- #1
Greg Egan
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?
>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 realize 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.
>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?
>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 realize 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.