbody emergent spectrum is the usual additional physical assumption, while the stated scattering-only model does not uniquely predict an emergent spectrum or its peak.

The coefficient $j_{\nu }$ is power per unit volume, frequency and solid angle. For uniform, optically thin, isotropic emission from $V=4\pi R^3/3$, the spectral luminosity is $L_{\nu }=4\pi V{j}_{\nu }$. With photon energy $\epsilon =h\nu$, conservation of energy and the change of spectral variable giveThusThe $4\pi$ outside $V$ integrates over emitted solid angle; $1/\epsilon$ changes radiated energy to a photon count, and $1/h$ is the Jacobian determinant from frequency to energy.

With $h$ in $\mathrm{J}\mathrm{s}$ and $\epsilon$ in joules, this formula produces ${\mathrm{J}}^{-1}{\mathrm{s}}^{-1}$. To compare with counts per $\mathrm{keV}$, multiply by $1.602176634\times 10^{-16}J/keV$. Equivalently, the conversion cancels the same joule-per-$\mathrm{keV}$ factor in $\epsilon$: one can use the numerical $\epsilon$ in $\mathrm{keV}$ in the $1/\epsilon$ factor, retain $h$ in $\mathrm{J}\mathrm{s}$, and express $k_{B}T$ in $\mathrm{keV}$ in the exponential. The radius in the volume must still be converted to metres because $j_0$ is per ${\mathrm{m}}^3$.

Put $x=h\nu /(k_{B}T)=hc/(\lambda k_{B}T)$. In the long-wavelength limit $x\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ in Planck's law gives the Rayleigh-Jeans law,This is the classical equipartition theorem result: each electromagnetic mode carries mean energy $k_{B}T$. Extending it to arbitrarily high frequency would make the total energy density diverge, the ultraviolet catastrophe.

In the short-wavelength limit $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximation,Large photon energies are thermally suppressed, removing the classical high-frequency divergence.

If the requested “form” is expressed per unit wavelength, the change of variables includes its Jacobian determinant: $U_{\lambda }=U_{\nu }|d\nu /d\lambda |=U_{\nu }c/{\lambda }^2$. ThusSubstituting $\nu =c/\lambda$ without the Jacobian determinant would give an incorrect spectral density.

Independent increments give $B_s\sim N(0,s)$ independent of $B_t-B_s\sim N(0,t-s)$. Their joint probability density function at $(x,v)$ is the productThe change of variables $(x,v)\mapsto (x,y)=(x,x+v)$ has Jacobian determinant one. Therefore the joint probability density function of $(B_s,B_t)$ isIn particular these coordinates are correlated, with covariance $s$, although the first pair has independent coordinates.

Normalize the probability density function on $x\ge 0$:For $y=\sqrt{x}\ge 0$, the change of variables is $x=y^2$, $dx=2ydy$. Conservation of probability gives the transformed probability density functionIt is a Rayleigh distribution with scale $\sigma =1/\sqrt{2\alpha }$. Its derivative is $g^{\prime }(y)=2\alpha e^{-\alpha y^2}(1-2\alpha y^2)$, so it rises from zero, has one maximum, and then decays:The expected value follows by putting $z=\sqrt{\alpha }y$:Here ${\int }_0^{\infty }{z}^2{e}^{-z^2}dz=\sqrt{\pi }/4$, obtained from the Gaussian integral by integration by parts. In particular $⟨\sqrt{x}⟩\ne \sqrt{⟨x⟩}$, and the mode changes under a nonlinear variable transformation because a density includes its Jacobian determinant.

For integers $k,l$,Thus reducing coordinates modulo one gives a well-defined function on the torus ${\mathbb{R}}^2/{\mathbb{Z}}^2$. Its determinant is one and its inverse matrix is the integer matrix $\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$, so it induces a bijection of the torus. The Jacobian determinant one makes this bijection area-preserving.

Its eigenvalues solve ${\lambda }^2-3\lambda +1=0$, givingThere are therefore expanding and contracting eigendirections and no eigenvalue on the unit circle: the function is a hyperbolic toral automorphism. The Lyapunov exponents per iterate areThey sum to zero, as required by area preservation.

The four vertices of the unit square map to $(0,0),(2,1),(3,2),(1,1)$, the vertices of an area-one parallelogram. To reduce it modulo one, cut it by the integer grid. For each piece in the cell $[j,j+1]\times [k,k+1]$, translate by $(-j,-k)$. Up to boundary identifications these translated pieces partition the unit square, because the inverse integer matrix gives exactly one torus preimage to every point. The colored pieces in the original diagram show the same pieces before and after these translations.

For the Fibonacci identity, at $n=1$ the claimed matrix is $\left(\begin{array}{cc}{F}_2& F_1\\ F_1& F_0\end{array}\right)=A$. If the identity holds at $n$, multiplication on the right by $A$ givesEach equality uses $F_{j+1}=F_j+F_{j-1}$, with the specified indexing $F_0=F_1=1$. This proves the formula for every positive integer $n$.

A fixed point satisfies $(A-I)v\in {\mathbb{Z}}^2$. Since $A-I=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has an inverse integer matrix, this forces $v\in {\mathbb{Z}}^2$, so the origin is the only fixed point modulo one.

For period dividing two,The second congruence gives $y=-3x(\mathrm{mod}1)$; inserting it into the first gives $5x=0(\mathrm{mod}1)$. Thus the complete list isThe first is the fixed origin. Direct application of $A$ exchanges the other four in exactly two pairs:This both verifies the periodic orbits and excludes additional ones.

Finally, if $v$ is fixed by $F^p$, then $(A^p-I)v=m$ for an integer vector $m$. The eigenvalues of $A^p-I$ are ${\lambda }_{+}^p-1$ and ${\lambda }_{-}^p-1$, neither zero. Hence its inverse exists and has rational entries by the adjugate matrix formula for an integer matrix. Therefore $v=(A^p-I{)}^{-1}m$ has rational coordinates. The derivative of the return map is $A^p$, with eigenvalues ${\lambda }_{+}^p>1$ and $0<{\lambda }_{-}^p<1$. Every periodic point is a hyperbolic saddle fixed point of a map and is unstable, although perturbations in its stable eigendirection contract.

For this change of variables,and $u_{xx}+u_{yy}=0=v_{xx}+v_{yy}$. The second derivative chain rule therefore givesThus the transformation rule for the Laplacian isFor $f=x+y$, its Laplacian is zero. At every point with $(x,y)\ne (0,0)$, division by $4(x^2+y^2)$ provesThis conclusion applies to each local inverse branch of the change of variables: its Jacobian determinant is $4(x^2+y^2)$, so such a branch exists away from the origin by the inverse function theorem. A single globally defined $g$ cannot represent $x+y$ on the entire plane, since $(x,y)$ and $(-x,-y)$ give the same $(u,v)$ but generally opposite values of $x+y$. Nor can a differentiable $g$ represent this $f$ in a neighborhood of the origin: the multivariable chain rule there would give $f_x=f_y=0$, contradicting $f_x=f_y=1$. The requested harmonic function conclusion is therefore a local statement on nonsingular branches.^[ref]

The divergence of the vector field generating a flow on phase space. Its time integral determines the Jacobian determinant of the characteristic flow map: $d\log |\det D{\Phi }_t|/dt=({\nabla }_z\cdot b)({\Phi }_t,t)$. A Hamiltonian vector field has zero phase-space divergence in canonical coordinates.

The Jacobian determinant of a differentiable characteristic flow map measures its change of volume. The variational equation and the Liouville formula for a fundamental matrix give the displayed formula for a complete smooth field $b$. For kinetic advection with $b=(v,F)$, its divergence is ${\nabla }_v\cdot F$. A zero divergence preserves phase-space Lebesgue measure.

For polar coordinates $x=r\cos \theta$, $y=r\sin \theta$ with $r\ge 0$, the Jacobian determinant is $r$. The change of variables formula therefore gives the area element $dA=rdrd\theta$. Integrating over the radial interval $r_1(\theta )\le r\le r_2(\theta )$ gives $[r_2(\theta {)}^2-r_1(\theta {)}^2]/2$ per unit angle. Bounds must describe the intended region without counting it more than once.

Put $I=I(1)$. The nonnegative integrand allows Tonelli's theorem to write the square as a double integral:In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant supplies the area factor $r$:Since $I>0$, take the positive square root to get $I=\sqrt{\pi }$. The computation proves finiteness as well as the value. For an explicit limiting argument, first integrate over disks of radius $R$; their integrals are $\pi (1-e^{-R^2})$, which increase to $\pi$ as $R\to \infty$. ^[ref]

The geometric surprise is that radial shells get longer as $r$ increases, even while the Gaussian function gets smaller. Its central height is largest at zero, but the radial contribution $2\pi r{e}^{-r^2}$ vanishes there and peaks at $r=1/\sqrt{2}$. Differentiating gives $2\pi e^{-r^2}(1-2r^2)$, which changes sign at that radius. Keeping the Jacobian determinant is what makes the one-dimensional integral and the plane geometry agree.

For independent standard normal random variables $X,Y$, put $X=R\cos \Theta$, $Y=R\sin \Theta$, with $R\ge 0$ and $0\le \Theta <2\pi$. The polar coordinates Jacobian determinant is $r$, so the joint probability density is $r{e}^{-r^2/2}/(2\pi )$ on $r>0$. It factorizes into a Rayleigh distribution density for $R$ and a continuous uniform distribution density for $\Theta$. Consequently the radius and angle are independent. The change of variables for probability densities $S=R^2$ gives $f_S(s)=e^{-s/2}/2$ for $s>0$, the exponential distribution of rate $1/2$, also the chi-squared distribution with two degrees of freedom. Measurable transformation preserves independence, giving $\Theta \perp S$. The angle at $R=0$ is arbitrary on an event of probability zero.

Thus a sector with angular length $\ell$ has probability $\ell /(2\pi )$, independently of any radial restriction. For instance $Y>|X|$ occupies a sector of angle $\pi /2$ and has probability $1/4$.

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ is its maximal spin component. Let $\theta$ be the angle of the emitted electron momentum to that axis and let $\beta =v/c$. Use $\mathcal{P}$ for the parity operator, so it is not confused with $P$.

For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization vector $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe factor $\beta$ is the conventional separation of the speed dependence; rotational symmetry alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-spin ensemble can also have alignment that is not specified by its vector nuclear polarization alone.

The unit-integral condition follows without any model of a detector:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

Under full spatial reflection, $\mathbf{p}\mapsto -\mathbf{p}$ but $\mathbf{J}\mapsto \mathbf{J}$. The latter follows already for orbital angular momentum from $\mathbf{r}\times \mathbf{p}\mapsto (-\mathbf{r})\times (-\mathbf{p})=\mathbf{r}\times \mathbf{p}$; spin has the same axial vector transformation. Hence $\mathbf{P}\cdot \hat{\mathbf{p}}$ is a pseudoscalar, and $\cos \theta \mapsto -\cos \theta$. If parity conservation held for this model, equal probabilities for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.

Integrating over the two hemispheres gives a useful original check:Choose $A=-0.8$, $P=0.5$, and $\beta =0.75$, so $a=-0.3$. Then the forward and backward hemisphere probabilities are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the Wu experiment. At a single energy, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.

To update a positive parameter, a symmetric additive proposal distribution can be generated on the whole real line. Reject nonpositive proposals; accept positive ones using the target density ratio. The off-diagonal proposal remains symmetric. Repeatedly redrawing until the proposal is positive instead truncates and renormalizes the proposal differently at different current states, so a Hastings correction is then required. A symmetric walk in $\log v$ is another valid method, but its transformed target includes the Jacobian determinant factor $v$.

If $(X,Y)$ has a joint probability density $f$, the product has the displayed probability density function, interpreted almost everywhere. For fixed $x\ne 0$, substitute $z=xy$ in the integral defining $\mathbb{P}(XY\in A)$; the absolute Jacobian determinant is $|x|$. The Tonelli theorem then exchanges the nonnegative integrals. The line $x=0$ has two-dimensional Lebesgue measure zero, so its omission creates no probability mass. A density may be infinite at exceptional points without creating an atom.

For a regular model with a suitable ancillary statistic $a$, the displayed expression approximates the conditional density of the maximum-likelihood estimator. The normalizing factor includes $(2\pi {)}^{-p/2}$ for a $p$-dimensional estimator, depending on convention. The observed information is evaluated at the sample's own fitted value. In a pure scale model the factor is proportional to $1/\hat{\theta }$; this Jacobian determinant can make the normalized approximation coincide with the exact conditional density.

For independent samples of size $n$ from exponential distributions with rates $\lambda$ and $\psi \lambda$, the maximum-likelihood estimators are $\hat{\lambda }=n/S$ and $\hat{\psi }=S/T$, where $S$ and $T$ are the respective sample sums. Their normalized P-star approximation is exact:Indeed, $S$ and $T$ have independent gamma distributions with shapes $n$ and respective rates $\lambda ,\psi \lambda$. The Jacobian determinant of their inverse transformation is $n^2/({\hat{\psi }}^2{\hat{\lambda }}^3)$, giving the displayed density. Alternatively the square root of the determinant of the fitted observed information is $n/(\hat{\psi }\hat{\lambda })$; multiplying by the exponential of the fitted log-likelihood difference gives the same shape. Its normalizing constant in the P-star approximation is $c_n=n^{2n-1}{e}^{-2n}/\Gamma (n{)}^2$. Integrating over $\hat{\lambda }$ givesso $\hat{\psi }/\psi$ has the F-distribution with degrees of freedom $(2n,2n)$.

On $x>0$, the linear partial differential equation $u_{xx}-x^2{u}_{yy}-u_x/x=0$ has characteristic curves $y\pm x^2/2=\text{constant}$. The change of variables $\xi =x^2-2y$, $\eta =x^2+2y$ has Jacobian determinant $8x\ne 0$. The chain rule givesThus its canonical form of a second-order partial differential equation is $U_{\xi \eta }=0$, and its general twice differentiable solution is $u=f(x^2-2y)+g(x^2+2y)$. The first derivative term exactly cancels the derivatives of the coordinate coefficients.

More generally, for positive differentiable $a(x)$, choose $s(x)$ with $s^{\prime }(x)=a(x)$. Then the same cancellation givesThe characteristic coordinates are $s\pm y$. The quadratic example is $a(x)=x$, $s=x^2/2$, with both coordinates multiplied by two.

A reflection acting as $(A,B)\mapsto (A,-B)$ forces an equivariant vector field to have $\dot{A}$ even and $\dot{B}$ odd in $B$. With the zero solution preserved, its quadratic normal form is $\dot{A}={\lambda }_{1}A+a_1{A}^2+a_2{B}^2$, $\dot{B}={\lambda }_{2}B+a_{3}AB$. For positive coefficients, every mixed equilibrium with $B\ne 0$ has Jacobian determinant $-2a_2{a}_3{B}^2<0$, hence is a saddle equilibrium. The two invariant-axis equilibria meet in a transcritical bifurcation, and the mixed branches meet them in pitchfork bifurcations.