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In 3D, I can calculate the total force due to gravity acting on a point on the surface of the unit sphere of constant density, where I choose units so that all physical constants (as well as the density of the sphere) is 1:

$$F = 4\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} \tfrac{x+1}{\left[(x+1)^2+y^2+z^2\right]^{3/2}} dz\, dy\, dx = \frac{16\pi}{3}.$$

This force agrees with what we get if we treat the sphere as a point particle at the sphere's center of gravity with lumped mass $\frac{4\pi}{3}$. So far so good.

But now if I try to calculate the force of gravity of a point on the boundary of the unit disk in 2D, I get infinity:

$$F = 2\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \tfrac{x+1}{\left[(x+1)^2+y^2\right]^{3/2}} dy\, dx = \infty?$$ Intuitively, if I think of the total force as the sum of contributions from nested circular rings around the point of interest, the circumference of the rings scales like $r$, the force density like $\frac{1}{r^2}$, so the contribution of each ring scales like $\frac{1}{r}$ which diverges as $r\to 0$.

What is going on here? How is the gravity potential derived? Is it an accident that in 3D, $\frac{G m_1 m_2}{r}$ is harmonic on the punctured space $\mathbb{R}^3 \backslash\{0\}$? Is the "right" gravity potential in 2D something like $G m_1 m_2 \log r$? If so, why, and isn't it a paradox if point masses in 2D orbit according to a different law than coplanar point masses in 3D?

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You're right that if you take Newton's law of gravity as is and apply it to a 2D universe, you'll get an infinite result. So you do need to use a modified theory in two dimensions, or indeed in any number of dimensions other than three.

The proper way to do this is using general relativity, and if you apply GR to 2+1D spacetime, you get something that looks basically nothing like gravity as we know it. In particular, space is only distorted or "curved" where there is actually mass, unlike our universe where the distortion extends beyond the region that actually contains the mass. Because that distortion is what we recognize as gravity, in a 2+1D world there would be no gravitational attraction. The presence of mass would cause some geometrical oddities, but there would be no force acting between separated masses. (I can put in some math if you like.)

Before GR was invented, on the other hand, physicists would have tried to generalize Newtonian gravity to other numbers of dimensions using Gauss's law for gravitation, which is exactly equivalent to Gauss's law for electric fields. As you may know, the electric version of Gauss's law states that the surface integral of electric field over a closed surface is proportional to the enclosed charge, so analogously, the surface integral of the gravitational field over a closed surface is proportional to the enclosed mass.

$$\int \vec{g}\cdot\mathrm{d}\vec{A} = -4\pi GM$$

In 3D, for a sphere enclosing a point mass, this gives you $(4\pi r^2\hat{r})\cdot(g\hat{r}) = -4\pi GM$, or $\vec{F}_g = m\vec{g} = -\frac{GMm}{r^2}\hat{r}$, which is recognizable as Newton's law of gravitation. In 2D space, you instead get $(2\pi r\hat{r})\cdot(g\hat{r}) = -4\pi GM$, which leads to

$$\vec{F}_{g\text{(2D)}} = m\vec{g} = -\frac{2GMm}{r}\hat{r}$$

So to do your integral with 2D Newtonian gravity, you would replace the exponent $\frac{3}{2}$ in the denominator of the integral with $1$, as well as changing the constant (but I'll skip that, since you've set it to 1 anyway).

$$F = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \tfrac{x+1}{(x+1)^2+y^2} \mathrm{d}y\, \mathrm{d}x = 2\pi$$

I may have done that integral wrong, but the point is, the answer is finite. You could also use Gauss's law directly to see it: a unit circle filled with material of unit surface density encloses a mass $\pi$, so the 2D Newtonian gravitational acceleration at the surface would be $-\frac{2GM}{r} = -2\pi G$.

From this modified force, you could determine that the potential energy would indeed have to have a logarithmic form, just using $U = -\int \vec{F}\cdot\mathrm{d}\vec{r}$.

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"Because that distortion is what we recognize as gravity, in a 2+1D world there would be no gravitational attraction. The presence of mass would cause some geometrical oddities, but there would be no force acting between separated masses. (I can put in some math if you like.)" Could you put in the math, please? – Leos Ondra Oct 19 '12 at 9:41
I'll try to do that when I have a chance, but I won't have time for a couple days at least. – David Z Oct 19 '12 at 13:58
Could I politely ask for the math or a hint how it should look like? – Leos Ondra Nov 23 '12 at 9:39
@Leos yes, I haven't forgotten, I've just been so busy with other things that I haven't gotten to it. It's not quite a trivial calculation. I promise I will do this as soon as I have time though. – David Z Nov 24 '12 at 16:42
I have found the math in this paper: – Leos Ondra Dec 4 '12 at 15:04

The usual definition of gravity in different dimension makes it so that if space is 2d, then Newtonian gravity goes as 1/r not 1/r^2 (as David Zaslavsky points out, GR in 2+1 dimensions is even weirder--- gravity doesn't exert long range forces in 2d, but it creates point conical deficits. I will ignore GR in this answer), so that the potential is logarithmic. This is because the proper 2d 1/r gravity is harmonic on the punctured plane, just like 3d gravity. The orbits are not ellipses anymore, they aren't closed curves, just radially oscillations are with a different period than the angular oscillations. Also, all orbits are bound. This is not what you are asking about, but you should know it.

What you are doing is using the 3d potential for a mass density which has been squashed into a plane. The divergence you see is due to the fact that you have an infinite density on the plane--- you have a massive plate of thickness $\epsilon$ with a mass density per unit area, so an infinite mass density per unit volume.

Away from the circle edge of the plate, when you are close to the plate, it looks like an infinite planar mass density. In this case, the near-plate solution for the potential is that it looks like the absolute value function $|a|$ where a is the perpendicular distance from the infinite plate. This is the same as in electrostatics--- the solution for the gravitational field for an infinite plate with constant mass density is by Gauss's law. You have a gravitational field pointing toward the plate which is finite at the plate, but which has a finite jump at the plate, and the jump magnitude is proportional to the density per unit area. This means that the potential has a discontinuous first derivative, but it isn't infinite at the plate.

But right near the circular edge, the near plate solution is different. In the limit that you are very close to the edge, you are looking at a half-infinite plate with a finite mass density. This scaling limit can be solved using 2-d methods. If you put the plate in the z-x plane, and stretching along the negative x-axis, the potential is a function only of y and x, not of z, so there is no electric field along the z-direction. The electric field is a harmonic function away from the negative x axis, and it is a 2-d harmonic function. This means that G_x + iG_y is the real part of a holomorphic function (G is the gravitational field).

For a point mass, G_x + i G_y turns out to be the holomorphic (except for one point) function $1/z$. So each pole is the location of a mass. The finite mass density along the negative x-axis means that the harmonic function has a constant residue density along the negative x-axis. The residue density is the cut-discontinuity (as explained in this answer: Correlation function which has branch cut in momentum space ), so you find that the solution is the log function:

$$ E_x + i E_y = \log(z) $$

This is indeed divergent at z=0, but not on the plate. This explains why your integral is divergent, and tells you the asymptotic form of the divergence.

This divergence at the edge of a flat plate is not a surprise. It is even worse when you have a thin line, then the divergence is as 1/r. Plates have diverging fields, and thin needles have even more diverging fields. This is the principle of the lightning rod, you make a thin line at zero potential to cause strong electric fields which ionize and discharge the air. This is not quite the same, because your plate is not at equal potential, but it's the same general principle.

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"GR in 2+1 dimensions is even weirder--- gravity doesn't exert long range forces in 2d, but it creates point conical deficits." Do you mean that a point mass create conical singularity and the space remains euclidean around it? – Leos Ondra Oct 19 '12 at 9:43
@LeosOndra: Yes, exactly. in 3 dimensions (of any signature, so 2+1 too) the Ricci tensor and the Riemann tensor have the same number of independent components, so saying Ricci is zero means Riemann is zero. – Ron Maimon Oct 19 '12 at 13:08

What is going on here? How is the gravity potential derived?

In the context of non-relativistic gravity, simply look to the fundamental solution for Laplace's equation. From the Wikipedia article "Fundamental solution":

Laplace equation

For the Laplace equation,

$[-\nabla^2] \Phi(\mathbf{x},\mathbf{x}') = \delta(\mathbf{x}-\mathbf{x}')$

the fundamental solutions in two and three dimensions are

$\Phi_{2D}(\mathbf{x},\mathbf{x}')= -\frac{1}{2\pi}\ln|\mathbf{x}-\mathbf{x}'|,\quad \Phi_{3D}(\mathbf{x},\mathbf{x}')= \frac{1}{4\pi|\mathbf{x}-\mathbf{x}'|}$

So yes, in 2D, the potential is logarithmic.

If so, why, and isn't it a paradox if point masses in 2D orbit according to a different law than coplanar point masses in 3D?

It's not a paradox; the force law between the two point masses is different in 2D from what it is in 3D. Consider a purely radial orbit. Would you consider it a paradox that colinear point masses in 3D don't follow a 1D force law?

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If their velocities are all also colinear, then yes, I would have expected so... – user2617 Jun 23 '12 at 17:54
This is not exactly the question--- the question is why a disk has diverging field on its boundary. – Ron Maimon Jun 23 '12 at 18:43
It's likely I'm confused but, as it seems to me, there a number of questions including but not limited to "how is the 2D gravity potential derived?". Moreover, it seems likely to me that the OP isn't clear on the difference of working a 2D problem in a 3D space and working a 2D problem in a 2D space. – Alfred Centauri Jun 23 '12 at 20:37

Inverse gravity has consequences for black holes. Answers #2 and #3 above detail why and how 2D gravity can actually exist. In a spinning black hole where mass-energy has collapsed to a singularity, only the z direction has collapsed. Because all matter/energy entering a black hole must initially have an orbital velocity component, all such material must orbit at faster and faster speeds, approaching infinity, as the singularity is approached. Such material must orbit in concert. Initially, that is, it must orbit all in the same direction so that as its orbital radius shrinks the black hole can be thought of as if it was rotating faster and faster. As it shrinks it acts more and more like a single rotating body. Rotating faster than c below the event horizon, the nature of spacetime itself must change. Collapse in the z direction means that mass-energy-spacetime becomes unified to form, not a ring singularity a la Kerr, but to an indefinitely broad disk singularity that is singular only in the z direction and infinite in the xy directions. So, this mass-energy-spacetime entity exerts a 2D gravitational field because the spacetime component of its nature can ignore the event horizon.

It must become asymptotically flat as it approaches the origin. It must also become asymptotically flat as it approaches infinity. This is the symmetry of a hyperbola, a gravitational field that declines as 1/r. Einstein calculated G for the 1/r^2 gravity field but he did not bother with the 1/r case.

But, using v = (G*M/r*)^1/2, where r* = the unit vector of r for dimensional integrity, and using the M-sigma relation that shows a plot of M versus sigma, the standard deviation of velocities of bulge stars in a galaxy chaotically orbiting a supermassive black hole, we can empirically determine G*. It is about 2 x 10^-32 m^3 kg^-1 s^-2 .

The consequences of this are profound. A gravitational field that falls off more slowly than the classical Newtonian field accounts for not only the M-sigma effect, it accounts for the anomalous velocity dispersion and all the other phenomena associated with Dark Matter. This hyperbolic (1/r) supermassive black hole gravitational field IS Dark Matter.

The 1/r gravitational potential energy profile follows a logarithmic relation while the 1/r^2 potential energy follows a hyperbolic profile. When they are placed on the same scale and plotted in natural or geometric units, they can be superposed so that the origins and the values at r = 1 coincide. These equivalent potential energies can be thought of as depicting the potential energy profiles of the primordial black hole, the Inflaton Particle in Alan Guth's highly excited inflaton field of the false vacuum, which is just an excited 2D gravitational field.

As the time dependent quantum collapse of the inflaton field donates its potential energy to the unexcited "ground state" gravitational field, there comes a time when they are both equal to 1. Thereafter, the collapsing excited gravitational field begins to re-exert itself and collapses from higher and higher energies into the ground state field. The result is that objects in the universe begin to kinematically move farther and farther apart at an accelerating rate.

So, the hyperbolic gravitational field can not only explain all the phenomena of Dark Matter, by extension it explains all the phenomena of Dark Energy. The hyperbolic gravitational field IS Dark Energy too.

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I thought also for years at the Newton formula of gravitation. For the intensity of gravitational field I did a differential calculation and I obtained logarithmic law.

There are 3 cases:

  • on points placed on the surface of the spherical body is available Newton's formula
  • on points placed at infinite distance is available Newton's formula
  • on points placed at intermediary or, finite distances it is available the logarithmic distribution for the Intensity of gravitational field.

Gamma= k*M/L^2*ln(1-R1/L)

Practically for the Newton formula it is added the additional term ln(1-R1/L)

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protected by Qmechanic Sep 9 '15 at 9:58

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