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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 <abergman-5F7D70.18103430052006@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 energy-momentum 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 energy-momentum' as a sort of ADM-like 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 energy-momentum 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
G-connections 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 Chern-Simons 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.
Greg Egan

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 group-valued 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 group-valued momentum corresponds to a
vector-valued 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 energy-momenta 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 energy-momenta to have values independent of these
choices, but I *do* expect the energy-momenta to give me *some*
gauge-independent 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 energy-momentum
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 "180-degree 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 right-handed orthonormal basis for the tangent space at B, putting e_t
= normalised (u1+u2), e_x = normalised (u1-u2), 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 vector-valued 4-momenta 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 4-momentum M u3 for
the outgoing particle with a non-zero, +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 parallel-transport tangent vectors
from the three world lines to my base point B, those parallel-transported
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 parallel-transport 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.

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).
Greg Egan

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 group-valued 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 (r-vt)) = 2pi ((1-alpha) r + alpha v t) where v is the speed at which the ring is expanding, alpha = mu / sqrt(1-v^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 (r-vt)) dphi^2 for r > vt. Inside the ring we can just use ordinary flat-spacetime polar coordinates. Global issues ============= alpha < 1 --------- On each surface t=constant the circumference is well-behaved, 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(1-mu^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 / (alpha-1) 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(1-v'^2) and a new coordinate: r' = (2v - (v+v') alpha') t / (1-alpha') - 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 + (v-v') t / (1-alpha') 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 laboratory-scale 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 -(1-alpha) | | v alpha (1-alpha) 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 = (alpha-1)/(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(1-v^2) we find that: g(n,n) = 1 - ( (mu-sqrt(1-v^2)) / (mu v) )^2 and that n changes from timelike to spacelike when: v = v_tach = |(1-mu^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 non-zero eigenvalue of T^2. There is only one, its eigenspace being the plane orthogonal to n, and it's equal to: lambda = - (alpha^2 (1-v^2) - 2 alpha + 1) or, making the substitution for alpha in terms of mu and v: lambda = - (mu^2 - 2 mu/sqrt(1-v^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 |mu-1| to zero. For the case mu < 1, the rotation takes the -ve sign, starting at -2pi (1-mu) 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 (mu-1) 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 (mu-1) 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 group-valued momentum changes with v ought to be similar ...
Greg Egan

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 2-particle collision in 2+1 gravity. It's mind-bogglingly 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 face-to-face 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 | |
| | | |
| | | |
............... ...............
Greg Egan

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.
Greg Egan

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
tachyon-tachyon and tachyon-luxon collisions (which yield either tachyons
or luxons as the outgoing particle, never tardyons) involve *coplanar*
vector-valued 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.
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

<DWCantrell@sigmaxi.org> wrote:
 Quote: Dik.Winter@cwi.nl (Dik T. Winter) wrote: Before Bach there were piano's that were not well-tempered. Perhaps something got lost in "translation". Of course, before Bach there were keyboard instruments (Klavier, in the _general_ sense) which were not well-tempered. 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
Hendrik van Hees

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 well-tempered. 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

--
Hendrik van Hees Texas A&M University
Phone: +1 979/845-1411 Cyclotron Institute, MS-3366
Fax: +1 979/845-1899 College Station, TX 77843-3366
http://theory.gsi.de/~vanhees/faq mailto:hees@comp.tamu.edu

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