Greg Egan had some nice questions about this "group-valued momentum" business, which I was able to answer
. However, I think there's still some work to be done to dig out the full meaning of this concept - some calculations, thought experiments, and so on.
John Baez wrote:
> The really cool part is the relation between the Lie algebra
> element p and the group element exp(p). Originally we thought
> of p as momentum - but there's a sense in which exp(p) is the
> momentum that really counts!
Would it be correct to assume that the ordinary tangent vector p still
transforms in the usual way? In other words, suppose I'm living in a
2+1 dimensional universe, and there's a point particle with rest mass m
and hence energy-momentum vector in its rest frame of p=m e_0. If I
cross its world line with a certain relative velocity, there's an
element g of SO(2,1) which tells me how to map the particle's tangent
space to my own. Would I measure the particle's energy-momentum to be
p'=gp? (e.g. if I used the particle to do work in my own rest frame)
Would there still be no upper bound on the total energy, i.e. by making
our relative velocity close enough to c, I could measure the particle's
kinetic energy to be as high as I wished?
I guess I'm trying to clarify whether the usual Lorentz transformation
of the tangent space has somehow been completely invalidated for
extreme boosts, or whether it's just a matter of there being a second
definition of "momentum" (defined in terms of the Hamiltonian) which
transforms differently and is the appropriate thing to consider in
gravitational contexts.
In other words, does the cut-off mass apply only to the deficit angle,
and do boosts still allow me to measure (by non-gravitational means)
arbitrarily large energies (at least in the classical theory)?
Greg Egan wrote:
>
John Baez wrote:
>>The really cool part is the relation between the Lie algebra
>>element p and the group element exp(p). Originally we thought
>>of p as momentum - but there's a sense in which exp(p) is the
>>momentum that really counts!
>Would it be correct to assume that the ordinary tangent vector p
>still transforms in the usual way?
Hi! Yes, it would.
>In other words, suppose I'm living in a 2+1 dimensional universe,
>and there's a point particle with rest mass m and hence
>energy-momentum vector in its rest frame of p=m e_0. If I
>cross its world line with a certain relative velocity, there's
>an element g of SO(2,1) which tells me how to map the particle's
>tangent space to my own. Would I measure the particle's
>energy-momentum to be p'=gp? (e.g. if I used the particle to
>do work in my own rest frame) Would there still be no upper
>bound on the total energy, i.e. by making our relative velocity
>close enough to c, I could measure the particle's kinetic energy
>to be as high as I wished?
To understand this, it's good to think of the momenta as
elements of the Lie algebra so(2,1) - it's crucial to the
game.
Then, if you have momentum p, and I zip past you, so you
appear transformed by some element g of the Lorentz group
SO(2,1), I'll see your momentum as
p' = g p g^{-1}
This is just another way of writing the usual formula for
Lorentz transforms in 3d Minkowski space. No new physics
so far, just a clever mathematical formalism.
But when we turn on gravity, letting Newton's constant k
be nonzero, we should instead think of momentum as group-valued,
via
h = exp(kp)
and similarly
h' = exp(kp')
Different choices of p now map to the same choice of h.
In particular, a particle of a certain large mass - the
Planck mass- will turn out to act just like a particle
of zero mass!
So, if we agree to work with h instead of p, we are now
doing new physics. This is even more obvious when we decide
to multiply momenta instead of adding them, since multiplication
in SO(2,1) is noncommutative!
But, if we transform our group-valued momentum in the correct
way:
h' = ghg^{-1}
this will be completely compatible with our previous transformation
law for vector-valued momentum!
>I guess I'm trying to clarify whether the usual Lorentz transformation
>of the tangent space has somehow been completely invalidated for
>extreme boosts, or whether it's just a matter of there being a second
>definition of "momentum" (defined in terms of the Hamiltonian) which
>transforms differently and is the appropriate thing to consider in
>gravitational contexts.
Good question! Amazingly, the usual Lorentz transformations still
work EXACTLY - even though the rule for adding momentum is new (now
it's multiplication in the group). We're just taking exp(kp) instead
of p as the "physical" aspect of momentum.
This effectively puts an upper limit on mass, since as
we keep increasing the mass of a particle, eventually it "loops
around" SO(2,1) and act exactly like a particle of zero mass.
But, it doesn't exactly put an upper bound on energy-momentum,
since SO(2,1) is noncompact. Of course energy and momentum don't
take real values anymore, so one must be a bit careful with this
"upper bound" talk.
>In other words, does the cut-off mass apply only to the deficit
>angle, and do boosts still allow me to measure (by non-gravitational
>means) arbitrarily large energies (at least in the classical theory)?
There's some sense in which energy-momenta can be arbitrarily
large. That's because the space of energy-momenta, namely SO(2,1),
is noncompact. Maybe you can figure out some more intuitive way
to express this.
