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Observer energy, spatial momentum and mass shell of a boosted test body

Example

Keep the setting of The relative-velocity decomposition of a boost, checked in an explicit tetrad: the flat metric g on R4, the tetrad (E0,E1,E2,E3), the observer n=E0, the parameters β∈(−1,1), γ=(1−β2)−1/2, c>0, and the future unit timelike vector u=γ(E0+βE1) with g(u,u)=−1 and −g(n,u)=γ. Let m>0 and set

P:=mc u.

Then P is future timelike, its rest mass is m(P)=m, and relative to the observer n its energy and spatial momentum are

En(P)=mc2γ,qn(P)=mcγβ E1,

so that g(qn(P),qn(P))=m2c2γ2β2 and the mass shell En(P)2=m2c4+c2g(qn(P),qn(P)) holds. This is the standard boosted test-body kinematics in the abstract observer split.

Facts & Assumptions

Given: The flat metric g and tetrad (EA) of the boosted-frame example, the observer n=E0, parameters β,γ,c, the vector u=γ(E0+βE1), a mass m>0, and the vector P=mcu.

[L1]

u is future unit timelike, −g(n,u)=γ, hn(u)=γβE1, and g(E1,E1)=1 with g(n,E1)=0 (The relative-velocity decomposition of a boost, checked in an explicit tetrad).

[L2]

For P∈TpM the observer energy and spatial momentum are En(P)=−c g(P,n) and qn(P)=hn(P); for g(P,P)≤0 the rest mass is m(P)=−g(P,P)/c≥0, and if m(P)>0 then U(P)=P/m(P) satisfies g(U(P),U(P))=−c2 (Observer-relative energy, spatial momentum and rest mass of a tangent vector).

[L3]

If P is future timelike with rest mass m, then with u=P/(mc) and γ=−g(n,u) one has En(P)=mc2γ, qn(P)=mc hn(u), g(qn(P),qn(P))=m2c2(γ2−1) and En(P)2=m2c4+c2g(qn(P),qn(P)) (Positivity, relative-velocity identities and the mass shell of observer-relative energy and momentum).

[L4]

For a>0 the nonnegative square root satisfies (a)2=a and a>0 (Rational powers ar of a positive base).

Verification

technique · direct
1.1L1L2L4algebra

Since g(u,u)=−1 by [L1], we get g(P,P)=m2c2g(u,u)=−m2c2, so −g(P,P)=m2c2>0; by [L4] and [L2] the rest mass is m(P)=m2c2/c=m. In particular P=mc u is future timelike and P/(m(P)c)=u.

2.1step 1.1L1L2algebra

By [L2] and [L1], En(P)=−c g(mcu,n)=−mc2g(u,n)=mc2γ because g(u,n)=−g(n,u)=−γ; and qn(P)=hn(P)=mc hn(u)=mcγβE1.

3.1step 2.1L1L3algebra∎

Using g(qn(P),qn(P))=m2c2γ2β2 and γ2(1−β2)=1, we compute En(P)2−c2g(qn(P),qn(P))=m2c4γ2(1−β2)=m2c4, which is the mass shell relation of [L3]; equivalently En(P)2=m2c4+c2g(qn(P),qn(P)). The energy is positive because m,c,γ>0.

Remarks

The example computes the same numbers as the elementary boost formula in an inertial frame, but only through the observer split of a single tetrad: the energy mc2γ and momentum mcγβ are vector and scalar data in TpM and its dual, not coordinate components. No measurement or empirical claim is made; the physical reading of En and qn belongs to the framework's units and postulates.

Depends on

Dependency tree · 2 levels

Sources