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Worked radar round trip: emission, reflection and reception at exact rational parameters

Physical scope

Exact SI radar assignments for an inertial observer in complete vacuum under the ideal-clock and ideal-light assumptions of the radar protocol, with available ideal instantaneous reflectors and a clock calibrated to zero at the chart origin. Coordinates and rest distances have unit metre, T and proper/radar time have unit second, and c has unit metre per second. No finite-apparatus, material-medium, empirical-measurement or uncertainty claim is made.

Example

Adopt the ideal-clock, ideal-light, available-reflector and SI premises of The operational radar protocol of an ideal observer, and calibrate the observer's ideal clock to read $τ = 0$ at $x^{0} = 0$ . In an inertial chart $(x^{0}, x^{1}, x^{2}, x^{3})$ of the affine Minkowski model with metric $diag (- 1, 1, 1, 1)$ let the observer be at rest at the spatial origin, so that its worldline is $x^{1} = x^{2} = x^{3} = 0$ , let $T > 0$ , and let $q : = x^{- 1} (c T, \frac{3}{5} c T, 0, 0)$ , an event of chart time $T$ and spatial coordinate $(3 c T / 5, 0, 0)$ . With $e_{-} : = x^{- 1} (\frac{2}{5} c T, 0, 0, 0)$ and $e_{+} : = x^{- 1} (\frac{8}{5} c T, 0, 0, 0)$ one has $q - e_{-} = (\frac{3}{5} c T, \frac{3}{5} c T, 0, 0)$ and $e_{+} - q = (\frac{3}{5} c T, - \frac{3}{5} c T, 0, 0)$ ; both displacements have $η$ -square $- \frac{9}{25} c^{2} T^{2} + \frac{9}{25} c^{2} T^{2} = 0$ and positive time component, hence are future null, so the radar protocol applies with emission $e_{-}$ , reflection $q$ and reception $e_{+}$ (The operational radar protocol of an ideal observer, The radar round trip of an inertial observer halves at the midpoint of the reception interval). Since the observer is inertial, its proper time between the events with temporal coordinates $λ_{-} = \frac{2}{5} c T$ and $λ_{+} = \frac{8}{5} c T$ (chart times $2 T / 5$ and $8 T / 5$ ) is $(λ_{+} - λ_{-}) / c$ with the stated clock calibration, so the radar assignments are $τ_{r a d} (q) = \frac{1}{2} (\frac{2}{5} T + \frac{8}{5} T) = T$ , the chart time of $q$ , and $ρ_{r a d} (q) = \frac{c}{2} (\frac{8}{5} T - \frac{2}{5} T) = \frac{3}{5} c T$ , the chart distance of $q$ from the observer. The same computation for $q^{'} : = x^{- 1} (c T, \frac{4}{5} c T, 0, 0)$ gives $e_{-}^{'} = x^{- 1} (\frac{1}{5} c T, 0, 0, 0)$ , $e_{+}^{'} = x^{- 1} (\frac{9}{5} c T, 0, 0, 0)$ , radar time $T$ and radar distance $\frac{4}{5} c T$ . Hence the two events $q$ and $q^{'}$ of equal chart time are radar-simultaneous, with radar distances equal to their chart distances, in agreement with The radar round trip of an inertial observer halves at the midpoint of the reception interval.

Facts & Assumptions

Given: The ideal-clock, ideal-light, reflector and SI premises of [L4], with the clock calibrated by $τ (0) = 0$ , an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with chart $x$ (Inertial frames and inertial charts of the affine Minkowski model), $T > 0$ , the observer worldline $W = {x^{- 1} (λ, 0, 0, 0) : λ \in R}$ at rest at the spatial origin, the events $q = x^{- 1} (c T, \frac{3}{5} c T, 0, 0)$ , $e_{-} = x^{- 1} (\frac{2}{5} c T, 0, 0, 0)$ , $e_{+} = x^{- 1} (\frac{8}{5} c T, 0, 0, 0)$ and $q^{'} = x^{- 1} (c T, \frac{4}{5} c T, 0, 0)$ , $e_{-}^{'} = x^{- 1} (\frac{1}{5} c T, 0, 0, 0)$ , $e_{+}^{'} = x^{- 1} (\frac{9}{5} c T, 0, 0, 0)$ (Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L1]

Chart and frame: $x^{- 1} (λ, r_{1}, r_{2}, r_{3}) = o + λ e_{0} + r_{1} e_{1} + r_{2} e_{2} + r_{3} e_{3}$ , the coordinate basis is $η$ -orthonormal with $η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ , and the spatial part of a displacement is measured by the Euclidean norm of The Euclidean spatial structure and cross product of an inertial frame (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L2]

Future null criterion: a nonzero vector $v = v^{0} e_{0} + v^{1} e_{1} + v^{2} e_{2} + v^{3} e_{3}$ is future null exactly when $η (v, v) = 0$ and $v^{0} > 0$ ; indeed the closure of the future cone is ${v : v^{0} \geq ∣ v_{v e c} ∣}$ in the coordinates of the frame, and on the null cone $v^{0} = ∣ v_{v e c} ∣$ , so for $v \neq 0$ the condition $v^{0} > 0$ is exactly $v \in cl (C^{+})$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space, Time orientation, future and past timelike/null/causal vectors, and the closed cones, The timelike set has exactly two convex components, described in orthonormal coordinates).

[L3]

Radar midpoint lemma: for a future unit $n$ , events $e_{\pm} = o + λ_{\pm} n$ with $λ_{-} < λ_{+}$ and an event $q$ , the conditions " $q - e_{-}$ future null and $e_{+} - q$ future null" hold if and only if $q = o + \frac{λ_{-} + λ_{+}}{2} n + u$ with $u \in n^{⊥}$ and $∣ u ∣ = (λ_{+} - λ_{-}) / 2$ ; in that case the midpoint is $m = \frac{e_{-} + e_{+}}{2}$ , the event $q$ lies in the slice $H (m, n)$ , and its rest distance in that slice is $(λ_{+} - λ_{-}) / 2$ (The radar round trip of an inertial observer halves at the midpoint of the reception interval, Simultaneity slices and rest distance relative to a future unit timelike vector).

[L4]

Radar protocol: an observer with proper-time function $τ$ assigns to a reflection event $q$ reached from $e_{-} = x (s_{-})$ and connected to $e_{+} = x (s_{+})$ by future null segments the values $τ_{r a d} (q) = \frac{1}{2} (τ (s_{-}) + τ (s_{+}))$ and $ρ_{r a d} (q) = \frac{c}{2} (τ (s_{+}) - τ (s_{-}))$ (The operational radar protocol of an ideal observer).

[L5]

Proper time of an inertial observer: for the straight worldline $λ \mapsto o + λ e_{0}$ the proper-time function satisfies $τ^{'} (λ) = c^{- 1} \sqrt{- η (e_{0}, e_{0})} = 1 / c$ , hence $τ (λ) = λ / c$ under the stated calibration (Proper time along a future timelike worldline, Inertial frames and inertial charts of the affine Minkowski model).

Verification

technique · direct

1.1L1L2L3algebra

The events of the statement are $e_{-} = o + \frac{2}{5} c T e_{0}$ , $e_{+} = o + \frac{8}{5} c T e_{0}$ , $q = o + c T e_{0} + \frac{3}{5} c T e_{1}$ and $m : = x^{- 1} (c T, 0, 0, 0) = o + c T e_{0}$ by [L1]; the observer direction is $n : = e_{0}$ , future unit. With $u : = \frac{3}{5} c T e_{1} \in n^{⊥}$ one has $q = m + u$ and $∣ u ∣ = \frac{3}{5} c T = \frac{1}{2} (\frac{8}{5} c T - \frac{2}{5} c T) = (λ_{+} - λ_{-}) / 2$ , so condition (ii) of [L3] holds; by the implication (ii) $\Rightarrow$ (i) of [L3] the displacements $q - e_{-}$ and $e_{+} - q$ are future null and the radar protocol applies with emission $e_{-}$ , reflection $q$ and reception $e_{+}$ . Directly, $q - e_{-} = \frac{3}{5} c T e_{0} + \frac{3}{5} c T e_{1}$ and $e_{+} - q = \frac{3}{5} c T e_{0} - \frac{3}{5} c T e_{1}$ by [L1], and $η (q - e_{-}, q - e_{-}) = - \frac{9}{25} c^{2} T^{2} + \frac{9}{25} c^{2} T^{2} = 0$ with time component $\frac{3}{5} c T > 0$ , so $q - e_{-}$ is future null by [L2], and likewise $η (e_{+} - q, e_{+} - q) = 0$ with time component $\frac{3}{5} c T > 0$ , so $e_{+} - q$ is future null.

1.2L1L3L4L5algebra

Radar assignments for $q$ : the proper-time function of the observer along $W$ is $τ (λ) = λ / c$ under the stated calibration by [L5], so the emission event $e_{-}$ has parameter $λ_{-} = \frac{2}{5} c T$ and clock reading $τ (e_{-}) = \frac{2}{5} T$ , and the reception event $e_{+}$ has parameter $λ_{+} = \frac{8}{5} c T$ and reading $τ (e_{+}) = \frac{8}{5} T$ . By [L4], $τ_{r a d} (q) = \frac{1}{2} (\frac{2}{5} T + \frac{8}{5} T) = T$ , the chart time $x^{0} (q) / c$ , and $ρ_{r a d} (q) = \frac{c}{2} \cdot \frac{6}{5} T = \frac{3}{5} c T$ . The chart distance of $q$ from the observer is the Euclidean norm of the spatial coordinate difference $(3 c T / 5, 0, 0)$ , namely $\frac{3}{5} c T$ , so $ρ_{r a d} (q)$ equals the chart distance; and by [L3] the same number is the rest distance $d_{n} (m, q)$ in the slice $H (m, n)$ through the midpoint $m = x^{- 1} (c T, 0, 0, 0)$ .

1.3L1L2L3L4L5algebra

The same computation for $q^{'} = o + c T e_{0} + \frac{4}{5} c T e_{1}$ : here $q^{'} - e_{-}^{'} = \frac{4}{5} c T e_{0} + \frac{4}{5} c T e_{1}$ and $e_{+}^{'} - q^{'} = \frac{4}{5} c T e_{0} - \frac{4}{5} c T e_{1}$ by [L1], both with $η$ -square $- \frac{16}{25} c^{2} T^{2} + \frac{16}{25} c^{2} T^{2} = 0$ and time component $\frac{4}{5} c T > 0$ , hence future null by [L2]; equivalently $q^{'} = m^{'} + u^{'}$ with $m^{'} = x^{- 1} (c T, 0, 0, 0) = m$ and $u^{'} = \frac{4}{5} c T e_{1}$ , where $∣ u^{'} ∣ = \frac{4}{5} c T = \frac{1}{2} (\frac{9}{5} c T - \frac{1}{5} c T)$ satisfies condition (ii) of [L3]. Since $τ (e_{-}^{'}) = \frac{1}{5} T$ and $τ (e_{+}^{'}) = \frac{9}{5} T$ , [L4] gives $τ_{r a d} (q^{'}) = \frac{1}{2} (\frac{1}{5} T + \frac{9}{5} T) = T$ and $ρ_{r a d} (q^{'}) = \frac{c}{2} \cdot \frac{8}{5} T = \frac{4}{5} c T$ , the chart distance of $q^{'}$ ; the midpoint of the $q^{'}$ round trip is again $m$ .

2.1step 1.1step 1.2step 1.3∎

Conclusion: by steps 1.1 and 1.3 the two round trips have the same midpoint $m = m^{'} = x^{- 1} (c T, 0, 0, 0)$ , and the assigned radar times agree, $τ_{r a d} (q) = τ_{r a d} (q^{'}) = T$ ; so $q$ and $q^{'}$ are radar-simultaneous although they are distinct events with different spatial coordinates, and each lies in the slice $H (m, n)$ with the rest vector $u = \frac{3}{5} c T e_{1}$ or $u^{'} = \frac{4}{5} c T e_{1}$ by [L3]. Their radar distances $\frac{3}{5} c T$ and $\frac{4}{5} c T$ equal their chart distances and their rest distances in that slice, while the reception intervals differ, $\frac{6}{5} c T$ for $q$ and $\frac{8}{5} c T$ for $q^{'}$ , reflecting the different rest distances from the same midpoint. This is the exact rational instance of [L3]: the reflection events on one slice need not be reflections for the same emission-reception pair, and the radar assignment reproduces the chart time and chart distance for each of them.

Remarks

Exact arithmetic only. All parameters are rational multiples of $c T$ ; the nullity of the two segments and the two radar assignments are exact evaluations, and no numerical approximation or measurement model is involved.
Clock calibration. The stated calibration fixes the additive proper-time constant to zero and hence gives the displayed radar time $T$ . A different common clock offset would add that offset to both radar times while leaving their equality and both radar distances unchanged.
One midpoint, different reflection radii. The two reflection events $q$ and $q^{'}$ lie on the same slice $H (m, n)$ but at different rest distances $\frac{3}{5} c T$ and $\frac{4}{5} c T$ from the midpoint $m$ ; the lemma's statement that admissible reflections for one fixed emission-reception pair differ only in the direction of $u$ is not being applied to the pair ${q, q^{'}}$ , which has two different reception intervals, but each event separately is checked against its own pair by [L3].

Worked radar round trip: emission, reflection and reception at exact rational parameters

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

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