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Worked frame data of a future null wave covector and its boost

Example

This worked example computes the frame data of a future null wave covector and its passive boost. In an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ let $ϖ > 0$ and $s = (0, 3 / 5, 4 / 5)$ , and let $k$ be the covector with frame components $(k_{0}, k_{1}, k_{2}, k_{3}) = (- ϖ, 0, 3 ϖ / 5, 4 ϖ / 5)$ . Then:

(a) $∣ s ∣ = 1$ and $k$ is a future null wave covector; its raised vector is $k^{♯} = ϖ (e_{0} + (3 / 5) e_{2} + (4 / 5) e_{3})$ and $k (k^{♯}) = 0$ .

(b) Let $x^{'}$ be the inertial chart with $x^{'} = B (β) x + b$ and $β = (3 / 5) e_{1}$ . The frame- $x^{'}$ data of the same covector is $(ϖ^{'}, s^{'})$ with $ϖ^{'} / ϖ = γ (1 - β \cdot s) = (5 / 4) (1 - 0) = 5 / 4$ and $s^{'} = (- 3 / 5, 12 / 25, 16 / 25)$ .

(c) Applying the transformation with $β$ replaced by $- β$ to $(ϖ^{'}, s^{'})$ returns $(ϖ, s)$ : $ϖ^{''} / ϖ^{'} = (5 / 4) (16 / 25) = 4 / 5$ , so $ϖ^{''} = ϖ$ , and $s^{''} = (0, 3 / 5, 4 / 5)$ .

The verification is exact rational algebra; no physical interpretation of the data is used, and no choice principle is used.

Facts & Assumptions

Given: An inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ , a real $ϖ > 0$ , $s = (0, 3 / 5, 4 / 5)$ , and the covector $k$ with components $(k_{0}, k_{1}, k_{2}, k_{3}) = (- ϖ, 0, 3 ϖ / 5, 4 ϖ / 5)$ .

[L1]

If a covector has frame components $(- ϖ, ϖ s_{1}, ϖ s_{2}, ϖ s_{3})$ with $ϖ > 0$ and $∣ s ∣ = 1$ , then it is a future null wave covector; its raised vector is $k^{♯} = ϖ (e_{0} + s_{1} e_{1} + s_{2} e_{2} + s_{3} e_{3})$ with $k (k^{♯}) = η (k^{♯}, k^{♯}) = 0$ (Frame data of a future null wave covector: the unique pair $(ϖ, s)$ with $∣ s ∣ = 1$ , clauses (d) and (e), The metric pairing is a linear bijection to the dual, and the raised covector in an orthonormal frame).

[L2]

For inertial charts with $x^{'} = B (β) x + b$ , $∣ β ∣ < 1$ , and a future null wave covector with frame- $x$ data $(ϖ, s)$ , the frame- $x^{'}$ data satisfies $ϖ^{'} = γ (1 - β \cdot s) ϖ$ , $s_{i}^{'} = [s_{i} + \frac{γ - 1}{∣ β ∣^{2}} (β \cdot s) β_{i} - γ β_{i}] / [γ (1 - β \cdot s)]$ for $β \neq 0$ , and in the axis case $β = b e_{1}$ : $s_{1}^{'} = (s_{1} - b) / (1 - b s_{1})$ , $s_{2}^{'} = s_{2} / [γ (1 - b s_{1})]$ , $s_{3}^{'} = s_{3} / [γ (1 - b s_{1})]$ ; applying the transformation with $- β$ returns the original data (Boost transformation of a future null wave covector and of its frame data, clauses (ii), (iii), (v)).

[L3]

$B (β)$ is the passive boost of The passive Lorentz boost with parameter beta with $γ = (1 - ∣ β ∣^{2})^{- 1 / 2}$ , and $B (β)$ is proper orthochronous with $B (- β) = B (β)^{- 1}$ (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)); the frame is an inertial frame with an inertial chart (Inertial frames and inertial charts of the affine Minkowski model).

Verification

technique · direct

1.1L1givenalgebra

Clause (a). Here $ϖ s = (0, 3 ϖ / 5, 4 ϖ / 5) = (k_{1}, k_{2}, k_{3})$ and $k_{0} = - ϖ$ , so the components of $k$ are exactly $(- ϖ, ϖ s)$ with $ϖ > 0$ ; and $∣ s ∣^{2} = 0 + (3 / 5)^{2} + (4 / 5)^{2} = 9 / 25 + 16 / 25 = 1$ . By [L1] the covector $k$ is a future null wave covector with $k^{♯} = ϖ (e_{0} + (3 / 5) e_{2} + (4 / 5) e_{3})$ , and $k (k^{♯}) = η (k^{♯}, k^{♯}) = - ϖ^{2} + ϖ^{2} (0 + 9 / 25 + 16 / 25) = 0$ .

2.1step 1.1L2L3algebra

Clause (b). With $β = (3 / 5) e_{1}$ one has $∣ β ∣ = 3 / 5$ and $γ = (1 - 9 / 25)^{- 1 / 2} = (16 / 25)^{- 1 / 2} = 5 / 4$ [L3, algebra], while $β \cdot s = (3 / 5) \cdot 0 + 0 \cdot 3 / 5 + 0 \cdot 4 / 5 = 0$ . By [L2] the frame- $x^{'}$ data is $ϖ^{'} = γ (1 - β \cdot s) ϖ = (5 / 4) (1 - 0) ϖ = 5 ϖ / 4$ , and the axis formulas with $b = 3 / 5$ , $s_{1} = 0$ , $γ = 5 / 4$ give $s_{1}^{'} = (0 - 3 / 5) / (1 - 0) = - 3 / 5$ , $s_{2}^{'} = (3 / 5) / ((5 / 4) (1 - 0)) = (3 / 5) \cdot (4 / 5) = 12 / 25$ and $s_{3}^{'} = (4 / 5) \cdot (4 / 5) = 16 / 25$ . As a check, $∣ s^{'} ∣^{2} = 9 / 25 + 144 / 625 + 256 / 625 = 9 / 25 + 16 / 25 = 1$ , in agreement with $∣ s^{'} ∣ = 1$ in [L2].

3.1step 2.1L2L3algebra∎

Clause (c). Apply [L2] to the chart pair $(x^{'}, x)$ , whose transition has parameter $- β = - (3 / 5) e_{1}$ , and to the data $(ϖ^{'}, s^{'})$ of step 2.1. Since $β \cdot s^{'} = (3 / 5) (- 3 / 5) = - 9 / 25$ , the factor is $γ (1 - (- β) \cdot s^{'}) = (5 / 4) (1 + β \cdot s^{'}) = (5 / 4) (1 - 9 / 25) = (5 / 4) (16 / 25) = 4 / 5$ , so $ϖ^{''} = (4 / 5) ϖ^{'} = (4 / 5) (5 / 4) ϖ = ϖ$ . The axis formulas with parameter $b = - 3 / 5$ give $s_{1}^{''} = (s_{1}^{'} - (- 3 / 5)) / (1 - (- 3 / 5) s_{1}^{'}) = (- 3 / 5 + 3 / 5) / (1 - 9 / 25) = 0$ , $s_{2}^{''} = s_{2}^{'} / [γ (1 - (- 3 / 5) s_{1}^{'})] = (12 / 25) / [(5 / 4) (1 - 9 / 25)] = (12 / 25) / [(5 / 4) (16 / 25)] = (12 / 25) / (4 / 5) = 3 / 5$ and $s_{3}^{''} = (16 / 25) / (4 / 5) = 4 / 5$ . Hence the round trip returns $ϖ^{''} = ϖ$ and $s^{''} = (0, 3 / 5, 4 / 5)$ , in agreement with clause (v) of [L2]. This proves (a), (b) and (c); all steps are exact rational arithmetic, no physical interpretation of the data was used, and no choice principle was used.

Worked frame data of a future null wave covector and its boost

Example

Facts & Assumptions

Verification

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