Get ready to have your eyes opened! I'm going to show you where the 1/2 went (and where it still resides). This will be new to you (no matter who you are). In fact, we're going to make this entertaining for you. I will pose the answer as an exercise. You will have to solve it to find what you need.
Exercise:
Assume αv² < 1 and m ≠ 0, and let u = v²/(1 + √(1 - αv²)).
Prove that if L = mu, H = Mu and M = m + αH, then M = m/√(1 - αv²), H = pv - L, where p = Mv, H = p²/(m+M) and p² - 2MH + αH² = 0
Calculus bonus: Prove that p = ∂L/∂v.
Comments:
Interpretations should be obvious, if you are familiar with the letter names. L,H,m,M,p,v are respectively named Lagrangian, Hamiltonian, rest mass, moving mass, momentum, velocity. The extra coordinate u has no name; but plays a central role in all of this.
The magic parameter α distinguishes different types of kinematics from one another. The geometry associated with it has, as its invariants, the following:
dt² - α(dx² + dy² + dz²), ∇² - α (∂/∂t)²
For α > 0, you have relativity with an underlying Minkowski geometry in which the invariant speed is given as c = 1/√α. Upon substitution into the formulae above, you arrive at the usual expressions for the associated terms that any Relativist should be familiar with.
For α = 0, you have an underlying non-relativistic framework (associated with a flat spacetime Newton-Cartan geometry). As you can see, the extra coordinate u reduces to your familiar ½ v². That's where it is.
For α < 0, these would be the expressions that apply in a Euclidean form of the kinematics, where the underlying geometry has 4 dimensions of space and none of time.
Heuristically, you can think of the Legendre transform from L to H as a kind of "renormalization" in which one first adds the mass equivalent αmu of the "bare" kinetic energy mu to the mass m; then the mass equivalent
α(αmu)u of the bare kinetic energy (αmu)u of that extra mass; then the mass equivalent α(α(αmu)u)u of the bare kinetic energy (α(αmu)u)u of the second contribution to the mass, and so on ad infinitum, to arrive at the total moving mass M:
m + αmu + α(αmu)u + α(α(αmu)mu)mu + ⋯ = m/(1 - αu) = m/√(1 - αv²) = M
(the last two equalities are what the exercise was, ultimately, to show!)
The corresponding kinetic energy Mu is the Hamiltonian H.
Underlying geometry;
Notice the mass shell invariant has 5 components (if you treat the momentum p as a 3-vector), not 4. The associated signature is 4+1, independent of what α is. Divide out M² from p² - 2MH + αH², and you get the invariant v² - 2u + αu². In addition, you also have the linear invariants M - αH and, associated with this, 1 - αu.
If, in place of u, you write -du/dt (I prefer it with the other sign), and express the velocity v as dr/dt, then you may (upon multiplying by dt²) rewrite this as a line element: dx² + dy² + dz² + 2dtdu - αdu²; and linear invariant ds = dt + αdu. For relativity, setting the line element to 0 (that is, confining one to the 5D light cone for this geometry) reduces ds to the proper time and the line element to the Minkowski metric.
This still has meaning, even when α = 0 ... and is closely linked to the natural 5-dimensional representation of the Galile group and its central extension, the Bargmann group. For α ≠ 0, it gives you the relativistic version of this interpretation.
The unification of Mikowski, Newton-Cartan and Euclidean geometries here is entirely analogous to the unification of Euclidean, spherical and hyperbolic geometries into projective geometry.
Symmetries:
Treating the momentum, now, as a vector p↑, then under infinitesimal rotations ω↑, infinitesimal boosts υ↑, the momentum p↑, moving mass M and kinetic energy H transform together as a 5-vector Δ(H,p↑,M) = (-υ↑·p↑, ω↑×p↑ - υ↑ M, -αυ↑·p↑). From this, you can derive a finite transform by exponentiating it (H,p↑,M) → exp(sΔ) (H,p↑,M) = (1 + sΔ + s²/2! Δ² + s³/3! Δ³ + ⋯)(H,p↑,M).
This leads to another exercise that generalizes this a bit further still...
Exercise:
Define the following operations
Δ (dr↑, dt, du) = (ω↑×dr↑ - β υ↑ dt, -αυ↑·dr↑, υ↑·dr↑)
Δ (H, p↑, M) = (-αυ↑·p↑, ω↑×p↑ - υ↑ M, -αβυ↑·p↑)
on 3-vectors p↑ = (p₁,p₂,p₃), dr↑ = (dx,dy,dz) and scalars dt, where ()×() and ()·() respectively denote the cross product and dot product.
Express, in closed form, exp(sΔ)(dr↑, dt, du) (where exp(sΔ) = 1 + sΔ + s²/2! Δ² + s³/3! Δ³ + ⋯) in terms of θ↑ = sω↑ and w↑ = sυ↑.
You may use the following functions
C(λ,x) = 1 + λ x²/2! + λ² x⁴/4! + ⋯
S(λ,x) = x + λ x³/3! + λ² x⁵/5! + ⋯
E(λ,x) = x²/2! + λ x⁴/4! + λ² x⁶/6! + ⋯
in your answer [note: the E function is ultimately where the v²/2 and its relativistic analogues stem from]; and assume the following properties
C(λ,0) = 1, S(λ,0) = 0, E(λ,0) = 0
C(λ,x) - λE(λ,x) = 1
S(λ,x)² - 2E(λ,x) C(λ,x) + λE(λ,x)² = 0
C(λ,x)² - λS(λ,x)² = 1
Prove the operator identity
Δ²(Δ² + ω² - αβυ²) = αβ(ω↑·υ↑)²
For αβ ≠ 0, express the answer also in terms of
v↑ = w↑/w S(αβ,w)/C(αβ,w)
(where w = |w↑|) restricted to vectors v↑ where αβv² < 1, and show that
C(λ,x)² = 1/(1 - αβv²).
Prove that the following are invariants under these transforms:
dx² + dy² + dz² + 2βdtdu + αβdu², dt + α du, β(p₁² + p₂² + p₃²) - 2MH + αH², M - αH
and, from these, derive the reduced 4D invariants
α(dx² + dy² + dz²) - β dt², M² - αβ(p₁² + p₂² + p₃²)
Discuss each of the cases αβ > 0, αβ < 0, (α = 0 & β ≠ 0), β = 0. For αβ > 0 what's the invariant velocity in terms of α and β; and what does the case β = 0 correspond to?
Related References:
Bargmann structures and Newton-Cartan Theory
C. Duval, G. Burdet, H. P. Künzle, and M. Perrin
Physical Review D, Volume 31, Number 8, 15 April 1985
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.31.1841
Although their 5D geometric formulation (corresponding to my β = 1 cases above) is tailored for Newton-Cartan geometry and non-relativistic gravity; the underlying framework is general and provides a nearly seamless unifying framework that also includes 4D General Relativity (embedded in the 5D geometric representation just described) as a special case. The "almost" part is that there is still an obstruction that prevents a smooth continuous transform from α > 0 → α = 0.
Possible Kinematics
Henri Bacry, Jean-Marc Lévy-Leblond
Journal of Mathematical Physics 9, 1605 (1968);
https://aip.scitation.org/doi/10.1063/1.1664490
They provide a general classification of all possibilities according with general assumptions ... but arbitrarily exclude the Euclideanized cases (αβ < 0) - despite the utility that they have in field theory! Their notation's different than mine, because everything I laid out here was independently developed (and much cleaner) and only later cross-fitted with their work.
Kinematical Lie algebras via deformation theory
José M. Figueroa-O'Farrill
https://arxiv.org/abs/1711.06111
An even bigger classification that derives everything as a deformation of the "static" group (which corresponds to the α = 0, β = 0 above). I did the same thing several years before this, but didn't take it as far as they did (they allow the central charge -- the invariant M - αH in my notation above) to have non-zero Lie brackets -- something I specifically excluded in my version.
