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Hendrik van Hees science forum addict
Joined: 15 May 2005
Posts: 60

Posted: Fri Jun 23, 2006 3:22 am Post subject:
Re: physics or mathematics of musical scales. (was This Week's



David W.Cantrell wrote:
Quote:  Perhaps something got lost in "translation". Of course, before Bach
there were keyboard instruments (Klavier, in the _general_ sense)
which were not welltempered. That was your point. But the piano _per
se_ did not exist before Bach. It was invented during his lifetime
(spec. in 1709) by Cristofori, who originally called it "gravicembalo
col pian e forte". IIRC, it is a matter of some speculation whether
Bach ever actually encountered a pianoforte.

As far as I know, he infact tried some pianofortes, and he was not
pleased at all.

Hendrik van Hees Texas A&M University
Phone: +1 979/8451411 Cyclotron Institute, MS3366
Fax: +1 979/8451899 College Station, TX 778433366
http://theory.gsi.de/~vanhees/faq mailto:hees@comp.tamu.edu 

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David W. Cantrell science forum Guru
Joined: 02 May 2005
Posts: 352

Posted: Thu Jun 22, 2006 6:43 am Post subject:
Re: physics or mathematics of musical scales. (was This Week's



In article <20060620125902.679$57@newsreader.com>, "David W.Cantrell"
<DWCantrell@sigmaxi.org> wrote:
Quote:  Dik.Winter@cwi.nl (Dik T. Winter) wrote:
Before Bach there were piano's that were not welltempered.
Perhaps something got lost in "translation". Of course, before Bach
there were keyboard instruments (Klavier, in the _general_ sense) which
were not welltempered. That was your point. But the piano _per se_ did
not exist before Bach. It was invented during his lifetime (spec. in
1709) by Cristofori, who originally called it "gravicembalo col pian e
forte". IIRC, it is a matter of some speculation whether Bach ever
actually encountered a pianoforte.

I didn't remember correctly, as pointed out by Gene. My retraction appeared
yesterday in sci.physics and sci.math, but not here. Let me correct that
now:
"Gene Ward Smith" <genewardsmith@gmail.com> wrote:
Quote:  Silbermann took up making early fortepianos, and Silbermann of course
knew Bach. The story as I understand it is that Silbermann showed Bach
one of his early pianos, and Bach didn't like it, but liked later
versions and even promoted them. Do you have a cite for the
controversy, by any chance?

My response:
I apologize. My memory just needed some refreshing, let's say. I have no
reason to suppose that <http://en.wikipedia.org/wiki/Gottfried_Silbermann>
is inaccurate. So Bach must have seen one, and he liked Silbermann's later
instruments better. But I doubt that he liked them well enough to buy one
for himself. There were quite a few instruments inventoried at the time of
his death (including IIRC more than one Lautenwerk) but no pianoforte.
David 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Thu Jun 22, 2006 6:42 am Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



I wrote:
Quote:  One way to see why the products of some rotations will be boosts is as
follows.
Suppose that for a pair of linearly independent vectors u and v in R^3 we
can find an element of O(2,1) with the following properties:
* it preserves both u and v
* its determinant is 1
* its square is the identity
If we can find such an element, we'll call it ref(u,v).
Now, if we pick three linearly independent vectors u, v and w so that
ref(u,v), ref(u,w) and ref(v,w) all exist, we can construct the following
elements of SO(2,1):
g1 = ref(u,v) ref(v,w) which preserves v
g2 = ref(v,w) ref(u,w) which preserves w
g3 = ref(u,v) ref(u,w) which preserves u
and we have g3 = g1.g2 because ref(v,w)^2 = I.
So if we pick a spacelike u and a timelike v and w, we'll have a boost
that's equal to a product of rotations.
The one question that remains is, when can we find ref(u,v)? This is
trivial in O(3), but O(2,1) is trickier. Generically we can construct a
normal n to the plane spanned by u and v:
n^a = g^{ab} eps_{bcd} u^c v^d
and if it's not a null vector we can construct ref(u,v) as the projector
into the plane minus the projector onto n. But if n is null, I don't
think ref(u,v) exists.

I think the only way the normal to the plane can be null is if the plane
is tangent to the light cone, in which case it will contain a single null
ray (which is itself normal to the plane), and all the other vectors in
it will be spacelike.
So if (but not only if) u and v are timelike, ref(u,v) will exist.
What's the significance of this construction failing if two of the
vectors lie on a tangent plane to the light cone? Well, if u and v are
*not* on such a plane, then you can choose w to be almost anywhere:
anywhere such that (u,w) and (v,w) also don't lie on tangent planes to
the light cone. In other words, given such generic u and v, you can find
SO(2,1) elements g_u and g_v that preserve them, *and* such that the
eigenvector of g_u g_v lies *almost* anywhere in R^3.
However, if u and v lie on a tangent plane to the light cone, then I
think the eigenvector of (g_u g_v) will always lie on that same tangent
plane, for all g_u and g_v that preserve u and v respectively.
The upshot of this for collisions in 2+1 gravity would be that certain
tachyontachyon and tachyonluxon collisions (which yield either tachyons
or luxons as the outgoing particle, never tardyons) involve *coplanar*
vectorvalued momenta.
What I'm claiming amounts to the statement that for each null ray n there
is a subgroup H(n) of SO(2,1), consisting of those elements whose
eigenvectors are normal to n.
So the simplest way to check this would be to look for a subalgebra of
so(2,1) consisting of elements whose null space lies in the plane normal
to n.
If we take the example of n being the null ray generated by (1,1,0), then
so(2,1) elements of the form:
 0 a b 
 a 0 b 
 b b 0 
have their null space generated by (b,b,a), which is always normal to
(1,1,0). And this set turns out to be closed under the Lie bracket, with
[e_a,e_b]=e_b (where by e_a I mean the element above with a=1, b=0, and
by e_b vice versa).
So it looks like those subgroups of H(n) do exist. But they won't be
Abelian! So those weird coplanar collisions in 2+1 gravity still won't
obey the old vector addition law. 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Thu Jun 22, 2006 6:41 am Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



John Baez wrote:
Quote:  So one thing you're saying
is that a product of rotations in two different rest frames can be
a boost??

One way to see why the products of some rotations will be boosts is as
follows.
Suppose that for a pair of linearly independent vectors u and v in R^3 we
can find an element of O(2,1) with the following properties:
* it preserves both u and v
* its determinant is 1
* its square is the identity
If we can find such an element, we'll call it ref(u,v).
Now, if we pick three linearly independent vectors u, v and w so that
ref(u,v), ref(u,w) and ref(v,w) all exist, we can construct the following
elements of SO(2,1):
g1 = ref(u,v) ref(v,w) which preserves v
g2 = ref(v,w) ref(u,w) which preserves w
g3 = ref(u,v) ref(u,w) which preserves u
and we have g3 = g1.g2 because ref(v,w)^2 = I.
So if we pick a spacelike u and a timelike v and w, we'll have a boost
that's equal to a product of rotations.
The one question that remains is, when can we find ref(u,v)? This is
trivial in O(3), but O(2,1) is trickier. Generically we can construct a
normal n to the plane spanned by u and v:
n^a = g^{ab} eps_{bcd} u^c v^d
and if it's not a null vector we can construct ref(u,v) as the projector
into the plane minus the projector onto n. But if n is null, I don't
think ref(u,v) exists. 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Mon Jun 19, 2006 8:51 pm Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



I wrote:
Quote:  I just figured out how to construct the complete spacetime for a general
2particle collision in 2+1 gravity. It's mindbogglingly simple.

D'oh! It *is* simple, but I messed up a crucial detail. And the reasons
stretch back a long time, to when I stupidly calculated "g1.g2" for the
holonomy of going around loop 1 and then loop 2, when of course that
should be "g2.g1".
The upshot is that when two particles collide, observers on either side
of the "plane" of the collision will think that the outgoing particle is
coming *towards* them, not moving away from them.
So you really do need to get these ordering conventions right in 2+1
gravity, or you won't know when a particle is coming to smack you in the
face.
Quote:  Take one copy of Minkowski spacetime. Draw in two world lines W1 and W2
for the two incoming particles, coming in from the past and meeting and
terminating at some point C. Then draw in the world line W3 for the
outgoing particle, starting at C and heading off in to the future. Don't
make these lines all coplanar (unless you're interested in a trivial
limiting case).

The above is all OK.
Quote:  Throw away all of the spacetime that lies *inside* the
"infinite tetrahedron" with its vertex at C and the three world lines as
its edges, leaving behind a kind of "concave infinite tetrahedron".
Now take a mirror image of that "concave infinite tetrahedron", and glue
it to the original along all three congruent faces.

Don't use the concave tetrahedron defined by the three world lines, use
the convex one. You still glue it facetoface to its mirror image.
Everything works now. This recipe gives angular deficits around the
incoming particles' world lines (the concave one didn't, it gave angular
excesses) and it gives what I now realise is the correct behaviour for
the geodesics of observers on either side of the collision: it makes
them collide!
A slice that cuts through both tetrahedra along the outgoing particle's
world line will look like this, where G and G' are geodesics through
observers at B and B'.
........ ........
 \ / 
 \ / 
G'  \ W3 W3 /  G
 \ / 
 \ / 
   
B' +   + B
  L L  
   
   
............... ............... 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Sat Jun 17, 2006 3:24 pm Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



John Baez wrote:
Quote:  In article <1149228612.361089.97090@j55g2000cwa.googlegroups.com>,
Greg Egan wrote:

[snip discussion of a collision between two particles in 2+1 gravity,
assuming that they combine to form a single new particle, and what
happens if we measure the groupvalued momentum of the outgoing particle
from two points, B and B', on either side of the plane of symmetry of the
problem.
In both cases, the total groupvalued momentum corresponds to a
vectorvalued momentum that seems to imply that the particle is moving
away from the observer.]
Quote:  So, are the geodesics through B and B' being made to diverge? Or were
they never really parallel to begin with? Or is there some reason why
these questions themselves are nonsensical?
I'm not completely sure, but I'm *hoping* the problem is this:
You're expecting that questions about the energymomenta of particles
have answers independent of the choice of "observer" B and little
loops encircling the particles... and they don't!

I'm not expecting the energymomenta to have values independent of these
choices, but I *do* expect the energymomenta to give me *some*
gaugeindependent physical information about what's happening in relation
to the various objects in the system  and this implies being able to
translate between and reconcile several observer's measurements.
I mean, I expect to be able to tell from a particle's energymomentum
whether it's going to come and smack me in the face, or whether the
distance between us is increasing. (I know that "distance" is a subtle
concept here, because there are different homotopy classes for the paths
that connect any two things.)
Let's consider this collision again. Two particles of equal rest mass m
collide, and combine into a third particle. Let's call the world lines
of the ingoing particles W1 and W2, and that of the outgoing particle W3.
I'm pretty sure there are three isometries of this spacetime:
(1) a kind of "reflection" F1 in the "plane" of the collision, which
preserves every individual point on all three world lines; we'll say
P1={x  F1(x)=x}.
(2) another "reflection" F2 that preserves W3, but exchanges W1 and W2;
we'll say P2={x  F2(x)=x}.
(3) a kind of "180degree rotation" R that exchanges W1 and W2, and
preserves W3; we'll say L={x  R(x)=x}. L should be the intersection of
P1 and P2.
I've used lots of words in quotes here because I don't mean them to be
taken literally, but I still think they give a useful sense of what's
going on.
Now let's choose B to lie somewhere on P2, the "plane" that divides the
incoming particles from each other. We parallel transport tangent
vectors from W1 and W2 to B along paths that don't cross either P2 or P1,
and get two timelike vectors in B's tangent space, u1 and u2. We choose
a righthanded orthonormal basis for the tangent space at B, putting e_t
= normalised (u1+u2), e_x = normalised (u1u2), and e_y orthogonal to
both and pointing towards L, by which I mean that the shortest spacelike
geodesic lying wholly in P2 that runs from B to L has a tangent vector at
B whose y component is +ve.
With this construction, W1 is on the right if I'm standing at B looking
out along e_y. The vectorvalued 4momenta m u1 and m u2 have zero y
components, and if we follow through all the holonomy computations using
the scheme I outlined previously, we end up with a 4momentum M u3 for
the outgoing particle with a nonzero, +ve y component. I would hope
that if I parallel transport a tangent vector to W3 to B along a path
lying wholly in P2, it would agree with u3.
My initial surprise was that when I paralleltransport tangent vectors
from the three world lines to my base point B, those paralleltransported
vectors are not coplanar. On reflection, though, that really shouldn't
surprise me. If you take a cone and draw two lines on its surface where
it intersects a plane of symmetry, then paralleltransport tangent
vectors from those lines to some common point and compare them, they
won't be colinear.
If we apply the isometry R to everything in sight, we'll get some
B'=R(B), images of all the paths we've used, and a map between the
tangent spaces, R*:T(B)>T(B), which all fit together to give a
description at B' that is identical to that at B.
So, both observers measure W3 as moving away from them, if their ideas of
"standing still" are represented by e_t and e_t'=R*(e_t) respectively.
Having thought about this some more, I *think* the spacetime geometry on
P2 can be formed by gluing together these two pieces:
................... ...................
 / \ 
 / \ 
G'  / W3 W3 \  G
 / \ 
 / \ 
   
B' +   + B
  L L  
   
   
............... ...............
Here G is a timelike geodesic through B with a tangent vector there of
e_t, and G'=R(G).
This would seem to make sense of the measurements at both B and B'. If
this is the correct solution, then the outgoing particle really *is*
moving away from both observers, and after the collision they're moving
away from each other (at least as measured in a certain reasonable way). 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Sat Jun 17, 2006 3:24 pm Post subject:
Re: 2+1 gravity (was This Week's Finds 232)



I previously described some numerical calculations of the total
groupvalued momentum of N particles of equal rest mass and speed, and a
uniform angular distribution of velocities.
I've updated the plots at the end of:
http://gregegan.customer.netspace.net.au/GR2plus1/GR2plus1.htm
to make the results clearer. They previously showed the trace of the
SL(2,R) holonomy, which doesn't really give enough information to see
what's going on.
It turns out to be quite easy to get the same results analytically.
Suppose we have a "ring of dust" of uniform density, expanding into empty
space. Some straightforward Lorentzian geometry shows that, if we excise
wedges around each particle's world line proportional to their mass, and
take the continuum limit, the circumference of a circle outside the ring
will be:
C(r) = 2pi (r  alpha (rvt))
= 2pi ((1alpha) r + alpha v t)
where v is the speed at which the ring is expanding,
alpha = mu / sqrt(1v^2),
mu = M / m_max,
M is the scalar sum of rest mass in the ring, and
m_max is the particle mass that gives an angular
deficit of 2pi; in units where G=c=1, m_max=1/4.
So the metric for the vacuum outside the ring will be:
ds^2 = dt^2 + dr^2 + (r  alpha (rvt)) dphi^2
for r > vt. Inside the ring we can just use ordinary flatspacetime
polar coordinates.
Global issues
=============
alpha < 1

On each surface t=constant the circumference is wellbehaved, increasing
with r, so the space outside the ring can be infinite. At t=0 we ought
to be able to join this metric to a similar one describing a contracting
ring for t < 0:
ds^2 = dt^2 + dr^2 + (r  alpha (r+vt)) dphi^2
alpha = 1

The circumference is C = 2pi alpha v t, independent of r. The space
outside the ring is still infinite for t>0, but there's a singularity at
t=0.
If mu <= 1, alpha = 1 occurs when v is equal to:
v_crit = sqrt(1mu^2)
whereas if mu > 1, we have alpha > 1 for all v.
alpha > 1

On each surface t=constant, the circumference decreases with r, so we
need to restrict r to values less than:
r_crit = alpha vt / (alpha1)
where C(r_crit)=0.
If we allow all values r < r_crit then there will be a singularity at
r=r_crit. It might be possible to "plug" this with a particle sometimes,
but I think not in general.
Alternatively, at least for 1 < alpha < 2, and v > 0, we can splice this
solution onto another expanding ring! If we choose:
alpha' = 2  alpha
v' = any value between 0 and v
mu' = alpha' sqrt(1v'^2)
and a new coordinate:
r' = (2v  (v+v') alpha') t / (1alpha')  r
then there is a second expanding dust ring at r'=v't. The reason we need
v' < v is that the r coordinate corresponding to r'=v't is:
vt + (vv') t / (1alpha')
so this is necessary to put the second dust ring outside the first (or
both rings outside each other).
The complete solution is topologically like an FRW expanding sphere, with
a singularity at t=0.
The message from all this seems to be that, although it might not be
unphysical nonsense to talk about alpha (and even mu) exceeding 1, it's
an issue of cosmology, rather than the kind of thing that might arise by
varying conditions in a laboratoryscale experiment.
Holonomy
========
The covariant derivative in the phi direction, anywhere outside the ring,
acts on the normalised coordinate vector fields as the following element
T of so(2,1):
 0 0 v alpha 
T =  0 0 (1alpha) 
 v alpha (1alpha) 0 
This is independent of t and r, as would be expected.
The loop holonomy will then be exp(2pi T), but we don't actually need to
perform this exponentiation.
First, note that T annihilates the vector:
n = (alpha1)/(v alpha) e_t + e_r
So n will be preserved by the holonomy, i.e. the holonomy will either be
a rotation around this vector, if it's timelike, or a boost orthogonal to
it, if it's spacelike. If we substitute:
alpha = mu / sqrt(1v^2)
we find that:
g(n,n) = 1  ( (musqrt(1v^2)) / (mu v) )^2
and that n changes from timelike to spacelike when:
v = v_tach = (1mu^2)/(1+mu^2)
Note that prohibiting values of alpha greater than 1 does *not* prevent
the transition to a tachyonic holonomy. For example, if mu=1/2, we have
alpha = 1 at v = v_crit:
v_crit = sqrt(3)/2 = 0.866, but
v_tach = 3/5 = 0.6
The easiest way to quantify the rotation/boost is to find the nonzero
eigenvalue of T^2. There is only one, its eigenspace being the plane
orthogonal to n, and it's equal to:
lambda =  (alpha^2 (1v^2)  2 alpha + 1)
or, making the substitution for alpha in terms of mu and v:
lambda =  (mu^2  2 mu/sqrt(1v^2) + 1)
When v=v_tach, lambda=0.
If the holonomy is a rotation, it will be by an angle:
+/2pi sqrt(lambda)
and if it's a boost, it will have a rapidity of:
+/2pi sqrt(lambda)
where we need to look at T and check some orientations to get the sign
correct for particular cases.
As v increases from 0 to v_tach, sqrt(lambda) falls monotonically from
mu1 to zero.
For the case mu < 1, the rotation takes the ve sign, starting at 2pi
(1mu) and shrinking in magnitude to zero as v increases, which is
equivalent to a +ve rotation by 2pi mu steadily *increasing* towards 2pi
as v approaches v_tach.
For mu > 1, the rotation takes the +ve sign, starting at 2pi (mu1) and
shrinking to zero. So the rotation, and hence the equivalent rest mass
of the total system, *decreases* as v increases.
For mu > 2, again the rotation takes a +ve sign, and its starting value
2pi (mu1) is greater than 2pi, so its value mod 2pi will exhibit one or
more discontinuous jumps as it crosses multiples of 2pi on its way down
to zero.
Whatever the value of mu, as v crosses v_tach lambda hits zero, becomes
positive, and increases monotonically. So the tachyonic mass of the
system starts at zero at v=v_tach and increases without bound.
My ultimate aim with all this was to understand what would happen in 2+1
gravity if you fired a relativistic particle at ever greater speeds into
a container of gas. An expanding ring of dust might not look much like a
gas held at constant volume, but the way its total groupvalued momentum
changes with v ought to be similar ... 

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John Baez science forum Guru Wannabe
Joined: 01 May 2005
Posts: 220

Posted: Fri Jun 16, 2006 10:36 pm Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



In article <abergman5F7D70.18103430052006@geraldo.cc.utexas.edu>,
Aaron Bergman <abergman@physics.utexas.edu> wrote:
Quote:  If one calculates for a while, one sees that g is a function of
the traditionally defined energymomentum of the particle, say p:
g = exp(kp)
where k is Newton's constant times something like 4pi.
I'm sorry, but I'm still confused. Should I think of this 'traditionally
defined energymomentum' as a sort of ADMlike thing in 2+1D?

Yeah.
More to the point: however you want to do it, you can figure out the
solution of GR in 2+1 dimensions that's the gravitational field
of a static point particle of mass m. Then you can boost it to
get the solution corresponding to a point particle of energymomentum p.
Then you can work out that this solution only depends on exp(kp)!
And then you can work out that exp(kp) is more useful than p:
Quote:  I'm sure you can define all the things as you say, but I'm not quite
sure that this is physically significant.

I doubt gravity in 2+1 dimensions is physically significant!
But, I know that the way I'm talking about it is a useful way
to talk about it. If you've got particles moving around in
2+1 dimensions, you've got a spacetime that's flat except along
their worldlines. And if you're trying to characterize a
spacetime that's flat except along certain worldlines, you'll
inevitably be led to study the holonomies of loops that wrap
around these worldlines  after all, the moduli space of flat
Gconnections on a spacetime M is hom(pi_1(M),G)/G, where the
homomorphism from pi_1(M) to G is obtained by studying holonomies
around loops.
And then, it turns out that the conservation laws which hold when
particles collide are most simply described in this language!
Quote:  Does this just translate to a holonomy in the related CS theory?

I'm using the BF description of gravity, where you've got a
Lorentz connection A and a triad field e. You can translate
this into a ChernSimons description, where the only field is
a Poincare group connection, built by combining A and e.
But, that's not what I've been talking about: all the holonomies
I'm talking about take values in the Lorentz group, not the Poincare
group. 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Thu Jun 15, 2006 4:39 pm Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



I wrote:
Quote:  [I]magine a cloud of particles with world lines starting from the
origin of Minkowski space with all possible velocities, spreading out at
all angles. By cutting out wedges around their world lines, we ought to
be able to get one of the expanding hyperboloid FRW solutions, where each
spacelike slice at time t is isometric to the unit hyperboloid multiplied
by sqrt(1+2C) t, and the world lines of all the particles are normal to
every spacelike slice.
In hyperbolic geometry the area and circumference of a circle increase
with distance faster than in the Euclidean plane, but as we move out
through our cloud of particles in Minkowski space with wedges cut out,
we're *losing* area and circumference. So all the hyperboloids in the
original Minkowski space get flattened out. I guess the trick is to find
a way to make sure that their geometry is still that of a hyperboloid, but
a slightly larger, hence flatter, one ... by a factor of sqrt(1+2C). But
I haven't been able to figure out yet how to get the particles distributed
in the original Minkowski spacetime in such a way that everything works
out nicely.

Amazingly enough, this does work! The geometry of the hyperboloids in
Minkowski spacetime is changed, by the excision of wedges, to exactly
that of the corresponding FRWsolution spacelike slices.
Calculations follow, for anyone interested in the grungy details ...
If a deficit angle A originates around a particle of velocity v on the
unit hyperboloid, then the amount of the azimuthal coordinate that this
steals further out at velocity w is, to first order in A:
A (wv) / ( w sqrt(1v^2) )
Rephrasing this in terms of geodesic distances along the unit hyperboloid
from the "pole" at (1,0,0), if the distances are r for v and q for w, we
get:
A (cosh(q)  sinh(q)/tanh(r))
To find T(r), the total angular deficit at distance r, we need to
integrate the deficit angles for all the matter out to r, weighted by
this adjustment factor. We can't actually know the amount of matter at
distance q without knowing the circumference of a circle at distance q
... which depends on the angular deficit. This leads to an integral
equation, but we don't need to solve it from first principles, because
the guess that the geometry remains hyperbolic but just changes radius
turns out to be correct.
On a hyperboloid of radius a, the circumference of a circle of radius r
is:
c_a(r) = 2pi a sinh(r/a)
On the unit hyperboloid with a deficit angle T(r), the circumference is:
c_{1,d}(r) = (2pi  T(r)) sinh(r)
If the deficit angle T(r) is to mimic a change of hyperboloid radius from
1 to a, making these two circumference formulas equal, we must have:
T(r) = 2pi (1  a sinh(r/a) / sinh(r) )
If we integrate a matter distribution based on a constant density rho, we
get:
T(r) = integral_{0,r} 8pi rho c_a(q) (cosh(q)  sinh(q)/tanh(r)) dq
If we evaluate this, then put a=sqrt(1+2C), rho=C/(4pi(1+2C))  where C
is a constant due to conservation of mass, discussed in my previous post
on the FRW solutions  then these two ways of computing T(r) are in
perfect agreement.
Of course, although our particle cloud is homogeneous on the FRW
spacelike slices, it won't be homogeneous on the *uncut* hyperboloids in
Minkowski spacetime. 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Wed Jun 14, 2006 7:01 am Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



I've worked out the FRW dust solutions in 2+1 gravity, that is the
homogeneous, isotropic solutions where the universe contains a continuum
of matter that approximates the presence of pressurefree particles at
the same density everywhere.
As in 3+1 gravity there can be negative, zero, or positive curvature for
the spacelike slices (i.e. hyperboloids, planes or spheres), but the time
evolution is extremely simple: either the solution is static (which is
only possible for spherical spatial geometry or zero density), or it is
expanding at a constant rate.
The metric is:
ds^2 = dt^2 + R(t)^2 dw^2
where dw^2 is the metric on the unit hyperboloid, plane, or unit sphere.
If we put k=1,0,1 for these alternatives, the relationship between R(t)
and the density of matter is:
R'(t)^2 + k = 8 pi rho R(t)^2
while the lack of pressure gives:
R''(t) = 0
For spherical geometry, the radius of the sphere is R(t), so the total
amount of matter present is M = 4 pi rho R(t)^2, and:
R'(t)^2 + 1 = 2 M
This implies that a static solution is possible iff:
M = 1/2 (in units where G=c=1).
In these units, the angular deficit associated with a mass of M is 8piM,
so this static solution has a total angular deficit of 4pi, in agreement
with the finite particle polyhedron solutions Ralph Hartley has described.
The only other *homogeneous, isotropic* static solution is the vacuum
solution. If we put rho=0 in the planar FRW solution, we get
R(t)=constant, and that's obviously Minkowski space, but if we put rho=0
in the hyperboloid solution we get R(t)=t, and the solution is the
interior of the light cone in Minkowski space, i.e. a portion of the same
vacuum solution.
For the nonstatic solutions, we note that
4 pi rho R(t)^2 = some constant C
by conservation of mass; for the spherical geometry we have C=M, the
total mass in the universe, but for the other geometries where there's an
infinite amount of matter in the universe C will still be constant. So
we have:
R(t) = sqrt(2C  k) t
rho(t) = C / [4pi (2Ck) t^2 ]
It'd be nice to be able to relate this to various discrete particle
solutions. I previously posted a solution with a ring of n particles
moving away from the origin with equal speed v in a symmetrical manner,
and noted that if we look at what's happening on the unit hyperboloid in
Minkowski space, the angular deficit associated with each particle
becomes much less at hyperbolic infinity, i.e. if we cut out a wedge by
arranging two planes to intersect along each particle's world line, the
sum of the angles between these planes can be much more than 2pi because
they take much smaller "bites" out of the light cone than the angles they
subtend around the world lines.
Specifically, if the angle of the "bite" is B, it turns out that for
small deficit angles A around the world line we have:
B = A sqrt( (1v) / (1+v) )
Now, imagine a cloud of particles with world lines starting from the
origin of Minkowski space with all possible velocities, spreading out at
all angles. By cutting out wedges around their world lines, we ought to
be able to get one of the expanding hyperboloid FRW solutions, where each
spacelike slice at time t is isometric to the unit hyperboloid multiplied
by sqrt(1+2C) t, and the world lines of all the particles are normal to
every spacelike slice.
In hyperbolic geometry the area and circumference of a circle increase
with distance faster than in the Euclidean plane, but as we move out
through our cloud of particles in Minkowski space with wedges cut out,
we're *losing* area and circumference. So all the hyperboloids in the
original Minkowski space get flattened out. I guess the trick is to find
a way to make sure that their geometry is still that of a hyperboloid,
but a slightly larger, hence flatter, one ... by a factor of sqrt(1+2C).
But I haven't been able to figure out yet how to get the particles
distributed in the original Minkowski spacetime in such a way that
everything works out nicely. 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Wed Jun 14, 2006 7:00 am Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



I've worked out the FRW dust solutions in 2+1 gravity, that is the
homogeneous, isotropic solutions where the universe contains a continuum
of matter that approximates the presence of pressurefree particles at
the same density everywhere.
As in 3+1 gravity there can be negative, zero, or positive curvature for
the spacelike slices (i.e. hyperboloids, planes or spheres), but the time
evolution is extremely simple: either the solution is static (which is
only possible for spherical spatial geometry or zero density), or it is
expanding at a constant rate.
The metric is:
ds^2 = dt^2 + R(t)^2 dw^2
where dw^2 is the metric on the unit hyperboloid, plane, or unit sphere.
If we put k=1,0,1 for these alternatives, the relationship between R(t)
and the density of matter is:
R'(t)^2 + k = 8 pi rho R(t)^2
while the lack of pressure gives:
R''(t) = 0
For spherical geometry, the radius of the sphere is R(t), so the total
amount of matter present is M = 4 pi rho R(t)^2, and:
R'(t)^2 + 1 = 2 M
This implies that a static solution is possible iff:
M = 1/2 (in units where G=c=1).
In these units, the angular deficit associated with a mass of M is 8piM,
so this static solution has a total angular deficit of 4pi, in agreement
with the finite particle polyhedron solutions Ralph Hartley has described.
The only other *homogeneous, isotropic* static solution is the vacuum
solution. If we put rho=0 in the planar FRW solution, we get
R(t)=constant, and that's obviously Minkowski space, but if we put rho=0
in the hyperboloid solution we get R(t)=t, and the solution is the
interior of the light cone in Minkowski space, i.e. a portion of the same
vacuum solution.
For the nonstatic solutions, we note that
4 pi rho R(t)^2 = some constant C
by conservation of mass; for the spherical geometry we have C=M, the
total mass in the universe, but for the other geometries where there's an
infinite amount of matter in the universe C will still be constant. So
we have:
R(t) = sqrt(2C  k) t
rho(t) = C / [4pi (2Ck) t^2 ]
It'd be nice to be able to relate this to various discrete particle
solutions. I previously posted a solution with a ring of n particles
moving away from the origin with equal speed v in a symmetrical manner,
and noted that if we look at what's happening on the unit hyperboloid in
Minkowski space, the angular deficit associated with each particle
becomes much less at hyperbolic infinity, i.e. if we cut out a wedge by
arranging two planes to intersect along each particle's world line, the
sum of the angles between these planes can be much more than 2pi because
they take much smaller "bites" out of the light cone than the angles they
subtend around the world lines.
Specifically, if the angle of the "bite" is B, it turns out that for
small deficit angles A around the world line we have:
B = A sqrt( (1v) / (1+v) )
Now, imagine a cloud of particles with world lines starting from the
origin of Minkowski space with all possible velocities, spreading out at
all angles. By cutting out wedges around their world lines, we ought to
be able to get one of the expanding hyperboloid FRW solutions, where each
spacelike slice at time t is isometric to the unit hyperboloid multiplied
by sqrt(1+2C) t, and the world lines of all the particles are normal to
every spacelike slice.
In hyperbolic geometry the area and circumference of a circle increase
with distance faster than in the Euclidean plane, but as we move out
through our cloud of particles in Minkowski space with wedges cut out,
we're *losing* area and circumference. So all the hyperboloids in the
original Minkowski space get flattened out. I guess the trick is to find
a way to make sure that their geometry is still that of a hyperboloid,
but a slightly larger, hence flatter, one ... by a factor of sqrt(1+2C).
But I haven't been able to figure out yet how to get the particles
distributed in the original Minkowski spacetime in such a way that
everything works out nicely. 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Wed Jun 14, 2006 6:59 am Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



Ralph Hartley wrote:
Quote:  Consider a polyhedron inscribed in a sphere of radius 1, centered at the
origin. Let the surface of the polyhedron inherit the metric from R^3
(which will be flat except at the vertexes).
For any point p other than the origin, let p_1 be the intersection of
the polyhedron with the ray from the origin through p. Let t(p) = p/p_1.
The metric (on R^3O) ds^2 = dt(p)^2 + dp_1^2 is flat except at the
rays from the origin through the vertexes, and any [space]like surface has
total deficit 4Pi.

That's a really elegant construction, but (at least in the static case) I
think you can get rid of your Big Bang at t=0. Just take the Cartesian
product of the real line R with the polyhedron, with its inherited
metric, and put ds^2 = dt^2 + dp_1^2, where now p_1 is the projection
from the Cartesian pair (t,p_1).
This would generalise to "polyhedra" formed by triangulating any compact
2manifold and putting flat metrics on the triangles. The total deficit
in the general case will be 2pi*chi, where chi is the Euler
characteristic of the manifold. 

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jhnrmsdn@yahoo.co.uk science forum addict
Joined: 25 Mar 2006
Posts: 68


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Ralph Hartley science forum addict
Joined: 07 May 2005
Posts: 92

Posted: Tue Jun 13, 2006 5:29 pm Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



Greg Egan wrote:
Quote:  Ralph Hartley wrote:
Consider a polyhedron inscribed in a sphere of radius 1, centered at the
origin. Let the surface of the polyhedron inherit the metric from R^3
(which will be flat except at the vertexes).
For any point p other than the origin, let p_1 be the intersection of
the polyhedron with the ray from the origin through p. Let t(p) = p/p_1.
The metric (on R^3O) ds^2 = dt(p)^2 + dp_1^2 is flat except at the
rays from the origin through the vertexes, and any [space]like surface has
total deficit 4Pi.
That's a really elegant construction, but (at least in the static case) I
think you can get rid of your Big Bang at t=0.

Yes, I realized, just after sending, that the area is constant, and the
space is just R x Polyhedron. The sphere was not needed either, any
convex polyhedron would do.
I'm pretty sure that the edges of the polyhedron are purely artifacts of
the construction as well. The space is flat except at the vertexes.
Given a reference frame on one face, you can extend it to the whole
space in many different ways, one for each way of cutting the polyhedron
up and spreading it out flat. How many there are depends on what you
count as a "different" cutting.
Quote:  This would generalise to "polyhedra" formed by triangulating any compact
2manifold and putting flat metrics on the triangles. The total deficit
in the general case will be 2pi*chi, where chi is the Euler
characteristic of the manifold.

If you require the manifold to be orientable, and don't allow negative
masses, I think the only new case is a flat torus (which does have some
parameters).
If you don't require that it be compact (and allow some infinite
triangles), you get the cases with deficit between 0 and 2Pi.
Lets see, how many more static solutions are there? If you allow
timelike loops, quite a few more I think.
Ralph Hartley 

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Greg Egan science forum addict
Joined: 01 May 2005
Posts: 75

Posted: Mon Jun 12, 2006 10:40 pm Post subject:
Re: This Week's Finds in Mathematical Physics (Week 232)



In article <20060610152605.CF2D11310F5@mail.netspace.net.au>, Greg Egan
<gregegan@netspace.net.au> wrote:
Quote:  Ralph Hartley wrote:

[snip]
Quote:  An *arbitrary* spacelike surface can have curvature anywhere, not just
at the particles.
Good point! Your comment made me think of the hyperbolic plane, and the
opportunity it gives for the sums of angles in triangles to be less than
2pi.
[snip description of a 2particle model with spacetime foliated by 
hyperboloids]
Quote:  [W]e can't
extend things back into the indefinite past, though it might be possible
to get around that with some further tricks.

I can't see how to extend this solution to infinite ve time, but why not
make a virtue out of necessity and consider a family of Big Bang style
solutions which start from a singularity?
If we pick an origin in Minkowski spacetime, we can take the topological
interior of the forward light cone of the origin and declare that this
set, minus some wedges excluded along particle world lines, will be our
entire solution.
If we have a collection of world lines starting from the origin (which
itself is excluded), we then cut out wedges around them and identify the
opposite faces to give these particles various masses. If we were using
all of Minkowski spacetime for our solution, we would need to be sure
that the wedge cut out around one worldline never overlapped with that of
any other particle, but in this case instead we only have to be sure that
there's no overlap *inside the light cone*.
That allows the particle masses to be greater than they otherwise would
be, and this is what enabled the extra mass in the construction I
described in my previous post, where two particles each with deficit
angles approaching 2pi are possible.
The same construction extends naturally to a symmetrical configuration of
N particles, each of mass m. If we arrange their worldlines
symmetrically around the origin of our chosen coordinate system, the
maximum possible value of m will occur when two wedge planes associated
with two neighbouring particles meet precisely on the light cone. Take
one particle's world line to lie on the xt plane, and to be generated by
the timelike vector:
u = (1,v,0)
The two wedge planes for this world line will pass through null vectors
that are halfway to this particle's clockwise and counterclockwise
neighbours. These null vectors are:
q1 = (1, cos pi/N, sin pi/N)
q2 = (1, cos pi/N, sin pi/N)
Normals to these planes are:
n1 = (v, 1, (vcos(pi/N))/sin(pi/N) )
n2 = (v, 1, (vcos(pi/N))/sin(pi/N) )
c = n1.n2 / n1n2 =
(cos(pi/N)^2  2) v^2 + 2 cos(pi/N) v  cos(2pi/N)

(cos(pi/N) v  1)^2
The deficit angle associated with each particle is then:
A = pi  arccos(c) v < cos(pi/N)
pi + arccos(c) v >= cos(pi/N)
where 0 <= arccos <= pi.
Between v=0 and v=1 (where these endpoints should be taken only as
limiting cases, not values that can actually be achieved):
v c A
  
0 cos(2pi/N) 2pi/N
cos(pi/N) 1 pi
1 1 2pi
So in the highvelocity limit v>1 each particle has a deficit angle
approaching 2pi.
For 0 < v < cos(pi/N), in the manyparticle limit N>infinity we have a
(maximum) total deficit angle of:
N A > 2pi sqrt((1+v)/(1v))
This means that if we want a system with a scalar sum of rest mass equal
to M (which, with a certain choice of units, means a total deficit angle
of 2M), there will be a minimum velocity required in these models of:
v_min = (M^2  pi^2) / (M^2 + pi^2)
For, say, M=4.5pi, v_min = 0.905. Interestingly, this comes at a point
where the total holonomy has ceased to be cyclic in v; see the plots at
the end of:
http://gregegan.customer.netspace.net.au/GR2plus1/GR2plus1.htm 

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