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Coordinate Hamiltonian and normal-observer energy: lapse and shift

Physical scope

Conditional energy identities in a supplied temporal chart; SR is used only as a flat consistency specialization, not as a curved premise.

Statement

Under the hypotheses of A curved temporal chart gives a local reduced Hamiltonian, all lapse, shift and spatial-metric coefficients in the reduced expressions below are evaluated at the spacetime event $(c t, q)$ , as in the temporal-shell supplier; $n$ is the normal at that same event. The future unit normal $n = N^{- 1} (\partial_{0} - β^{i} \partial_{i})$ measures positive particle energy

$E_{n} = - c p (n) = c κ,$

while the chart's reduced Hamiltonian is $H = c (N κ - β^{i} p_{i})$ . The normal-frame spatial velocity is $w^{i} = ({\dot{q}}^{i} / c + β^{i}) / N = h^{i j} p_{j} / κ$ , with $h_{i j} w^{i} w^{j} < 1$ for $m > 0$ and $= 1$ for a nonzero massless state. Write $∣ β ∣_{h} : = \sqrt{h_{i j} β^{i} β^{j}}$ and $∣ p ∣_{h^{- 1}} : = \sqrt{h^{i j} p_{i} p_{j}}$ . If $β^{2} : = h_{i j} β^{i} β^{j} < N^{2}$ , then $H > 0$ . If $β^{2} > N^{2}$ , $H$ can be negative even though $E_{n} > 0$ . Along a Hamiltonian solution, $d H / d t = \partial_{t} H$ ; explicit time independence is sufficient for its conservation. On a connected solution interval, $H (t, q (t), p (t))$ is constant if and only if $\partial_{t} H (t, q (t), p (t)) = 0$ throughout that interval.

Facts & Assumptions

Given: A lapse-shift temporal chart, positive $N, h$ , a future shell state, and a solution of its reduced equations.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Assume countable choice ACω, meaning every sequence of nonempty sets has a choice function; no full Axiom of Choice is assumed.

[F1]

Lorentz background, particle quantities and parameter units: $p$ is a covector and an observer is a future unit timelike vector.

[F2]

Temporal chart with lapse, shift and positive slice metric: $n = N^{- 1} (\partial_{0} - β^{i} \partial_{i})$ , $g (n, n) = - 1$ , and $h$ is positive definite.

[F3]

A curved temporal chart gives a local reduced Hamiltonian: κ, shell root and $H = c (N κ - β^{i} p_{i})$ as displayed.

[F4]

Energy, momentum, invariant mass and four-force of relativistic particles, and necessary collision thresholds: in the flat specialization, energy is the time component paired with the inertial observer; this supplier is not assumed on curved backgrounds.

[F5]

Free massive test-particle action: the massive branch is an adopted square-root test-particle model.

[F6]

Separate massless shell action: the massless branch is a separate model with nonzero future-null momentum.

Proof

1.1givenA1F1F2F3algebra

Use the inherited smooth bundle structure under [A1] and the quantity and temporal-shell conventions [F1,F2,F3]. Evaluate the covector on the normal: $p (n) = N^{- 1} (p_{0} - β^{i} p_{i}) = - κ$ by [F3]. Thus $E_{n} = - c p (n) = c κ > 0$ , while multiplying the shell root by $- c$ gives the chart Hamiltonian stated in [F3].

1.2givenF2F3F5F6algebra

Differentiate $H = c (N κ - β^{i} p_{i})$ with respect to $p_{i}$ , keeping $t, q$ fixed. Since $\partial κ / \partial p_{i} = h^{i j} p_{j} / κ$ , Hamilton's equation gives ${\dot{q}}^{i} / c = N h^{i j} p_{j} / κ - β^{i}$ , so $w^{i} = h^{i j} p_{j} / κ$ . Its squared $h$ -norm is $h^{i j} p_{i} p_{j} / (m^{2} c^{2} + h^{i j} p_{i} p_{j})$ , strictly below one for $m > 0$ and equal to one for $m = 0$ with nonzero spatial momentum.

2.1givenF2F3step 1.1algebra

Cauchy–Schwarz in the positive definite metric $h$ gives $β^{i} p_{i} \leq ∣ β ∣_{h} ∣ p ∣_{h^{- 1}}$ , and $κ \geq ∣ p ∣_{h^{- 1}}$ . If $β^{2} < N^{2}$ , then $∣ β ∣_{h} < N$ and $N κ - ∣ β ∣_{h} ∣ p ∣_{h^{- 1}} > 0$ unless $p = 0$ ; for $p = 0, m > 0$ , $N κ = N m c > 0$ as well, while the massless case excludes $p = 0$ . Hence $H > 0$ . If $β^{2} > N^{2}$ , choose $p_{i}$ a positive multiple of the metric dual of $β$ and let that multiple grow. Then $β^{i} p_{i} = ∣ β ∣_{h} ∣ p ∣_{h^{- 1}}$ and $H / c = N \sqrt{m^{2} c^{2} + ∣ p ∣_{h^{- 1}}^{2}} - ∣ β ∣_{h} ∣ p ∣_{h^{- 1}}$ is negative for sufficiently large $∣ p ∣_{h^{- 1}}$ , because its ratio to $∣ p ∣_{h^{- 1}}$ tends to $N - ∣ β ∣_{h} < 0$ . Step 1.1 still gives $E_{n} = c κ > 0$ .

2.2F3step 1.2algebra

Along a solution, $d H / d t = H_{t} + H_{q^{i}} {\dot{q}}^{i} + H_{p_{i}} {\dot{p}}_{i} = H_{t} + H_{q^{i}} H_{p_{i}} - H_{p_{i}} H_{q^{i}} = H_{t}$ . Thus explicit time independence suffices for conservation. More precisely, the value of $H$ along a connected solution interval is constant exactly when $H_{t}$ vanishes along that solution, by differentiating a constant in one direction and the mean value theorem in the other. This does not require $H_{t}$ to vanish at states off that solution.

3.1F4step 1.1∎

In a flat inertial chart with $N = 1, β = 0, h = δ$ , one has $H = c κ = E_{n}$ , agreeing with [F4]; this is the only use of the SR supplier.

Depends on

Dependency tree · 2 levels

Sources

Eric Gourgoulhon, 3+1 Formalism and Bases of Numerical Relativity, complete lecture notes (arXiv:gr-qc/0703035)