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Mass shell inversion, observer work and variable mass

Physical scope

Conditional kinematic identities for a massive point particle of constant positive mass in one inertial chart of flat Minkowski spacetime, plus a separate variable-mass ansatz; no force law or variable-mass dynamics is adopted.

Statement

Let I⊆R be a nonempty open interval, and let x:I→A be a C2 future-timelike worldline in the flat Minkowski model of Particle spacetime, observers and worldlines. Use one inertial frame and parameterize the curve by that chart's time coordinate t, so its coordinates are (ct,x(t)) (Inertial frames and inertial charts of the affine Minkowski model). Identify the observer rest space with R3 by the frame's spatial orthonormal basis and use its Euclidean dot product and norm (The Euclidean spatial structure and cross product of an inertial frame). Write v=dx/dt, γ=(1−∣v∣2/c2)−1/2, proper time τ, and U=dx/dτ. In the constant-mass sector of Adopted massive-particle momentum, observer measurement and isolated-collision conservation, let m>0 be constant and P=mU. In this chart write P=(E/c,p), where E and p are the observer energy and spatial momentum of Mechanical momentum and observer energy. Then ∣v∣<c, E=γmc2>0, p=γmv, and

E2=m2c4+c2∣p∣2,v=c2pE.

Define the three-force by F:=dp/dt and the four-force by K:=dP/dτ in this flat inertial chart. In SI units m is in kilograms, c and v in metres per second, E in joules, p in kilogram metres per second, and F,K in newtons. Then

dEdt=F⋅v,K=γ(F⋅vc,F).

Separately, let Iτ⊆R be a nonempty open proper-time interval, let m:Iτ→(0,∞) be C1, and let U:Iτ→V be a C1 future timelike vector field with η(U,U)=−c2. Measure m in kilograms and dm/dτ in kilograms per second, and set P(τ):=m(τ)U(τ). With K:=dP/dτ, the kinematic ansatz satisfies

η(P,K)=−m dmdτ c2.

The last identity is only a differentiated mass-shell relation; it prescribes no evolution for m. None of these conclusions assumes a force law.

Facts & Assumptions

Given: A C2 future-timelike worldline on an open interval, one inertial chart with x0=ct, c>0, and either a constant mass m>0 in the adopted massive sector or the separately stated C1 fields m(τ)>0 and U(τ) with P=mU and η(U,U)=−c2.

[A1]

The constant-mass sector of Adopted massive-particle momentum, observer measurement and isolated-collision conservation adopts m>0 constant and P=mU for massive particles. This theorem uses no force equation from that postulate.

[F1]

Mechanical momentum and observer energy defines En(P)=−cη(P,n) and qn(P)=P+η(P,n)n; in the inertial chart with time-axis observer n=e0, P=(E/c,p).

[F2]

Norms of four-velocity and four-acceleration, and the observer rest-space decomposition gives ∣v∣<c, U=γ(c,v) with γ=(1−∣v∣2/c2)−1/2, η(U,U)=−c2, and dτ/dt=1/γ in the stated chart.

[F3]

Particle spacetime, observers and worldlines fixes the C2 future-timelike curve domain, the distinction between an observer and a chart, and the SI convention x0=ct with t in seconds and spatial coordinates in metres.

[F4]

Inertial frames and inertial charts of the affine Minkowski model and The Euclidean spatial structure and cross product of an inertial frame: the chosen chart has a future unit time-axis observer and an orthonormal Euclidean spatial basis; the coordinates of qn(P) are the spatial components of P.

Proof

technique · direct differentiation and component decomposition
1.1A1F1F2F3F4algebra

By [A1] and [F2], P=mU=(γmc,γmv), and [F4] identifies these with the time and spatial components in the chosen frame. The definitions in [F1] give E=γmc2 and p=γmv. Hence E2−c2∣p∣2=γ2m2c4(1−∣v∣2/c2)=m2c4, and γm>0 gives E>0.

2.1step 1.1algebra

Dividing the spatial identity p=γmv by E=γmc2 yields p/E=v/c2, and therefore v=c2p/E. The division is valid because E>0 by step 1.1.

3.1step 1.1step 2.1F4algebra

Differentiating the mass shell of step 1.1 with respect to t gives 2E dE/dt=2c2p⋅F. Substituting p/E=v/c2 from step 2.1 yields dE/dt=F⋅v.

4.1F2step 3.1algebra

In inertial coordinates, dP/dt=(c−1dE/dt,F), and [F2] gives d/dτ=(dt/dτ)d/dt=γ d/dt. Substituting the work identity of step 3.1 proves K=γ(F⋅v/c,F). This is a kinematic identity, not a force law.

5.1F2F3algebra∎

For the separate variable-mass ansatz, bilinearity and η(U,U)=−c2 give η(P,P)=m(τ)2η(U,U)=−m(τ)2c2. Since m and U are C1, P is differentiable and K=dP/dτ exists. Differentiating with respect to τ gives 2η(P,K)=−2m(dm/dτ)c2, so η(P,K)=−m(dm/dτ)c2. No equation for m was used.

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