Abstract
When considering gravity and electromagnetic fields (Coulomb potential), it is believed that the force acting on an object is inversely proportional to the square of the distance:
\[ \small r = \sqrt{x^2+y^2+z^2}. \]
However, this distance is clearly defined in Euclidean space, and is probably different from the properties of actual space. If we infer the properties of global space based on the special theory of relativity, it is assumed to be a three-dimensional sphere, so distances should actually be measured on a three-dimensional sphere. In addition, gravity and electromagnetic forces are thought of as attractive (repulsive) forces acting in Euclidean space, but in reality they must be thought of as forces acting in a curved space (three-dimensional sphere). Let us consider the properties of how distance on a sphere differs from distance in Euclidean space.
Distance on a One-dimensional Sphere
Let us consider the distance between two points on a one-dimensional sphere (circumference) \(\small S^1\) and find the distance on the circumference between two points \(\small (x_1,y_1),(x_2,y_2)\) that satisfy
\[ \small \begin{align*} &x_1^2+y_1^2 = r \\ &x_2^2+y_2^2 = r. \end{align*} \]
If we consider it in Euclidean space \(\small R^2\), it is:
\[ \small d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}, \]
but since this distance must pass through a space that does not exist on a one-dimensional sphere, it cannot be used as is. Instead, you would have to calculate the distance traveled by going all the way around the circumference. When this is expressed as:
\[ \small \begin{align*} &x_1 = r \cos \theta_1 \\ &y_1 = r \sin \theta_1 \\ &x_2 = r \cos \theta_2 \\ &y_2 = r \sin \theta_2 \end{align*} \]
using polar coordinates, it corresponds to the length of a circular arc, so it can be calculated as:
\[ \small d = r |\theta_1-\theta_2|. \]
Since
\[ \small \cos(\theta_1-\theta_2) = \cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_1 \]
holds, the above distance is expressed in coordinates as:
\[ \small d = r \cos^{-1} \frac{x_1x_2+y_1y_2}{r^2}. \]
This would be the definition of distance on a one-dimensional sphere.
As an approximation, we calculate this distance when \(\small x_1,x_2\) is negligibly small compared to \(\small y_1,y_2\). By rearranging the equation, it can be expressed as:
\[ \small d = r\cos^{-1} \frac{x_1x_2+\sqrt{r^2-x_1^2}\sqrt{r^2-x_2^2}}{r^2}, \]
but by approximating it as:
\[ \small \sqrt{r^2-x_1^2} \approx r-\frac{x_1^2}{2r}, \]
we get
\[ \small x_1x_2+\sqrt{r^2-x_1^2}\sqrt{r^2-x_2^2} \approx r^2-\frac{1}{2}(x_1-x_2)^2+\frac{x_1^2x_2^2}{4r^2}. \]
If we ignore the last term,
\[ \small d \approx r\cos^{-1} \frac{r^2-\frac{1}{2}(x_1-x_2)^2}{r^2} \]
can be obtained. Using Taylor expansion, we get
\[ \small \cos \frac{d}{r} \approx 1-\frac{1}{2}\frac{d^2}{r^2} \approx 1- \frac{1}{2}\frac{(x_1-x_2)^2}{r^2}, \]
which induces
\[ \small d \approx |x_1-x_2|, \]
and this can be approximated as a distance in one-dimensional Euclidean space \(\small R\). Let us perform a similar calculation in two dimensions.
Distance on a Two-dimensional Sphere
The shortest distance (geodesic) on a two-dimensional sphere is the length of a circular arc on a plane that passes through two points and the origin, and is called the great-circular distance. This problem often appears as a kind of exercise in Riemannian geometry. Using polar coordinates:
\[ \small \begin{align*} &x=r\sin\theta\cos \varphi \\ &y=r\sin\theta\sin \varphi \\ &z = r\cos \theta, \end{align*} \]
let us express two points as \(\small (\theta_1,\varphi_1)\),\(\small (\theta_2,\varphi_2)\) and find the shortest curve connecting these two points. …No, I’ll give up after all. Apparently, using a method called spherical trigonometry, this can be calculated as:
\[ \small d = r\cos^{-1} (\sin\theta_1\sin\theta_2\cos(\varphi_1-\varphi_2)+\cos\theta_1\cos\theta_2). \]
As in the one dimension, if we note that the distance between two points that satisfy
\[ \small \begin{align*} &x_1^2+y_1^2+z_1^2 = r^2 \\ &x_2^2+y_2^2+z_2^2 = r^2 \end{align*} \]
is
\[ \small \cos(\varphi_1-\varphi_2) = \cos \varphi_1 \cos \varphi_2+\sin \varphi_1\sin \varphi_2, \]
it can be calculated as:
\[ \small d = r \cos^{-1} \frac{x_1x_2+y_1y_2+z_1z_2}{r^2}. \]
It is likely that this can be extended to the \(\small n\)-dimensional sphere and calculated as:
\[ \small d_n = r \cos^{-1} \frac{\sum_{k=1}^nx^{(k)}_1x^{(k)}_2}{r^2}. \]
Distance in the Central Potential
Since actual space is assumed to be three-dimensional (although some may disagree), the issue is distance on a three-dimensional sphere. Since
\[ \small x^2+y^2+z^2+s^2=c^2t^2 \]
holds from the theory of special relativity, even though it is called a three-dimensional sphere, it is inferred that space is actually distributed almost entirely along coordinate axes \(\small s\approx ct\) that we do not recognize as space. In other words, the space we perceive \(\small (x,y,z)\) is negligibly small compared to the value of the coordinate axis \(\small s\).
As an approximation, if \(\small (x_1,y_1,z_1)\) and \(\small (x_2,y_2,z_2)\) are small enough to be ignored compared to \(\small s_1,s_2\), we calculate the distance of the 3-sphere as follows:
\[ \small d = r\cos^{-1} \frac{x_1x_2+y_1y_2+z_1z_2+\sqrt{r^2-x_1^2-y_1^2-z_1^2}\sqrt{r^2-x_2^2-y_2^2-z_2^2}}{r^2}. \]
If we approximate it like
\[ \small \sqrt{r^2-x_1^2-y_1^2-z_1^2} \approx r-\frac{x_1^2+y_2^2+z_1^2}{2r}, \]
we can obtain
\[ \small \begin{align*} &x_1x_2+y_1y_2+z_1z_2+\sqrt{r^2-x_1^2-y_1^2-z_1^2}\sqrt{r^2-x_2^2-y_2^2-z_2^2} \\ &\quad \approx r^2-\frac{1}{2}\left((x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2 \right)+\frac{(x_1^2+y_1^2+z_1^2)(x_2^2+y_2^2+z_2^2)}{4r^2}. \end{align*} \]
If we ignore the last term,
\[ \small d \approx r\cos^{-1} \frac{r^2-\frac{1}{2}\left((x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2 \right)}{r^2} \]
can be obtained. Using Taylor expansion, we get
\[ \small \cos \frac{d}{r} \approx 1-\frac{1}{2}\frac{d^2}{r^2} \approx 1- \frac{1}{2}\frac{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}{r^2} \]
which induces
\[ \small d \approx \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2} \]
and it can be approximated as a distance in three-dimensional Euclidean space \(\small R^3\).
Considering this way, it is possible that the distances used in gravity and the Coulomb potential are not actually distances in Euclidean space, but are approximations of distances on the three-dimensional sphere. If we could determine the distance at which these inverse square laws are broken, it might be possible to estimate the radius of the three-sphere that represents the entire space. Well, if it’s not broken, it’s possible that the radius is actually infinite…In this case, even if the space has the properties of a three-dimensional sphere, the entire space may be a different shape. By the way, I don’t know much about this, but when it comes to Coulomb potential, the inverse square law is not clearly violated even when the unit is quantum (electron), and it seems that it cannot be confirmed experimentally for distances shorter than that. Gravity is thought to be a weaker force than electromagnetic force, so it may be something that cannot be measured anyway. The nuclear force (strong force) seems a bit suspicious, but that is beyond the understanding of the author at present…
Appendix: Geodesics on the Two-dimensional Sphere (Incomplete)
In Section 3, I gave up on finding the geodesic on the two-dimensional sphere because I tried to do it halfway but gave up. I’ll just write down what I’ve done so far.
In the polar form of the two-dimensional sphere, the distance traveled in an infinitesimal time \(\small dt\) is:
\[ \small \left(\frac{ds}{dt}\right)^2 = \left(\frac{dr}{dt}\right)^2+r^2\left(\frac{d\theta}{dt}\right)^2+r^2\sin^2\theta \left(\frac{d\varphi}{dt}\right)^2 \]
from the line element:
\[ \small ds^2 = dr^2+r^2d\theta^2+r^2\sin^2\theta d\varphi^2 \]
in polar coordinates. Since \(\small dr/dt=0\) on a sphere, the length of the curve can be expressed as
\[ \small S[\theta,\varphi] = \int_{(\theta_1,\varphi_1)}^{(\theta_2,\varphi_2)} ds = r\int_{(\theta_1,\varphi_1)}^{(\theta_2,\varphi_2)} \sqrt{\left(\frac{d\theta}{dt}\right)^2+\sin^2\theta\left(\frac{d\varphi}{dt}\right)^2}dt. \]
The problem of finding the shortest distance on a two-dimensional sphere can be formulated as a variational problem to find the path \(\small \theta(t),\varphi(t)\) that minimizes this value. Let define
\[ \small F(\theta,\varphi,\dot{\theta},\dot{\varphi},t)=\sqrt{\dot{\theta}^2+\sin^2\theta\dot{\varphi}^2} \]
and solve
\[ \small \begin{align*} &\frac{\partial}{\partial \theta} F(\theta,\varphi,\dot{\theta},\dot{\varphi},t) = \frac{d}{dt}\left(\frac{\partial }{\partial \dot{\theta}} F(\theta,\varphi,\dot{\theta},\dot{\varphi},t)\right) \\ &\frac{\partial}{\partial \varphi} F(\theta,\varphi,\dot{\theta},\dot{\varphi},t) = \frac{d}{dt}\left(\frac{\partial }{\partial \dot{\varphi}} F(\theta,\varphi,\dot{\theta},\dot{\varphi},t)\right) \end{align*} \]
from the Euler-Lagrange equations. By calculating it, we get
\[ \small \begin{align*} &\frac{d}{dt}\frac{\dot{\theta}}{\sqrt{\dot{\theta}^2+\sin^2\theta\dot{\varphi}^2}}=\frac{\sin\theta\cos\theta \dot{\varphi}^2}{\sqrt{\dot{\theta}^2+\sin^2\theta\dot{\varphi}^2}} \\ &\frac{d}{dt}\frac{\sin^2\theta \dot{\varphi}}{\sqrt{\dot{\theta}^2+\sin^2\theta\dot{\varphi}^2}}=0. \end{align*} \]
After integrating the second equation and squaring both sides, we get
\[ \small \dot{\varphi} = \frac{C_1\dot\theta}{\sin\theta \sqrt{\sin^2\theta-C_1^2}}. \]
Here, \(\small C_1\) is an integral constant. From this equation,
\[ \small \frac{d\varphi}{d\theta} = \frac{\dot{\varphi}}{\dot{\theta}} = \frac{C_1}{\sin\theta \sqrt{\sin^2\theta-C_1^2}} \]
holds true. It seems that this integral can be calculated and is:
\[ \small \varphi(\theta) = \cos^{-1}\left(\frac{C_1}{\sqrt{1-C_1^2}}\cot\theta\right)+C_2. \]
If we differentiate it while keeping in mind that
\[ \small \begin{align*} &\frac{d\cos^{-1}x}{dx} = -\frac{1}{\sqrt{1-x^2}} \\ &\frac{d\cot x}{dx} = -\frac{1}{\sin^2x}, \end{align*} \]
it can be confirmed that the result is:
\[ \small \begin{align*} \frac{d\varphi}{d\theta} &= \left(-\frac{1}{\sqrt{1-\frac{C_1^2}{1-C_1^2}\frac{\cos^2 \theta}{\sin^2\theta}}}\right)\left(-\frac{C_1}{\sqrt{1-C_1^2}}\frac{1}{\sin^2\theta} \right) \\ & = \frac{C_1}{\sin\theta \sqrt{\sin^2\theta-C_1^2}} \end{align*} \]
and that it indeed matches. Therefore, rearranging the equation from
\[ \small \frac{C_1}{\sqrt{1-C_1^2}}\cot\theta=\cos(\varphi-C_2) = \cos\varphi\cos C_2+\sin\varphi\sin C_2 \]
gives us
\[ \small \cos\theta = \frac{\cos C_2\sqrt{1-C_1^2}}{C_1} \sin \theta \cos\varphi+\frac{\sin C_2\sqrt{1-C_1^2}}{C_1} \sin \theta \sin\varphi. \]
By multiplying both sides by \(\small r\), we can replace it with the equation of a plane passing through the origin
\[\small z = \frac{\cos C_2\sqrt{1-C_1^2}}{C_1} x+\frac{\sin C_2\sqrt{1-C_1^2}}{C_1} y. \]
In other words, the path on this plane is the condition for a geodesic line. Since the plane passing through the origin and \(\small (x_1,y_1,z_1)\),\(\small (x_2,y_2,z_2)\) is:
\[ \small z = \frac{z_1y_2-z_2y_1}{x_1y_2-x_2y_1}x+\frac{x_1z_2-x_2z_1}{x_1y_2-x_2y_1}y, \]
we need to determine the integral constant so that it matches this equation. Finally, the geodesic line is a curve of the common coordinates of
\[ \small \begin{align*} &z = \frac{z_1y_2-z_2y_1}{x_1y_2-x_2y_1}x+\frac{x_1z_2-x_2z_1}{x_1y_2-x_2y_1}y \\ &x^2+y^2+z^2=r^2, \end{align*} \]
and we need to calculate the length of the intervals \(\small (x_1,y_1,z_1)\),\(\small (x_2,y_2,z_2)\). At this point it became too much of a hassle and I gave up…
If \(\small z_1=z_2=0\), then it is the same as a one-dimensional sphere, so the distance is:
\[ \small d = r\cos^{-1} \frac{x_1x_2+y_1y_2}{r^2}. \]
In addition, since this must be spherically symmetric (For example, if \(\small x_1=x_2=0\),
\[ \small d = r\cos^{-1} \frac{y_1y_2+z_1z_2}{r^2} \]
must hold true.), it is easy to guess that
\[ \small d = r\cos^{-1} \frac{x_1x_2+y_1y_2+z_1z_2}{r^2} \]
holds true after all, but it seems troublesome to perform a precise calculation.
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