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Differentiate the position vector with respect to the three coordinates:Their pairwise dot products vanish. For example, ${\mathbf{r}}_{\rho }\cdot {\mathbf{r}}_u=-\sinh \rho \cosh \rho \sin u\cos u+\sinh \rho \cosh \rho \sin u\cos u=0$. Thus these are orthogonal coordinates wherever the coordinate tangents are nonzero. Their scale factors of orthogonal coordinates areThe absolute Jacobian determinant is their product, givingThe ordered coordinate system is left handed: ${\mathbf{r}}_{\rho }\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)=-h_{\rho }{h}_u{h}_v$. The volume measure uses the absolute value. The axis and the focal circle contain coordinate degeneracies, as expected for oblate spheroidal coordinates. This is the hyperbolic parametrization of oblate spheroidal coordinates. The local TeX has an OCR substitution ${\sinh }^{2}u$ in the volume formula; the original PDF correctly prints ${\sin }^{2}u$.

Integrating the Dirac delta distribution in $x_3$ reduces the inertia integral to the planar lamina. Use $u=x_1-x_2$, $v=x_2$, so $x_1=u+v$ and the parallelogram becomes the square $-D\le u,v\le D$, with unit Jacobian determinant. The mass is $M=4\sigma D^2$, and symmetry gives a centre of mass at the origin.

Odd rectangular integrals vanish. The required second moments areThe mixed moments involving $x_3$ vanish. Hence the inertia tensor isThe off-diagonal entries are minus the product moment, not the product moment itself. This is the unit-shear case of the inertia tensor of a sheared rectangular lamina.

The Jacobian determinant is the oriented scalar triple product of the three coordinate basis vectors:Here ${\mathbf{h}}_u\times {\mathbf{h}}_v=(u^2+v^2)(-\sin \varphi ,\cos \varphi ,0)$, soThe sign is nonnegative on the stated range, and the absolute Jacobian determinant also equals $h_u{h}_v{h}_{\varphi }$ as expected for orthogonal coordinates.

Since $\mathbf{F}=\nabla \Omega$, its divergence is the Laplacian of $\Omega$:The divergence theorem changes the outward flux through the whole closed boundary into $I={\int }_V(6x+4z)dV$. Rotational symmetry makes the $x$ term vanish. With $z=(u^2-v^2)/2$ and the Jacobian determinant from part (c),Both the curved face and the top disk are included in this surface integral through the divergence theorem.

Under a proper rotation matrix $R$, the product of coordinates transforms as a rank-$n$ tensor. The unit ball is unchanged by the substitution $\mathbf{y}=R\mathbf{x}$, whose Jacobian determinant is one. HenceThus $S$ is an isotropic tensor. An isotropic second-rank tensor has the form $S_{ij}=A{\delta }_{ij}$. Taking the trace and using spherical coordinates,Therefore the isotropic second moment of a ball isAn isotropic third-rank tensor has the form $B{ϵ}_{ijk}$, but $S_{ijk}$ is symmetric in every pair of indices, forcing $B=0$. Alternatively, $\mathbf{x}\mapsto -\mathbf{x}$ makes its integrand odd on the ball. Thus

A marginal density is obtained by integrating out the other variable: $\boxed{f_X(x)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dy}$. Here independence gives $f_{X,Y}=e^{-(x^2+y^2)/2}/(2\pi )$.

The polar transformation has absolute Jacobian determinant $r$. Consequentlywith density zero outside this range. Integrating over the angle gives $\boxed{f_R(r)=r{e}^{-r^2/2}}$, the Rayleigh distribution density. Integrating over radius, using ${\int }_0^{\infty }r{e}^{-r^2/2}dr=1$, gives $\boxed{f_{\Theta }(\theta )=1/(2\pi )}$ on the angle interval. The joint density factors as the product of these marginals, so $\boxed{R\text{ and }\Theta \text{ are independent}}$. The arbitrary angle assigned at $R=0$ is irrelevant because that event has probability zero.

The annulus is open and hence Lebesgue measurable; multiplying its indicator by the continuous Gaussian gives a measurable function. It is bounded by the indicator of $[-b,b{]}^2$, whose Lebesgue integral is $4b^2$. Thus it is Lebesgue integrable by comparison.

Use the change of variables formula with polar coordinates $x=r\cos \theta$, $y=r\sin \theta$. Removing one radial cut makes this a smooth one-to-one coordinate map; the cut is a null set and does not affect the integral. The absolute Jacobian determinant is $r$. By Fubini theorem,The antiderivative is $-e^{-r^2/2}$, which also makes the sign of the answer transparent.

The inverse transformation is $x=2u+v$, $y=v$, with absolute Jacobian determinant two. The original positivity constraints become $v>0$ and $2u+v>0$. ThusIntegrating out $v$ givesThis is the centred Laplace distribution with scale one.

The inverse transformation is $x=uv$, $y=u(1-v)$, with support $u>0$, $0<v<1$. Its absolute Jacobian determinant isThe original joint probability density function, by independence, is ${\lambda }^2{e}^{-\lambda (x+y)}$ on the positive quadrant. The change of variables formula therefore givesIntegrating over $v$ givesthe gamma distribution of shape two and rate $\lambda$. Integrating over $u$ gives $f_V(v)=1$ on $(0,1)$. The factorization also shows that $U$ and $V$ are independent.

For sigma-finite measures, Tonelli's theorem states that the double integral of a nonnegative measurable function equals either iterated integral, allowing $+\infty$. Fubini's theorem states that if a measurable function is absolutely integrable, its sections are integrable almost everywhere, both iterated integrals exist, and they equal the double integral.

Assign any value at $(0,0)$, which does not affect Lebesgue integrability. Near that corner set $s=x+y$ and $t=(x-y)/(x+y)$. Then $x=s(1+t)/2$, $y=s(1-t)/2$, and the absolute Jacobian determinant is $s/2$. On $0<s<1$, the full range is $-1<t<1$, soThe remaining part of the unit square has $x+y\ge 1$ and is harmless. The improper power integral is finite precisely when $\alpha <3$. Under the given restriction,For those values, the change $(x,y)\mapsto (y,x)$ preserves the square and changes the sign of $F$. Absolute integrability justifies this change of variables, givingFor $\alpha \ge 3$ the positive and negative parts each diverge, so symmetry does not define a Lebesgue integral.

Independence means $\mathrm{Pr}(X\in A,Y\in B)=\mathrm{Pr}(X\in A)\mathrm{Pr}(Y\in B)$ for all Borel sets $A,B$. For variables with a joint density this is equivalent to factorization of that density into the two marginal densities almost everywhere.

The inverse transformation is $x=uv$, $y=(1-u)v$, on $0<u<1$, $v>0$. Its absolute Jacobian determinant is $v$. Multiplying the independent gamma densities by it givesThe density is zero outside that support. Integrating in $v$ with the gamma integral givesIt is a beta density. Its normalization also proves the beta integral ${\int }_0^1{u}^{\alpha -1}(1-u{)}^{\beta -1}du=\Gamma (\alpha )\Gamma (\beta )/\Gamma (\alpha +\beta )$. The joint density consequently factors as this normalized beta density timesThus $V$ is gamma with shape $\alpha +\beta$ and rate $\lambda$, and $U,V$ are independent.

Finally $X=UV$. All relevant first moments are finite, and independence gives $\mathbb{E}X=\mathbb{E}U\mathbb{E}V$, while $\mathbb{E}V=\mathbb{E}X+\mathbb{E}Y>0$. HenceThis identity follows from the particular beta-gamma independence here; a general ratio of random variables would not permit moving expectation through division.

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open image, with continuously differentiable inverse. Equivalently one may specify a one-to-one continuously differentiable map whose Jacobian determinant is everywhere nonzero; the inverse function theorem gives the corresponding smooth inverse on its image.

Away from the null event $X=0$, the inverse transformation is $x=u$, $y=uv$. Its Jacobian determinant is $u$, so its absolute value is $|u|$. Independence therefore gives the joint densityA value at $u=0$ may be assigned arbitrarily without changing the distribution.

Successive interarrival times $W_1=T_1$ and $W_2=T_2-T_1$ are independent exponential variables of rate $\lambda$. Their joint density is ${\lambda }^2{e}^{-\lambda (w_1+w_2)}$ for $w_1,w_2>0$. The transformation $t_1=w_1$, $t_2=w_1+w_2$ has Jacobian determinant one, givingIntegrating over $0<t_1<t_2<\infty$ gives one, as a joint density should.

The variable initial energy $E_0$ replaces $m_{\pi }$ in the kinematic expression before differentiation. The charged-lepton mass remains fixed:Thus the required Jacobian determinant in this one-dimensional change of variables isEquivalently, $E_0=\sqrt{p^2+m_{\ell }^2}+p$ gives $d{E}_0/dp=1+p/E_{\ell }$, so $dp/d{E}_0=E_{\ell }/(E_{\ell }+p)=E_{\ell }/m_{\pi }$, the same result. It is dimensionless in natural units.

Treat the electron and electron antineutrino as effectively massless particles in the centre-of-momentum frame, retain only the charged current, and use the Fermi interaction: $m_e^2\ll s\ll M_W^2$. Take ${\theta }^{\ast }$ between incoming and outgoing electron antineutrino directions. The V-A interaction selects a right-helicity electron antineutrino and a left-helicity electron. Their opposite incoming directions give total angular momentum projection one. The outgoing pair has the same projection along its own axis. The Wigner D-matrix element is $d_{11}^1({\theta }^{\ast })=(1+\cos {\theta }^{\ast })/2$, so the squared helicity amplitude is proportional to $(1+\cos {\theta }^{\ast }{)}^2$. The inactive initial electron helicity is included in the unpolarized normalization supplied by the paper.

The boost from the centre-of-momentum frame to the laboratory frame has $\beta =E/(E+m_e)$, while the incoming electron antineutrino has $E_{\ast }=(s-m_e^2)/(2\sqrt{s})=m_{e}E/\sqrt{s}$ and $s=m_e^2+2m_{e}E$. ThusThe requested identity therefore uses $E\gg m_e$; it is not exact at finite target mass. With $c_{\ast }=\cos {\theta }^{\ast }$, the absolute Jacobian determinant is $|dy/d{c}_{\ast }|=1/2$, givingThe suppression at backward scattering follows from angular momentum conservation: rotating projection $+1$ through $\pi$ cannot leave projection $+1$ along the original axis. Replacing $1+c_{\ast }$ by $2(1-y)$ gives the supplied $(1-y{)}^2$ dependence.

Write $s_{ij}=M_{ij}^2$ and factor the three-body phase space through the pair $q=p_1+p_3$, with $s=q^2$:A scalar parent permits integration over the overall orientation. In its rest frame let the pair have momentum magnitude $p$, and in the pair rest frame let particle 3 have magnitude $p^{\ast }$. After integrating the azimuth of its direction, the two-body factors giveIn that pair rest frame, particle 2 has momentum magnitude $M_{0}p/\sqrt{s}$, so $|d{s}_{23}/d\cos {\theta }^{\ast }|=2M_{0}p{p}^{\ast }/\sqrt{s}$. The Jacobian cancels both momentum factors, leavingThe physical domain is bounded by the requirement that the relative cosine lie between $-1$ and $1$. The phase-space density is uniform within that domain; the event density is uniform only for a constant squared amplitude. Scalar parent spin alone does not forbid resonance-dependent amplitudes. A factor $1/(s_{13}-m_R^2+i{m}_R{\Gamma }_R)$ is a concrete nonuniform example. For two identical pions, a conventional overall $1/2!$ compensates integration over both labelings without changing the local flatness of phase space.^[ref]

Neglecting the electron mass in the centre-of-mass kinematics, take $k=(E^{\ast },0,0,E^{\ast })$, $p=(E^{\ast },0,0,-E^{\ast })$, and $k^{\prime }=(E^{\ast },E^{\ast }\sin {\theta }^{\star },0,E^{\ast }\cos {\theta }^{\star })$. Then $p\cdot k=2E^{\ast 2}$ and $p\cdot k^{\prime }=E^{\ast 2}(1+\cos {\theta }^{\star })$. HenceThe constant Jacobian $|d\cos {\theta }^{\star }/dy|=2$ givesIncluding both Standard Model chiral channels in the contact approximation instead gives $d\sigma /dy\propto g_L^2(1-y{)}^2+g_R^2$. The right-electron term is flat in $y$.

In the electron-rest laboratory frame $p=(m_e,0)$, the invariant definition immediately givesThis last equality is exact. With electron mass retained in the centre-of-mass relation, the angular formula becomes $y=(s-m_e^2)(1-\cos {\theta }^{\star })/(2s)$, approaching the boxed massless relation at high energy.

With relativistically normalized one-particle states, Fermi's golden rule becomesIn the parent rest frame, the three-momentum delta function fixes ${\mathbf{p}}_2=-{\mathbf{p}}_1$. The energy delta function fixes the common magnitude $p^{\ast }$; its radial Jacobian converts the remaining phase-space factor intoOnly one momentum magnitude remains because the two final momenta are constrained, not independent. Explicitly,For leptonic W boson decay both final masses are small compared with $m_W$, including the tau mass at the precision requested, so $p^{\ast }\simeq m_W/2$.

At a hadron collider, select $W\to \ell \nu$ events and reconstruct the transverse massusing missing transverse momentum for the neutrino. Fit the shape, especially the region above the nominal Jacobian edge, with finite-width resonance templates convolved with detector resolution, recoil modelling and backgrounds. The Breit-Wigner width controls this off-shell tail. Calibrated momentum and missing-momentum resolutions must be separated from the physical width. This is a lineshape measurement; the lifetime is far too short for a direct flight-distance measurement.