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Greg Egan
science forum addict


Joined: 01 May 2005
Posts: 75

PostPosted: Fri Jun 09, 2006 7:17 am    Post subject: Re: This Week's Finds in Mathematical Physics (Week 232) Reply with quote

In article <4486EF88.1010902@aic.nrl.navy.mil>, Ralph Hartley
<hartley@aic.nrl.navy.mil> wrote:

Quote:
Greg Egan wrote:
However, if M exceeds the maximum particle mass, then there is cyclic
behaviour in v, with the total holonomy being a rotation by an angle
that cycles back and forth between 0 and 2pi ... and then only
becomes a boost after going through several cycles.

I've put some plots on this page:

http://gregegan.customer.netspace.net.au/GR2plus1/GR2plus1.htm

Two comments, minor and major:

Minor:
Your graphs look suspiciously like they go through *infinitely* many
cycles, like sin(1/x) near x=0. Do you trust your program to be
numerically stable?

With lower values of M it's clear that there is a finite number of
cycles. Basically as you increase M the plots, which start out monotonic
decreasing, smoothly concertina to develop cyclic sections with an
increasing number of cycles before the final plunge.

I've lowered the values of M plotted on the web page to make this clearer.

Quote:
Major:
You seem to assume a simple topology. For small M that's fine, for each
particle you cut out a wedge and glue the sides together.

Looking just at a spacelike slice, each particle is the tip of a cone.

But there is a limit to how many wedges you can cut out of a plane, and
still have the topology of a plane. If the deficit angle is 2Pi the
plane closes up into a sphere.

If the deficit angle is more than 2Pi then it will become disconnected.

So at *least* we can say that if the total Mass is greater than 2Pi the
space cannot have any connected spacelike slices.

Does the whole space become disconnected, or does it just not have
connected spatial slices?

Good point. I did wonder about this, but I clearly haven't given it
enough thought. I'll have to check the literature more carefully; I
expect someone has analysed this issue.

In fact, when there's relative motion of the particles the problem seems
to become more acute, because for a spacelike slice in the centre-of-mass
frame the angular deficits that need to be accommodated are all larger.
In the limit of N->infinity, we need to have v < sqrt(1-(M/Pi)^2) to
avoid this problem.

However, that restriction doesn't seem to rule out a tachyonic total
momentum. It's clear that for N=2 there is no limit on v, because the
worst the relativistic effects can do is make each angular deficit in the
c.o.m. frame tend towards pi. For large N, my numeric results show a
tachyonic transition well before the total angular deficit in the c.o.m.
frame reaches 2pi.
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Greg Egan
science forum addict


Joined: 01 May 2005
Posts: 75

PostPosted: Fri Jun 09, 2006 7:17 am    Post subject: Re: This Week's Finds in Mathematical Physics (Week 232) Reply with quote

I wrote:

Quote:
In article <4486EF88.1010902@aic.nrl.navy.mil>, Ralph Hartley
hartley@aic.nrl.navy.mil> wrote:

[snip]

Quote:
[T]here is a limit to how many wedges you can cut out of a plane, and
still have the topology of a plane. If the deficit angle is 2Pi the
plane closes up into a sphere.

If the deficit angle is more than 2Pi then it will become disconnected.

Good point. I did wonder about this, but I clearly haven't given it
enough thought. I'll have to check the literature more carefully; I
expect someone has analysed this issue.

One tricky way around this would be to allow some particles of "negative
mass", i.e. with negative deficit angles. That way the total deficit
angle in a spacelike slice could be limited to 2 pi, but you could still
analyse a group of particles whose collective deficit angle would exceed
2 pi.

For example:

to infinity
^ ^
A | | B
| |
| |
| |
. 3 4 .
/ \
/ \
/ \
1 . . 2
\ /
\ /
\ /
. 3 4 .
| |
| |
| |
A | | B
v v
to infinity

Take the interior of this diagram as flat space, and identify the pairs
of lines that run between 1-3, 3-A, 2-4, and 4-B. The points marked
1,2,3 and 4 are singularities, while A and B are just marked to clarify
the identification of the edges.

The angular deficit around the points 1 and 2 individually both exceed
pi, and as a group their total deficit exceeds 2pi. The negative
deficits around 3 and 4 mean that the total angular deficit of this
connected spacelike slice does not exceed 2pi.

Of course, there might be good reasons to rule out these negative mass
particles as unphysical.
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Ralph Hartley
science forum addict


Joined: 07 May 2005
Posts: 92

PostPosted: Fri Jun 09, 2006 7:17 am    Post subject: Re: This Week's Finds in Mathematical Physics (Week 232) Reply with quote

Greg Egan wrote:
Quote:
hartley@aic.nrl.navy.mil> wrote:
[T]here is a limit to how many wedges you can cut out of a plane, and
still have the topology of a plane. If the deficit angle is 2Pi the
plane closes up into a sphere.

If the deficit angle is more than 2Pi then it will become disconnected.

Oops, I was off by a factor of 2. A sphere actually has a total deficit
of 4Pi. I think that gives "Big Bang" solutions.

An example of a surface with deficit 2Pi is an infinite prism truncated
on one end. I still think 2Pi is where you start seeing unavoidable
topology changes.

If the total deficit is 2Pi < M < 4Pi then there have to be additional
particles, bringing the total to 4Pi (excluding negative mass as noted
below). Proof: Consider curve surrounding particles with total deficit
M>2Pi. The curve is *concave*, it gets shorter as it gets farther away.
If the region outside the curve were flat, it could be shrunk to a point
without changing its holonomy, which is not zero unless M is a multiple
of 2Pi.

This is based on a static picture. If the velocities are large there may
be relativistic corrections.

Quote:
One tricky way around this would be to allow some particles of "negative
mass", i.e. with negative deficit angles. That way the total deficit
angle in a spacelike slice could be limited to 2 pi, but you could still
analyse a group of particles whose collective deficit angle would exceed
2 pi.

Very tricky! With more points you can get any total deficit. Also, you
can have any total deficit if your space has a boundary (which will be
concave if M>2Pi).

I'm not sure this helps you with collisions though. As the positive
curvature particles approach each other the other required particles
have to move with them (at least I think they do). You end up with a
total deficit involved in the collision of less that 2Pi (or exactly 4Pi).

Ralph Hartley
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markwh04@yahoo.com
science forum Guru Wannabe


Joined: 12 Sep 2005
Posts: 137

PostPosted: Fri Jun 09, 2006 7:18 am    Post subject: Breaking the Light Barrier & Tachyon/Synchron Mechanics. Reply with quote

Joe Jakarta wrote:
Quote:
Is the c-barrier a kinematic sound barrier, i.e. a shock-wave of
space-time?

No. It's (to put it in layman's terms) the boundary between space and
time. Trajectories representing motions at velocities greater than c
don't represent motion in time at all, but play the analogous role to
the "synchron" of Galilean physics. [Note 1]

In Poincare' relativity, in contrast to Galilean relativity, the past
and future of an event depend not only on when it is, but where. The
past of an event is a sphere (and its inside) that shrinks down to the
place and time of the event at light speed. The future of the event is
a sphere (and its inside) that emanates from the place and time of the
event, growing at light speed. In Poincare' relativity, light speed is
absolute, while infinity is a relative speed -- in fact, any speed
greater than light speed is infinity with respect to another given
reference.

In Galilean relativity, the past and future of an event are independent
of its location. It comprises all of space up to the very instant of
the event, the past being before, the future after. The events that are
neither past nor future, relative to the fixed event, exist only along
a 3-dimensional space -- the "plane of simuntaneity". In Galilean
relativity, no finite speed is absolute, the absolute speed is
infinite, and the plane of simultaneity represents all the possible
destinations of a synchron emanating from the place and time of the
event.

In Poincare' relativity, there is a wedge of positive time interval
between the past and future of any event and in totality they comprise
a 4-dimensional subspace, rather than just 3. In particular, relative
to an event taking place at the Earth, there will be about a 3 second
interval of the moon's timeline which is neither before nor after than
event; about a 16 or 17 minute interval of the sun's existence
likewise, and about an 8 1/2 year interval of the existence of the
nearest star.

Instead of a plane of simultaneity, you now have what's called the
"absolute elsewhere". And instead of synchrons of Galilean physics, you
now have TWO separate classes of particles: the Luxons and Tachyons
[Note 2], each adopting some of the properties synchrons had.

Your body is made of tardions [Note 3], so you're not going anywhere
faster than light, since there is no known way to transfer energy from
tardions to tachyons, and (in fact) no known tachyons.

If tachyons did exist, their corresponding Dirac equation would have a
pseudoscalar in the Lagrangian, instead of the scalar for mass. In
general, the mass shell constraint for the enveloping
Galilei/Poincare'/Euclidean spacetime symmetry group is
P^2 - 2MH + a H^2 = constant
where a is the family parameter (a = (1/c)^2 for Poincare', a = 0 for
Galilei, a < 0 for Euclid); and you have a 3rd Casimir invariant
M - aH = constant.
To write down a Dirac equation, requires taking the square root of the
expression on the left
alpha.P + delta M + epsilon H
which, in turn, requires an algebra of the form
{ alpha^i, alpha^j } = delta^{ij}
{ alpha^i, delta } = { alpha^i, epsilon } = 0
delta^2 = 0, epsilon^2 = -a, { delta, epsilon
} = -2.
Outside the Galilean sector, a is non-zero, and since one also has
(epsilon + a delta)^2 = a,
with
{ epsilon, epsilon + a delta } = 0,
then the resulting algebra corresponds to a Clifford algebra of a
metric with the signature (+,+,+,+,-) -- which is just the complex
Dirac algebra. The two extra parameters are naturally interpreted as
beta and (beta gamma_5). Which one is which depends on the sign of a.

For a = 0, one has for a fixed parameter V, beta_{+/-} = 2^{-1/2}
(delta/V -/+ epsilon/V). This, too, produces an algebra isomorphic to
the complex Dirac algebra.

Galilean particles thus satisfy an appropriate version of the Dirac
equation. So the Dirac equation is not relativistic, per se.

What you'll ultimately find for the tachyonic sector is an equation of
the form
gamma^m P_m - gamma_5 pi = 0
where pi is the "asymptotic momentum" of the tachyon [Note 3].

All the relations to follow, and those above, emerge from a
comprehensive analysis of the combined Galilean, Poincare', Euclidean
spacetime symmetry group, as a result of carrying out a "Wigner"
classification on the irreducible representations of the general group.
Mathematical details may be found under

"Wigner Classification for Galilei/Poincare'/Euclid"
currently listed under http://federation.g3z.com/Physics/Index.htm

Note 1:
A synchron is an action-at-a-distance particle which exists everywhere
along a line at an instant. It has a fixed momentum and kinetic energy,
but 0 mass, and represents an instantaneous transfer of momentum across
a distance in space. In effect, Newtonian gravity is mediated by
synchrons.

In a corresponding quantum theory, there would be no Heisenberg
relation between the X coordinate and x-component p_X of momentum,
where X is the direction of travel. Instead, like time does in quantum
theory, X would just be a parameter for the trajectory. Instead, there
would be a time-energy uncertainty relation [t,H] = i h-bar. The time
operator t represents the time of the synchron's existence.

Note 2:
Luxons and Tachyons are particles that travel at or faster than light.
Both are massless and have an "asymptotic momentum". Luxons move at
light speed and have 0 asymptotic momentum. Their "relativistic" mass
(M), kinetic energy (H) and momentum (P) are related by P = Mc, H =
Mc^2.

Tachyons move at a speed greater than light or at infinite speed. Like
sychrons, they have an asymptotic momentum, Pi, which would represent
the momentum of the particle in a frame of reference in which its speed
is infinite. Their "relativistic" mass M, momentum P and kinetic energy
H are given by the relations
P = Pi v/sqrt(v^2 - c^2), M = Pi/sqrt(v^2 - c^2),
and
H = Pi c^2/sqrt(v^2 - c^2).

Note 3:
Tardions move at a speed less than light speed or are at rest and have
associated with them a non-zero "rest mass", m, which represents the
value of M in a frame in which they are at rest.

In Galilean relativity, their "relativistic" mass M is fixed as m, the
kinetic energy H and momentum P are given by
M = m, P = mv, H = 1/2 mv^2.

In Euclidean relativity (where time is a spatial dimension), they would
be given by the relations
M = m/sqrt(1 + (v/C)^2), P = mv/sqrt(1 + (v/C)^2),
H = mv^2/(1 + sqrt(1 + (V/C)^2))
with a "total energy", E, given by
E = -mC^2/sqrt(1 + (v/C)^2)
The parameter C would play the Euclidean analogue of light speed,
providing the conversion of spatial and time units.

In Poincare' relativity, they would be given by the relations
M = m/sqrt(1 - (v/c)^2), P = mv/sqrt(1 - (v/c)^2)
H = mv^2/(1 + sqrt(1 - (v/c)^2))
and a "total energy", E, given by
E = mc^2/sqrt(1 - (v/c)^2).
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Joe Rongen
science forum beginner


Joined: 02 Jun 2005
Posts: 15

PostPosted: Sun Jun 11, 2006 1:28 am    Post subject: Re: Breaking the Light Barrier & Tachyon/Synchron Mechanics. Reply with quote

"Rock Brentwood" <markwh04@yahoo.com> wrote in message
news:1149803772.032456.318270@i39g2000cwa.googlegroups.com...
Quote:
Joe Jakarta wrote:
Is the c-barrier a kinematic sound barrier, i.e. a shock-wave of
space-time?

No. It's (to put it in layman's terms) the boundary between space and
time. Trajectories representing motions at velocities greater than c
don't represent motion in time at all, but play the analogous role to
the "synchron" of Galilean physics. [Note 1]

[snip]

Quote:
Note 1:
A synchron is an action-at-a-distance particle which exists everywhere
along a line at an instant. It has a fixed momentum and kinetic
energy, but 0 mass, and represents an instantaneous transfer of
momentum across a distance in space. In effect, Newtonian gravity
is mediated by synchrons.


A report in "Science" 12 May, 2006, Vol.312, Issue 5775,
page 892 - 894 maybe of interest. It explains the behavior
of a recent experimental setup that appears counterintuitive.

"Simultaneous Negative Phase and Group
Velocity of Light in a Metamaterial."

"We investigated the propagation of femtosecond laser pulses
through a metamaterial that has a negative index of refraction
for wavelengths around 1.5 micrometer. From the interference
fringes of a Michelson interferometer with and without the sample,
we directly inferred the phase time delay. From the pulse-envelope
shift, we determined the group time delay. In a special region,
phase and group velocity are negative simultaneously. This means
that both the carrier wave and the pulse envelope peak of the output
pulse appear at the rear side of the sample before their input pulse
counterparts have entered the front side of the sample."
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Ralph Hartley
science forum addict


Joined: 07 May 2005
Posts: 92

PostPosted: Sun Jun 11, 2006 11:05 am    Post subject: Re: This Week's Finds in Mathematical Physics (Week 232) Reply with quote

Ralph Hartley wrote:
Quote:
[T]here is a limit to how many wedges you can cut out of a plane, and
still have the topology of a plane. If the deficit angle is 4Pi [corrected]
the plane closes up into a sphere.
If the deficit angle is more than 2Pi then it will become disconnected.

After a little more thought, I don't think this is as bad as it seems.

It just means that in 2+1 (as in 3+1) GR, there may not be any globally
defined reference frames, relative to which you can define surfaces of
constant time. If you follow a path with holonomy that has a boost
component, then your concept of what constitutes "constant time" will
inevitably change.

An *arbitrary* spacelike surface can have curvature anywhere, not just
at the particles.

It is true that an isolated particle has a (local) rest frame, and that
the surfaces of constant time in that frame are (locally) cones, but for
a collection of particles there is not in general a well defined "center
of mass" frame, nor is there always a spacelike slice that is flat
except at the particles.

There is no problem defining the group valued momentum for a particle,
as long as you specify both a loop, a base point and a coordinate frame
at the base point. You also need to be careful defining the "ordinary"
velocity. A reference frame is not enough. You need a reference frame at
a some chosen point, *and* a path from that point to the particle.

There is still the question of how the distribution of particles
interacts with the topology of the whole 3D space. I don't know about
that, the classification of manifolds is a *bit* harder in 3D than in 2D.

Ralph Hartley
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Greg Egan
science forum addict


Joined: 01 May 2005
Posts: 75

PostPosted: Sun Jun 11, 2006 11:05 am    Post subject: Re: This Week's Finds in Mathematical Physics (Week 232) Reply with quote

Ralph Hartley wrote:

Quote:
There is no problem defining the group valued momentum for a particle,
as long as you specify both a loop, a base point and a coordinate frame
at the base point. You also need to be careful defining the "ordinary"
velocity. A reference frame is not enough. You need a reference frame at
a some chosen point, *and* a path from that point to the particle.

By specifying a path and a loop, do you mean homotopy classes of these
things?

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.

I'm not sure if the following construction is the kind of thing you had
in mind ...

Consider the unit timelike hyperboloid in Minkowski space,
-t^2+x^2+y^2=-1. Suppose we have a particle moving with speed v e_x in
some reference frame, so its world line lies in the xt plane, and
punctures the hyperboloid at:

u = (1, v, 0) / sqrt(1-v^2)

Now consider the two planes with unit spacelike normals n1 and n2:

n1 = (v, 1, v)
n2 = (v, 1, -v)
n1.n2 = 1 - 2v^2 (Lorentzian dot product)

Note that n1.u = n2.u = 0, so the intersection of these two planes is the
world line of our particle.

These two planes intersect the unit hyperboloid along the curves:

y = +/- (v^2 - (1 - v^2) x^2) / (2vx)
t = (v^2 + (1 + v^2) x^2) / (2vx)

which can be confirmed by checking that (t,x,y) with the above
substitutions is always a unit timelike vector, and is orthogonal to n1
or n2 respectively.

Both curves are asymptotic to the yt plane, with y->+/-infinity as x->0.
So these two hyperbolas meet in a cusp on our particle's worldline, at u,
and then spread out as they approach the yt plane. Their projection into
the xy plane will be something like this:

y
^
|.
|.
| .
| .
| .
|______.____ x
| .
| .
| .
|.
|.

The Lorentz transformation that rotates around the particle's worldline
and carries n2 into -n1 will carry the bottom curve into the top curve,
counterclockwise around this diagram. So if we identify the bottom curve
with the top one this way, we will have an angular deficit of
pi+arccos(1-2v^2) associated with this particle (choosing the branch 0 <
arccos < pi).

We can turn this pair of curves into a pair of surfaces meeting in a cusp
along the world line by linearly rescaling everything by a factor lambda
over some range of positive values. The same rotation around the world
line will take one surface into the other, and although the angular
deficit will be greater than pi the excised wedge will never encroach
into the region x<=0. (We have to stick to +ve lambda, so we can't
extend things back into the indefinite past, though it might be possible
to get around that with some further tricks.)

We can then use a mirror-reversed version of the same construction to add
a second particle. We don't have to use the same value for v, and we
could displace the origin by some vector if we liked. And of course it's
not compulsory to make either rotation exactly pi+arccos(1-2v^2), that's
just an upper bound.

I realise that the boundary in the past is a bit messy, but I don't have
the energy to try to fix that up just yet.
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Ralph Hartley
science forum addict


Joined: 07 May 2005
Posts: 92

PostPosted: Mon Jun 12, 2006 9:20 pm    Post subject: Re: This Week's Finds in Mathematical Physics (Week 232) Reply with quote

Greg Egan wrote:
Quote:
Ralph Hartley wrote:

There is no problem defining the group valued momentum for a particle,
as long as you specify both a loop, a base point and a coordinate frame
at the base point. You also need to be careful defining the "ordinary"
velocity. A reference frame is not enough. You need a reference frame at
a some chosen point, *and* a path from that point to the particle.

By specifying a path and a loop, do you mean homotopy classes of these
things?

Yes.

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.

I'm not sure if the following construction is the kind of thing you had
in mind ...

I didn't really have any construction in mind. With one exception, I am
not convinced that it *is* possible to have a (connected) manifold with
total deficits > 2Pi. I had just noticed that my proof that there aren't
had a hole in it.

The one exception is a big bang with a total deficit of *exactly* 4Pi.

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^3-O) ds^2 = -dt(p)^2 + dp_1^2 is flat except at the
rays from the origin through the vertexes, and any timelike surface has
total deficit 4Pi.

Any loop divides the surface into two parts, either of which can be
viewed as "inside". The holonomy around the loop can only have one
value, and should be the sum of the deficits of the points it encloses.
This only works if holonomy is modulo 2Pi.

This solution is static, but there should be dynamic variants as well.
They shouldn't be too complicated in principle, but I would have to
abandon pencil and paper and start programing to figure them out (which
I don't have time to do).

I would also have to do a great deal more reading than I have so far.

Quote:
We can turn this pair of curves into a pair of surfaces meeting in a cusp
along the world line by linearly rescaling everything by a factor lambda
over some range of positive values.

I'm afraid you lost me about there.

Quote:
The same rotation around the world
line will take one surface into the other, and although the angular
deficit will be greater than pi the excised wedge will never encroach
into the region x<=0. (We have to stick to +ve lambda, so we can't
extend things back into the indefinite past, though it might be possible
to get around that with some further tricks.)

It is possible that you have a piece of the solution I outlined above.

Quote:
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?

I'm not sure if what you describe is the same as mine.

Quote:
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.

It looks like your construction has a boundary, other than the origin.
If so, then it can't be the same as my big bang, but it could be a piece
of it (e.g. with a hole cut out).

Ralph Hartley
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Greg Egan
science forum addict


Joined: 01 May 2005
Posts: 75

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

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 2-particle 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, (v-cos(pi/N))/sin(pi/N) )
n2 = (v, 1, -(v-cos(pi/N))/sin(pi/N) )

c = n1.n2 / |n1||n2| =

(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 high-velocity limit v->1 each particle has a deficit angle
approaching 2pi.

For 0 < v < cos(pi/N), in the many-particle limit N->infinity we have a
(maximum) total deficit angle of:

N A -> 2pi sqrt((1+v)/(1-v))

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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Ralph Hartley
science forum addict


Joined: 07 May 2005
Posts: 92

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

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^3-O) 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
2-manifold 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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jhnrmsdn@yahoo.co.uk
science forum addict


Joined: 25 Mar 2006
Posts: 68

PostPosted: Wed Jun 14, 2006 3:18 am    Post subject: Re: This Week's Finds in Mathematical Physics (Week 234) Reply with quote

John Baez wrote:
Quote:

Also available at http://math.ucr.edu/home/baez/week234.html

June 12, 2006
This Week's Finds in Mathematical Physics (Week 234)
John Baez

Today I'd like to talk about the math of music - including
torsors, orbifolds, and maybe even Mathieu groups

How about topoi? Being a category theory buff, you may be
interested in a work called "The Topos of Music", discussed
on sci.math a couple of months ago:

http://groups.google.com/group/sci.math/browse_frm/thread/74260e4d010f4b5/b0161928089c6c1b?lnk=st&q=&rnum=33#b0161928089c6c1b


Cheers

John R Ramsden
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Greg Egan
science forum addict


Joined: 01 May 2005
Posts: 75

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

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^3-O) 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
2-manifold 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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Greg Egan
science forum addict


Joined: 01 May 2005
Posts: 75

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

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 pressure-free 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 non-static 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 (2C-k) 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( (1-v) / (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

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

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 pressure-free 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 non-static 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 (2C-k) 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( (1-v) / (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

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

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 FRW-solution 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 (w-v) / ( w sqrt(1-v^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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