# Gravity on a torus

1. Aug 27, 2007

### Gerard Westendorp

I was playing with a 2D model with point particles and gravitational
attraction. To avoid particles going off to infinity, I thought I might
turn the 2D space into a topological torus: connect x=L = x=0 and y=L to
y=0.

But now a problem occurs: What is the distance between 2 points?

Because of the periodic boundary, you get for the distance (s):

s^2 = (x2-x1+i*L)^2 + (y2-y1+j*L)^2

You could set i = j = 0, but that would destroy translation symmetry, it
would make the coordinates at 0 and L physically different from other
points.

So there are a multitude of distances, a grid of them described by the
integers (i,j).

But that does seem a bit weird. Any comments on this?

Gerard

2. Aug 30, 2007

### ebunn@lfa221051.richmond.edu

In article <46cfe90b$0$240\$e4fe514c@news.xs4all.nl>,
Gerard Westendorp <westy31@xs4all.nl> wrote:
>I was playing with a 2D model with point particles and gravitational
>attraction. To avoid particles going off to infinity, I thought I might
>turn the 2D space into a topological torus: connect x=L = x=0 and y=L to
>y=0.
>
>But now a problem occurs: What is the distance between 2 points?
>
>Because of the periodic boundary, you get for the distance (s):
>
> s^2 = (x2-x1+i*L)^2 + (y2-y1+j*L)^2
>
>You could set i = j = 0, but that would destroy translation symmetry, it
>would make the coordinates at 0 and L physically different from other
>points.
>
>So there are a multitude of distances, a grid of them described by the
>integers (i,j).
>
>But that does seem a bit weird. Any comments on this?

This is kind of a funny situation. The most natural-seeming thing to do is
to take all of the different values of (i,j) as "legitimate" distances.
That is, assume that any given particle is attracted to any other particle
at all of the other particle's locations in this hall of mirrors. But
if you do this, you find that the force on any given particle is a divergent
sum.

I suppose one way to get around this would be to give the graviton a mass --
that is, to replace the usual gravitational potential G m1 m2 / r with
a Yukawa potential (G m1 m2 / r) exp(- lambda r) for some parameter lambda.
That would tame the divergence, but it'd mean you were talking about a
physically different theory.

Whether that's what you should do or not depends largely on what you're
planning to use this model for, I suppose.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

3. Aug 30, 2007

### Alexey Popov

Gerard Westendorp wrote:

> I was playing with a 2D model with point particles and gravitational
> attraction. To avoid particles going off to infinity, I thought I might
> turn the 2D space into a topological torus: connect x=L = x=0 and y=L to
> y=0.
>
> But now a problem occurs: What is the distance between 2 points?

I think you must try to solve Poisson equation on the torus.

4. Aug 30, 2007

### Gerard Westendorp

Gerard Westendorp wrote:
> I was playing with a 2D model with point particles and gravitational
> attraction. To avoid particles going off to infinity, I thought I might
> turn the 2D space into a topological torus: connect x=L = x=0 and y=L to
> y=0.
>
> But now a problem occurs: What is the distance between 2 points?
>
> Because of the periodic boundary, you get for the distance (s):
>
> s^2 = (x2-x1+i*L)^2 + (y2-y1+j*L)^2
>
> You could set i = j = 0, but that would destroy translation symmetry, it
> would make the coordinates at 0 and L physically different from other
> points.
>
> So there are a multitude of distances, a grid of them described by the
> integers (i,j).
>
> But that does seem a bit weird. Any comments on this?

Think I get it now.

The particles simply see each other particle, plus all the infinite
number of images of of them. In 2D (ie gravity force varies as 1/r) this
would be a problem, because the series 1/x + 1/(x+L) + 1/(x+2L) + ...
doesn't converge. But with the inverse square law it is OK.

Gerard

5. Oct 28, 2007

### pirillo

This discussion is funny to me because apparently all the
string theory guys have solved this already---since they
speak of the description of gravity in an alternative world
say 2 space dimensions,
where one dimension is compactified and they claim to know
what the gravitational potential would be in such a case.

I wonder what their criterion is for what the potential of gravity
should be in such a case, since they need such a criterion to make
the claims they do.

I wonder if one of the guys "in the know" would read this and tell us
how their argument works.

6. Oct 29, 2007

### torre@cc.usu.edu

> This discussion is funny to me because apparently all the
> string theory guys have solved this already---since they
> speak of the description of gravity in an alternative world
> say 2 space dimensions,
> where one dimension is compactified and they claim to know
> what the gravitational potential would be in such a case.
>

I don't think string theory is meant to work as a gravity theory in
2-d. One is instead building a gravity theory, say, in ten dimensions
using
(quantum) fields in 2-d.

> I think you must try to solve Poisson equation on the torus.

This would seem to be exactly what is needed. A complication is that
a necessary condition for the Poisson equation (with a well-behaved,
non-negative mass density) to have a solution on the torus is that
the mass density vanishes everywhere!

To see this, just integrate the equation over the torus and use the
divergence theorem. This implies the total mass vanishes. Since the
mass density is non-negative, it must vanish everywhere.

charlie torre

7. Oct 31, 2007

### Igor Khavkine

On 2007-10-28, pirillo <ultraman2002@hotmail.com> wrote:
> This discussion is funny to me because apparently all the
> string theory guys have solved this already---since they
> speak of the description of gravity in an alternative world
> say 2 space dimensions,
> where one dimension is compactified and they claim to know
> what the gravitational potential would be in such a case.

Sure, string theorists sometimes include gravity on the 2D worldsheets
of strings, but most of the time it doesn't have much effect. The way
that gravity is defined in other than 4 dimensions, nowadays, is using
the Einstein-Hilbert action of general relativity (which is the integral
of the Ricci curvature scalar over space-time). In 2 and 3 dimensions,
this version of gravity behaves very differently from what you get by
reducing Newtonian gravity to 2 or 3 dimensions (which is described by
the 1 or 2 dimensional Poisson equation for the gravitational
potential).

In 2 and 3 dimensions, general relativity is in a sense trivial.
Einstein's field equations are R = T, where R is the Ricci curvature
tensor and T is the energy-momentum tensor. In 2 and 3 dimension, the
full Riemann curvature tensor can be altebraically related to the Ricci
tensor R. So, in the absence of matter, R = 0 implies the Riemann
curvature is also zero, which in turn implies that the metric is flat. A
flat metric implies no acceleration due to gravity. This is very
different from lower dimensional Newtonian gravity. Solving the Poisson
equation in in 1 or 2 dimensions still produces a non-trivial
gravitational potential.

For higher dimensional extensions of gravity, Newtonian equations still
work in the weak field limit. However, if some of these higher
dimensions are compact, then the shape behavior of the gravitational
potential as a function of the compactified coordinates strongly depends
on what kind of sources are used for matter. I'm not sure if point
sources or other kinds of sources are used in those cases. It is
possible that some matter distributions are not allowed, because there
will exist no solution to Einstein's equations compatible with them.
One would have to look at this issue in more detail.

Hope this helps.

Igor

8. Nov 4, 2007

### pirillo

To Igor and the others,

see
a misinterpretation of what I meant. I never said that the 2dimensions
(Torus) I was talking about, are the dimensions of the string world
sheet.
I'm saying the torus here is a toy model for --space--of spacetime
itself
E.g., a simpler version of R^{10} with one dimension made into a
circle.
Regardless, due to many of the same concerns many of the posters
before me expressed. I do not know how the write the potential for a
Gauge (EM--like) theory in such a compactified space. But since
people like Lisa Randall have in their book that "at long distances,
gravitation on a cylinder behaves a certain way and differently when
the
distance is small compared to the zize of the compact dimension--
as evidenced by the potential" eg somethis that at long distances
behaves
like r^2 will behave as r^4 at shoter distances etc. How do they get
this potential? Is it some local interaction index(to extract
potential As in QCD)
or is it a classical potential that one puts on the cylinder (say,
emanating
from a source).

9. Nov 9, 2007

### Igor Khavkine

On 2007-11-04, pirillo <ultraman2002@hotmail.com> wrote:

> Regardless, due to many of the same concerns many of the posters
> before me expressed. I do not know how the write the potential for a
> Gauge (EM--like) theory in such a compactified space. But since
> people like Lisa Randall have in their book that "at long distances,
> gravitation on a cylinder behaves a certain way and differently when
> the
> distance is small compared to the zize of the compact dimension--
> as evidenced by the potential" eg somethis that at long distances
> behaves
> like r^2 will behave as r^4 at shoter distances etc. How do they get
> this potential? Is it some local interaction index(to extract
> potential As in QCD)
> or is it a classical potential that one puts on the cylinder (say,
> emanating
> from a source).

The behavior of the potential that Lisa Randall and other string
theorists allude to in this context is just the classical Newtonian one
(there are arguments for why the more general general relativistic
situation reduces to this one).

To get the Newtonian potential for a stationary point mass, you discard
the time axis and solve the Poisson equation in the spatial directions
for the point mass distribution. In R^n, for n > 2, the solution is
proportional to 1/r^(n-2) (n=3 gives the usual 1/r potential). If you
solve for the Newtonian potential on a space where all dimensions except
three have been compactified, that is on R^3 x C^n, then the full
expression for the potential will be different. Here, C^n is some
compact manifold, it could be a circle (n=1), an n-torus (a product of n
circles), an n-sphere, or something more exotic. However, its behavior
very close and very far from the source is easy to obtain. If you are
very close to the source, much closer than the diameter of C^n, then the
local behavior of the potential is exactly the same as in the R^(3+n)
case: the potential diverges as 1/r^(1+n). On the other hand, if you are
very far from the source, the spatial dependence of the potential is
very weak in the compactified directions. Assuming that the potential is
constant in these directions, the potential becomes the same as for R^3:
the long distance decay of the potential will go as 1/r.

If the compactified dimensions are there and are very small, it makes
sense that we would only see the 1/r behavior on ordinary length scales.
However, when two particles would come very close to each other, they
would feel the stronger 1/r^(1+n) short range gravitational potential.
This hypothesis has been used to test for the existence of large
(centimeter or millimeter scale) extra dimensions, so far with negative
results (http://www.npl.washington.edu/eotwash/).

Hope this helps.

Igor

10. Nov 9, 2007

### Igor Khavkine

On 2007-11-04, pirillo <ultraman2002@hotmail.com> wrote:

> Regardless, due to many of the same concerns many of the posters
> before me expressed. I do not know how the write the potential for a
> Gauge (EM--like) theory in such a compactified space. But since
> people like Lisa Randall have in their book that "at long distances,
> gravitation on a cylinder behaves a certain way and differently when
> the
> distance is small compared to the zize of the compact dimension--
> as evidenced by the potential" eg somethis that at long distances
> behaves
> like r^2 will behave as r^4 at shoter distances etc. How do they get
> this potential? Is it some local interaction index(to extract
> potential As in QCD)
> or is it a classical potential that one puts on the cylinder (say,
> emanating
> from a source).

The behavior of the potential that Lisa Randall and other string
theorists allude to in this context is just the classical Newtonian one
(there are arguments for why the more general general relativistic
situation reduces to this one).

To get the Newtonian potential for a stationary point mass, you discard
the time axis and solve the Poisson equation in the spatial directions
for the point mass distribution. In R^n, for n > 2, the solution is
proportional to 1/r^(n-2) (n=3 gives the usual 1/r potential). If you
solve for the Newtonian potential on a space where all dimensions except
three have been compactified, that is on R^3 x C^n, then the full
expression for the potential will be different. Here, C^n is some
compact manifold, it could be a circle (n=1), an n-torus (a product of n
circles), an n-sphere, or something more exotic. However, its behavior
very close and very far from the source is easy to obtain. If you are
very close to the source, much closer than the diameter of C^n, then the
local behavior of the potential is exactly the same as in the R^(3+n)
case: the potential diverges as 1/r^(1+n). On the other hand, if you are
very far from the source, the spatial dependence of the potential is
very weak in the compactified directions. Assuming that the potential is
constant in these directions, the potential becomes the same as for R^3:
the long distance decay of the potential will go as 1/r.

If the compactified dimensions are there and are very small, it makes
sense that we would only see the 1/r behavior on ordinary length scales.
However, when two particles would come very close to each other, they
would feel the stronger 1/r^(1+n) short range gravitational potential.
This hypothesis has been used to test for the existence of large
(centimeter or millimeter scale) extra dimensions, so far with negative
results (http://www.npl.washington.edu/eotwash/).

Hope this helps.

Igor

11. Dec 4, 2007

### pirillo

Igor said:

" However, its behavior very close and very far from the source is
easy to
obtain. If you are very close to the source, much closer than the
diameter
of C^n, then the local behavior of the potential is exactly the same
as in the R^(3+n)
case: the potential diverges as 1/r^(1+n). On the other hand, if you
are
very far from the source, the spatial dependence of the potential is
very weak in the compactified directions. Assuming that the potential
is
constant in these directions, the potential becomes the same as for
R^3:
the long distance decay of the potential will go as 1/r. "

Igor , how could one justify this claim? In other words, roughly,
whats your
argument that very close to the source the potential behaves like
that?
Do you know The greens function for the torus explicitly? I've heard
people talk
of field lines but I dont know how to argue using field lines or if
such arguments
are very precise.

12. Dec 5, 2007

### Igor Khavkine

Pirillo, you may want to watch out for spurious line breaks in your
posts. As you can see, they are somewhat hard to read as they appear. If
you post through Google Groups, you should either insert your own breaks
(with up to 71 characters per line) or not insert any breaks (except in
between paragraphs) and let Google do the wrapping for you.

On 2007-12-03, pirillo <ultraman2002@hotmail.com> wrote:
> Igor said:
>
> " However, its behavior very close and very far from the source is
> easy to
> obtain. If you are very close to the source, much closer than the
> diameter
> of C^n, then the local behavior of the potential is exactly the same
> as in the R^(3+n)
> case: the potential diverges as 1/r^(1+n). On the other hand, if you
> are
> very far from the source, the spatial dependence of the potential is
> very weak in the compactified directions. Assuming that the potential
> is
> constant in these directions, the potential becomes the same as for
> R^3:
> the long distance decay of the potential will go as 1/r. "
>
> Igor , how could one justify this claim? In other words, roughly,
> whats your
> argument that very close to the source the potential behaves like
> that?
> Do you know The greens function for the torus explicitly? I've heard
> people talk
> of field lines but I dont know how to argue using field lines or if
> such arguments
> are very precise.

The argument is very simple. You do not need to know the torus Green
function explicitly. Although, it might be a good exercisefor you to sit
down with your favorite PDE book, find an explicit expression for it,
and check that these arguments give correct results.

If you pick the usual Cartesian coordinates on the non-compact as well
as the toroidal coordinates, then the Laplacian takes the same form as
in R^(3+n). Now, place a point charge somewhere and try to solve the
corresponding Poisson equation in its neighborhood. In a small
neighborhood around the charge, you can use spherical coordinates. An
arbitrary solution can be decomposed into spherical harmonics
(eigenfunctions of the angular part of the Laplacian; this is standard
separation of variables). Each harmonic will have a corresponding
r-dependent coefficient. From physical reasoning, all harmonics but the
lowest one (the spherically symmetric one) can be eliminated or
discarded as negligible: higher harmonics correspond to dipole,
qudruple, etc. charge sources, but we stipulate a monopole charge. It
remains to obtain the r-dependent coefficient by solving the radial
Laplace equation. The most divergent part of the solution corresponds
precisely to the well known 1/r^(1+n) solution for uncompactified space,
with possibly many subleading terms. The basic lession is that to obtain
the divergent part of the Green function, we need only consider a tiny
neighborhood around the source charge, so that the global space-time
structure and topology does not matter.

When considering the Green function far from the source, again, we are
looking for a solution that is spherically symmetric in the non-compact
dimensions and that tends to zero at infinity. It makes sense to further
assume that the gradient of the solution goes to zero as well. So, far
enough from the source, the gradient of the Green function becomes much
smaller than the inverse radius of the compactified dimensions. This
basically means that under these conditions the Green function is
essentially constant in the compactified coordinates. Using this
assumption in the Poisson equation drops the dimensionality of the
problem from 3+n to 3. But the solution in that case is again well
known, its the 1/r Newtonian potential.

Hope this helps.

Igor

13. Dec 5, 2007

### Igor Khavkine

Pirillo, you may want to watch out for spurious line breaks in your
posts. As you can see, they are somewhat hard to read as they appear. If
you post through Google Groups, you should either insert your own breaks
(with up to 71 characters per line) or not insert any breaks (except in
between paragraphs) and let Google do the wrapping for you.

On 2007-12-03, pirillo <ultraman2002@hotmail.com> wrote:
> Igor said:
>
> " However, its behavior very close and very far from the source is
> easy to
> obtain. If you are very close to the source, much closer than the
> diameter
> of C^n, then the local behavior of the potential is exactly the same
> as in the R^(3+n)
> case: the potential diverges as 1/r^(1+n). On the other hand, if you
> are
> very far from the source, the spatial dependence of the potential is
> very weak in the compactified directions. Assuming that the potential
> is
> constant in these directions, the potential becomes the same as for
> R^3:
> the long distance decay of the potential will go as 1/r. "
>
> Igor , how could one justify this claim? In other words, roughly,
> whats your
> argument that very close to the source the potential behaves like
> that?
> Do you know The greens function for the torus explicitly? I've heard
> people talk
> of field lines but I dont know how to argue using field lines or if
> such arguments
> are very precise.

The argument is very simple. You do not need to know the torus Green
function explicitly. Although, it might be a good exercisefor you to sit
down with your favorite PDE book, find an explicit expression for it,
and check that these arguments give correct results.

If you pick the usual Cartesian coordinates on the non-compact as well
as the toroidal coordinates, then the Laplacian takes the same form as
in R^(3+n). Now, place a point charge somewhere and try to solve the
corresponding Poisson equation in its neighborhood. In a small
neighborhood around the charge, you can use spherical coordinates. An
arbitrary solution can be decomposed into spherical harmonics
(eigenfunctions of the angular part of the Laplacian; this is standard
separation of variables). Each harmonic will have a corresponding
r-dependent coefficient. From physical reasoning, all harmonics but the
lowest one (the spherically symmetric one) can be eliminated or
discarded as negligible: higher harmonics correspond to dipole,
qudruple, etc. charge sources, but we stipulate a monopole charge. It
remains to obtain the r-dependent coefficient by solving the radial
Laplace equation. The most divergent part of the solution corresponds
precisely to the well known 1/r^(1+n) solution for uncompactified space,
with possibly many subleading terms. The basic lession is that to obtain
the divergent part of the Green function, we need only consider a tiny
neighborhood around the source charge, so that the global space-time
structure and topology does not matter.

When considering the Green function far from the source, again, we are
looking for a solution that is spherically symmetric in the non-compact
dimensions and that tends to zero at infinity. It makes sense to further
assume that the gradient of the solution goes to zero as well. So, far
enough from the source, the gradient of the Green function becomes much
smaller than the inverse radius of the compactified dimensions. This
basically means that under these conditions the Green function is
essentially constant in the compactified coordinates. Using this
assumption in the Poisson equation drops the dimensionality of the
problem from 3+n to 3. But the solution in that case is again well
known, its the 1/r Newtonian potential.

Hope this helps.

Igor

14. Dec 11, 2007

### jafrosty

distances

In your construct, the torus is occupying the central 'region'?

2d is going to give you some major problems, in the 4 dimensions that we are used to living, the torus exerts it's influence. That is what the torus is.