Isotropic projection fractions

Geometric preparation

Let $v$ be a uniformly distributed unit vector in a $D$-dimensional Euclidean spatial space, and $P$ the orthogonal projector onto a fixed $q$-dimensional subspace, $0<q<D$. The squared projected fraction is

$$F=\|Pv\|^2\in[0,1].$$

Uniformity is a preparation assumption. In a scattering application the interaction, incoming state, and detector acceptance determine whether that assumption is justified. The subspaces here are ordinary spatial directions within the single nested physical manifold.

Distribution and moments

Take independent standard real Gaussian coordinates $z_1,\ldots,z_D$. Their joint density is rotation invariant, so $v=z/\|z\|$ is uniform on the unit sphere. In a basis adapted to $P$, put

$$X=\sum_{j=1}^q z_j^2,\qquad Y=\sum_{j=q+1}^D z_j^2,\qquad F=\frac{X}{X+Y}.$$

$X$ and $Y$ are independent gamma variables of common scale $2$ and shapes $a=q/2$, $b=(D-q)/2$. Change variables to $F$ and $S=X+Y$, whose Jacobian is $S$. Their density factors into a gamma density for $S$ and

$$\boxed{f_F(t)=\frac{t^{a-1}(1-t)^{b-1}}{B(a,b)}, \quad 0<t<1,\quad B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.}$$

Integration gives

$$\mathbb E[F^k]=\frac{(q/2)_k}{(D/2)_k},\qquad \mathbb E[F]=\frac qD,\qquad \operatorname{Var}(F)=\frac{2q(D-q)}{D^2(D+2)}.$$

Here $(a)_k=a(a+1)\cdots(a+k-1)$. The beta-integral formula also holds for real $k>-q/2$. Thus a magnitude fraction has $\mathbb E[\sqrt F]=B(a+1/2,b)/B(a,b)$, which generally differs from $\sqrt{q/D}$. The endpoint cases $q=0,D$ are deterministic.

Beam-relative example in six ordinary spatial dimensions

Suppose a beam direction lies within the observer's three-dimensional spatial subspace of an electron's six dimensions. Its orthogonal spatial complement has $D=5$ dimensions. Two of them lie in the observer's transverse plane and three are perpendicular to the observer's subspace. If the transverse direction is isotropic in that five-dimensional space, the fraction perpendicular to the observer's subspace has

$$F\sim\operatorname{Beta}(3/2,1),\qquad \mathbb E[F]=\frac35,\qquad \operatorname{Var}(F)=\frac{12}{175},\qquad \Pr(F\leq t)=t^{3/2}.$$

The electron has six equivalent ordinary spatial dimensions. The grouping above is relative to the beam and observer, with no inherent electron split. The $3/5$ result concerns squared transverse momentum under the specified preparation. It supplies neither a scattering rate nor an energy threshold. Opposing momentum components can balance within a multi-particle event even when each has a nonzero perpendicular component.

Projected mass shell

If a particle satisfies the relativistic mass shell in the containing spatial sector, decompose its momentum relative to the observer as $p=p_{\parallel}+p_{\perp}$. Orthogonality gives

$$E^2=c^2(|p_{\parallel}|^2+|p_{\perp}|^2)+m^2c^4.$$

An observer reconstruction using the full particle energy $E$ and only $p_{\parallel}$ would therefore assign

$$M_{\rm app}^2=\frac{E^2}{c^4}-\frac{|p_{\parallel}|^2}{c^2} =m^2+\frac{|p_{\perp}|^2}{c^2}.$$

The qualification that $E$ is the full particle energy matters: a partial energy deposit has its own detector model. This equation is kinematic; whether a process populates $p_{\perp}$ requires its interaction.

Verification and related

spectral_analysis_checks.py checks beta moments by quadrature, the isotropic Gaussian construction, the special cumulative distribution, and the projected mass shell under rotations and changes of observer frame.

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