The chord-length probability density of the regular octahedron

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by  Salvino Ciccariello

The chord-length probability density of the
regular octahedron
Salvino Ciccariello
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Universit`a di Padova,Dipartimento di Fisica G. Galilei
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2 Via Marzolo 8, I-35131 Padova,Italy
[email protected]
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F January 16, 2014
revised February 8, 2014
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]
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p Abstract
-
h
t The chord length probability density of the regular octahedron is ex-
a
m plicitly evaluated throughout its full range of distances by separating
it into three contributions respectively due to the pairs of facets op-
[
posite to each other or sharing an edge or a vertex.
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1 Introduction
Small-angle scattering1 3 (SAS) is a powerful tool to get information on the
−
interface size andshapeofasample. Infact, theobserved scattering intensity
is the square modulus of the Fourier transform (FT) of n(r), the scattering
density of the sample, or, equivalently, the FT of the convolution of n(r) by
itself. In the SAS realm n(r) can fairly be approximated by a discrete value
function that takes only two values in most of the cases. This approximation
implies that the sample is either a bi-continuous or a particulate system.
Confining ourselves to the second case, if one further assumes that particles
have the same shape and size, that are isotropically distributed and that
their number density is small (conditions fairly met in the case of biological
samples), thenthescatteringdensityisproportionaltotheFToftheisotropic
correlation function of the particle defined as
γ(r) (1/4πV) dωˆ ρ (r )ρ (r +rωˆ)dv . (1)
V 1 V 1 1
≡
Z Z
Here ρ (r) denotes the characteristic function of set (having volume V)
V
V
occupied by the particle (i.e. it is defined as being equal to one inside the
particle and to zero elsewhere). Further, the inner integral is performed over
(or the all space) and the outer integral over all the directions of the unit
V
vector ωˆ.
As first pointed out by Debye et al.4, the right hand side of (1), suitably
scaled, can be interpreted as the probability density that by randomly toss-
ing a stick of length r both ends of the stick fall within the particle. This
remark shows the stochastic meaning of γ(r). Consequently, SAS theory is
intimately related to stochastic geometry5 as well as to integral geometry6
that aims to get general properties investigating suitable integrals over the
particle volume or surface. Interestingly, the derivatives of γ(r) can be ex-
pressed as integrals over the particle surface7,8 and, in the case of convex
particles, the first and second derivatives (after being appropriately scaled)
can be interpreted as the probability densities for respectively finding the
stick with one end or with both ends on the particle surface. The investiga-
tions of these integral relations yield some general result as the Porod9 and
the Kirste-Porod10 relations as well as the particle surface features that yield
discontinuous γ′′(r)s11 13.
−
These considerations illustrate the importance of determining the explicit
′ ′′
expressions of γ(r), γ (r) or γ (r) for specific particle shapes. In fact, these
expressions were obtained in the cases of spheres2, cubes14, cylinders15, right
parallelepipeds16, rotational ellipsoids17 and regular tetrahedrons18.
2
C
E
F
B
A
D
Figure 1: The regular octahedron.
This paper goes one step further since it determines the chord-length proba-
bility density (r) of another Platonic solid: the regular octahedron (see Fig.
C
1).
2 Basic integral expression
In order to evaluate (r) one starts from the general integral expression of
γ′′(r) obtained by CicCcariello et al.7, namely
1
′′
γ (r) = dωˆ dS dS (νˆ (r ) ωˆ)(νˆ (r ) ωˆ)δ(r +rωˆ r ). (2)
1 2 1 1 2 2 1 2
−4πV · · −
Z ZS1 ZS2
Here S1 = S = S and V respectively denote the surface and volume of
2
the octahedron, νˆ (r ) [νˆ (r )] the unit normal (pointing outwardly to the
1 1 2 2
octahedron) to the infinitesimal surface element dS [dS ] of S, located at
1 2
the point with position vector r [r ]. The Dirac function δ( ), present in (2),
1 2
·
requires that the distance between dS and dS be equal to r( 0) since ωˆ
1 2
≥
denotes a unit vector which, according to the leftmost integral of (2), ranges
over all the possible directions. Thus, at fixed dS , the integrals over dωˆ and
1
dS amounts to integrating over the curve resulting from the intersection of
2
the sphere of radius r and centered at dS with S.
1
To explicitly evaluate expression (2) it is convenient to account for the fact
that S is formed by eight triangular faces. Thus, expression (2) reduces to
a sum of terms having still the form of (2) but with the important changes
3
that integration domains S and S are two facets of the octahedron. Clearly
1 2
the only cases where S and S are different is important, otherwise the
1 2
correspondent integrand vanishes because ωˆ is perpendicular both to νˆ and
1
νˆ . Then, for each couple (S , S ), the facets share an edge or a vertex or, in
2 1 2
the negative case, they lie oppositely and the associated planes are parallel.
In the last case, for conciseness, the facets will be said parallel. For each
couple (S , S ), expression (2) defines a scalar r-function. Hence, whatever
1 2
′′
S and S , expression (2) takes the same form, denoted by γ (r), if S
1 2 E 1
and S share an edge. The same happens if S and S share a vertex or
2 1 2
′′
are parallel and the corresponding integrals will be denoted by γ (r) and
V
′′
γ (r), respectively. Thus, it results that
P
′′ ′′ ′′ ′′
γ (r) = 24γ (r)+24γ (r) +8γ (r), (3)
E V P
′′ ′′ ′′ ′′
and the evaluation of γ (r) reduces to that of γ (r), γ (r) and γ (r).
E V P
The evaluations will be carried out in the following three sections assuming
that the edge length ℓ of the octahedrons be equal to one. This assumption
′′
is by no way restrictive. In fact, if one respectively denotes by γ (r) and
ℓ
′′
γ (r) the functions relevant to the octahedrons with edges equal to ℓ and 1,
one has γ′′(r) = (1/ℓ2)γ′′(r/ℓ).
ℓ
3 Evaluation of γ ′′(r)
E
′′
The geometrical configuration relevant to γ (r) is shown in Fig. 2. Surfaces
E
S and S respectively are the regular triangles ABC and ABD with unit
1 2
sides. It is now put
β = BAˆC = π/3, α = DOˆC = arccos( 1/3), α = π α, (4)
c
− −
√3 2
H = OC = OD = , h = D D = Hsin α = , DC = √2. (5)
0 c
2 3 | |
r
The minimax distances11 are
d = h, d = H, d = 1 and d = √2. (6)
1 2 3 4
In terms of the axes shown in Fig. 2, one has
dS = dxdz and dS = dY dZ. (7)
1 2
The integration domain of dS (i.e. the triangle ABC) is defined by the
1
inequalities
0 < x < H and L (x) < z < L (x), (8)
E E
−
4
Y
y
D
‡
B
Α
Do O Β ¯ C x
A
z=Z
Figure 2: Cartesian frames used for evaluating the chord-length probability
density (CLPD) contribution due to a couple of plane facets with a common
edge.
and that of dS by
2
0 < Y < H and L (Y) < Z < L (Y), (9)
E E
−
with
1
L (x) z xctgβ = xctgβ. (10)
E A
≡ − 2 −
Further νˆ = (0, 1,0) and νˆ = ( sinα,cosα,0). With ωˆ = (ωˆ ,ωˆ ,ωˆ ) =
1 2 x y z
− −
(cosϕ sinθ, sinϕ sinθ, cosθ) one finds that
νˆ ωˆ = sinθ sinϕ and νˆ ωˆ = sinθ sin(α ϕ). (11)
1 2
· − · − −
Since rωˆ joins a point of ABC to a point of ABD, variable ϕ varies within the
′
interval [α, π]. Then, putting ϕ ϕ +(α+π)/2, itfollowsthat (π α)/2 <
′ ≡ ′ − −
ϕ < (π α)/2 and, hereafter denoting ϕ as ϕ for notational simplicity, one
−
gets
A (θ,ϕ) (νˆ ωˆ)(νˆ ωˆ) sinθ = sin3θ cos(ϕ+α/2) cos(ϕ α/2), (12)
E 1 2
≡ · · − −
In terms of the present variables the Dirac function requires that
Z = Z¯(r,z,θ) z +r cosθ,
≡
Y = Y¯(r,θ,ϕ) r cos(ϕ+α/2)sin(θ)/sinα, (13)
≡
¯ ¯
x = x¯(r,θ,ϕ) Y(r,θ,ϕ)cosα+r sin(ϕ+α/2) sinθ = Y(r,θ, ϕ).
≡ −
5
Thex, Y andZ integralsareimmediatelyperformedand, takingintoaccount
that the Jacobian related to the Y variable is 1/sinα, one finds that
γ ′′(r) = 1 πdθ π−2α A (θ,ϕ)Θ (x¯,Y¯)dϕ LE(x¯) Θ (Z¯)dz,
E E E z
−4πV sinα Z0 Z−π−2α Z−LE(x¯)
(14)
where function Θ (x¯,Y¯), depending on θ and ϕ, is equal to one if x¯ and
E
¯ ¯
Y obey inequalities (8a) and (9a) and to zero elsewhere. Similarly Θ (Z) is
z
equal to one or zero depending on whether Z¯ obeys inequalities (9b) or not.
According to (13a) Z¯ linearly depends on z. Then, the above z-integral is
equal to
F (r,θ,ϕ) min[L (x¯), L (Y¯) rcosθ] max[ L (x¯), L (Y¯) rcosθ] =
E E E E E
≡ − − − − −
min[L (x¯), L (Y¯) rcosθ]+min[L (x¯), L (Y¯)+rcosθ] (15)
E E E E
−
if the above quantity is positive otherwise it is equal to zero. It is now
observed that each of the following transformations θ (π θ) and ϕ
ϕ interchanges x¯ with Y¯ and leaves expressions (12)→and −(15) invarian→t.
−
Thus, (14) becomes
1 π/2 (π α)/2
′′ −
γ (r) = dθ A (θ,ϕ)F (r,θ,ϕ)dϕ, (16)
E E E
−πV sinα
Z0 Z0
with the constraints
¯
0 < x¯(r,θ,ϕ) < H, 0 < Y(r,θ,ϕ) < H, F (r,θ,ϕ) > 0 (17)
E
that will generally reduce the integration domain
0 < θ < π/2, 0 < ϕ < (π α)/2 (18)
E
D ≡ { − }
toasmaller one. Alltheconstraints mustexplicitly bereducedinordertoget
′′
the explicit expression of γ (r). To this aim we start from the constraints
E
implicit in definition (15). Putting
µ (θ,ϕ) 2ctgβ sinθ cos(ϕ α/2)/sinα (19)
± ≡ ±
one easily shows that F (r,θ,ϕ) becomes
E
r
F (r,θ,ϕ) = 1 max[µ +cosθ, µ cosθ]+max[µ cosθ, µ +cosθ] .
E + +
−2 −− − −
(20)
(cid:0) (cid:1)
The condition (µ + cosθ) > (µ cosθ) is fulfilled within the sub-
+ E
− − D
domain
¯
0 < ϕ < (π α)/2, 0 < θ < θ (ϕ) arcotg(sinϕ) , (21)
E,A E
D ≡ { − ≡ }
6
and (µ +cosθ) > (µ cosθ) throughout . Then, from equations (20)
+ E
− − D
and (4) follows that F (r,θ,ϕ) takes the forms
E
sinθ cosϕ
F (r,θ,ϕ) 1 r cosθ+ if (θ,ϕ) , (22)
E,A E,A
≡ − √2 ∈ D
(cid:0) (cid:1)
and
F (r,θ,ϕ) 1 r 3/2 sinθ cos(ϕ α/2) if (θ,ϕ) , (23)
E,B E,B
≡ − − ∈ D
p
being defined as the complementary domain of in .
E,B E,A E
D D D
3.1 Constraints and integration domains
Before proceeding to the integral evaluation one must still reduce inequalities
(17a), (17b) and (17c). The last requires that F > 0 within and that
E,A E,A
D
F > 0within . Theinequalitiesdependonr andthesamewillhappen
E,B E,B
D
for the associated domains that will be determined confining ourselves to
E
and to the non-trivial r-domain [0, √2] of γ′′(r). D
For inequalities (17a) one finds that the associated domain (r) is
E,x
D
(r) = if 0 < r < √3/2, (24)
E,x E
D D
(r) = (r) (r) if √3/2 < r < √2, (25)
E,x E,x,1 E,x,2
D D ∪D
with
¯
(r) 0 < θ < π/2, 0 < ϕ < φ (r) , (26)
E,x,1 E,x
D ≡ { }
¯ ¯
(r) φ (r) < ϕ < (π α)/2, 0 < θ < θ (r,ϕ) , (27)
E,x,2 E,x E,x
D ≡ { − }
and
¯
φ (r) α/2 arcos( 2/3 r), (28)
E,x
≡ −
¯
θE,x(r,ϕ) arcsin 2/3p[rcos(cid:14)(ϕ α/2)] . (29)
≡ −
(cid:0)p (cid:14) (cid:1)
Inequalities (17b) are fulfilled throughout if 0 < r < √2.
E
D
Inequality F (r,θ,ϕ) > 0 coincides with x¯(r,θ,ϕ) < H and is therefore
E,B
obeyed throughout .
DE,FB ≡ DE,x
The reduction of the last inequality F (r,θ,ϕ) > 0 yields the domain
E,A
(r) defined as
DE,FA
(r) if 0 < r < 2/3, (30)
DE,FA,a ≡ DE,A
p
7
to
(r) 0 < θ < (r), 0 < ϕ < (π α)/2 ,
E,A,1 E,1
D ≡ { L − }
(r) = (r) (r) < θ < (r), φ¯ (r,θ) < ϕ < (π α)/2
DE,FA,b DE,A,2 ≡ {LE,1 LE,2 E,A − }
 (r) (r) < θ < π/2, 0 < ϕ < (π α)/2 ,
E,A,3 E,2
D ≡ {L − }
(31)

if 2/3 < r< √3 2, to
p (cid:14)(r),
E,A,1
D
′ ¯
(r) (r) < θ < (r), φ (r,θ) < ϕ < (π α)/2 ,
(r) = D E,A,2 ≡ {LE,1 LE,3 E,A − }
DE,FA,c D′E,A,3(r) ≡ {LE,4(r) < θ < LE,2(r), φ¯E,A(r,θ) < ϕ < (π −α)/2},
′
(r) (r) < θ < π/2, 0 < ϕ < (π α)/2
E,A,4 E,2
D ≡ {L − }
 (32)


if √3 2 < r < 1, and to
(cid:14) ′′ ¯
(r) (r) < θ < (r), φ (r,θ) < ϕ < (π α)/2 ,
E,A,1 E,4 E,2 E,A
(r) = D ≡ {L L − }
DE,FA,d ( ′′E,A,2(r) E,2(r) < θ < π/2, 0 < ϕ < (π α)/2
D ≡ {L − }
(33)
if 1 < r < √2. The involved (r) functions are defined as follows
..
L
¯ ¯
(r) θ (r) α/2, (r) π α/2 θ (r) (34)
E,1 E,A,1 E,2 E,A,1
L ≡ − L ≡ − −
¯ ¯
(r) θ (r) π/3, (r) 2π/3 θ (r) (35)
E,3 E,A,2 E,4 E,A,2
L ≡ − L ≡ −
with
¯
θ (r) arcsin( 2/3 r), (36)
E,A,1
≡
θ¯ (r) arcsin(p√3 2(cid:14)r), (37)
E,A,2
≡
φ¯ (r,θ) arccos √2(1 rco(cid:14)sθ) (rsinθ) . (38)
E,A
≡ −
(cid:0) (cid:14) (cid:1)
′′
3.2 γ (r) expression
E
′′
The γ (r) expression is given by the sum of the following two expressions
E
1
′′
γ (r) A (θ,ϕ)F (r,θ,ϕ)dθdϕ (39)
E,A E E,A
≡ −πV sinα
Z¯E,A(r)
D
1
′′
γ (r) A (θ,ϕ)F (r,θ,ϕ)dθdϕ (40)
E,B E E,B
≡ −πV sinα
Z¯E,B(r)
D
with
¯ (r) (r) (r) (41)
DE,A ≡ DE,FA ∩DE,x ∩DE,A
¯ (r) (r). (42)
E,B E,B E,x
D ≡ D ∩D
8
j j
0.6 0.6
aL bL
0.5 DE,B=DB 0.5 DE,A DE,B
0.4 0.4
0.3 DE,A=DA GR 0.3 GFA
0.2 0.2 GR
0.1 0.1
0.0 Θ0.0 Θ
0.0 0.5 1.0 1.5 0.0 0.5 1.0 1.5
j j
0.6 0.6
cL dL
G
0.5 X 0.5
V
0.4 A D 0.4 DE,B¯
E, E,B
0.3D 0.3
0.2 A 0.2
0.1 GFA GFA DE, 0.1 DE,A®
0.0 Θ0.0 Θ
0.0 0.5 1.0 1.5 0.0 0.5 1.0 1.5
Figure 3: The four panels shows the shapes of the integration domains
¯ (r)and ¯ (r)inthefourr-ranges: a = [0, 2/3], b = [ 2/3, √3 2],
E,A E,B
D D
c = [√3 2, 1] and d = [1, √2]. The ¯ (r) domains are bounded by thick
DE,A p p (cid:14)
blue lines and the ¯ (r)s by red continuous ones. The dotted thick curve
E,B
(cid:14) D
Γ separates domain from and is defined by Eq. (21). The bell
R DE,FA DE,FB
shaped curve Γ is related to the condition F > 0 [see Eq.s (31)-(33)]
FA E,A
and curve Γ to the boundary of [see Eq.s (24)-(29)].
x E,x
D
The shapes of these domains are shown in Fig. 3 in the four ranges of
distances a, b, c and d, defined inthecaption. The evaluationof theintegrals
is a long task that was made possible by the MATHEMATICA software. By
9
′′
these results the final γ (r) expression is
E
2√2 π +α (18+(7√3 9)π)r
′′
γ (r) − − , (43)
E,a
≡ 8π − 96√2π
1 1 4+√2(π +α) (18 (9 13√3)π)r
′′
γ (r) + − − , (44)
E,b
≡ 12√6r3 − 2√6r 8√2π − 96√2 π
1
γ ′′(r) 4√3π 36(4+√2α)r3+3(18 9π +10√3π)r4 +
E,c
≡ −288√2πr3 − − −
(cid:20)
9r2 7
6(25r2 6)R (r)+8√3arcsin − +
− 34 2R3 (r)
23
h i
1 r
96√3r2arcsin 72√2r3arcsin (45)
2R (r) − √3R (r) −
23 23
h i h i
√3(27 90r2 +96r4 34r6 +2r8)
18√3r4arcsin − − ,
2r7R (r)
23 (cid:21)
h i
1
γ ′′(r) 16√2+8√2(3+√3π)r2 12πr3 +
E,d ≡ −192πr3 − −
(cid:20)
3√2(4√3 3)πr4 4√2(5r2 2)R (r)
11
− − − −
4+4r2 7r4 R (r)
24r3arcsin − +36√2 r4arcsin 11
R4 (r) r −
23
h i h i
1+3R (r)
8√6r2(2+3r2)arcsin 11 . (46)
2R (r)
23 (cid:21)
h i
Here suffices a, b, c and d specify the r-range where the expressions apply
and the following definitions have also been used
R (r) = √r2 1, R (r) = √3r2 2, R (r) = √4r2 3. (47)
11 23 34
− − −
4 Evaluation of γ ′′(r)
V
One passes now to evaluate the CLPD relevant to a pair of facets sharing a
vertex. The configuration is shown in Fig. 4 that also shows the Cartesian
framesusedtoworkouttheCLPDexpression. Theevaluationproceedsalong
the same route described in section 3. The integration domains S = ABC
1
and S = CEF are defined by the inequalities
2
0 < x < H, L (x) < z < L (x) (48)
V V
−
and
0 < Y < H, L (Y) < Z < L (Y) (49)
V V
−
10