A path-connected region is multiply connected when some closed path cannot contract to a point within the region. An annulus and the complement of a solid torus are examples. The accessible region in the ideal magnetic Aharonov-Bohm effect has this obstruction. Such a region is not a simply connected space.

For orbital angular momentum $\ell =1$ and spin $s=1/2$, considerThe quantum number $m_j=m_{\ell }+m_s$ is conserved. The stretched states with $m_j=\pm 3/2$ have energies $K\pm 2a$. The remaining eigenvalues follow from two $2\times 2$ blocks with off-diagonal entries $\sqrt{2}K$:The subscripts denote $m_j=\pm 1/2$. For $K>0$, the zero-field multiplets are $j=3/2$ at $K$ and $j=1/2$ at $-2K$. Their weak-field Zeeman splitting is $g_{j}a{m}_j$, with Landé g-factors $4/3$ and $2/3$, respectively. When $a\gg |K|$, the Paschen-Back effect gives approximate product states $(m_{\ell },m_s)$ and energies $a(m_{\ell }+2m_s)+2K{m}_{\ell }{m}_s$. Mixing within a fixed $m_j$ causes an avoided crossing; different $m_j$ sectors can cross because the Hamiltonian does not couple them.

The particle flux of neutrinos through a specified surface; an incident-energy distribution may also be specified.

In diffusion-limited cluster aggregation, mobile clusters undergoing Brownian motion attach with probability close to one when they encounter another cluster. Irreversible first-contact sticking tends to trap particles in exposed outer branches, producing open aggregates. In the common three-dimensional isotropic cluster model, the mass fractal dimension is roughly $1.8$. This is different from depositing individual random walkers onto one fixed cluster. Rotational motion, anisotropic attraction and bond restructuring can change the morphology.

In reaction-limited cluster aggregation, encounters occur frequently compared with successful attachment: an activation or interaction barrier gives a small sticking probability. Repeated unsuccessful approaches allow clusters to sample more relative configurations before bonding, commonly producing denser aggregates than diffusion-limited cluster aggregation. Typical three-dimensional isotropic irreversible cluster models give mass fractal dimension around $2.1$. This structural contrast does not require previously formed bonds to break and restructure; it can arise from unsuccessful encounters alone. ^[ref]

Poisson's equation is $\Delta u=f$, where $\Delta$ is the Laplace operator and $f$ a prescribed source. It generalizes the Laplace equation, recovered when $f=0$. Boundary conditions are needed to determine a unique solution; Neumann boundary conditions can leave an additive constant.

For electrostatics in vacuum, the convention is $\Delta V=-\rho /{ϵ}_0$. For Newtonian gravity, $\Delta \Phi =4\pi G\rho$. The opposite signs reflect the specified force and potential conventions. A simple mathematical example is $u(x)=x^2/2$ on $[0,1]$, which solves $u^{\prime \prime }=1$ with endpoint values zero and $1/2$. On a periodic domain the source must have zero mean, by integrating the equation.

For a state of integer total angular momentum $J$, natural parity means $\pi =(-1{)}^J$: $0^{+},1^{-},2^{+},3^{-},\dots$. The other sign is called unnatural parity. The terminology is useful when a spin-zero positive-parity parent emits a spin-zero positive-parity particle: angular momentum conservation fixes relative orbital angular momentum $L=J$, and parity conservation then forces natural parity for the daughter.

For a prime number $p$ and integers $0\le k\le n$, the p-adic valuation of the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ equals the number of arithmetic carries in the base-$p$ addition of $k$ and $n-k$.^[ref] Indeed, subtract the two denominator factorial valuations in the Legendre formula:Each arithmetic carry accounts for one unit of this quotient by subadditivity of the base-p digit sum. In particular, the binomial coefficient is divisible by $p$ exactly when at least one carry occurs.

For $n=50,k=17,p=3$, the factorial counts giveThe matching base-p expansions are $17=(0122{)}_3$ and $33=(1020{)}_3$. Their units add to two; the next column gives $2+2=4=1+3$, sending one carry into the nines column. That column then gives $1+0+1=2$, and no further carry occurs. The result is $(1212{)}_3=50$. The base-p digit sums check the same result: $(5+3-6)/(3-1)=1$. Consequently $\left(\genfrac{}{}{0ex}{}{50}{17}\right)$ is divisible by three but not by nine.