Abstract
In a cold, dilute gas the force between two particles is represented by a single number g. It is written g = 4πħ²a/m, where a is the scattering length and m the mass of a particle. This paper counts what the 4π in this formula is, and looks at what disappears and what remains when the mass is set to 4π. No new theorem or law is claimed.
First, g is the flow going out of the target. The force summed with the weight of the zero-energy solution ψ can be rewritten, by Gauss’s theorem, as the flow passing outward through a distant sphere. The distant solution has the form 1 − a/r, so the flow is the full solid angle 4π multiplied by a. For six targets (steps, Gaussians, an attractive well) the ratio of the summed quantity to 4πħ²a/m was $\mathbf{1.000000000}$ in every case. Inside the target the flow has not yet grown (for a Gaussian target, at the sphere of radius 1 it is $\mathbf{0.398254166}$ of its final value), and once outside the target it no longer depends on the radius.
Second, the force summed by itself agrees only when it is weak. The ratio of the true scattering length to the one fixed by summing the force without the weight of the solution (the Born approximation) is $\mathbf{0.996016125}$ when the strength κR of a step target is 0.1, $\mathbf{0.715217532}$ when it is 1 and $\mathbf{0.027000000}$ when it is 10. Towards a hard sphere the summed force grows without bound, while the scattering length settles to the radius of the target and g stays finite.
Third, setting the mass to 4π gives g = ħ²a. g·m/(ħ²a) is $\mathbf{12.566370614}$ however ħ and m are chosen, and with m = 4π the ratio g/(ħ²a) is $\mathbf{1.000000000}$. But the 4π in the numerator is a solid angle, which has no unit, and the 4π in the denominator is a mass placed there by a choice of unit. They are different things, and they merely cancel as numbers.
Fourth, and this is the core of the paper: a formula that does not contain the mass keeps its 4π. The length over which the wave of the gas recovers from a disturbance (the healing length) is 1/√(8πna), where n is the number density. Neither the mass nor Planck’s constant appears in it. With the numbers assumed here (n = 10²⁰ m⁻³, a = 100 Bohr radii) it is $2.742076$×10⁻⁷ m, $\mathbf{1.272759}$ times the spacing of the particles. Reducing the mass to a quarter multiplies g and the speed of sound by $\mathbf{4.000000}$ but leaves the healing length at $\mathbf{1.000000}$ times its value. However the mass is chosen, this 4π cannot be moved.
Fifth, when the dimension is changed, 4π becomes (d−2) times the surface area of the unit sphere. In d dimensions g is (d−2)×S×ħ²/m multiplied by a to the power d−2, with S the surface area of the unit sphere in d dimensions. The coefficient is 4π, 4π², 8π², 4π³ from three dimensions upward, and for a step target it agreed with the numerical value with a ratio of $\mathbf{1.000000000}$. Even when the mass is taken to be the surface area of the unit sphere, a factor of 2 remains in four dimensions, 3 in five and 4 in six. Only in three dimensions does nothing remain.
Sixth, ħ²/m plays the part of a diffusion coefficient. Paper 249 showed that the rate at which a sphere captures diffusing molecules is 4π times the diffusion coefficient times the radius, and that this 4π is the solid angle of Gauss’s law. g = 4π×(ħ²/m)×a has the same form, and the equation outside the target is the same Laplace equation.
To be candid, the scattering length of 100 Bohr radii is an assumed number, not a measured value for a particular atom. That the healing length does not depend on the mass follows from rearranging the formulas and has not been checked by measurement here.
Keywords: scattering length, coupling constant, Gauss’s theorem, solid angle, healing length, Born approximation, choice of unit.
Scope (what is not claimed)
No new theorem or law is claimed. The scattering length, the replacement of the force by a contact force with g = 4πħ²a/m (Huang and Yang 1957), the Born approximation, and the mean-field equation with its healing length are all known. Setting the mass to 4π is not called a fact about nature. It is a choice of the unit of mass. Only the appearance of the formulas changes; nothing measurable changes. Cases where the scattering length diverges are not treated. The neighbourhood of a resonance, the appearance of a bound state and three-particle effects are left out. Nothing beyond the mean field is treated. Corrections from quantum fluctuations and the shift of the condensation temperature by interactions are left out. No value for a particular atom is used. The scattering length of 100 Bohr radii is an assumed number; only the mass is that of rubidium 87, the same value as in the check of Paper 663. No claim is made that gases exist in higher dimensions. The formula is only listed against the dimension. Relation to earlier papers: Paper 663 showed that the condensation of an ideal Bose gas needs no interaction, and stated in its scope note that the scattering length and the mean-field equation are left out and only the ideal gas is considered. The present paper sits in that seat of the scope note and counts the scattering length and the mean-field coupling (no earlier occupant; it does not sit on the shift of the condensation temperature). Paper 249 showed that the 4π in the rate at which a sphere captures diffusing molecules comes from the same equation as the 4π of a spreading field, and counted the coefficients 4π, 4π², 8π² in d dimensions and the absence of steady absorption in two dimensions. Here it is counted that the same coefficients appear in the quantum coupling, and that even with a force inside the target the coupling is fixed by the outward flow alone. Paper 02 showed that integrating an inverse-square density over the unit ball gives a mass of 4π, the solid angle itself. Here it is seen that putting that mass into the formula for g cancels, as a number, the 4π that counted the flow. Paper 60 showed that the 4π in E = 4πc² is not a solid angle but a mass placed there by a choice of unit. Here too the 4π entering the denominator is the placed mass, a different thing from the solid angle in the numerator. Paper 912 showed that taking the mass as the surface area of the unit sphere in d dimensions makes the coefficient of temperature the volume of the unit ball in every dimension. Here it is seen that with the same choice a factor d−2 remains in g. Paper 1081 counted that the long-wavelength cross section of a hard sphere is the full solid angle 4π multiplied by a². Here the same distant form 1 − a/r gives a flow equal to the solid angle 4π multiplied by a. What is added: counting g as an integral weighted by the solution for six targets, tabulating the difference from the force summed by itself, separating the 4π that disappears when the mass is set to 4π from the 4π that remains because there is no mass, and changing the dimension to count that the choice of mass can remove everything only in three dimensions.
1. The question
In a cold, dilute gas the particles collide slowly. The fine shape of the force between them then becomes invisible, and only one length $a$ (the scattering length) remains. In formulas for many particles the force is condensed further into a single number $g$.
$$g = \frac{4\pi\hbar^2 a}{m}.$$$m$ is the mass of a particle and $\hbar$ is Planck’s constant divided by $2\pi$. Two things are counted about this formula. One is where the $4\pi$ comes from. The other is what happens when the choice of mass used throughout this corpus, $m=4\pi$, is put in. It is immediate that $g=\hbar^2a$ results. What we want to know is whether one may then say that the $4\pi$ has disappeared.
2. $g$ is the flow going out of the target
Consider the relative motion of two particles of the same mass $m$. The reduced mass is $m/2$, so the coefficient of the kinetic energy is $\hbar^2/m$. The wave $\psi$ at zero energy satisfies
$$-\frac{\hbar^2}{m}\nabla^2\psi + V\psi = 0 .$$Outside the range of the force $V$, $\nabla^2\psi=0$ and the solution has the form $\psi = 1 - a/r$. This $a$ is the scattering length. Defining $g$ as “the force summed with the weight $\psi$”, Gauss’s theorem gives
$$g = \int V\psi\,d^3r = \frac{\hbar^2}{m}\int \nabla^2\psi\,d^3r = \frac{\hbar^2}{m}\oint \nabla\psi\cdot d\mathbf{S} = \frac{\hbar^2}{m}\times 4\pi r^2\times\frac{a}{r^2} = \frac{4\pi\hbar^2a}{m}.$$The $4\pi$ is the solid angle that appears when the flow passing outward through a distant sphere is summed over all directions.
This was checked without the formula. For six targets the equation was solved numerically to obtain the scattering length, and separately $\int V\psi d^3r$ was summed ($\hbar=1$, $m=1$, target size $R=1$).
| Target | $a/R$ | $\int V\psi d^3r$ | $4\pi\hbar^2a/m$ | Ratio |
|---|---|---|---|---|
| Step ($\kappa R=0.5$) | $+0.075765685$ | $+0.952099684$ | $+0.952099684$ | $1.000000000$ |
| Step ($\kappa R=2$) | $+0.517986210$ | $+6.509206688$ | $+6.509206688$ | $1.000000000$ |
| Step ($\kappa R=10$) | $+0.900000000$ | $+11.309733558$ | $+11.309733558$ | $1.000000000$ |
| Gaussian (height 1) | $+0.328721131$ | $+4.130831562$ | $+4.130831562$ | $1.000000000$ |
| Gaussian (height 5) | $+0.834791447$ | $+10.490298714$ | $+10.490298714$ | $1.000000000$ |
| Well ($\kappa R=1$, attractive) | $-0.557407725$ | $-7.004592051$ | $-7.004592051$ | $1.000000000$ |
A step is a target of constant height $V_0$ inside the radius $R$, with strength $\kappa=\sqrt{mV_0}/\hbar$. The largest departure of the ratio from 1 was $4.582\times10^{-13}$. The step and the well have closed forms ($a/R=1-\tanh(\kappa R)/(\kappa R)$ and $1-\tan(\kappa R)/(\kappa R)$), and the numerical values differed from them by at most $5.632\times10^{-13}$. For the attractive well $a$ is negative and the flow is inward.
Where the flow is settled was also examined. For the Gaussian target (height 1), the flow through the sphere of radius $r$ divided by its final value $4\pi a$ was $0.071891158$ at $r=0.5$, $0.398254166$ at $r=1$, $0.947110488$ at $r=2$, $0.999999351$ at $r=4$, and $1.000000000$ at $r=6$ and $r=10$. The flow grows inside the target and no longer depends on the radius once outside. However intricate the inside of the target, all that is seen from outside is the total flow, and that is condensed into the single length $a$.
3. The force summed by itself is not enough
There is an approximation that fixes $g$ by the force summed by itself, $\int V d^3r$, without the weight $\psi$ (the Born approximation). Writing the resulting scattering length as $a_{\rm B}$, a step target has $a_{\rm B}/R=(\kappa R)^2/3$.
| $\kappa R$ | $a/R$ (exact) | $a_{\rm B}/R$ | $a/a_{\rm B}$ |
|---|---|---|---|
| $0.1$ | $0.003320054$ | $0.003333$ | $0.996016125$ |
| $0.3$ | $0.028957958$ | $0.030000$ | $0.965265283$ |
| $1$ | $0.238405844$ | $0.333333$ | $0.715217532$ |
| $3$ | $0.668315082$ | $3.000000$ | $0.222771694$ |
| $10$ | $0.900000000$ | $33.333333$ | $0.027000000$ |
| $100$ | $0.990000000$ | $3333.333333$ | $0.000297000$ |
For a weak target the two agree. The departure is proportional to $(\kappa R)^2$, with a coefficient of $0.398387$ at $\kappa R=0.1$ (it tends to $2/5$). For a weak Gaussian target (height $0.001$) too, the ratio of the numerical scattering length to the Born value was $0.999647$. For a strong target they differ greatly. At $\kappa R=100$ the length fixed by the summed force is $3333.333333$ times the radius, while the true scattering length is $0.990000000$ times the radius. In the hard-sphere limit the summed force is infinite and the scattering length is the radius itself.
The reason is that the wave cannot enter a strong target. The force in a place the wave cannot enter is not counted, because the weight $\psi$ is close to 0 there. $g$ counts not the total amount of force but how much the wave has been pushed aside. Paper 1081 counted that the long-wavelength cross section of a hard sphere is $4\pi a^2$, the full solid angle $4\pi$ multiplied by $a^2$. The cross section sums the squared amplitude of the distant form $1-a/r$ over all directions, and $g$ sums the flow of the same form over all directions. In both, the $4\pi$ is the full solid angle.
4. Setting the mass to $4\pi$
Changing the units does not change $g\cdot m/(\hbar^2a)$. For a step target ($\kappa R=2$) the equation was solved again with four choices of $\hbar$ and $m$.
| $\hbar$ | $m$ | $g\cdot m/(\hbar^2a)$ | $g/(\hbar^2a)$ |
|---|---|---|---|
| $1$ | $1$ | $12.566370614$ | $12.566370614$ |
| $1$ | $4\pi$ | $12.566370614$ | $1.000000000$ |
| $0.5$ | $3$ | $12.566370614$ | $4.188790205$ |
| $2$ | $4\pi$ | $12.566370614$ | $1.000000000$ |
Setting the mass to $4\pi$ gives $g=\hbar^2a$. Paper 02 showed that integrating an inverse-square density over the unit ball gives a mass of $4\pi$, that is, that there is exactly one density for which the mass is the same number as the solid angle itself. Putting that mass into the formula for $g$ cancels the $4\pi$ counted by Gauss’s theorem.
But the two $4\pi$ must not be read as the same thing. Paper 60 showed that the $4\pi$ in $E=4\pi c^2$ is not a solid angle but a mass placed there by a choice of unit. The same holds in the formula for $g$. The $4\pi$ in the numerator is a solid angle, which has no unit, and the $4\pi$ in the denominator is a number placed there by choosing the unit of mass. That they cancel as numbers is due to the choice of unit; the solid angle has not gone from the formula. To check this, it suffices to list formulas in which the mass enters differently.
| Quantity | Usual form | Form with $m=4\pi$ | How the mass enters |
|---|---|---|---|
| Coupling $g$ | $4\pi\hbar^2a/m$ | $\hbar^2a$ | once in the denominator |
| Interaction energy $gn$ | $4\pi\hbar^2an/m$ | $\hbar^2an$ | once in the denominator |
| Born amplitude (identical particles) | $-(m/4\pi\hbar^2)\int V$ | $-(1/\hbar^2)\int V$ | once in the numerator |
| Born amplitude (fixed target) | $-(m/2\pi\hbar^2)\int V$ | $-(2/\hbar^2)\int V$ | once in the numerator |
| Speed of sound | $(\hbar/m)\sqrt{4\pi na}$ | $\hbar\sqrt{na/4\pi}$ | once in the denominator, $4\pi$ under the root |
| Healing length | $1/\sqrt{8\pi na}$ | $1/\sqrt{8\pi na}$ | not at all |
| Cross section (distinguishable) | $4\pi a^2$ | $4\pi a^2$ | not at all |
In numbers, with $\hbar=1$, $g/a$ is $12.566370614$ for $m=1$ and $1.000000000$ for $m=4\pi$. The Born coefficient for a fixed target, $m/(2\pi\hbar^2)$, goes from $0.159154943$ at $m=1$ to $2.000000000$ at $m=4\pi$, and the coefficient for identical particles, $m/(4\pi\hbar^2)$, goes from $0.079577472$ to $1.000000000$. The speed of sound, with $\hbar=1$, $n=1$ and $a=0.01$, is $0.354490770$ for $m=1$ and $0.028209479$ for $m=4\pi$, the latter agreeing with $\hbar\sqrt{na/4\pi}$. The $4\pi$ becomes invisible only in formulas where the $4\pi$ of the solid angle and the mass face each other, one each, across a division.
5. A formula without the mass keeps its $4\pi$
A gas in which many particles occupy the same wave $\Psi$ is described by the mean-field equation
$$-\frac{\hbar^2}{2m}\nabla^2\Psi + g|\Psi|^2\Psi = \mu\Psi$$($\mu=gn$, with $n$ the number density). The length over which a wave held at 0 beside a wall returns to the original density is called the healing length. Defining it as the length $\xi$ at which the kinetic energy $\hbar^2/(2m\xi^2)$ equals the interaction energy $gn$,
$$\xi = \frac{\hbar}{\sqrt{2mgn}} = \frac{1}{\sqrt{8\pi na}},$$and both $\hbar$ and $m$ drop out. The $m$ in the denominator of $g$ cancels the $m$ in the denominator of the kinetic energy. With $\hbar=1$, $n=1$ and $a=0.01$, and $\hbar$ and $m$ chosen in the four ways of the previous section, $g$ moves through $0.125663706$, $0.010000000$, $0.010471976$ and $0.040000000$, while the healing length was $1.994711402$ in all four.
To see actual magnitudes, numbers were assumed. The number density is $10^{20}$ m$^{-3}$ as in Paper 663, the mass is that of rubidium 87 ($86.909180527$ u), and the scattering length is assumed to be 100 Bohr radii, $5.291772105\times10^{-9}$ m.
| Quantity | Value |
|---|---|
| Coupling $g$ | $5.124465\times10^{-51}$ J m$^3$ |
| Interaction energy $gn$ (as a temperature) | $37.1164$ nK |
| Speed of sound $\sqrt{gn/m}$ | $1.884373$ mm/s |
| Healing length $\xi=1/\sqrt{8\pi na}$ | $2.742076\times10^{-7}$ m |
| Width of the wall $\sqrt{2} \xi=1/\sqrt{4\pi na}$ | $3.877881\times10^{-7}$ m |
| Spacing of the particles $n^{-1/3}$ | $2.154435\times10^{-7}$ m |
| Healing length $\div$ spacing | $1.272759$ |
| Gas parameter $na^3$ | $1.481847\times10^{-5}$ |
For a hypothetical particle with only the mass reduced to a quarter, $g$ becomes $4.000000$ times and the speed of sound also $4.000000$ times, but the healing length stays at $1.000000$ times. In a hypothetical world with $\hbar$ doubled, $g$ becomes $4.000000$ times and the speed of sound $2.000000$ times, and the healing length again stays at $1.000000$ times. What fixes the healing length is only the density, the scattering length and the $4\pi$ of the solid angle. However the unit of mass is chosen, this $4\pi$ cannot be moved.
The 2 in $8\pi$ comes from the $1/2$ of the kinetic energy. The shape of the wave beside a wall is $\tanh(x/L)$ with width $L=\sqrt{2} \xi=1/\sqrt{4\pi na}$. Inserting this shape into the equation left a remainder of $2.183\times10^{-8}$ for $L=\sqrt{2} \xi$ and, as a control, $0.384900$ for $L=\xi$. In the convention that defines the healing length by the width of the wall, the formula is $1/\sqrt{4\pi na}$ and no 2 appears. Of the $8\pi$, what moves with the convention is the 2; the $4\pi$ does not move.
6. Changing the dimension
In $d$ dimensions the solution outside the range of the force is $\psi=1-(a/r)^{d-2}$. Using Gauss’s theorem in the same way,
$$g = \int V\psi\,d^dr = \frac{\hbar^2}{m}\,S_{d-1}r^{d-1}\times\frac{(d-2)\,a^{d-2}}{r^{d-1}} = \frac{(d-2)\,S_{d-1}\,\hbar^2a^{d-2}}{m}.$$$S_{d-1}$ is the surface area of the unit sphere in $d$ dimensions; in three dimensions it is $4\pi$. This was checked numerically for a step target ($\kappa R=2$).
| $d$ | $(a/R)^{d-2}$ | $\int V\psi d^dr$ | Coefficient $(d-2)S_{d-1}$ | In terms of $\pi$ |
|---|---|---|---|---|
| 3 | $0.517986209962$ | $6.509206688$ | $12.566370614$ | $4\pi$ |
| 4 | $0.302225342036$ | $11.931378264$ | $39.478417604$ | $4\pi^2$ |
| 5 | $0.194027918909$ | $15.319830419$ | $78.956835209$ | $8\pi^2$ |
| 6 | $0.133745146555$ | $16.587756075$ | $124.025106721$ | $4\pi^3$ |
The ratio of the summed value to the formula was $1.000000000$ in all four. $(a/R)^{d-2}$ agreed with the closed form written with Bessel functions to within $2.450\times10^{-13}$. The coefficients $4\pi$, $4\pi^2$ and $8\pi^2$ are the same values that Paper 249 counted as the flow of diffusion.
Besides the solid angle $S_{d-1}$, the coefficient carries a factor $d-2$. Paper 912 showed that taking the mass as the surface area $S_{d-1}$ of the unit sphere in $d$ dimensions makes the coefficient of temperature the volume of the unit ball in every dimension. Making the same choice for $g$ gives the following.
| $d$ | $m=S_{d-1}$ | $g/(\hbar^2a^{d-2})$ |
|---|---|---|
| 3 | $12.566370614$ | $1.000000$ |
| 4 | $19.739208802$ | $2.000000$ |
| 5 | $26.318945070$ | $3.000000$ |
| 6 | $31.006276680$ | $4.000000$ |
The solid angle goes, but $d-2$ remains. The form $g=\hbar^2a$ with nothing left over arises when two things coincide: the mass has been set to $4\pi$, and in three dimensions $d-2$ is 1. $d-2$ is the number that appears on differentiating the power $r^{-(d-2)}$ of the distant solution, and it has the same origin as the fact that the power of an inverse-square force is one less than the dimension.
In two dimensions $d-2$ is 0 and a formula of this form does not stand. The outside solution becomes $\ln(r/a)$, and the flow is $2\pi\hbar^2/m$ whatever $a$ is. For a step target ($\kappa R=2$), $a/R=0.488427753$ and the flow was $6.283185307$. In two dimensions the strength of the force enters not through a power of the length $a$ but through a logarithm. This is the same place where Paper 249 counted that there is no steady absorption in two dimensions.
7. $\hbar^2/m$ plays the part of a diffusion coefficient
Paper 249 showed that the rate at which a sphere of radius $a$ captures diffusing molecules is $4\pi Da$ ($D$ the diffusion coefficient), and that this $4\pi$ comes from the same equation as the solid angle of Gauss’s law. Placing the formula of this paper beside it,
$$\text{capture rate} = 4\pi\times D\times a,\qquad g = 4\pi\times\frac{\hbar^2}{m}\times a,$$the two have the same form. This is not a coincidence. Outside the target, the equation of the zero-energy wave and the equation of steady diffusion are the same Laplace equation. In diffusion the concentration is 0 on the surface of the sphere, and for a hard sphere the wave is 0 on the surface, so the boundary condition is the same as well. The only difference is whether the coefficient multiplying the flow is $D$ or $\hbar^2/m$.
There is also a difference. The sphere of diffusion is a target that absorbs molecules at its surface. The target of this paper has a force inside, and the wave goes in. That $g$ is nevertheless fixed by the outward flow alone is because Gauss’s theorem replaces the integral inside the target by an integral over a sphere outside. What happens inside the target shows up only in how far the scattering length $a$ departs from the radius $R$.
Paper 663 showed that the condensation of an ideal Bose gas needs no interaction, and stated in its scope note that the scattering length and the mean-field equation are left out. The present paper sits in that seat of the scope note and has looked at what the single number $g$ representing the interaction counts. Whether condensation happens does not depend on $g$ (Paper 663). What $g$ fixes is the stiffness of the gas after it has happened, the speed of sound and the healing length.
Closing
Closing: the 4π in g = 4πħ²a/m is the solid angle obtained by summing the flow out of the target over all directions. The force summed with the weight of the zero-energy solution becomes, by Gauss’s theorem, the flow through a distant sphere, and for six targets its ratio to 4πħ²a/m was $\mathbf{1.000000000}$. The force summed by itself agrees only for weak targets; at a step strength of 10 the ratio falls to $\mathbf{0.027000000}$. Setting the mass to 4π gives g = ħ²a, but what cancels is a solid angle against a placed mass, and the two are different things. The healing length 1/√(8πna), which contains no mass, keeps its 4π however the mass is chosen. Of the 8π, it is the 2 that moves with the convention. When the dimension is changed the coefficient is (d−2) times the surface area of the unit sphere; even with the mass taken as that surface area d−2 remains, and nothing remains only in three dimensions. To be candid, setting the mass to 4π does not remove the solid angle from the formulas. It becomes invisible only in formulas where the solid angle and the mass face each other one to one. The separator is whether the mass appears in the formula.
References
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- Yamagishi, Y. (2026). Taking $m$ as the Area of the Unit Sphere Makes the Coefficient of Temperature the Volume of the Unit Ball in Every Dimension [Paper 912]. YOI: https://www.yamagishiyuuki.com/yoi/912; DOI: 10.5281/zenodo.22804990
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Reference
Yamagishi, Y. (2026). The 4π in g = 4πħ²a/m Is a Solid Angle, and Setting m = 4π Turns It into g = ħ²a [Paper 1082]. YOI: https://www.yamagishiyuuki.com/yoi/1082
ORCID: 0009-0006-8587-3450
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The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 1082. This paper is a preprint and has not been peer reviewed.