Jim Shao
In Newtonian mechanics, the gravitational force \(\vec{F}\) exerted on a particle of gravitational mass \(m\) by gravitational mass \(M\) is:
$$\vec{F} = -\dfrac{GMm}{r^3}\,\vec{r} \tag{1}$$Where \(G\) is the gravitational constant, \(r\) is the distance between the two particles, and \(\vec{r}\) is the direction vector from \(M\) to \(m\).
The potential energy \(U\) of particle \(m\) in the gravitational field of \(M\) is:
$$ U(r)=-\dfrac{GMm}{r} \tag{2}$$In general relativity, in the Schwarzshild spherically symmetric center gravitational field \(M\), the potential energy of particle \(m\) has an additional term compared to equation (2), because the metric of space-time is no longer flat,
$$ U(r)=-\dfrac{GMm}{r} - \dfrac{GML^2}{mc^2r^3} \tag{3}$$Where \(L\) is the angular momentum, \(L=rmv_\perp\), \(v_\perp\) is the velocity component of particle \(m\) perpendicular to the direction vector \(\vec{r}\), and \(c\) is the speed of light.
Therefore, the gravitational force exerted on \(m\) by \(M\) can be calculated as:
$$\begin{aligned} \vec{F} &= -\dfrac{\partial U(r)}{\partial r}\,\hat{r}_0 \\ &= -\dfrac{GMm}{r^3}\,\vec{r} \;-\; \dfrac{3v_\perp^2GMm}{c^2r^3}\,\vec{r} \end{aligned} \tag{4}$$According to Newton's second law, the equation of motion for a particle with inertial mass \(m\) is:
$$\vec{F} = m\vec{a} = m\,\dfrac{d^2\vec{r}}{dt^2} \tag{5}$$Let (5) = (4), if we consider that the gravitational mass of particle \(m\) is the same as its inertial mass, after eliminating \(m\) from both sides, we can obtain its second-order differential equation of motion:
$$\dfrac{d^2\vec{r}}{dt^2} = -\left(1 + \dfrac{3v_\perp^2}{c^2}\right)\dfrac{GM}{r^3}\,\vec{r} \tag{6}$$In a two-dimensional rectangular coordinate system, according to equation (6), we may write down the equations of motion in \(x\)- and \(y\)-direction for particles \(M\) and \(m\) respectively.
After non-dimensionalization and order-reduction, yield eight first-order equations. Then, using the Runge-Kutta numerical method, the instantaneous coordinates and velocity of each particle are calculated, and their trajectories are plotted.
Discussion 1
In 1859, French astronomer Leverrier discovered that even considering the perturbations of the other seven planets, and even including the effect of the Solar oblateness, the observed precession of Mercury's orbit around the Sun was still 43 arcseconds per century more than expected.
In 1915, Einstein published his theory of general relativity. After revising Newton's equations for calculating gravitational motion, he could accurately calculate this missing 43 arcseconds.
For over a century, the explanations of general relativity and the analytical solutions to its field equations have largely consisted of theoretical and mathematical derivations. Many readers, faced with complex formulas and strange symbols, struggle to understand them, often becoming intimidated and frustrated, then attributing simply to "space-time curvature" without further understanding.
However, precession is ultimately a problem of classical mechanics. This paper attempts to qualitatively, intuitively, and vividly explain some concepts of general relativity through force analysis and numerical solutions.
Figure 1 shows that if the Sun \(M\) had only one planet \(m\), according to Newtonian mechanics, \(m\) would orbit the Sun stably in an elliptical orbit without precession.
Figure 1.
However, according to general relativity, the force on a particle moving in a gravitational field has one more term than in Newtonian equation (1), as shown in the equation (4).
Our numerical calculations show that it is this tiny force, proportional to \(\dfrac{v_\perp^2}{c^2}\), that causes the precession of the planet's orbit around the Sun, as shown in Figure 2.
Figure 2.
In the case of Mercury, due to its proximity to the Sun and the high eccentricity of its elliptical orbit, its perihelion radius \(r\) becomes very small, and \(v_\perp\) is at its maximum. Therefore, near the immense mass of the Sun, this tiny precession, originating from general relativistic effects, becomes directly observable.
Discussion 2
With the concept of "force" and through simulations and comparisons, this long-unexplained "redundant" precession seems easier to understand.
However, attentive readers may not be satisfied and will ask, "Where does this 'force' come from?"
Indeed, this extra gravitational term is quite strange; it is related to \(v_\perp\). If the particle \(m\) is stationary in the field \(M\), or only moves along \(\vec{r}\) direction, this force does not exist. Why is this?
To understand this, we must return to the concept of "space-time curvature."
Imagine a train traveling along a curve. Due to inertia, the wheels press against the outer rails, while the outer rails, fixed to the ground, exert back a constraint force on the wheels. It is this constraint that allows the train to move along the curve.
Similarly, when a particle moves in a strong gravitational field, if space-time is flat, according to Newtonian mechanics, the gravitational force acting on the particle perfectly balances the centrifugal force caused by its inertia, allowing it to move precisely along a quadratic curve within the gravitational field.
However, according to general relativity, space-time is a property of matter, and a gravitational field is a form of matter with curved space-time. The non-radial motion of a particle within it "compresses" the surrounding space-time ©¤the "railway"¡ªand space-time itself responds with the same "force," constraining the particle's motion. This "constraint" from the gravitational field's space-time manifests as the precession of the elliptical orbit itself.
In short, finding the "force" from the "curvature of space-time" might be the key to understanding general relativity.
Written on March 22, 2026.
Relevant Reading£º
Jim Shao£º"The Three-body Problem" (April 28, 2024)
Jim Shao£º"Entropy and State" (October 17, 2023)
Jim Shao£º"Langton's Ants" (August, 2017)