# probability of a line not crossing spheres

This question in geometric probability was inspired by a problem in radiation physics: What is the probability that a "ray" hits an atom in a crystal, or rather the probability that the ray goes through the lattice without collision?

We will idealize a "ray" to be a straight line and the atoms to be identical size non-penetrable perfect spheres of common radius $$r$$. Of course, the answer will also depend on the size of the crystal and its shape but first let me ask of the special case when it is infinite in all direction. It is easy to find the condition of non-collision for an idealized lattice but my question is more complicated as it involves two stages of randomization.

(1) From one fixed unit cell we take a point $$\mathcal {C} = \mathbf{c}$$ and a cone of unit length direction vectors $$\hat{\mathbf{a}}$$ pointing from $$\mathcal C$$ so that the area of their projected patch on a unit sphere is uniform. (This is a model of a radiation source whose emission is uniform with a cone; if another prior distribution is more convenient that would be grate, too.)

(2) Now displace each spheres that are centered in the lattice points $$\mathbf{m}$$ by a random amount $$\mathbf{p}_m$$ so that they are centered $$\mathbf{m}+\mathbf{p}_m$$. Let us assume, for simplicity, that the displacements are $$\mathbf{p}_m$$ i.i.d random variates from some common distribution but one can imagine correlated displacements, too. This would be the simplest model of thermal vibrations so the distribution of $$\mathbf{p}_m$$ reflects the temperature of the crystal, and just assume that these are $$\mathcal{N}(0,\sigma^2)$$ for some given $$\sigma$$.

For further simplification I would just like to get an estimate for a randomized snapshot, in other words the ray moves infinitely fast through the crystal and the random samples are static while the ray passes through.

My question: How does the non-collision probability depend on $$\sigma$$ and $$r$$?

Take a cubic integer lattice at points $$\mathbf{m} = (m_1, m_2, m_3)$$, $$m_k \in \mathbb{Z}$$, and at each lattice point, place a sphere of radius $$r<1$$. Draw a line $$\mathcal {L}[\mathbf{c}, \hat{\mathbf {a}}]$$ in the direction of the unit vector $$\hat{\mathbf {a}}=(\mathrm{cos}\alpha_1, \mathrm{cos}\alpha_2,\mathrm{cos}\alpha_3)$$ through a point $$\mathbf{c}$$.

The distance $$d_m$$ of the lattice point $$\mathbf{m}$$ to the line $$\mathcal {L}[\mathbf{c}, \hat{\mathbf {a}}]$$ is $$d_m=|\mathbf{m-c}|\mathrm{sin}(\beta_m)$$ where $$\beta_m$$ is defined as the angle between $$(\mathbf{m-c})$$ and the line $$\mathcal {L}[\mathbf{c}, \hat{\mathbf {a}}]$$, so that $$(\mathbf{m-c})\cdot \hat{\mathbf {a}} = |\mathbf{m-c}|\mathrm{cos}(\beta_m)$$.

The line $$\mathcal {L}$$ will not cross the sphere centered at $$\mathbf{m}$$ if and only if $$r, that is, $$r<|\mathbf{m-c}|\mathrm{sin}(\beta_m)$$ which can be written as $$r^2<|\mathbf{m-c}|^2-|(\mathbf{m-c})\cdot \hat{\mathbf {a}}|^2\tag{1}\label{1}$$or expressed in coordinates $$r^2<\sum_k (m_k-c_k)^2- \big|\sum_k(m_k-c_k)\mathrm{cos}(\alpha_k)\big|^2 \tag{2}\label{2}$$

Given some range of the directions $$\hat {\mathbf{a}}$$ and starting points $$\mathbf{c}$$ of the line $$\mathcal {L}[\mathbf{c}, \hat{\mathbf {a}}]$$ by $$\alpha'_\ell < \alpha_\ell <\alpha''_\ell$$ with $$\ell=1,2$$ and $$\mathrm{cos}^2(\alpha_1)+\mathrm{cos}^2(\alpha_2)+\mathrm{cos}^2(\alpha_3)=1$$, and $$0 with $$k=1,2,3$$ I would like to know what is the $$probability$$ of non-crossing for all $$\mathbf{m}$$ if we are given some distribution over the ranges of the uniform random variates $$c_k$$ and the uniform random $$direction$$ defined by the range of the angles $$\alpha_{\ell}$$.

Next stage is to randomize the centers with $$\mathbf{p}_m$$ and then ask what is the non-collision probability?

You mention the context of radiation physics. In that field an “interaction cross section,” which has units of area, is typically measured in “barns,” where $$\rm 1\,barn=(10\,fm\times10\,fm)=10^{-28}\,m^2$$ is substantially larger than the physical size of a nucleus. However the distance between scattering centers is the atomic scale, $$\rm1\,Å=10^{-10}\,m$$.
Let’s consider a lattice with number density $$n$$ scatterers per unit volume, where each scatterer has an interaction cross section $$\sigma$$. Consider a thin slice of the material with thickness $$\ell$$. The number of scatterers per unit area is $$n\ell$$, so the fraction of the slice which is occluded is $$\sigma n\ell$$ and the probability your ray reaches the other side of the slice is $$(1-\sigma n \ell)$$. In a symmetric material with $$n\approx\rm1\,Å^{-3}$$ a single layer would have $$\ell\approx\rm1\,Å$$, so barn-scale interactions give an interaction probability $$\sigma n \ell \approx 10^{-8}$$ for each layer.
If you traverse many slices with total thickness $$L=N\ell$$, the transmission probability is the product of the transmission probabilities for each thin layer:
$$T(L) = \lim_{N\to\infty}\left(1 - \frac{\sigma n L}{N}\right)^N = e^{-\sigma n L}$$
• Thank you. I do see that $nuclear$ interaction must be insignificant but I was asking about $ionizing$ interaction specifically; to my limited understanding the size of the atom is then a more important parameter than that of the nucleus. For Si the lattice spacing is ~540pm, nearest neighbor distance ~235pm (300K) while an atom's covalent radius is ~111pm. I was wondering if the crystal vibrations were smaller (say, cryogenically cooled) would there be measurable reduction of the composite (effective) cross section of the $whole$ crystal in ionizing radiation? – hyportnex Apr 23 at 17:03
• This is how you compute significant interactions: just extend to $L > 1/n\sigma$. For electrons in silicon the scale of the total cross section is $a_0^2$, where $a_0 \approx 50\rm\,pm$ is the Bohr radius. (Source.) If $1/n\ell$ is roughly the square of the lattice spacing, then $a_0^2 n\ell \approx 1/100$ and this argument is still relevant. Furthermore, at high energies most ionization radiation scatters in the forward direction and doesn't leave the beam, so questions about opacity would want to consider somewhat less than the total cross section. – rob Apr 23 at 19:55