Abstract
When something is sent at a hard sphere of radius a, the size of the target (the cross section) takes four values depending on the setting. For point particles it is πa², for a wave of short wavelength 2πa², for a wave of long wavelength 4πa², and for two identical bosons 8πa². This paper obtains all four from one calculation (the partial-wave sum) and counts where each of the factors 2, 4 and 8 comes from. No new theorem or law is claimed.
First, the πa² for particles is the area of a circle and, at the same time, 4π×(a/2)². A sphere that reflects like a mirror scatters particles with the same strength in every direction. When the area that is hit is sorted into ten bins by the cosine of the outgoing direction, every bin receives $\mathbf{0.100000}$, and the strength per solid angle is a²/4 everywhere.
Second, for a wave the value moves with the ratio of wavelength to radius. With k the wavenumber and a the radius, the cross section in units of πa² is $\mathbf{3.999999}$ at ka = 0.001, $\mathbf{3.382438}$ at ka = 1 and $\mathbf{2.006806}$ at ka = 5000. The approach to 2 goes as the −2/3 power of ka, with a coefficient of $\mathbf{1.985245}$ at ka = 1000 and $\mathbf{1.989941}$ at ka = 5000.
Third, the 4 at long wavelength is the full solid angle 4π multiplied by a². At long wavelength the wave is scattered equally in all directions, a² per solid angle. That is $\mathbf{3.999999}$ times the a²/4 of the particle case, a factor of 2 in amplitude. 4π×a² is the same number as the surface area of a sphere of radius a, and the fact that the surface of a sphere is four times its great circle (Archimedes) shows up here as the ratio of 4πa² to πa².
Fourth, the 2 at short wavelength is πa² of reflection plus πa² of shadow. At ka = 200 the amount scattered into the backward half is $\mathbf{0.500065}$ times πa², and away from the forward direction the strength per direction is between $\mathbf{1.000042}$ and $\mathbf{1.003550}$ times the a²/4 of the particle case: the same reflection as for particles is still there. The rest sits in a narrow forward peak whose strength straight ahead is $\mathbf{42443.168}$ times that of the reflection. This peak is the wave needed to make the shadow behind the sphere, and it has the same shape as the diffraction by a disc of the same size.
Fifth, the 8 for identical particles is the solid angle of a hemisphere, 2π, multiplied by (2a)². When the two particles cannot be told apart, the wave scattered to the right and the wave scattered to the left are added before squaring. At long wavelength the result is $\mathbf{7.999997}$ for bosons and $2.666663$×10⁻¹² for fermions with aligned spins, with no intermediate value between these and the $\mathbf{3.999999}$ of distinguishable particles. At ka = 1000 both are back within 0.1 percent of the distinguishable value.
Sixth, and this is the core of the paper: when the dimension is changed, the factors of 2 remain and the 4 does not. Repeating the calculation in two to six dimensions, the short-wavelength value tends to twice the frontal area in every dimension (from $\mathbf{2.009960}$ to $\mathbf{2.049127}$ at ka = 1000), and identical bosons at long wavelength give $\mathbf{2.000000}$ times the distinguishable value in every dimension. But the long-wavelength value itself is proportional to the wavenumber to the power d−3: it goes to 0 in four or more dimensions and keeps growing in two. Only in three dimensions does it settle to a number independent of wavelength, and that number is the same as the surface area of the sphere.
To be candid, one check failed once. Before measuring, the short-wavelength value in each dimension had been set to lie “between 2.00 and 2.03”, but the measured value rises with dimension and was 2.049127 in six dimensions. Reading the failure showed that the coefficient of the approach to 2 is nearly proportional to d−1 (divided by d−1 it runs from $\mathbf{0.497983}$ to $\mathbf{0.491265}$).
Keywords: scattering cross section, hard sphere, partial waves, shadow scattering, identical particles, dimension, solid angle.
Scope (what is not claimed)
No new theorem or law is claimed. The partial-wave solution for a hard sphere, the 4πa² at long wavelength, the 2πa² at short wavelength with its forward shadow (known as the extinction paradox), the symmetrisation of the wave for identical particles, and the partial-wave expansion in higher dimensions are all known. Only a hard sphere is considered. Attractive targets, resonances, bound states and soft walls are left out. Only a scalar wave is considered. The polarisation of electromagnetic waves and the material of the target (conductor or dielectric) are left out. Only one spin case is considered. Fermions are counted only with aligned spins, and the weighting for a beam of mixed spin directions is not treated. The coefficient of the approach to 2 is not compared with the literature. A paper that treats the short-wavelength correction (Wu 1956) is cited, but its value has not been checked here, and only the measured numbers are listed. 4πa² is not said to be the same thing as the surface area of the sphere. The two numbers agree only in three dimensions; in other dimensions the long-wavelength cross section moves with the wavenumber. No claim is made that higher-dimensional spheres exist in the world. The d-dimensional formula is assembled from the standard expansion and checked by numerical integration; it is not put forward as a derivation. Relation to earlier papers: Paper 02 showed that integrating an inverse-square density over the unit ball gives a mass of 4π, and that moving the exponent of the density gives a staircase 2π, 4π, 8π. Here the same three numbers line up as cross sections, but what is moved is not the exponent of the density: it is the wavelength, and whether the particles can be told apart. The same numbers lining up is not written as a common root. Paper 546 showed that an aperture and its complement make the same diffraction pattern only away from the forward direction. Here it is counted that the πa² belonging to the shadow of the sphere sits in that forward direction. Paper 498 showed that the entropy of mixing is continuous in composition but jumps, with no intermediate value, in whether the species are the same. Here the cross section jumps in the same way. Paper 335 showed that a round trip changes the exponent of the received strength, and stated in its scope note that it does not enter into computing the reflection cross section from shape or material and treats the cross section as a given number. The present paper is placed beside that seat in the scope note. Restricted to a scalar wave and a hard sphere, it counts that the cross section defined from the backward strength is 4πa² at long wavelength and πa² at short wavelength (no earlier occupant of this item was found). Paper 300 showed that whether two things share a root is decidable. Here the factors 2, 4 and 8 are sorted by what is moved (wavelength, distinguishability, dimension). What is added: obtaining the four cross sections from one calculation and sorting each π into the area of a circle or a solid angle, splitting the 2 at short wavelength into reflection and shadow by numbers, changing the dimension to separate the factors of 2 that remain from the 4 that does not, measuring that the coefficient of the approach to 2 is nearly proportional to d−1, and recording a mistake in measuring.
1. The question
Take a hard sphere of radius $a$. When something is sent at it, the amount that is hit and deflected, expressed as an area perpendicular to the flow, is called the cross section. Naively this should be the area of the sphere seen from the front, $\pi a^2$. Textbooks, however, give four values, one for each setting.
| Setting | Cross section | Multiple of $\pi a^2$ (computed here) | Reading |
|---|---|---|---|
| Particles (classical) | $\pi a^2$ | $1.000000$ | Area of a circle; also $4\pi\times(a/2)^2$ |
| Wave, short wavelength ($ka=5000$) | $2\pi a^2$ | $2.006806$ | $\pi a^2$ of reflection and $\pi a^2$ of shadow |
| Wave, long wavelength ($ka=0.001$) | $4\pi a^2$ | $3.999999$ | Full solid angle $4\pi$ times $a^2$ |
| Identical bosons, long wavelength | $8\pi a^2$ | $7.999997$ | Solid angle of a hemisphere $2\pi$ times $(2a)^2$ |
$k$ is the wavenumber ($2\pi$ divided by the wavelength); the smaller $ka$, the longer the wavelength compared with the sphere. The values double from 1 to 2 to 4 to 8, but do the three doublings happen for the same reason? This paper obtains the four from one calculation and counts, in turn, where each factor comes from. At the end the dimension of space is changed, to see which factors remain and which do not.
2. Particles: a sphere that reflects like a mirror
Let a particle fly in on a straight line and bounce off the surface like light off a mirror. With $b$ the distance from the centre line (the impact parameter) and $\theta$ the angle between the outgoing and the incoming direction, $b = a\cos(\theta/2)$. The amount scattered per solid angle is then
$$\frac{d\sigma}{d\Omega} = \frac{b}{\sin\theta}\left|\frac{db}{d\theta}\right| = \frac{a^2}{4},$$which does not depend on the angle. The sphere scatters particles with the same strength in every direction.
This was checked without the formula. The disc that is hit was divided into 400,000 rings of equal area; for each ring the direction was folded back at the surface normal, and the cosine of the outgoing direction was sorted into ten bins. Every bin received $0.100000$ of the area, and the strength per solid angle was $0.250000$ (in units of $a^2$) at both its smallest and its largest. As a control, a wrong sphere whose deflection angle is taken proportional to $b$ gave bin fractions that differ from $0.1$ by up to $0.090000$.
So $\pi a^2$ can be read in two ways. One is the area of a circle; the other is “the full solid angle $4\pi$ times the strength per direction $(a/2)^2$”. The second reading is the one that connects to the wave case.
3. Waves: the partial-wave sum
For a wave, the wave is split into components of angular momentum $l$ (partial waves), and the shift of phase is found for each. For a hard sphere the phase shift $\delta_l$ is fixed by one condition, that the wave vanish on the surface.
$$\tan\delta_l = \frac{j_l(ka)}{n_l(ka)},\qquad \sigma = \frac{4\pi}{k^2}\sum_{l=0}^{\infty}(2l+1)\sin^2\delta_l .$$$j_l$ and $n_l$ are the spherical Bessel and Neumann functions. The $l=0$ component has the closed form $\delta_0=-ka$. The numerical $\sin^2\delta_0$ and $\sin^2(ka)$ were both $0.087332192545$ at $ka=0.3$ and both $0.879843956429$ at $ka=7.5$.
| $ka$ | $\sigma/(\pi a^2)$ | $(\sigma/(\pi a^2)-2)\times(ka)^{2/3}$ |
|---|---|---|
| $0.001$ | $3.999999$ | |
| $0.1$ | $3.986816$ | |
| $1$ | $3.382438$ | |
| $2$ | $2.999087$ | |
| $10$ | $2.397229$ | $1.843772$ |
| $100$ | $2.090949$ | $1.959443$ |
| $1000$ | $2.019852$ | $1.985245$ |
| $5000$ | $2.006806$ | $1.989941$ |
The value tends to 4 on the long-wavelength side and to 2 on the short-wavelength side. The approach to 2 is slow: at $ka=5000$ there is still $0.006806$ left. Multiplying the remainder by $(ka)^{2/3}$ gives a nearly constant number that rises slowly with $ka$, $1.985245$ and then $1.989941$. The short-wavelength correction is treated by Wu (1956); the present paper does not compare with that value.
The sum was checked in two ways. Doubling the number of $l$ did not change the values at $ka=10$ and $ka=1000$. And without using the sum, integrating the squared modulus of the scattering amplitude $f(\theta)$ over all directions gave $3.7410195846$ at $ka=0.5$, $2.6023763997$ at $ka=5$ and $2.1988198855$ at $ka=30$, in agreement with the sum to ten digits.
4. The 4 at long wavelength: $a^2$ in every direction
At $ka=0.001$ the strength per direction $|f(\theta)|^2$, in units of $a^2$, was $0.999998$, $0.999999$, $1.000000$, $1.000001$ and $1.000002$ from backward (180 degrees) to forward (0 degrees). The wave is scattered equally in all directions, $a^2$ per solid angle. The full solid angle is $4\pi$, so the cross section is $4\pi\times a^2$.
For particles, too, the scattering did not depend on direction. The only difference is the strength per direction: $a^2/4$ for particles and $a^2$ for the wave, a ratio of $3.999999$. In amplitude, the $a/2$ of the particle case becomes $a$, a factor of 2. The 4 in $4\pi a^2$ is the square of this factor of 2.
$4\pi a^2$ is the same number as the surface area of a sphere of radius $a$. That the surface of a sphere is four times the area of its great circle was shown by Archimedes (Heath 1897). Here that factor of four appears as the ratio of the long-wavelength cross section to the particle cross section. Whether the agreement is more than a coincidence cannot be told without changing the dimension. That is counted in Section 7.
The departure from 4 was also counted. The $l=0$ component alone gives $4\sin^2(ka)/(ka)^2$, and the rest is carried by the $l=1$ component. That rest tends to $(4/3)(ka)^4$; the ratio was $0.953884$ at $ka=0.2$, $0.988121$ at $ka=0.1$ and $0.997008$ at $ka=0.05$.
5. The 2 at short wavelength: reflection and shadow
A wave of short wavelength should behave like particles, and yet the cross section tends not to $\pi a^2$ but to $2\pi a^2$. What doubles it was counted by sorting the scattering by direction.
| Quantity (in units of $\pi a^2$) | $ka=50$ | $ka=200$ | $ka=1000$ | For particles |
|---|---|---|---|---|
| Total | $2.142965$ | $2.057649$ | $2.019852$ | $1$ |
| Scattered into the backward half | $0.501001$ | $0.500065$ | $0.500003$ | $0.5$ |
| Scattered beyond 30 degrees | $0.963012$ | $0.937496$ | $0.933330$ | $0.933013$ |
| Total minus 1 | $1.142965$ | $1.057649$ | $1.019852$ | $0$ |
Away from the forward direction, the same reflection as for particles is still there. Dividing the strength per direction by the particle value $a^2/4$ gave, at $ka=200$, $1.003550$ at 60 degrees, $1.000398$ at 90 degrees, $1.000096$ at 120 degrees and $1.000042$ at 150 degrees (averaged over 2 degrees on either side to smooth the fine ripple). Summed over all directions this reflection is $\pi a^2$, the particle cross section itself.
The other $\pi a^2$ is in a narrow forward peak. The strength straight ahead is $2912.760$ times the strength of the reflection at $ka=50$, $42443.168$ times at $ka=200$ and $1020250.839$ times at $ka=1000$, that is, $1.165104$, $1.061079$ and $1.020251$ times $(ka)^2$. The width of the peak is about $1/(ka)$: the shorter the wavelength, the taller and narrower it is.
| Part of the forward peak inside | $ka=50$ | $ka=200$ | $ka=1000$ | Disc diffraction formula |
|---|---|---|---|---|
| the first dark ring | $0.915405$ | $0.864610$ | $0.846361$ | $0.837785$ |
| the second dark ring | $1.007354$ | $0.941189$ | $0.919569$ | $0.909931$ |
| the third dark ring | $1.052771$ | $0.972063$ | $0.947810$ | $0.937647$ |
The right-hand column is the fraction of light that falls inside each dark ring in the diffraction by a disc of radius $a$ (or by a round hole of the same size). The larger $ka$, the closer the forward peak of the sphere comes to these fractions. No wave reaches the region behind the sphere. That it does not reach there means that the sphere sends forward a wave that exactly cancels the incoming one. That wave has the shape of a wave cut out by the frontal outline of the sphere (a disc), and it carries a flow of $\pi a^2$. This is the shadow part, and it is not counted for particles. Particles count only “was the direction changed”, while a wave also counts “was the forward wave reduced”.
Paper 546 showed, for a one-dimensional slit, that an aperture and its complement make the same diffraction pattern only away from the forward direction. The shadow of the sphere is the amount that gathers in that forward direction.
There is also a convention that defines a cross section from the backward strength alone (it is used in detection). The strength straight back, $|f(\pi)|^2$, is treated as if it went out equally in all directions and is multiplied by $4\pi$.
| $ka$ | $4\pi|f(\pi)|^2/(\pi a^2)$ |
|---|---|
| $0.001$ | $3.999991$ |
| $0.1$ | $3.908170$ |
| $0.5$ | $2.404084$ |
| $1$ | $1.346059$ |
| $2$ | $1.151840$ |
| $5$ | $1.042768$ |
| $10$ | $1.011434$ |
| $50$ | $1.000498$ |
| $500$ | $1.000005$ |
This cross section is $4\pi a^2$ at long wavelength and $\pi a^2$ at short wavelength, and it does not include the shadow part. Paper 335 stated in its scope note that it does not enter into computing the reflection cross section from shape or material, and treats the cross section as a given number. The present paper is placed beside that seat in the scope note. What is counted here is only a scalar wave and a hard sphere; the polarisation of electromagnetic waves and the material are not included.
6. The 8 for identical particles
When the two particles are of the same kind and cannot be told apart, one cannot tell, in the centre-of-mass frame, whether a particle went to the angle $\theta$ or to the opposite angle $\pi-\theta$. The two amplitudes are therefore added before squaring. For bosons the sum is taken, and for fermions with aligned spins the difference. In terms of partial waves, only even $l$ remain for bosons and only odd $l$ for fermions.
$$\sigma_{\rm B} = \frac{8\pi}{k^2}\sum_{l\ \mathrm{even}}(2l+1)\sin^2\delta_l,\qquad \sigma_{\rm F} = \frac{8\pi}{k^2}\sum_{l\ \mathrm{odd}}(2l+1)\sin^2\delta_l .$$Here $a$ is the closest distance between the centres of the two particles.
| $ka$ | Bosons | Fermions | Distinguishable |
|---|---|---|---|
| $0.001$ | $7.999997$ | $2.666663\times10^{-12}$ | $3.999999$ |
| $0.1$ | $7.973369$ | $2.634988\times10^{-4}$ | $3.986816$ |
| $1$ | $5.676428$ | $1.088447$ | $3.382438$ |
| $2$ | $2.335018$ | $3.663157$ | $2.999087$ |
| $5$ | $2.412765$ | $2.791988$ | $2.602376$ |
| $10$ | $2.576638$ | $2.217819$ | $2.397229$ |
| $50$ | $2.122362$ | $2.163568$ | $2.142965$ |
| $1000$ | $2.021713$ | $2.017992$ | $2.019852$ |
At long wavelength, bosons give twice the distinguishable value and fermions go to 0. The fermion value tends to $(8/3)(ka)^4$; the ratio was $0.999880$ at $ka=0.01$. As the wavelength gets shorter the two values approach the distinguishable value while crossing above and below it, and at $ka=1000$ bosons are at $1.000921$ times and fermions at $0.999079$ times that value. Being indistinguishable matters only when the wavelength is long.
The reading of $8\pi a^2$ is as follows. At long wavelength the amplitude is $a$ in every direction. Adding two of them gives $2a$, and the strength per direction is $|f(\theta)+f(\pi-\theta)|^2=3.999999 a^2$. On the other hand $\theta$ and $\pi-\theta$ are the same event, so the directions to be summed are only the $2\pi$ of a hemisphere. $2\pi\times(2a)^2=8\pi a^2$, and the calculation gave $7.999997$. Integrating the symmetrised amplitude over a hemisphere gave, at $ka=0.5$, $7.3552313897$ for bosons and $0.1268077795$ for fermions, in agreement with the even and odd sums.
Paper 498 showed that the entropy of mixing is continuous in composition but jumps, with no intermediate value, in whether the species are the same. The cross section jumps in the same way. Between the $4\pi a^2$ of distinguishable particles and the $8\pi a^2$ or 0 of identical ones there is no intermediate value.
7. Changing the dimension
The factors found so far are: 2 from particles to a short-wavelength wave, 2 from short to long wavelength, and 2 from distinguishable particles to identical bosons. Whether the three are factors of the same nature is examined by changing the dimension $d$ of space.
In $d$ dimensions, too, the phase shift of a hard sphere is fixed by the condition that the wave vanish on the surface. With $\nu=(d-2)/2$,
$$\tan\delta_l = \frac{J_{l+\nu}(ka)}{Y_{l+\nu}(ka)},\qquad \sigma = \frac{4\,(2\pi)^{d-1}}{S_{d-1}\,k^{d-1}}\sum_{l=0}^{\infty}N(d,l)\sin^2\delta_l,\qquad N(d,l)=\frac{(2l+d-2)\,(l+d-3)!}{l!\,(d-2)!} .$$$S_{d-1}$ is the surface area of the unit sphere in $d$ dimensions and $N(d,l)$ is the number of states of angular momentum $l$; for $d=3$ it returns to $2l+1$. The formula is assembled from the standard expansion and was checked in three ways. The expansion of a plane wave in Gegenbauer polynomials differed from the direct value by at most $1.112\times10^{-15}$ in three to six dimensions. The squared modulus of the amplitude integrated over the sphere of dimension $d-1$ differed from the formula above by at most $4.565\times10^{-13}$ in the eight cases of three to six dimensions with $ka=0.7$ and $ka=3$. Setting $d=3$ gave a difference of $1.332\times10^{-15}$ from the formula of Section 3.
First, short wavelength. The cross section is divided by the frontal area $V_{d-1}a^{d-1}$, where $V_{d-1}$ is the volume of the unit ball one dimension lower.
| $d$ | $ka=50$ | $ka=200$ | $ka=1000$ | Coefficient of the approach $\div(d-1)$ |
|---|---|---|---|---|
| 2 | $2.073267$ | $2.029109$ | $2.009960$ | $0.497983$ |
| 3 | $2.142965$ | $2.057649$ | $2.019852$ | $0.496311$ |
| 4 | $2.208978$ | $2.085613$ | $2.029678$ | $0.494634$ |
| 5 | $2.271196$ | $2.112993$ | $2.039436$ | $0.492952$ |
| 6 | $2.329514$ | $2.139783$ | $2.049127$ | $0.491265$ |
In every dimension the value comes closer to 2 as $ka$ grows. Counting reflection plus shadow as a factor of 2 does not depend on the dimension. The right-hand column is $(\sigma/(2V_{d-1}a^{d-1})-1)\times(ka)^{2/3}$ at $ka=1000$, divided by $d-1$. Before dividing, the values are $0.497983$, $0.992622$, $1.483902$, $1.971809$ and $2.456327$, nearly proportional to $d-1$. Reading this as the radius of the sphere looking fatter by a fraction of order $(ka)^{-2/3}$ gives this proportionality, but the divided value drifts down from $0.497983$ to $0.491265$, so that reading does not exhaust it.
Next, long wavelength. The cross section is divided by the surface area of the sphere, $S_{d-1}a^{d-1}$.
| $d$ | $ka=0.1$ | $ka=0.01$ | $ka=0.001$ | Power of the wavenumber (measured) | $d-3$ |
|---|---|---|---|---|---|
| 2 | $1.892257$ | $6.345123$ | $3.032450\times10^{1}$ | $-0.679354$ | $-1$ |
| 3 | $9.967040\times10^{-1}$ | $9.999667\times10^{-1}$ | $9.999997\times10^{-1}$ | $-0.000014$ | $0$ |
| 4 | $1.522113\times10^{-1}$ | $1.569937\times10^{-2}$ | $1.570784\times10^{-3}$ | $+0.999766$ | $1$ |
| 5 | $9.881227\times10^{-3}$ | $9.998800\times10^{-5}$ | $9.999988\times10^{-7}$ | $+1.999948$ | $2$ |
| 6 | $3.900772\times10^{-4}$ | $3.926729\times10^{-7}$ | $3.926988\times10^{-10}$ | $+2.999971$ | $3$ |
Only three dimensions settle to 1. In four or more dimensions the long-wavelength cross section is proportional to the wavenumber to the power $d-3$ and goes to 0. In four dimensions it is $(\pi/2) ka$ times the surface area; the numerical coefficient was $1.570784$ ($\pi/2=1.570796$). In two dimensions, on the contrary, the cross section grows as the wavelength gets longer and cannot be written as a power (a logarithm enters; Adhikari 1986). Only in three dimensions does the long-wavelength cross section become a single number independent of wavelength, and that number is the surface area of the sphere.
Even so, the form “a solid angle is multiplied” does not depend on the dimension. At $ka=0.001$, in each of three to six dimensions the strength per direction is uniform (the ratio of largest to smallest is at most $1.000004$), and the cross section divided by the strength per direction was $1.000000$ times the surface area $S_{d-1}$ of the unit sphere. What moves is the strength per direction, which is proportional not to $a^{d-1}$ but to $k^{d-3}a^{2d-4}$. The cross section of identical bosons was $2.000000$ times that of distinguishable particles in each of three to six dimensions.
| Factor | Three dimensions | $d$ dimensions |
|---|---|---|
| Particles $\to$ short-wavelength wave | 2 | 2 (checked in two to six dimensions) |
| Distinguishable $\to$ identical bosons (long wavelength) | 2 | 2 (checked in three to six dimensions) |
| Short wavelength $\to$ long wavelength | 2 | not a single number (power $d-3$ of the wavenumber) |
| Surface area of the sphere $\div$ frontal area | 4 | $\pi$, 4, $3\pi/2$, $16/3$, $15\pi/8$ ($d=2$ to 6) |
The numerical values in the last row are $3.141592654$, $4.000000000$, $4.712388980$, $5.333333333$ and $5.890486225$. Of the three factors of 2, the 2 of the shadow and the 2 of indistinguishability remain in every dimension. The middle one, that is, the 4 at long wavelength, appears only in three dimensions.
8. The staircase of Paper 02, and a mistake in measuring
Paper 02 showed that integrating an inverse-square density over the unit ball gives a mass of $4\pi$, and that moving the exponent $n$ of the density gives $m(n)=4\pi/(3-n)$, which is $2\pi$, $4\pi$ and $8\pi$ for $n=1$, 2 and $5/2$. Numerically these are $6.283185307$, $12.566370614$ and $25.132741229$. The same three numbers line up in the cross sections of this paper, in units of $a^2$.
But what is moved is different. The three of Paper 02 come from moving the exponent of the density. The three here come from moving the wavelength, and whether the particles can be told apart. The contents differ as well. The $2\pi a^2$ here is two areas of a circle, and the $8\pi a^2$ is the solid angle of a hemisphere times the square of a doubled amplitude. The full solid angle itself appears only in the $4\pi a^2$ at long wavelength. Following the criterion of Paper 300, the same numbers lining up is not, by itself, written as a common root.
One check failed once. Before measuring, the short-wavelength value in each dimension had been set to lie “between 2.00 and 2.03 at $ka=1000$”. The measured value rises with dimension, from $2.009960$ in two dimensions to $2.049127$ in six, and four or more dimensions fell outside. Listing the values that fell outside showed that the distance from 2 is nearly proportional to $d-1$. The mistake was not in the formula but in measuring with an expectation set in advance.
Closing
Closing: a hard sphere of radius a has four cross sections, πa² for particles, 2πa² for a wave of short wavelength, 4πa² for a wave of long wavelength, and 8πa² for identical bosons. The πa² for particles is the area of a circle and also 4π×(a/2)², the strength per direction a²/4 summed over all directions. The 4πa² at long wavelength is the full solid angle 4π multiplied by a strength per direction that has become a²; the partial-wave calculation gave $\mathbf{3.999999}$ times πa². The 2πa² at short wavelength is the πa² of the same reflection as for particles plus the πa² of the shadow gathered in a narrow forward peak; at ka = 5000 it was $\mathbf{2.006806}$ times. The 8πa² for identical bosons is the solid angle of a hemisphere, 2π, multiplied by (2a)²; it was $\mathbf{7.999997}$ times. When the dimension is changed, the factor of 2 from the shadow and the factor of 2 from indistinguishability remain in every dimension, while the long-wavelength value is proportional to the wavenumber to the power d−3 and becomes independent of wavelength only in three dimensions. To be candid, the fact that 4πa² is the same number as the surface area of the sphere does not allow one to say that the surface does the scattering. What could be said is that this number appears only in three dimensions, and that the form in which a solid angle is multiplied does not depend on the dimension. The separator is the ratio of wavelength to radius, and whether the two particles can be told apart.
References
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Reference
Yamagishi, Y. (2026). A Hard Sphere Has Four Cross Sections, 1, 2, 4 and 8 Times πa², and the 4 Appears Only in Three Dimensions [Paper 1081]. YOI: https://www.yamagishiyuuki.com/yoi/1081
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 1081. This paper is a preprint and has not been peer reviewed.