7/28/2019 Symplectic Representation of the Tardion: Normal Forms

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Symplectic Representation of the Tardion Sector: Normal FormsThis is a reformatted, rewritten version of the article below, which I originally posted. In the process of discussing the

spin-orbit decomposition, a normal form for the symplectic leafs corresponding to the spin non-zero tardion is derived.Out of this also comes the Newton-Wigner position vector.

Mark Hopkins

Originally:

Re: The First Law and Angular Momentumsci.physics

Part 1: 2009 December 11 14:07:12 -0800 (PST)

Part 2: 2009 December 11 14:41:33 -0800 (PST)

Part 1Uncle Al , 2009 December 11 11:31

>The spinning perfect circle angular velocity is an artifact of the uniformly spinning circle itself. The angular

>velocity of a spinning disk, sphere, or solid object, is an artifact of the uniformly spinning disk, sphere, or solid. So

>we have least action consistent, single object, spin angular momentum.

>1) How do you know it is spinning? Translation is strictly POV.

On a somewhat related note, it may be of interest to note that spin, itself, is translation-invariant.

For ordinary systems, this comes straight out of the representations of the space-time symmetry group; namely, that the

Poisson bracket { }, 0=S P , where ( )1 2 3, ,S S S=S is spin and ( )1 2 3, ,P P P=P , the momentum, is the generator for

spatial translations. Also, { },H =S 0 , where H, the kinetic energy, is the generator of temporal translations.

One way of thinking of this is that the angular momentum contained in S has an infinite lever arm, attached to

asymptotic infinity. Then, Penroses famous cable construction can be directly used to explain the unusual 720-degree

symmetry for half-integer spin representations.

This is true both the relativistic and non-relativistic settings. It comes about by decomposing the rotation generator (i.e.

angular momentum) into = +J r P S , and the boost generator (i.e. mass moment) into ( ) ( )M a m M= + +K r P S ,

where M m H= + is the relativistic mass and m the rest mass. For relativity21 c = , for non-relativistic theory

0 = .

Under this decomposition, the following Poisson brackets

{ } { } { }

{ } { } { }

, 0, , , , 0,

, 0, , 0, , ,

= = =

= = =

r a r b r a P b a b P a P b

r a S b P a S b S a S b S a b

all follow from the fundamental bracket relations

{ } { } { } { } { } { }

{ } { } { } { } { } { }

, , , , , , , , , , , ,

, , , , , 0, , , , , , ,

M H H H

M M M

= = = = = =

= = = = = =

J a J b J a b K a K b J a b K a P b a b P 0 J 0 K P

J a K b K a b J a P b P a b P a P b P 0 J 0 K P

that govern space-time symmetries.

This holds irrespective of the paradigm divide between relativistic and non-relativistic theory; parameterized by , as

described above.

The brackets { }, 0=r S also mean that spin is independent of the momentum scale.

Part 2Spin can also be thought of as whats left over after taking out the translation-dependent part of J .

The Poisson brackets can also be represented as transformation properties on the generators, themselves; by writing an

infinitesimal transformation as

7/28/2019 Symplectic Representation of the Tardion: Normal Forms

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{ }_, H = + + J K P

where , , and are respectively infinitesimal rotations, boosts, spatial translations and time translations.

Then the quantities transform as

, , , , ,a M M H M = + + = + = = = J J K P K K J P P P P P

with J the Thomas precession term and P the mass deficit term that distinguish the relativistic from

non-relativistic form.

So, having this, you can then ask what the normal forms are. For ordinary systems ( 2 2M P> ), the spin S results

from transforming J to its minimum residual that is, by finding a lever arm ( R ) under which the infinitesimal

transform (by ) integrates out to

, , , ,M H H M M + + J J R P K K R P P

so as to minimize J .

Also: the r vector described above almostarises in this way. In fact, it arises by the following 3 step process:

(1) Carry out a finite spatial translation (i.e. center of mass). The result is J S , and the translation is by the leverarm = R r , where r is given above.

(2) Carry out a finite boost parallel to P , such that J J , K K and P 0 (i.e. the rest frame). Therequirement in K uniquely determines what translation to use in step (1).

(3) Carry out a finite spatial translation J J and K 0 (corrected center of mass).With the decomposition made explicit, the results of the steps are

(1) ( ), , , , , ,Mm M m M

= + +

+ +

p S p SJ K P r p S r p S p ,

(2) ( ), , , , , ,m M m M

=

+ +

p S p SJ K P S p S 0 ,

(3) ( ), , , ,m M

+

p SS 0 S 0 0 .

The resulting normal form is invariant under all transformations except rotations on axes parallel to S ; and arbitrary

time translations. So,

(a) its stationary,(b) it defines a unique position and(c) it defines a unique velocity.

Therefore, it is interpreted as a system in its rest frame with the center of mass at the given position, possessing a

residual angular momentum S .

This construction only works if 2 2M P> . For the case 2 2M P= , one has luxons (light-speed modes) and for2 2

M P< , faster-than-light modes. Then the normal forms and corresponding reductions are different. (In particular,

for tachyons, the reduction is to 0M , while P 0 yields the tachyon impulse as its square:2 2

P = ; after

the reduction to 0M ).