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The measured energy density of a fixed tensor depends on the observer congruence

Physical scope

A flat-space kinematic counterexample in the framework's reading: the energy density assigned by the observer split of a fixed smooth symmetric tensor depends on the chosen observer congruence, shown for a static dust tensor measured by a static and by a boosted congruence. The statement refuted is the observer independence of the measured energy density in the smooth branch; no dynamics, no conservation, no measurement and no empirical claim is involved.

Statement refuted

Statement refuted: "Let $U$ be a nonempty open subset of a time-oriented smooth Lorentzian four-manifold and let $T$ be a smooth symmetric covariant $2$ -tensor field on $U$ . If $n$ and $n^{'}$ are smooth observer congruences on $U$ , then the measured energy densities of The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress agree, $T_{a b} n^{a} n^{b} = T_{a b} n^{' a} n^{' b}$ on $U$ ."

Facts & Assumptions

Given: The flat Lorentzian four-manifold $(R^{4}, g)$ with identity chart, $g = diag (- 1, 1, 1, 1)$ and time orientation $N = \partial_{0}$ , with the metric-selected connection; the constant $ρ > 0$ , read as rest energy density in $J m^{- 3}$ with $ρ_{m} = ρ / c^{2}$ the SI rest-mass density; the dust model $(ρ, u)$ with $u = \partial_{0}$ and tensor $T^{a b} = ρ u^{a} u^{b}$ (The dust matter model: rest density, four-velocity, stress-energy and dust system); the number $β \in (0, 1)$ , $γ = (1 - β^{2})^{- 1 / 2}$ ; and the vector fields $n = \partial_{0}$ and $n^{'} = γ (\partial_{0} + β \partial_{1})$ .

[L1]

$n = \partial_{0}$ and $n^{'} = γ (\partial_{0} + β \partial_{1})$ are smooth future-directed unit timelike vector fields on $R^{4}$ , hence observer congruences: $g (n, n) = - 1$ , $g (n, \partial_{0}) = - 1$ , and $g (n^{'}, n^{'}) = γ^{2} (- 1 + β^{2}) = - 1$ , $g (n^{'}, \partial_{0}) = - γ < 0$ (Observer congruences, their observer curves and adapted charts, Observers, rest spaces and the spatial projection at an event, Timelike, null, spacelike and causal vectors of a Lorentz vector space, Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model).

[L2]

The measured energy density is $μ = T_{a b} n^{a} n^{b}$ (The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress clause (b), Physical reading of the observer split: measured energy density, energy flux and momentum density, and measured stress clause (a)).

[L3]

On the identity chart the metric equals its inverse, $η^{a b} = η_{a b} = diag (- 1, 1, 1, 1)$ , and $\partial_{0}^{a} = δ_{0}^{a}$ with $\partial_{0, a} = - δ_{a}^{0}$ (The inverse metric, musical isomorphisms and index raising, Smooth covariant 2-tensor fields on a smooth manifold).

Counterexample

1.1L1given

(Two congruences.) By [L1], $n$ and $n^{'}$ are both smooth observer congruences on $U : = R^{4}$ , and they are distinct because $n^{'}$ has the nonzero spatial component $γ β \partial_{1}$ .

1.2givenL3

(The tensor.) $T^{a b} = ρ u^{a} u^{b}$ is the tensor of the dust model $(ρ, \partial_{0})$ ; it is a smooth symmetric type $(2, 0)$ tensor field, equivalently $T_{a b} = ρ \partial_{0, a} \partial_{0, b}$ after lowering both indices with [L3].

2.1step 1.2L2L3algebra

(Energy density for the static congruence.) For $n = \partial_{0}$ , [L3] gives $u_{a} n^{a} = \partial_{0, a} δ_{0}^{a} = - 1$ , so $μ = T_{a b} n^{a} n^{b} = ρ (u_{a} n^{a})^{2} = ρ$ by [L2].

2.2step 1.2L1L2L3algebra

(Energy density for the boosted congruence.) For $n^{'}$ , [L1] and [L3] give $u_{a} n^{' a} = γ \partial_{0, a} (δ_{0}^{a} + β δ_{1}^{a}) = - γ$ . Hence $μ^{'} = T_{a b} n^{' a} n^{' b} = ρ (u_{a} n^{' a})^{2} = ρ γ^{2}$ by [L2], and $μ^{'} - μ = ρ (γ^{2} - 1) > 0$ because $β \in (0, 1)$ implies $γ > 1$ .

3.1step 2.1step 2.2algebra∎

(Conclusion.) The fixed smooth symmetric tensor $T$ assigns the energy density $ρ$ to the congruence $n$ and $ρ γ^{2}$ to the congruence $n^{'}$ , and these values differ since $γ > 1$ . The displayed statement is therefore refuted: in the framework's reading, measured energy density is observer dependent. This completes the counterexample.

Remarks

What the counterexample does not say. It does not contradict the algebraic decomposition of The orthogonal observer decomposition of a symmetric two-tensor and its uniqueness, which holds for each congruence separately; it shows that the measured density is not a property of $T$ alone. In the framework's reading this is the kinematic fact that a moving medium has a boosted energy density.
No empirical content. Both congruences are idealizations and no measurement is asserted; the comparison is between two exact model readings of one prescribed tensor.

The measured energy density of a fixed tensor depends on the observer congruence

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources