Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck pass
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Low-velocity expansions of observer energy and momentum with uniform error bounds

Statement

Let c>0 be a fixed real parameter, let (M,g,N) be a time-oriented smooth Lorentzian four-manifold, p∈M, n an observer at p and P∈TpM a future timelike vector with rest mass m:=m(P)>0 (Observer-relative energy, spatial momentum and rest mass of a tangent vector, Observers, rest spaces and the spatial projection at an event). Write U:=P/m, u:=U/c, γ:=−gp(n,u) and, as in clause 3 of Positivity, relative-velocity identities and the mass shell of observer-relative energy and momentum,

w:=c hn(u)γ∈n⊥,s:=gp(w,w).

For z∈n⊥ write ∣z∣n:=gp(z,z), the norm from the positive-definite restriction of gp to n⊥. Thus s=∣w∣n, and ∣qn(P)∣ in the displays below means ∣qn(P)∣n; qn(P)∈n⊥ by the observer decomposition.

Then s<c, γ=(1−s2/c2)−1/2,

En(P)=mc2(1−s2c2)−1/2,∣qn(P)∣=ms(1−s2c2)−1/2,

and for every 0≤v0<c, with z0:=v02/c2<1 and all s≤v0,

∣En(P)−mc2−12ms2∣≤38 (1−z0)−5/2 ms4c2,∣∣qn(P)∣−ms∣≤12 (1−z0)−3/2 ms3c2.

Facts & Assumptions

Given: A fixed real parameter c>0 and the Lorentz vector space data n,P of the statement, with m>0, U=P/m, u=U/c, γ=−gp(n,u), w=c hn(u)/γ and s=gp(w,w).

[L1]

For a future timelike P with m=m(P)>0: γ≥1, En(P)=mc2γ, qn(P)=mγw, gp(qn,qn)=m2c2(γ2−1) and En2=m2c4+c2gp(qn,qn) (Positivity, relative-velocity identities and the mass shell of observer-relative energy and momentum clause 3, Observer-relative energy, spatial momentum and rest mass of a tangent vector clauses (a)-(b)).

[L2]

Taylor's theorem with Lagrange remainder: if f is C2 on an interval containing 0 and z, then ∣f(z)−f(0)−f′(0)z∣≤12sup⁡0≤t≤z∣f′′(t)∣z2; if f is C1, then ∣f(z)−f(0)∣≤sup⁡0≤t≤z∣f′(t)∣z (The Lagrange and Cauchy forms of Taylor's remainder, On a convex open set, a uniform bound ∥Df(z)v∥2≤M∥v∥2 implies ∥f(y)−f(x)∥2≤M∥y−x∥2, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[L3]

For z<1, the base 1−z is positive; the rational power defined by Rational powers ar of a positive base at exponent −12 agrees with the corresponding real power (The exponential definition of real powers agrees with the existing rational powers). The real-power derivative formula (yα)′=αyα−1 for y>0, applied at α=−12 and then α=−32, together with the chain and scalar rules (Continuity and derivatives of positive-base real powers, The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c), Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0), gives f(z)=(1−z)−1/2∈C2((−∞,1)) and f′(z)=12(1−z)−3/2,f′′(z)=34(1−z)−5/2.

Proof

technique · direct; read $\gamma$ and the velocity from the mass-shell identities, then expand the resulting scalar functions with the Lagrange remainder
1.1L1algebra

(The relative speed and the Lorentz factor.) Combining qn=mγw with gp(qn,qn)=m2c2(γ2−1) from [L1] gives m2γ2s2=m2c2(γ2−1), hence γ2(c2−s2)=c2; since γ≥1>0 this reads γ=(1−s2/c2)−1/2, and the identity forces s<c as well as s2/c2=1−γ−2.

1.2L2L3algebra

(Taylor bounds for the Lorentz factor.) Let f(z):=(1−z)−1/2 on [0,z0] with z0<1. By [L3] and [L2], ∣f(z)−1−12z∣≤12sup⁡t∈[0,z0]∣f′′(t)∣z2=38(1−z0)−5/2z2,∣f(z)−1∣≤sup⁡t∈[0,z0]∣f′(t)∣z=12(1−z0)−3/2z.

2.1step 1.1L1algebra

(Energy and momentum formulas.) By [L1] and step 1.1, gp(qn,qn)=m2γ2s2 and γ>0, so ∣qn(P)∣=gp(qn,qn)=mγs; hence, with En=mc2γ, En(P)=mc2(1−s2/c2)−1/2 and ∣qn(P)∣=ms(1−s2/c2)−1/2.

3.1step 2.1step 1.2algebra

(Energy expansion.) Substituting z=s2/c2 in the first bound of step 1.2 into En=mc2f(z) of step 2.1 gives ∣En−mc2−12ms2∣≤mc2⋅38(1−z0)−5/2(s4/c4)=38(1−z0)−5/2ms4/c2.

3.2step 2.1step 1.2algebra

(Momentum expansion.) Likewise ∣qn∣=msf(z) by step 2.1, so ∣∣qn∣−ms∣=ms∣f(z)−1∣≤ms⋅12(1−z0)−3/2(s2/c2)=12(1−z0)−3/2ms3/c2.

4.1step 2.1step 3.1step 3.2given∎

(Conclusion.) Step 2.1 gives the identities of the statement (including s<c), and steps 3.1 and 3.2 give the two uniform error bounds for s≤v0<c.

Remarks

  • Restriction to a cone. The statement is a low-velocity expansion for s≤v0<c with constants blowing up as v0↑c; near the light cone the expansion is useless and no claim is made there.
  • No dynamics. The lemma is purely kinematic: it expands the observer-relative energy and momentum of a tangent vector. It asserts no conservation, no force law and no observation.

Depends on

Dependency tree · 2 levels

Sources