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Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum

Statement

Let (A,V,η,C+) be the oriented time-oriented affine Minkowski model (Oriented time-oriented affine Minkowski space (the spacetime primitives)), c>0, let n∈V be future unit timelike, η(n,n)=−1, and let En,qn,m,Tn be the quantities of Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum.

(a) If P is nonzero future causal, then En(P)>0, qn(P)∈n⊥ and

η(qn(P),qn(P))=En(P)2c2+η(P,P) ≥ 0.

Consequently: (i) if P is future null, then η(P,P)=0 and En(P)=c ∣qn(P)∣>0; (ii) if P is future timelike with invariant mass m(P), then η(P,P)=−m(P)2c2 and the mass-shell identity

En(P)2=m(P)2c4+c2 ∣qn(P)∣2

holds. The invariant mass m(P) depends only on the scalar η(P,P) and therefore on no observer n and no basis, while En and qn do depend on n.

(b) Let P=mU with m>0 constant, let U be future timelike with η(U,U)=−c2 (the four-velocity normalisation of Four-velocity and four-acceleration of a future timelike worldline), and put γ:=−η(U/c,n)≥1 and w:=c h(U/c)/γ∈n⊥, the relative velocity of Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(d) applied to the future unit vector u=U/c; equivalently w=h(U)/γ. Then

En(P)=mc2γ,Tn(P)=mc2(γ−1) ≥ 0,qn(P)=mγw,

and m(P)=m. Moreover Tn(P)=0 if and only if γ=1, that is U=cn, equivalently P=mcn (the particle is at rest relative to n). Finally ∣qn(P)∣=mγ∣w∣=mcγ2−1.

Facts & Assumptions

Given: the affine Minkowski model (A,V,η,C+); a real number c>0; a future unit timelike n∈V with η(n,n)=−1; clause (a): a nonzero P∈cl(C+); clause (b): a real m>0 and a future timelike U∈V with η(U,U)=−c2, with P=mU.

[F1]

The definition of the observer quantities (Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum, Definition): for P∈cl(C+) one has En(P)=−c η(P,n) and qn(P)=h(P)=P+η(P,n)n∈n⊥, and η(n,P)=−En(P)/c; for nonzero future causal P the invariant mass is m(P)=−η(P,P)/c; for P=mU with m>0 and η(U,U)=−c2 the kinetic energy is Tn(P)=En(P)−mc2; and ∣w∣=η(w,w) on n⊥.

[F2]

Norms of four-velocity and four-acceleration, and the observer rest-space decomposition, clauses (c) and (d): h is linear with image n⊥ and kernel span(n), the restriction of η to n⊥ is positive definite, and V=span(n)⊕n⊥; and for future unit timelike n,u one has, with γ:=−η(n,u) and w:=c h(u)/γ, the relations γ≥1, w∈n⊥, u=γ(n+w/c), ∣w∣2=c2(1−γ−2), and γ=1 if and only if u=n.

[F3]

Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space: η is a symmetric bilinear form on V, so η(v,w) is additive and homogeneous in each argument and η(v,w)=η(w,v); in particular polarisation gives η(v+w,v+w)=η(v,v)+2η(v,w)+η(w,w) and η(λv,μw)=λμ η(v,w) for real λ,μ.

[F4]

Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1): there is an ordered basis (e0,e1,e2,e3) of V with η(ea,eb)=diag(−1,1,1,1)ab.

[F5]

Existence of future-oriented inertial frames at every event with Inertial frames and inertial charts of the affine Minkowski model: in the oriented time-oriented affine Minkowski space there exists, for every event o, an inertial frame with origin o, and an inertial frame's basis is an η-orthonormal ordered basis whose first vector lies in C+; in particular such a basis of V exists.

[F6]

The timelike set has exactly two convex components, described in orthonormal coordinates: in every η-orthonormal ordered basis whose first vector is timelike, writing v=v0e0+vs with vs∈S:=span(e1,e2,e3) and ∣vs∣:=(v1)2+(v2)2+(v3)2, one has η(v,v)=−(v0)2+∣vs∣2, C+={v:v0>∣vs∣} (for the appropriate naming of the two components) and cl(C+)={v:v0≥∣vs∣}.

[F7]

The Euclidean spatial structure and cross product of an inertial frame: for an inertial frame, S=span(e1,e2,e3) equals e0⊥, the restriction δ of η to S is positive definite, and ∣w∣=δ(w,w) is a norm on S; in coordinates δ(vs,ws)=vs⋅ws=∑i=13viwi.

[F8]

The Cauchy-Schwarz inequality for finite sums: for real numbers ai,bi, ∣∑i=13aibi∣≤(∑i=13ai2)1/2(∑i=13bi2)1/2.

[F9]

Timelike, null, spacelike and causal vectors of a Lorentz vector space and Time orientation, future and past timelike/null/causal vectors, and the closed cones: a vector v is causal exactly when v≠0 and η(v,v)≤0 (it is timelike when η(v,v)<0 and null when v≠0 and η(v,v)=0), and C+ is a component of the timelike set with closed cone cl(C+).

[F10]

Rational powers ar of a positive base and Laws of rational exponents: for a,b>0 and rationals r,s the power ar is the positive real number defined there, with (ab)r=arbr, (ar)s=ars and a1/2 the unique nonnegative square root of a; in particular λ2=λ for λ>0 and γ1−γ−2=γ2−1 for γ≥1.

Proof

technique · fix orthonormal coordinates, prove the algebraic identity and positivity there, then specialise to $P=mU$ using the relative-velocity decomposition
1.1F1F2F3

By [F1], qn(P)=P+η(P,n)n∈n⊥, η(P,n)=−En(P)/c, m(P)2=−η(P,P)/c2 for nonzero future causal P, and Tn(P)=En(P)−mc2 in the setting of (b). Expanding with [F3] and η(n,n)=−1 gives η(qn(P),qn(P))=η(P,P)+2η(P,n)2+η(P,n)2η(n,n)=η(P,P)+η(P,n)2=η(P,P)+En(P)2/c2. The formula m(P)2=−η(P,P)/c2 involves only the scalar η(P,P), so m(P) depends on no observer and no basis, whereas En and qn are built from η(P,n) and h and depend on n.

1.2F3F4F5F6F7F8

Choose by [F4] and [F5] an η-orthonormal ordered basis (e0,e1,e2,e3) with e0∈C+; write v=v0e0+vs with vs∈S=span(e1,e2,e3) and ∣vs∣ as in [F6]. Bilinearity [F3] extends the quadratic form of [F6] to η(v,w)=−v0w0+δ(vs,ws) with δ(vs,ws)=vs⋅ws the Euclidean inner product of [F7]. By [F6], C+={v:v0>∣vs∣} and cl(C+)={v:v0≥∣vs∣}; hence n0>∣ns∣≥0 and P0≥∣Ps∣≥0, while η(n,n)=−1 gives (n0)2=1+∣ns∣2. By [F8], ∣Ps⋅ns∣≤∣Ps∣ ∣ns∣.

1.3F2F3F7

In the setting of (b), put u:=U/c. Since η(U,U)=−c2, bilinearity [F3] gives η(u,u)=c−2η(U,U)=−1, so u is future unit timelike, and [F2] clause (d) applies: with γ:=−η(n,u)≥1 and w:=c h(u)/γ one has w∈n⊥, u=γ(n+w/c), ∣w∣2=c2(1−γ−2), and γ=1 if and only if u=n; equivalently U=cn, and P=mU=mc n when γ=1.

2.1step 1.1step 1.2F8

Positivity of En: using [step 1.2], c−1En(P)=−η(P,n)=P0n0−Ps⋅ns, so c−1En(P)≥P0n0−∣Ps∣∣ns∣≥∣Ps∣n0−∣Ps∣∣ns∣=∣Ps∣(n0−∣ns∣)≥0, where the first inequality uses [F8], the second uses P0≥∣Ps∣ and n0>0, and the last uses n0>∣ns∣ of [step 1.2]; if ∣Ps∣>0 the last term is strictly positive, while if ∣Ps∣=0 then P≠0 forces P0>0 and c−1En(P)=P0n0>0, so En(P)>0 in both cases.

2.2step 1.1F1F2

Since qn(P)∈n⊥ by step 1.1 and the restriction of η to n⊥ is positive definite by [F2], η(qn(P),qn(P))=∣qn(P)∣2≥0. Combining this with the identity of step 1.1 gives the displayed equality and inequality of (a).

2.3step 1.1step 1.3F2F3

The energy and momentum of P=mU: by [step 1.1] and [step 1.3], En(P)=−c η(mU,n)=−mc2η(U/c,n)=mc2γ and qn(P)=h(mU)=m h(U)=mc h(u)=mγw, the last equality because h(u)=h(γ(n+w/c))=γ h(w/c)=γ w/c by linearity of h and h(n)=0 in [F2].

3.1step 2.1step 2.2F1F9

The null case: if P is future null then η(P,P)=0 by [F9], so [step 2.2] gives η(qn(P),qn(P))=En(P)2/c2>0 by [step 2.1], and since qn(P)∈n⊥ the norm of [F1] gives ∣qn(P)∣=η(qn(P),qn(P))=En(P)/c, that is En(P)=c ∣qn(P)∣>0.

3.2step 1.1step 2.2F1

The timelike case and the mass shell: if P is future timelike, then η(P,P)=−m(P)2c2 by the definition of m(P) in [F1]; substituting this into [step 2.2] and multiplying by c2 gives En(P)2=c2η(qn(P),qn(P))+m(P)2c4=c2∣qn(P)∣2+m(P)2c4, using ∣qn(P)∣2=η(qn(P),qn(P)) for qn(P)∈n⊥.

3.3step 1.3step 2.3F1

The kinetic energy: by [step 2.3] and Tn(P)=En(P)−mc2 of [F1], Tn(P)=mc2γ−mc2=mc2(γ−1)≥0 because γ≥1 in [step 1.3]; moreover Tn(P)=0 if and only if γ=1, which by [step 1.3] is equivalent to u=n, that is to U=cn, that is to P=mU=mcn, and conversely P=mcn gives γ=1 and Tn(P)=0.

3.4step 1.3step 2.3F10

The momentum magnitude: by [step 2.3] and [step 1.3], ∣qn(P)∣=mγ∣w∣=mγ c1−γ−2, and since γ≥1, [F10] gives γ1−γ−2=γ2(1−γ−2)=γ2−1, so ∣qn(P)∣=mcγ2−1.

4.1step 1.1step 2.1step 2.3step 3.1step 3.2step 3.3step 3.4F3∎

Collect the conclusions and evaluate the invariant mass of P=mU: by [step 1.1], m(P)2=−η(P,P)/c2, and for P=mU with η(U,U)=−c2 bilinearity [F3] gives η(P,P)=m2η(U,U)=−m2c2, so m(P)2=m2 and, since both m(P) and m are nonnegative and c>0, m(P)=m; together with [step 2.1] and [step 3.1] this proves clause (a) including (i), with [step 3.2] proving clause (a)(ii), and with [step 2.3], [step 3.3] and [step 3.4] it proves clause (b), the vanishing statement of (b) included by [step 3.3].

Depends on

Dependency tree · 2 levels

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