Gravity in the 2+1 dimensions

In a previous post, we point out in the d+1 dimensions, the Coulomb's law reads, \[
\varphi(r) = \frac{\Gamma(\frac{d-2}{2})}{4\pi^{\frac{d}{2}}}\frac{1}{r^{d-2}}.
\] Apparently, in the 2+1 dimensions, this formula is divergent. To see this, first note that $x^\epsilon = 1 + \epsilon \log x$, and $\Gamma(\epsilon) = 1/\epsilon - \gamma$, the potential reads, \[
\varphi(r) \triangleq \lim_{\epsilon\to 0^+} \frac{\Gamma(\frac{\epsilon}{2})}{4\pi^{\frac{2+\epsilon}{2}}}\frac{1}{r^{\epsilon}} = \frac{1}{2\pi}\log\frac{1}{r} + \frac{1}{2\pi\epsilon} - \frac{1}{4\pi}\log e^\gamma\pi
\] We can regularize the potential by adding an infinitely large constant such that it obtains a finite value at a given distance $r_0$: \[
\varphi_r(r) = \frac{1}{2\pi}\log\frac{r_0}{r}.
\] But still, the potential divergences at infinity, which sort of violates the spirit of "locality".

The interpretation of this mathematical phenomenon is that in the 2+1 dimensions, long range correlation can be created with infinitely small energy. A consequence of this is the Coleman's theorem, stating that there is no spontaneous symmetry breaking of the continuous symmetry in the 2 dimensions.

In terms of the gravity, this infrared divergence is obviously non-physical. The potential becomes infinitely large near the source and far away from the source. A plausible interpretation is therefore $\varphi = \text{constant}$, at least for the type of gravity as we understand (We are not claiming the Poisson equation has no non-trivial solution in 2 dimensions.).

Of course, with a strong field strength, Newton's law should be replace by Einstein's equation: \[
R_{\mu\nu}  =  8\pi G \Big( T_{\mu\nu} - T g_{\mu\nu} \Big).
\] Here $T = g^{\mu\nu} T_{\mu\nu}$. In the vacuum, this equation reduces to $R_{\mu\nu} = 0$.  But this does not necessarily mean the space-time curvature is zero. The space time curvature is ultimately determined by the Riemann tensor $R^\mu_{\nu\sigma\rho}$. Non-vanishing curvature represents the propagation of the gravity from the source inside the vacuum.

Riemann tensor has $D^2(D^2-1)/12$ independent components, where $D=d+1$ is the space-time dimensionality. In the 2+1 case, it has 6 independent components, whereas the Ricci tensor $R_{\mu\nu}$ has $(3 \times 3 +3)/2 = 6$ independent components as well. Thus, Riemann tensor is expressible through the Ricci tensor: \[
R_{\mu\nu\rho\sigma} = \big( g_{\mu\rho}R_{\nu\sigma} - (\rho \leftrightarrow \sigma) \big) - (\nu \leftrightarrow \rho) - \frac{1}{2} (g_{\mu\rho}g_{\nu\sigma} - g_{\mu\sigma}g_{\nu\rho}) R
\] Here $R = g^{\mu\nu}R_{\mu\nu}$. As a result, Riemann tensor, hence the space-time curvature, is zero in the vacuum, in the 2+1 dimensions. This means gravity cannot propagator through the free space in the 2+1 dimensions! Einstein's equation is a full pledged dynamical theory of gravity with Newton's gravitational theory as its weak field approximation. Now, Einstein's theory actually agrees with Newton's theory that in the 2+1 dimensions there is no gravity!

Applying the gravitational theory in the 3+1 dimensions into the 2+1 dimension is somewhat unjustified. After all, non-body knows wether these are the correct modeling. However Newton's law is not an ad loc construction. It follows, as the Coulomb's law, the Poisson equation $\nabla^2 \varphi(x) = 0$. Einstein's equation has a stronger built-in beauty in it: it acquires a geometric interpretation. So these interesting observation is not a major consequence of our gravitational theory per se. It is more closely related to the special topology of the low dimensions, which is also well known. In fact, in the 2 dimensions, topology is more central than the local operators we cherish much in the 3 dimensions.

The Gravity Train Revisited

Introduction 

The gravity train is a conceptual of design of transportation making use of the gravity of the planet, first proposed by 17th century scientist Robert Hooke in a letter to Issac Newton. It features a tunnel through the Earth.

Suppose a straight tunnel was drill through the Earth. A train operating solely on gravity starts from one end of the tunnel. Assuming there is no friction inside the tunnel, and the density of the Earth is uniform. The problem asks how long it takes for the train to arrive at the other end.

In the classic solution, the spin (rotation) of the Earth is not considered. (Strictly speaking, the rotation effect can be factorized in by using a locally measured gravitational acceleration $g$). The answer then is $t = \pi \sqrt{\frac{R_E}{g}} \simeq 42 \mathrm{min.}$, regardless of the exact location of the tunnel, where $R_E \simeq 6371 \mathrm{km}$ is the radius of the Earth, $g \simeq 9.8 \mathrm{m/s^2}$ is the gravitational acceleration. The rotation of the planet introduces a second time scale: $T_s \simeq 24 \mathrm{hr} \simeq 17 T_G$, where  $T_G = 2\pi \sqrt{\frac{R_E}{g}}$. The separation of the scales validates the neglecting of the Earth spin in the classical solution. Nevertheless, in this essay we want to look at the correction effect introduced by the spin of the Earth.

Fig. 1

General Settings

Let the center of the Earth be the origin of our coordinate system. The tunnel is characterized by the position of the center of the tunnel $\vec d$, and the orientation of the tunnel $\hat \tau$, $\vec d \cdot \hat \tau = 0$. Let $\vec{r} = \vec{d} + x \hat{\tau}$ be the position of the train, where $x$ is the distance of the train from the center of the tunnel, $\vec{v} = \dot x\hat\tau$ the velocity of the train. The spin of the Earth is described by an angular velocity vector $\vec\Omega$. Suppose $\Omega \cos \theta \equiv \vec \Omega \cdot \hat \tau$, $\Omega d \cos\phi = \vec\Omega \cdot \vec d$.

In the reference frame of the Earth (a non-inertial frame), there are four forces acting on the train: the gravity $\vec F_G = -\frac{GM_Em}{R^3_E} \vec r \equiv -mg \frac{\vec r}{R_E}$, the centrifugal force $\vec F_e = - m \vec\Omega \times (\vec\Omega \times \vec r)$, the Coriolis force $\vec F_c = 2 m \vec v \times \vec \Omega$, and the contact force. Since there is no friction, the contact force simply counter all forces in the direction that perpendicular to the tunnel. We only need to consider the force along the tunnel. $\hat \tau \cdot \vec F_G = -m g \frac{x}{R_E}$, $\hat \tau \cdot \vec F_e = m \Omega^2 x - m \Omega^2 (d \cos\phi + x \cos \theta)$, $\hat \tau \cdot \vec F_c = 0$. The total force along the tunnel is $F = -mg\frac{x}{R_E} + m \Omega^2 ( (1-\cos\theta)x - d\cos\phi)$.

The equation of motion thus reads, \[
\ddot x = \left( -\frac{g}{R_E} + \Omega^2 (1-\cos\theta) \right) x - \Omega^2 d \cos\phi
\] The solution is a harmonic oscillator with angular frequency $\omega = \frac{g}{R_E} - \Omega^2 (1-\cos\theta)$ or period $T = 2\pi \sqrt{\frac{R_E}{g - \Omega^2R_E(1-\cos\theta)}}$. The center of the harmonic oscillator is not in the middle of the tunnel. In stead, it is shifted away from there by $\Delta = \frac{\Omega^2dR_E\cos\phi}{g - \Omega^2 R_E(1-\cos\theta)}$.

To calculate the travel time, we need to consider two cases. If the train starts from the high altitude side, it arrives at the low altitude terminal in time $t = \pi \sqrt{\frac{R_E}{g - \Omega^2R_E(1-\cos\theta)}}\left(1 - \frac{1}{\pi}\arccos\frac{\sqrt{R_E^2-d^2}-\Delta}{\sqrt{R_E^2-d^2}+\Delta} \right)$. If the train starts from low altitude terminal (farther from the axis of the Earth) with zero velocity, it would never reach the high altitude side. Some minimum velocity is needed in order to arrive the other side, $v_{\min} = 2 \Omega \sqrt{\left( g - \Omega^2R_E (1-\cos\theta) \right) d\cos\phi} \sqrt[4]{1-\frac{d^2}{R^2_E}}$. by doing so, it also takes time $t$ to arrive at the destination.

Discussions

The spin of the Earth affects the problem in two ways. The correction in the period of the oscillator is about $\sim \frac{\Omega^2 R_E}{g}\sin^2\frac{\theta}{2} = 0.345\% \sin^2\frac{\theta}{2}$. The second correction is through the displacement of the center of the harmonic oscillator. We compare $\Delta$ and $l = \sqrt{R_E^2 - d^2}$ half the length of the tunnel. $\Delta \simeq \frac{\Omega^2 R_E}{g} \cos\phi d = 0.345371\% \cos\phi \sqrt{R_E^2 - l^2}$. Compare $\Delta_{\max}$ (with $\phi = 0$) and $l$ (see Fig. 2). The correction from $\Delta$ is only small for tunnels either of thousands of kilometers long or well oriented (i.e. choosing a good $\phi$). Thousands of kilometers long tunnel would be for transcontinental express.

Fig. 2, the numbers for the tunnel length $l$ and maximum shift of the harmonic oscillator center $\Delta_{\max}$ are in kilometers.

We have only considered the straight tunnels. It turns out, straight tunnels are not the fastest ones for gravity train. The fastest tunnels are the brachistochrone tunnels. The correction due to the Earth rotation can be considered accordingly.

Newton's Law of Gravity for Solar System Planets (visualization)

According to Newtonian gravity theory \[ v(r) =\sqrt{ \frac{G M}{r}} \] If the orbit is circular, the speed is simply proportional to $1/\sqrt{r}$. For general cases, however, after some derivation, the average speed, \[ \bar{v} = \frac{1}{2\pi}\int_0^{2\pi} \mathrm{d}\theta v(\theta) = \sqrt{\frac{GM}{a} } \frac{2 \mathrm{E}(\frac{2\epsilon}{1+\epsilon})}{\pi \sqrt{1-\epsilon}}, \] where $\mathrm{E}(z) = \int_0^{\frac{\pi}{2}} (1-z \sin^2\theta)^{1/2} \mathrm{d}\theta$ is the elliptic function. Note that $\mathrm{E}(0) = \frac{\pi}{2}$, restoring the circular motion result. So, the average speed is proportional to $\frac{1}{\sqrt{a}}$ where $a$ is the semi-major axis.

Fig. 1: the semi-major axis vs. average orbital speed for solar system planets in linear coordinates

Fig. 2: the semi-major axis vs. average orbital speed for solar system planets in logarithmic coordinates

The best fit of the slope gives 29.779763 km/s/AU, which is about the earth average orbital speed. Using the data of solar mass and gravitational constant, the average eccentricity is about 0.0195386. This is the absolute value.