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An isotropic two-photon decay of a massive parent with laboratory Lorentz factor $\gamma >1$ has the displayed normalized opening-angle probability density, with $\beta =\sqrt{1-{\gamma }^{-2}}$ and $2\arcsin (1/\gamma )<A<\pi$. To derive it in units $c=1$, the laboratory photon energies are $E_{1,2}=\gamma M(1\pm \beta z)/2$, where $z$ is uniform on $[-1,1]$. The diphoton invariant mass gives ${\sin }^2(A/2)=1/[{\gamma }^2(1-{\beta }^2{z}^2)]$. Apply the change of variables for probability densities to both signs of $z$. Its minimum-angle peak is an inverse-square-root singularity. At $\gamma =1$, the probability distribution is concentrated at $A=\pi$.

Immanuel Estermann, O. C. Simpson and Otto Stern used gravitational deflection to measure atomic speeds. Stern describes scanning a hot-wire surface ionization detector through a cesium beam and recording ion electric current.^[ref] Their 1947 paper reports potassium and cesium beams; a deficit of slow arrivals increased with oven pressure, supporting a source-slit collision explanation.^[ref]

First consider an original point-launch model, with horizontal speed $v>0$, zero initial vertical velocity, flight distance $L$, and uniform gravitational acceleration $g$. Downward displacement $y$ follows from flight time $t=L/v$:Slow particles fall farther. However, a detector scan counts arrivals per interval of height, not per interval of speed. Change of variables for probability densities supplies the Jacobian determinant:Here $g(v)$ denotes the effusive beam speed distribution; the unadorned constant $g$ in $C$ denotes gravitational acceleration. The integral of $p_y$ is one: substituting $u=b/y$ reduces it to ${\int }_0^{\infty }u{e}^{-u}du=1$. Its mode is $y=b/3$. Simply mapping the most probable arrival speed into a fall distance instead gives $2b/3$. The peaks differ because equal height bins cover unequal speed intervals.

For those chosen parameters, $b\simeq 176.54\mu \mathrm{m}$, so the detector-height probability density peaks near $58.85\mu \mathrm{m}$. A detector covering only $0<y<Y$ accepts the fractionAt $Y=0.80\mathrm{mm}$ this is about $97.9\%$, even before allowing for imperfect detection. Discarding the remainder changes the observed normalization.

Real slits transmit a range of launch heights and vertical velocities. Their geometrical response must be combined with the gravitational deflection and detector width; the point-launch formula is not an exact reduction of the historical apparatus. In the point-launch model, an initial vertical velocity $w$ would add $wL/v$ to $y$. Source collisions, detector efficiency and alignment can also distort a profile. Varying oven pressure helps distinguish such source losses from an incorrect thermal distribution; inferring a source temperature from the peak alone ignores both selection and the detector's coordinate transformation.

The inverse change of variables is $y_1=uv$, $y_2=(1-u)v$, on $0<u<1$, $v>0$. Its absolute Jacobian determinant isThe change of variables for probability densities therefore givesThis factorizes on a product domain into a unit uniform density in $u$ and a shape-two Gamma density in $v$. Thus $U$ and $V$ are independent, with $U\sim \mathrm{Uniform}(0,1)$ and $V\sim \mathrm{Gamma}(2,\lambda )$. This is the shape-$(1,1)$ case of beta-gamma independence.

Set $\alpha =\arctan c\in (0,\pi /2)$. In polar coordinates, the event is $\sin \Theta >c|\cos \Theta |$, the angular sectorBecause $\Theta$ has the continuous uniform distribution on an interval of length $2\pi$, the probability is the sector's angular length divided by $2\pi$:The symmetry of the standard normal distribution in $Y$ makes the lower sector $Y<-c|X|$ equally likely. Their disjoint union is $|Y/X|>c$, apart from $X=0$, an event of probability zero. Therefore the requested cumulative distribution function isThis is the unit-scale half-Cauchy distribution. At $z=0$ there is no atom, and as $z\to \infty$ the cumulative distribution function tends to one.

The left panel uses $c=1$, giving a sector of angle $\pi /2$ and probability $1/4$. The right panel displays the domain $u<v<2u$ for the next change of variables for probability densities; a vertical section through it determines the conditional range.

The joint probability density of the two independent standard normal random variables is $(2\pi {)}^{-1}{e}^{-(x^2+y^2)/2}$. The polar coordinates map has absolute Jacobian determinant $r$, so the change of variables for probability densities givesAt the origin the angle may be assigned arbitrarily: that point has probability zero. The joint probability density factorizes as a Rayleigh distribution density $r{e}^{-r^2/2}$ and a continuous uniform distribution density $1/(2\pi )$. Hence $R$ and $\Theta$ are independent.

For $u>0$, substitute $s=r^2/2$ to findTherefore $R^2\sim \mathrm{Exp}(1/2)$, equivalently ${\chi }_2^2$, and applying a measurable function to $R$ preserves its independence from $\Theta$. Explicitly, with $S=R^2$,Thus $\Theta \sim \mathrm{Unif}[0,2\pi )$ and $S\sim \mathrm{Exp}(1/2)$ independently. This polar decomposition of a planar standard normal vector separates the uniformly distributed direction from its random squared distance.

The inverse linear transformation isThe positive support $x>0,y>0$ is exactly $u>0$ and $u<v<2u$. The independence and rate-one exponential distributions give $f_{X,Y}(x,y)=e^{-x-y}$ there. Since $x+y=u$, the change of variables for probability densities yieldsAlthough the density's value depends only on $u$, its support does not factor as a product of fixed intervals. Thus this expression does not imply independence of $U$ and $V$.

Use the inverse change of variables for probability densitiesIts absolute Jacobian determinant isThereforeSince ${\int }_0^{\infty }w{e}^{-w}dw=1$, its marginal densities are $f_V(v)=1_{(0,1)}(v)$ and $f_W(w)=w{e}^{-w}{1}_{(0,\infty )}(w)$. The joint density factorizes as $f_V{f}_W$, so $V$ and $W$ are independent random variables, with $V$ uniform on $[0,1]$ and $W\sim \mathrm{Gamma}(2,1)$ in the shape-rate convention. This is beta-gamma independence with both shape parameters equal to one.

The support condition $-y<x<y$ forces $y>0$. As a check, the joint probability density integrates toFor any real $x$, the marginal density of $X$ isDividing the joint probability density by this marginal density gives the conditional density:Thus, conditionally on $X=x$, $Y-|x|$ has the rate-two exponential distribution.

For $U=X^2$, use the two inverse branches $x=\sqrt{u}$ and $x=-\sqrt{u}$. Each branch has absolute Jacobian determinant $1/(2\sqrt{u})$. Both lie in the original support exactly when $u>0$ and $y>\sqrt{u}$. The change of variables for probability densities therefore yieldsNo atom of a measure is created at $u=0$, since $\mathbb{P}(X=0)=0$. The value of the probability density function at this boundary may be assigned arbitrarily. Checking the integral over $0<u<y^2$ gives $4y{e}^{-2y}$, which integrates to one.

For a Borel set $A\subseteq \mathbb{R}$, the joint probability density givesFor each $x\ne 0$, substitute $z=xy$. The change of variables for probability densities contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional Lebesgue measure zero, so omitting it changes no probability. By the Tonelli theorem,The bracket is nonnegative and its integral over $\mathbb{R}$ is one. This proves the probability density of a product of random variables formula:As usual, a probability density function is determined only almost everywhere; exceptional infinite values are harmless.

For the independent random variables with the specified uniform distribution, $f(x,y)=b^{-2}$ on $(0,b{)}^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. ThereforeThe value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks normalization of a probability distribution: ${\int }_0^1-\log sds=1$. This is the product of independent uniform random variables law.

The inverse transformation of a random variable is $u=e^{-v}$, with absolute derivative $e^{-v}$. The condition $0<u<b$ becomes $v>-\log b$. The change of variables for probability densities givesEquivalently, $V_n+\log b$ has the exponential distribution of rate one. Applying a separate measurable transformation of a random variable to each independent $U_n$ preserves independence of random variables, so the $V_n$ are independent and identically distributed.

The inverse change of variables is $x=2u+v$, $y=v$, with absolute Jacobian determinant two. Positivity of $x,y$ givesThe change of variables for probability densities givesFor fixed $u$, integrate over $v>\max (0,-2u)$:Thus $\boxed{f_U(u)=\frac{1}{2}{e}^{-|u|}}$, the standard Laplace distribution.

The joint density of $X,Y$ is ${\lambda }^2{e}^{-\lambda (x+y)}$ for positive $x,y$. Inverting the transformation gives $x=uv$, $y=u(1-v)$, with absolute Jacobian determinant $u$ and domain $u>0$, $0<v<1$. The change of variables for probability densities givesThe joint density factors into the gamma density of $U$ and the uniform density of $V$ on $(0,1)$, so these two transformed variables are independent.

By conformal invariance of planar Brownian motion, $w(Z_{T_D})$ has the uniform boundary law from part (a), because the image process starts at zero. A disk automorphism extends to an orientation-preserving boundary diffeomorphism. Choose a continuous angular lift $\varphi :\mathbb{R}\to \mathbb{R}$ satisfyingThe principal-angle version may have a branch jump, which is harmless for the density. The change of variables for probability densities yields the exit-angle density ${\varphi }^{\prime }(\theta )/(2\pi )$. Differentiating the boundary relation givesTaking absolute values proves ${\varphi }^{\prime }(\theta )=|w^{\prime }(e^{i\theta })|$. ThereforeThe angular lift explains rigorously the derivative appearing in the printed hint despite the principal-angle branch cut.

In the center-of-momentum frame, four-momentum conservation gives two opposite photon momenta, each of energy $M_{\pi }/2$. Let $z=\cos {\theta }_{\ast }$, with the axis along the Lorentz boost from that frame to the laboratory. Reversing the axis specified in the source changes $z$ to $-z$ and has no effect on the following product. The Lorentz transformation of four-momentum givesWriting $s=\sin (A/2)$, b now impliesThe isotropy assumption is $dN=a|dz|$ on $-1\le z\le 1$. Each interior opening angle has two inverse branches, $z$ and $-z$, both of which must enter the change of variables for probability densities. The positive branch has derivativeConsequently the diphoton opening-angle distribution isIts support is $2\arcsin (1/\gamma )<A<\pi$ for $0<\beta <1$, and its integral is $2a$, the integral of the original angular probability density before normalization. Thus the normalized probability density is $(2a{)}^{-1}dN/dA$. Omitting one inverse branch would incorrectly halve the total count.

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 $1/4$.

If $X,Y$ are independent exponential random variables with one common rate $\lambda >0$, put $U=X+Y$. The change of variables for probability densities $(x,y)=(u-y,y)$ gives $f_{U,Y}(u,y)={\lambda }^2{e}^{-\lambda u}$ on $0<y<u$. Its marginal density is ${\lambda }^{2}u{e}^{-\lambda u}$, so the conditional probability density of $Y$ given $U=u>0$ is $1/u$ on $(0,u)$. This is the shape-$(1,1)$ case of beta-gamma independence. In particular $Y/U$ is uniform on $(0,1)$ and independent of $U$. Equality of the rates is essential: different rates leave an exponential tilt across the conditional interval.