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Worked simultaneity gap of the traveler's two momentary rest frames at the turn

Example

Fix reals $c > 0$ , $T > 0$ , $v = 3 c / 5$ and $γ = 5 / 4$ . This is a mathematical proper-time computation: the inertial chart $x$ has metric $diag (- 1, 1, 1, 1)$ and $x^{0} = c t$ ; the stay-at-home worldline is $y_{s t a y} (t) = x^{- 1} (c t, 0, 0, 0)$ for $t \in [0, 2 T]$ ; the traveler is $y_{t r a v e l} (t) = x^{- 1} (c t, v t, 0, 0)$ on $[0, T]$ and $y_{t r a v e l} (t) = x^{- 1} (c t, v (2 T - t), 0, 0)$ on $[T, 2 T]$ . All clock readings below mean the based mathematical proper-time values of these curves, initialized to zero at $t = 0$ (Proper time along a future timelike worldline, Total proper time of a piecewise C^1 future timelike worldline), so the stay-at-home reading equals $t$ ; the turn event is $r : = x^{- 1} (c T, 3 c T / 5, 0, 0)$ ; the two momentary inertial rest frames of the traveler are the charts $x^{'}$ and $x^{''}$ of the boosts with velocities $+ 3 c / 5$ and $- 3 c / 5$ , both with the same origin $p = x^{- 1} (0, 0, 0, 0)$ as $x$ . Explicitly their bases are $e_{a}^{'} = B (- \frac{3}{5} e_{1}) e_{a}$ and $e_{a}^{''} = B (\frac{3}{5} e_{1}) e_{a}$ for $a = 0, 1, 2, 3$ ; the two charts are defined from these bases and origin by Inertial frames and inertial charts of the affine Minkowski model.

(a) Outbound rest frame. The chart $x^{'}$ has $x^{' 0} = \frac{5}{4} (x^{0} - \frac{3}{5} x^{1})$ ; at $r$ this is $\frac{5}{4} (c T - \frac{3}{5} \cdot \frac{3 c T}{5}) = \frac{5}{4} (c T - \frac{9 c T}{25}) = \frac{5}{4} \cdot \frac{16 c T}{25} = \frac{4 c T}{5}$ , so the slice $x^{' 0} = \frac{4 c T}{5}$ through $r$ meets the stay-at-home worldline $x^{1} = 0$ at $\frac{5}{4} x^{0} = \frac{4 c T}{5}$ , that is $x^{0} = \frac{16 c T}{25}$ , i.e. coordinate time $t = \frac{16 T}{25} = T (1 - \frac{9}{25})$ .

(b) Inbound rest frame. The chart $x^{''}$ has $x^{'' 0} = \frac{5}{4} (x^{0} + \frac{3}{5} x^{1})$ ; at $r$ this is $\frac{5}{4} (c T + \frac{9 c T}{25}) = \frac{5}{4} \cdot \frac{34 c T}{25} = \frac{17 c T}{10}$ , so the slice $x^{'' 0} = \frac{17 c T}{10}$ through $r$ meets the stay-at-home worldline at $\frac{5}{4} x^{0} = \frac{17 c T}{10}$ , that is $x^{0} = \frac{34 c T}{25}$ , i.e. coordinate time $t = \frac{34 T}{25} = T (1 + \frac{9}{25})$ .

(c) The gap. The two momentary frames assign to the stay-at-home clock 'at the time of the turn' the two different readings $\frac{16 T}{25}$ and $\frac{34 T}{25}$ ; their difference is $\frac{18 T}{25} = \frac{2 T v^{2}}{c^{2}}$ , the simultaneity gap.

(d) Reconciliation. The traveler's own clock reads $\frac{4}{5} T = \frac{20 T}{25}$ at the turn and $\frac{8}{5} T = \frac{40 T}{25}$ at the reunion; during each leg his momentary frame assigns to the stay-at-home clock the advance $\frac{16 T}{25}$ (outbound from $0$ to $\frac{16 T}{25}$ ; inbound from $\frac{34 T}{25}$ to $2 T = \frac{50 T}{25}$ ); with the gap $\frac{18 T}{25}$ the total assignment is $\frac{16 T}{25} + \frac{18 T}{25} + \frac{16 T}{25} = \frac{50 T}{25} = 2 T$ , the stay-at-home reading at the reunion, while the traveler reads $\frac{8}{5} T = \frac{40 T}{25}$ ; the two readings differ by $\frac{2 T}{5} = \frac{10 T}{25}$ .

(e) The two rest frames are distinct. Their relative boost parameter is the collinear composition $\frac{2 \cdot (3 / 5)}{1 + 9 / 25} = \frac{6 / 5}{34 / 25} = \frac{15}{17} \neq 0$ , so the traveler's momentary inertial frame changes at the turn.

Facts & Assumptions

Given: An inertial frame with chart $x$ , metric $diag (- 1, 1, 1, 1)$ and $x^{0} = c t$ ; $T > 0$ , $v = 3 c / 5$ , $γ = 5 / 4$ ; the two explicitly parametrized worldlines and mathematical proper-time readings of the Example, the turn event $r = x^{- 1} (c T, 3 c T / 5, 0, 0)$ , the reunion event $q = x^{- 1} (2 c T, 0, 0, 0)$ , and the boost charts $x^{'}$ , $x^{''}$ of the lemma Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation for velocities $+ 3 c / 5$ and $- 3 c / 5$ , both with origin $p = x^{- 1} (0, 0, 0, 0)$ .

[L1]

Boost chart transitions (Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation): for inertial frames with a common origin whose bases are related by a boost of velocity $v$ with $∣ v ∣ < c$ along a common spatial direction one has $x^{' 0} = γ (x^{0} - \frac{v}{c} x^{1})$ , $x^{' 1} = γ (x^{1} - \frac{v}{c} x^{0})$ with $γ = (1 - v^{2} / c^{2})^{- 1 / 2}$ ; and along the worldline at rest in $x^{'}$ the clock rate is $d τ / d t = 1 / γ$ .

[L2]

Simultaneity slices (Simultaneity slices and rest distance relative to a future unit timelike vector, Inertial frames and inertial charts of the affine Minkowski model): for a future unit timelike vector $n$ the slice through an event $p$ is ${u : η (n, u - p) = 0}$ ; in a chart the slice of constant $x^{' 0}$ through an event consists of the events with that coordinate, and coordinate differences of events are the coordinates of their displacements.

[L3]

Boosts (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta), Linear and affine Lorentz maps, and the Lorentz group): for $∣ β ∣ < 1$ the passive boost $B (β e)$ is a proper orthochronous Lorentz map, so it maps inertial bases to inertial bases and its charts are inertial charts.

[L4]

Collinear composition (Collinear boost composition and the relativistic velocity-addition law): $B (β_{1} e) B (β_{2} e) = B (\frac{β_{1} + β_{2}}{1 + β_{1} β_{2}} e)$ .

[L5]

Proper times (Proper time along a future timelike worldline, Total proper time of a piecewise C^1 future timelike worldline): a straight leg with constant tangent $(c, \pm v, 0, 0)$ over a parameter interval of length $T$ has proper time $T / γ$ , and the two-leg total is $2 T / γ$ ; the stay-at-home worldline has tangent $(c, 0, 0, 0)$ , so its based proper time is $c^{- 1} \int_{0}^{t} c d u = t$ . This identifies the mathematical reading, without any physical-clock assumption.

[L6]

Arithmetic, roots and order (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives, Ordered field): all the rational evaluations below are exact; $(5 / 4)^{- 1} = 4 / 5$ and $(3 / 5)^{2} = 9 / 25$ .

Verification

technique · direct

1.1givenL1L3L5

By [L1] with $v / c = 3 / 5$ and $γ = 5 / 4$ the outbound chart is $x^{' 0} = \frac{5}{4} (x^{0} - \frac{3}{5} x^{1})$ and the inbound chart is $x^{'' 0} = \frac{5}{4} (x^{0} + \frac{3}{5} x^{1})$ ; both are inertial charts by [L3], and the stay-at-home worldline has $x^{1} = 0$ with clock reading the coordinate time $t = x^{0} / c$ by [L5].

2.1step 1.1L2L6algebra

For (a): the turn event has $x^{0} (r) = c T$ and $x^{1} (r) = 3 c T / 5$ , so $x^{' 0} (r) = \frac{5}{4} (c T - \frac{3}{5} \cdot \frac{3 c T}{5}) = \frac{5}{4} \cdot \frac{16 c T}{25} = \frac{4 c T}{5}$ ; the slice $x^{' 0} = \frac{4 c T}{5}$ meets $x^{1} = 0$ at $\frac{5}{4} x^{0} = \frac{4 c T}{5}$ , that is $x^{0} = \frac{16 c T}{25}$ and clock reading $t = \frac{16 T}{25}$ .

2.2step 1.1L2L6algebra

For (b): $x^{'' 0} (r) = \frac{5}{4} (c T + \frac{3}{5} \cdot \frac{3 c T}{5}) = \frac{5}{4} \cdot \frac{34 c T}{25} = \frac{17 c T}{10}$ ; the slice $x^{'' 0} = \frac{17 c T}{10}$ meets $x^{1} = 0$ at $\frac{5}{4} x^{0} = \frac{17 c T}{10}$ , that is $x^{0} = \frac{34 c T}{25}$ and clock reading $t = \frac{34 T}{25}$ .

2.3step 1.1L4L6algebra

For (e): the outbound frame is obtained from $x$ by a boost of dimensionless parameter $3 / 5$ and the inbound frame by one of parameter $- 3 / 5$ ; the boost from the inbound to the outbound momentary frame is therefore the collinear composition of parameters $3 / 5$ and $3 / 5$ in the composed description, whose parameter is $\frac{3 / 5 + 3 / 5}{1 + (3 / 5) (3 / 5)} = \frac{6 / 5}{34 / 25} = \frac{15}{17}$ by [L4]; since $\frac{15}{17} \neq 0$ , the two momentary inertial frames are distinct.

3.1step 2.1step 2.2L6algebra

For (c): the two readings differ by $\frac{34 T}{25} - \frac{16 T}{25} = \frac{18 T}{25}$ , and $\frac{2 T v^{2}}{c^{2}} = 2 T \cdot \frac{9}{25} = \frac{18 T}{25}$ , so the gap equals $\frac{2 T v^{2}}{c^{2}}$ .

4.1step 3.1L1L5algebra

For (d): by [L5] the traveler's clock reads $T / γ = \frac{4}{5} T = \frac{20 T}{25}$ at the turn and $2 T / γ = \frac{8}{5} T = \frac{40 T}{25}$ at the reunion; the outbound frame assigns to the stay-at-home clock the advance from $t = 0$ to $t = \frac{16 T}{25}$ , namely $\frac{16 T}{25}$ , the inbound frame assigns the advance from $t = \frac{34 T}{25}$ to $t = 2 T = \frac{50 T}{25}$ , again $\frac{16 T}{25}$ , so with the gap the total assigned reading is $\frac{16 T}{25} + \frac{18 T}{25} + \frac{16 T}{25} = \frac{50 T}{25} = 2 T$ , equal to the stay-at-home clock reading at the reunion, while the traveler reads $\frac{40 T}{25}$ ; the readings differ by $\frac{10 T}{25} = \frac{2 T}{5}$ .

5.1step 2.1step 2.2step 3.1step 4.1step 2.3L1L2L4L5∎

All the numbers of (a)-(e) are exact rational evaluations of the model's chart transitions, slices and proper-time values: the simultaneity gap of the traveler's two momentary frames at the turn is $\frac{18 T}{25} = \frac{2 T v^{2}}{c^{2}}$ , and the leg-by-leg assignment $\frac{16 T}{25} + \frac{18 T}{25} + \frac{16 T}{25} = 2 T$ reproduces the stay-at-home reading. The example makes no empirical claim; it displays a reconciliation of the traveler's momentary-frame accounts within the model.

Worked simultaneity gap of the traveler's two momentary rest frames at the turn

Example

Facts & Assumptions

Verification

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