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All Of The Kinematic Groups
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rockbr...@gmail.com
Jun 24, 2020, 1:15:22 AM6/24/20
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2020 June 17 23:55 sci.physics.research
https://groups.google.com/forum/#!original/sci.physics.research/kwEyJRGdUyQ/kX25BnfdBwAJ
Repeated here, using UTF-8.
On Friday, May 15, 2020 at 4:06:46 PM UTC-5, Nicolaas Vroom wrote:
> Today we understand group theory (which Einstein did not in 1905). We
> can use group theory to show that there can be only three transformation
> groups among inertial frames, and that just one of them, the Lorentz
> group [#], agrees with all the experiments.
14, not 3. There are 14 kinematic groups consistent with a general set of assumptions - a 3-parameter family of groups distinguished by the sign of each parameter, which I'll call here (α,β,κ), such that the (-α,-β,-κ) group is isomorphic to the (α,β,κ) group.
They are
Static: (α,β,κ) = (0,0,0)
Galilei: (α,κ) = (0,0), β ≠ 0
Carroll: (β,κ) = (0,0), α ≠ 0
Para-Galilei: (α,β) = (0,0), κ ≠ 0
Newton-Hooke(±): α = 0, ζ > 0 (+) or ζ < 0 (-)
Para-X: β = 0; λ > 0 (X = Poincare') or λ < 0 (X = Euclid)
X: κ = 0; γ > 0 (X = Poincare') or γ < 0 (X = Euclid-4D)
deSitter(±): γ > 0; λ > 0 (+) or λ < 0 (-)
Hyper-X: γ < 0; ζ > 0 (X = bolic), ζ < 0 (X = spherical)
where γ = α β and λ = α κ and ζ = β κ.
The Bacry Levi-Leblond (BLL) 1968 classification.
Roughly speaking: α = 0 corresponds to c = infinity, β = 0 corresponds to c = 0, with β/α = c², when α β > 0. The 3 groups where α β < 0 correspond to geometries with signature 4+0 and were excluded in the 1968 paper ... though they are relevant for geometries with signature-changing metrics.
Both the c = 0 and c = infinity cases can occur together: α = 0 and β = 0 (the Static and Carroll groups). Static and Carroll have the same central extension, even though one is trivial; Galilei has the Bargmann group as its central extension.
If the symmetries are denoted:
∙ Spatial translations P = (P₁,P₂,P₃)
∙ Spatial rotations J = (J₁,J₂,J₃)
∙ Time translations H
∙ Boosts K = (K₁,K₂,K₃)
then the groups are those with the Lie brackets:
∙ [Ja,Vb] = Vc, where V is (J,K,P), (a,b,c)=(1,2,3), (2,3,1) or (3,1,2)
∙ [J,H] = 0
∙ [Ka,Kb] = -γ Jc; [Pa,Pb] = λ Jc, with (a,b,c) as above
∙ [Ka,H] = β Pa, [Pa,H] = κ Ka
∙ [Ka,Pb] = α δ_{a,b} H where δ_{a,b} = 1 if a = b, 0 else
where γ = α β, λ = α κ. The combination ζ = β κ is also useful, but does not appear as a structure coefficient.
This set is derived from the assumptions:
∙ isotropy to fix all the J brackets as above
∙ consistency with the discrete transforms (J,K,P,H) → (J,aK,bP,abH) for time reversal (a,b)=(-,+) and parity reversal (a,b)=(+,-), which fixes the forms of the brackets,
∙ Jacobi identities (from which follow γ = α β, λ = α κ).
A slightly more general classification can be derived by seeking out all possible deformations of the (α,β,κ) = (0,0,0) case.
The dimensions of the various quantities (using M,L,T,V respectively for mass, length, time, speed) are:
α: T/VL, β: L/TV, κ: V/LT
γ: 1/VV, λ: 1/LL, ζ: 1/TT
J: MLV, K: ML, P: MV, H: MLV/T
The V dimension has to be treated as independent of L/T for consistency with β → 0, though it can be normalized to L/T for non-zero β by setting β = +1 or β = -1.
All of the kinematic groups have a central extension of the form:
[Ka,Pb] = M δ_{ab}, where M ≡ μ + α H and μ is the central charge.
Both M and μ have the dimension M. The central extension is only non-trivial when α = 0; it is "trivial" otherwise; i.e. the group, for non-zero α, splits into (J,K,P,H,μ) → (J,K,P,M) × (μ). In this case, E = M/α can be used in place of M. This corresponds to "total energy" in Relativity, while M corresponds to "relativistic mass".
For uniformity, however, it's necessary to use (J,K,P,H,M) or (J,K,P,H,μ); since E is ill-defined for α = 0.
Two of the groups coincide under central extensions, so the number then drops to 13.
The transformations in infinitesimal form are
∙ rotations: ω
∙ boosts: υ
∙ spatial translations: ε
∙ time translations: τ
∙ extra translation: ψ
which have dimensions:
ω: 1, υ: V, ε: L, τ: T, ψ: LV
Their actions on (J,K,P,H,M,μ) are all given by
δ X = [X, J.ω + K.υ + P.ε - H τ + μ ψ]
leading to the following infinitesimal transformations:
δ J = ω × J + υ × K + ε × P
δ K = ω × K - γ υ × J + ε M - β τ P
δ P = ω × P - υ M + λ ε × J - κ τ K
δ H = -β υ.P - κ ε.K
δ M = -γ υ.P - λ ε.K
δ μ = 0
where ()x() denotes vector cross-product and ().() vector dot product.
They can be integrated to finite form to see how each transforms under the respective symmetry.
Rotations: (J,K,P,H,M,μ) → (RJ,RK,RP,RH,RM,R μ)
where R is a rotation operator given by
RV = V + sin(theta) n × V + (1 - cos(theta)) n × (n × V),
where the unit vector n is the axis of rotation, theta the rotation angle
Time Translations by time-shift s:
(J,H,M,μ) → (J,H,M,μ)
(K,P) → (K - β s P)/r, (P - κ s K)/r)
where r = √(1 + ζ s²) and ζ s² > -1
For ζ < 0, the transform has an horizon at |s| = √(-1/ζ); time is hyperbolic. For ζ > 0, time is circular and the transform is actually a rotation and you can take either the + or - sign of the square root, √(1 + ζ s²).
Spatial Translations by shift-vector a:
(μ,J₀,K₁,P₀) → (μ,J₀,K₁,P₀)
(J₁,P₁) → ((J₁ + a × P₁)/r, (P₁ + λ a × J₁)/r)
(K₀,M → ((K₀ + a M)/r, (M - λ a.K₀)/r)
H → H - κ a.K₀/r + M/r κ a²/(1 + r)
where r = √(1 - λ a²) and λ a² < 1.
The components (J₀,K₀,P₀) are those parallel to ε, while (J₁,K₁,P₁) are perpendicular to ε.
J = J₀ + J₁, K = K₀ + K₁, P = P₀ + P₁
ε.V₁ = 0 = ε × V₀, where V is (J,K,P).
This has an horizon at a = 1/√(λ) if λ > 0; the spatial dimensions are hyperbolic if λ > 0, circular if λ < 0 and flat if λ = 0. For λ < 0, both signs of the square root can be used and the transformations are ordinary sinusoidal rotations.
Boosts by a velocity vector v:
(J₀,K₀,P₁,μ) → (J₀,K₀,P₁,μ)
(J₁,K₁) → ((J₁ + v × K₁)/r, (K₁ - γ v × J₁)/r)
(P₀,M) → ((P₀ - v M)/r, (M - γ v.P₀)/r)
H → H - β v.P/r + M/r β v²/(1 + r)
where r = √(1 - γ v²) and γ v² < 1.
This has an horizon at v = 1/√γ if γ > 0 and reduces to a circular rotation if γ < 0. For γ = 0, it reduces to the form:
(J,K) → (J + v × K, K)
(P,H,M,μ)→ (P - v M, H - β v.P + β v² M/2, M, μ)
If there is a "rest frame" (i.e. where P → 0), then in terms of the transformed values in that frame (P,H,M) = (0,U,m), one has
0 = (P₀ - v M)/r + P₁
m = (M - γ v.P₀)/r
U = H - β v.P/r + M/r β v²/(1 + r)
which, when inverted, yields:
P = vm/r, M = m/r, H = U + m/r β v²/(1 + r),
μ = M - α H = m - α U
which simultaneously generalizes both the non-relativistic and relativistic formulas for (relativistic) mass M, momentum P and kinetic energy H ... except that U is absent in relativity.
The appearance of an "internal energy" U term in the relativistic case is directly connected with the inclusion of the 11th generator μ. If setting U = 0, then both μ and the rest mass m coincide. Otherwise, μ would have to be given a separate name, e.g. "Intrinsic Mass".
To emphasize: the case U ≠ 0 and μ ≠ m is a GENERALIZATION of Special Relativity that is NOT equivalent to it, but only includes it as a special case!
There are, in the general cases of (α,β,κ), 3 invariants under the transforms:
The Intrinsic Mass: μ = M - α H
The Extended Mass Shell: Φ₂ = β P² - 2MH + α H² - κ K² + α β κ J²
The "Spin" shell: Φ₄ = W² - γ W₀² + λ W₄²
There is also the derived invariant:
The Mass Shell: μ² - α Φ₂ = M² - γ P² + λ K² - λ γ J²
The 5-vector (W₀, W = (W1,W2,W3), W₄) is the generalization of the Pauli-Lubanski vector, and is given by
W₀ = P.J, W = MJ + P × K, W₄ = K.J.
It transforms as
δ W₀ = -υ W - β τ W₄
δ W = ω × W - γ υ W₄ - λ ε W₄
δ W₄ = ε W - κ τ W₀
If there is a rest frame P = 0, then in it, it reduces to (W₀,W,W₄) → (0,mS,m² R.S).
This, along with the fact that there are 5 translation generators (P₁,P₂,P₃,H,μ) in the central extension, instead of 4, strongly suggests that the geometry naturally suited for this is 5 dimensional, not 4.
It's something one of us found a couple years ago.
The centrally extended members of the BLL family all have a uniform geometric coordinate representation with the following features:
∙ It is non-linear if κ ≠ 0
∙ It is 5 dimensional, not 4.
Coordinates: (t, u, x, y, z)
Differential Operators: denoted (T, U, X, Y, Z)
J = (z Y - y Z, x Z - z X, y X - x Y)
K = β t (X,Y,Z) + (x,y,z) (α T - U)
P = -a (X,Y,Z)
H = aT + bU
M = a (α T - U) = μ + α H
μ = -c U
where
a = √(1 + ζ s² - λ r²)
b = κ r²/(a + c)
c = √(1 + ζ s²) = a + α b
r² = x² + y² + z² + 2 β t u + γ u²
s = t + α u
It is linear if κ = 0 (with a = 1, b = 0, c = 1), non-linear otherwise. The non-linear case reflects an underlying curved geometry.
The dimensions of the coordinates are x,y,z: L, t: T, u: LV.
Both a and c play the role of "extra" coordinates in the associated conic sections:
a² + λ (x² + y² + z²) - ζ t² = 1
c² - ζ s² = 1
For the Static/Carroll groups (α = 0, β = 0), one has:
a = 1, b = κ r²/2, c = 1, r² = x² + y² + z², s² = t².
From this, you can integrate to find the action of the symmetry transforms on the coordinates. The transform (ψ) corresponding to (μ) will have a non-trivial effect.
To determine these - which I won't do here - you will need to calculuate the action of (J,K,P,H,M,μ) on (a,b,c); and these can be found from the following actions of (T,U,X,Y,Z) on (a,b,c):
a (T,U,X,Y,Z) a = κ (β t, 0, -α x, -α y, -α z)
ac (T,U,X,Y,Z) b = κ (β (au-bt), β as, cx, cy, cz)
c (T,U,X,Y,Z) c = κ (β s, γ s, 0, 0, 0)
https://groups.google.com/forum/#!original/sci.physics.research/kwEyJRGdUyQ/kX25BnfdBwAJ
Repeated here, using UTF-8.
On Friday, May 15, 2020 at 4:06:46 PM UTC-5, Nicolaas Vroom wrote:
> Today we understand group theory (which Einstein did not in 1905). We
> can use group theory to show that there can be only three transformation
> groups among inertial frames, and that just one of them, the Lorentz
> group [#], agrees with all the experiments.
14, not 3. There are 14 kinematic groups consistent with a general set of assumptions - a 3-parameter family of groups distinguished by the sign of each parameter, which I'll call here (α,β,κ), such that the (-α,-β,-κ) group is isomorphic to the (α,β,κ) group.
They are
Static: (α,β,κ) = (0,0,0)
Galilei: (α,κ) = (0,0), β ≠ 0
Carroll: (β,κ) = (0,0), α ≠ 0
Para-Galilei: (α,β) = (0,0), κ ≠ 0
Newton-Hooke(±): α = 0, ζ > 0 (+) or ζ < 0 (-)
Para-X: β = 0; λ > 0 (X = Poincare') or λ < 0 (X = Euclid)
X: κ = 0; γ > 0 (X = Poincare') or γ < 0 (X = Euclid-4D)
deSitter(±): γ > 0; λ > 0 (+) or λ < 0 (-)
Hyper-X: γ < 0; ζ > 0 (X = bolic), ζ < 0 (X = spherical)
where γ = α β and λ = α κ and ζ = β κ.
The Bacry Levi-Leblond (BLL) 1968 classification.
Roughly speaking: α = 0 corresponds to c = infinity, β = 0 corresponds to c = 0, with β/α = c², when α β > 0. The 3 groups where α β < 0 correspond to geometries with signature 4+0 and were excluded in the 1968 paper ... though they are relevant for geometries with signature-changing metrics.
Both the c = 0 and c = infinity cases can occur together: α = 0 and β = 0 (the Static and Carroll groups). Static and Carroll have the same central extension, even though one is trivial; Galilei has the Bargmann group as its central extension.
If the symmetries are denoted:
∙ Spatial translations P = (P₁,P₂,P₃)
∙ Spatial rotations J = (J₁,J₂,J₃)
∙ Time translations H
∙ Boosts K = (K₁,K₂,K₃)
then the groups are those with the Lie brackets:
∙ [Ja,Vb] = Vc, where V is (J,K,P), (a,b,c)=(1,2,3), (2,3,1) or (3,1,2)
∙ [J,H] = 0
∙ [Ka,Kb] = -γ Jc; [Pa,Pb] = λ Jc, with (a,b,c) as above
∙ [Ka,H] = β Pa, [Pa,H] = κ Ka
∙ [Ka,Pb] = α δ_{a,b} H where δ_{a,b} = 1 if a = b, 0 else
where γ = α β, λ = α κ. The combination ζ = β κ is also useful, but does not appear as a structure coefficient.
This set is derived from the assumptions:
∙ isotropy to fix all the J brackets as above
∙ consistency with the discrete transforms (J,K,P,H) → (J,aK,bP,abH) for time reversal (a,b)=(-,+) and parity reversal (a,b)=(+,-), which fixes the forms of the brackets,
∙ Jacobi identities (from which follow γ = α β, λ = α κ).
A slightly more general classification can be derived by seeking out all possible deformations of the (α,β,κ) = (0,0,0) case.
The dimensions of the various quantities (using M,L,T,V respectively for mass, length, time, speed) are:
α: T/VL, β: L/TV, κ: V/LT
γ: 1/VV, λ: 1/LL, ζ: 1/TT
J: MLV, K: ML, P: MV, H: MLV/T
The V dimension has to be treated as independent of L/T for consistency with β → 0, though it can be normalized to L/T for non-zero β by setting β = +1 or β = -1.
All of the kinematic groups have a central extension of the form:
[Ka,Pb] = M δ_{ab}, where M ≡ μ + α H and μ is the central charge.
Both M and μ have the dimension M. The central extension is only non-trivial when α = 0; it is "trivial" otherwise; i.e. the group, for non-zero α, splits into (J,K,P,H,μ) → (J,K,P,M) × (μ). In this case, E = M/α can be used in place of M. This corresponds to "total energy" in Relativity, while M corresponds to "relativistic mass".
For uniformity, however, it's necessary to use (J,K,P,H,M) or (J,K,P,H,μ); since E is ill-defined for α = 0.
Two of the groups coincide under central extensions, so the number then drops to 13.
The transformations in infinitesimal form are
∙ rotations: ω
∙ boosts: υ
∙ spatial translations: ε
∙ time translations: τ
∙ extra translation: ψ
which have dimensions:
ω: 1, υ: V, ε: L, τ: T, ψ: LV
Their actions on (J,K,P,H,M,μ) are all given by
δ X = [X, J.ω + K.υ + P.ε - H τ + μ ψ]
leading to the following infinitesimal transformations:
δ J = ω × J + υ × K + ε × P
δ K = ω × K - γ υ × J + ε M - β τ P
δ P = ω × P - υ M + λ ε × J - κ τ K
δ H = -β υ.P - κ ε.K
δ M = -γ υ.P - λ ε.K
δ μ = 0
where ()x() denotes vector cross-product and ().() vector dot product.
They can be integrated to finite form to see how each transforms under the respective symmetry.
Rotations: (J,K,P,H,M,μ) → (RJ,RK,RP,RH,RM,R μ)
where R is a rotation operator given by
RV = V + sin(theta) n × V + (1 - cos(theta)) n × (n × V),
where the unit vector n is the axis of rotation, theta the rotation angle
Time Translations by time-shift s:
(J,H,M,μ) → (J,H,M,μ)
(K,P) → (K - β s P)/r, (P - κ s K)/r)
where r = √(1 + ζ s²) and ζ s² > -1
For ζ < 0, the transform has an horizon at |s| = √(-1/ζ); time is hyperbolic. For ζ > 0, time is circular and the transform is actually a rotation and you can take either the + or - sign of the square root, √(1 + ζ s²).
Spatial Translations by shift-vector a:
(μ,J₀,K₁,P₀) → (μ,J₀,K₁,P₀)
(J₁,P₁) → ((J₁ + a × P₁)/r, (P₁ + λ a × J₁)/r)
(K₀,M → ((K₀ + a M)/r, (M - λ a.K₀)/r)
H → H - κ a.K₀/r + M/r κ a²/(1 + r)
where r = √(1 - λ a²) and λ a² < 1.
The components (J₀,K₀,P₀) are those parallel to ε, while (J₁,K₁,P₁) are perpendicular to ε.
J = J₀ + J₁, K = K₀ + K₁, P = P₀ + P₁
ε.V₁ = 0 = ε × V₀, where V is (J,K,P).
This has an horizon at a = 1/√(λ) if λ > 0; the spatial dimensions are hyperbolic if λ > 0, circular if λ < 0 and flat if λ = 0. For λ < 0, both signs of the square root can be used and the transformations are ordinary sinusoidal rotations.
Boosts by a velocity vector v:
(J₀,K₀,P₁,μ) → (J₀,K₀,P₁,μ)
(J₁,K₁) → ((J₁ + v × K₁)/r, (K₁ - γ v × J₁)/r)
(P₀,M) → ((P₀ - v M)/r, (M - γ v.P₀)/r)
H → H - β v.P/r + M/r β v²/(1 + r)
where r = √(1 - γ v²) and γ v² < 1.
This has an horizon at v = 1/√γ if γ > 0 and reduces to a circular rotation if γ < 0. For γ = 0, it reduces to the form:
(J,K) → (J + v × K, K)
(P,H,M,μ)→ (P - v M, H - β v.P + β v² M/2, M, μ)
If there is a "rest frame" (i.e. where P → 0), then in terms of the transformed values in that frame (P,H,M) = (0,U,m), one has
0 = (P₀ - v M)/r + P₁
m = (M - γ v.P₀)/r
U = H - β v.P/r + M/r β v²/(1 + r)
which, when inverted, yields:
P = vm/r, M = m/r, H = U + m/r β v²/(1 + r),
μ = M - α H = m - α U
which simultaneously generalizes both the non-relativistic and relativistic formulas for (relativistic) mass M, momentum P and kinetic energy H ... except that U is absent in relativity.
The appearance of an "internal energy" U term in the relativistic case is directly connected with the inclusion of the 11th generator μ. If setting U = 0, then both μ and the rest mass m coincide. Otherwise, μ would have to be given a separate name, e.g. "Intrinsic Mass".
To emphasize: the case U ≠ 0 and μ ≠ m is a GENERALIZATION of Special Relativity that is NOT equivalent to it, but only includes it as a special case!
There are, in the general cases of (α,β,κ), 3 invariants under the transforms:
The Intrinsic Mass: μ = M - α H
The Extended Mass Shell: Φ₂ = β P² - 2MH + α H² - κ K² + α β κ J²
The "Spin" shell: Φ₄ = W² - γ W₀² + λ W₄²
There is also the derived invariant:
The Mass Shell: μ² - α Φ₂ = M² - γ P² + λ K² - λ γ J²
The 5-vector (W₀, W = (W1,W2,W3), W₄) is the generalization of the Pauli-Lubanski vector, and is given by
W₀ = P.J, W = MJ + P × K, W₄ = K.J.
It transforms as
δ W₀ = -υ W - β τ W₄
δ W = ω × W - γ υ W₄ - λ ε W₄
δ W₄ = ε W - κ τ W₀
If there is a rest frame P = 0, then in it, it reduces to (W₀,W,W₄) → (0,mS,m² R.S).
This, along with the fact that there are 5 translation generators (P₁,P₂,P₃,H,μ) in the central extension, instead of 4, strongly suggests that the geometry naturally suited for this is 5 dimensional, not 4.
It's something one of us found a couple years ago.
The centrally extended members of the BLL family all have a uniform geometric coordinate representation with the following features:
∙ It is non-linear if κ ≠ 0
∙ It is 5 dimensional, not 4.
Coordinates: (t, u, x, y, z)
Differential Operators: denoted (T, U, X, Y, Z)
J = (z Y - y Z, x Z - z X, y X - x Y)
K = β t (X,Y,Z) + (x,y,z) (α T - U)
P = -a (X,Y,Z)
H = aT + bU
M = a (α T - U) = μ + α H
μ = -c U
where
a = √(1 + ζ s² - λ r²)
b = κ r²/(a + c)
c = √(1 + ζ s²) = a + α b
r² = x² + y² + z² + 2 β t u + γ u²
s = t + α u
It is linear if κ = 0 (with a = 1, b = 0, c = 1), non-linear otherwise. The non-linear case reflects an underlying curved geometry.
The dimensions of the coordinates are x,y,z: L, t: T, u: LV.
Both a and c play the role of "extra" coordinates in the associated conic sections:
a² + λ (x² + y² + z²) - ζ t² = 1
c² - ζ s² = 1
For the Static/Carroll groups (α = 0, β = 0), one has:
a = 1, b = κ r²/2, c = 1, r² = x² + y² + z², s² = t².
From this, you can integrate to find the action of the symmetry transforms on the coordinates. The transform (ψ) corresponding to (μ) will have a non-trivial effect.
To determine these - which I won't do here - you will need to calculuate the action of (J,K,P,H,M,μ) on (a,b,c); and these can be found from the following actions of (T,U,X,Y,Z) on (a,b,c):
a (T,U,X,Y,Z) a = κ (β t, 0, -α x, -α y, -α z)
ac (T,U,X,Y,Z) b = κ (β (au-bt), β as, cx, cy, cz)
c (T,U,X,Y,Z) c = κ (β s, γ s, 0, 0, 0)
rockbr...@gmail.com
Jun 24, 2020, 1:22:17 AM6/24/20
to
On Wednesday, June 24, 2020 at 12:15:22 AM UTC-5, rockbr...@gmail.com wrote:
> Rotations: (J,K,P,H,M,μ) → (RJ,RK,RP,RH,RM,R μ)
> where R is a rotation operator [...]
> Rotations: (J,K,P,H,M,μ) → (RJ,RK,RP,RH,RM,R μ)
(J,K,P,H,M,μ) → (RJ,RK,RP,H,M,μ)
The scalars H, M, μ are invariant under rotation.
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