Exercises — Simultaneity — relativity of simultaneity

What we test: whether spatial separation is what drives disagreement. Here $\Delta x=0$. The simultaneity gap needs a nonzero $\Delta x$ to act, but more directly these events are timelike (same place, real time gap → light easily connects them). Timelike order is preserved in all frames. So yes, same order for the rocket — the earlier event stays earlier. The rocket sees a longer gap (that's Time Dilation), but never a flip.

What we test: reading the sign without plugging numbers. Map onto Figure s01: A is the back B (at $x=0$), B is the front F (at $x=+500$). Look at the amber clock — the front clock lags, so in $S^{\prime }$ the front event B happens first. Check with the signed formula (event 2 $=$ B, event 1 $=$ A): $\Delta x=x_B-x_A=+500>0$, so $\Delta t^{\prime }=t_B^{\prime }-t_A^{\prime }=-\gamma v\Delta x/c^2<0$. A negative signed gap means $t_B^{\prime }<t_A^{\prime }$: B occurs earlier. ✅

Step 1 — get $\gamma$ (why: the formula needs it). $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=1.25$ Step 2 — plug into the signed gap formula. Let event 2 $=$ front (larger $x$), event 1 $=$ back. $\Delta x=+400\text{m}$ is the separation in the platform frame $S$ where events are simultaneous — exactly what the formula wants. $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(1.25)(0.6c)(400)}{c^2}=-\frac{300}{c}$ Step 3 — numbers. $\Delta t^{\prime }=-\frac{300}{3\times 10^8}=-1.0\times 10^{-6}\text{s},|\Delta t^{\prime }|=1.0\mu \text{s}$ The negative sign (amber clock in Figure s01 lags) says the front bolt is recorded first in the train frame; its magnitude is $1.0\mu \text{s}$.

Reasoning first: both $\gamma$ and $v$ increase, so the gap grows. $\gamma =1/\sqrt{1-0.64}=1/0.6=5/3\approx 1.6667$. $\Delta t^{\prime }=-\frac{(5/3)(0.8c)(400)}{c^2}=-\frac{(5/3)(0.8)(400)}{3\times 10^8}=-\frac{533.3}{3\times 10^8}\approx -1.78\times 10^{-6}\text{s}$ So $|\Delta t^{\prime }|\approx 1.78\mu \text{s}$ — larger than the $1.0\mu \text{s}$ at $0.6c$, as predicted. Sign still negative: front bolt first.

Why compute each separately: it shows the gap isn't a trick — it drops out of the full Lorentz Transformation $t^{\prime }=\gamma (t-vx/c^2)$. $\gamma =1/\sqrt{1-0.25}=1/\sqrt{0.75}=1.1547$. $t_A^{\prime }=\gamma \left(0-\frac{(0.5c)(0)}{c^2}\right)=0$ $t_B^{\prime }=\gamma \left(0-\frac{(0.5c)(600)}{c^2}\right)=-\gamma \frac{0.5\cdot 600}{c}=-1.1547\cdot \frac{300}{3\times 10^8}$ $t_B^{\prime }=-1.1547\times 1.0\times 10^{-6}=-1.15\times 10^{-6}\text{s}$ Difference: $\Delta t^{\prime }=t_B^{\prime }-t_A^{\prime }=-1.15\times 10^{-6}\text{s}$ (i.e. $|\Delta t^{\prime }|=1.15\mu \text{s}$). Negative → B (the leading event, the amber clock in Figure s01) happens before A in $S^{\prime }$. ✅ Matches the rule.

The subtlety: the flashes are simultaneous in the rod frame $S$, so $\Delta x$ in the formula must be the rest length (proper length) $L_0$, not the contracted ground-frame 300 m. Step 1 — undo length contraction to get $L_0$. Length contraction: $L=L_0/\gamma$, so $L_0=\gamma L$. $\gamma =1/\sqrt{1-0.64}=1/0.6=5/3$. $L_0=\frac{5}{3}\times 300=500\text{m}$ Step 2 — apply the signed gap formula. Take event 2 $=$ the flash at the front end (larger $x$, the direction of motion), event 1 $=$ the rear flash, so $\Delta x=+L_0=+500\text{m}$. $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(5/3)(0.8c)(500)}{c^2}=-\frac{(5/3)(0.8)(500)}{3\times 10^8}=-\frac{666.7}{3\times 10^8}$ $\Delta t^{\prime }\approx -2.22\times 10^{-6}\text{s},|\Delta t^{\prime }|=2.22\mu \text{s}$ Which flash first? The sign is negative, so $t_{\text{front}}^{\prime }<t_{\text{rear}}^{\prime }$: the front (leading) flash occurs first in the ground frame $S^{\prime }$, by $2.22\mu \text{s}$. This is the amber clock in Figure s01 again.

Frame bookkeeping (staying with the top-of-page convention). The convention is: $S$ is the frame where the two events are simultaneous. Here the nose- and tail-"same reading" events are simultaneous in the ship, so for this problem the ship plays the role of $S$ and the ground plays $S^{\prime }$, with the ground moving at $v=0.6c$ in the $-x$ direction relative to the ship (equivalently, the ship moves $+x$ past the ground). We only ever call "$S$" the simultaneous frame — that is exactly the top-of-page rule, applied consistently. $\gamma =1.25$, $\Delta x=L_0=120\text{m}$ (separation in $S$, the ship). $|\Delta t^{\prime }|=\frac{\gamma v\Delta x}{c^2}=\frac{(1.25)(0.6c)(120)}{c^2}=\frac{90}{c}=\frac{90}{3\times 10^8}=3.0\times 10^{-7}\text{s}=0.30\mu \text{s}$ Which is ahead? By "leading clocks lag" (amber clock in Figure s01), the nose (front, in the direction of ship motion) clock reads behind, so the tail clock is ahead by $0.30\mu \text{s}$ as read from the ground. ✅

(a) Classify (why: only spacelike events can flip order — see the tilted line in Figure s02 and Spacetime Diagrams). $\Delta x=900\text{m}$, $\Delta t=2\times 10^{-6}\text{s}$, so $c\Delta t=3\times 10^8\times 2\times 10^{-6}=600\text{m}$. Since $|\Delta x|=900>c\Delta t=600$, the events are spacelike — light can't get from one to the other. Order-flipping is allowed. (b) Full transform (both events have nonzero $t$, so use the general signed formula, not the simultaneous shortcut): $\gamma =1.25$. $\Delta t^{\prime }=\gamma \left(\Delta t-\frac{v\Delta x}{c^2}\right)=1.25\left(2\times 10^{-6}-\frac{(0.6c)(900)}{c^2}\right)$ $\frac{(0.6)(900)}{c}=\frac{540}{3\times 10^8}=1.8\times 10^{-6}\text{s}$ $\Delta t^{\prime }=1.25(2\times 10^{-6}-1.8\times 10^{-6})=1.25\times 0.2\times 10^{-6}=2.5\times 10^{-7}\text{s}$ Positive and small: B is still after A, no flip here. In Figure s02 the $S^{\prime }$ line of simultaneity hasn't yet tilted past B. Push $v$ higher and it can flip: the flip needs $v\Delta x/c^2>\Delta t$, i.e. $v>c^2\Delta t/\Delta x$. Numerically $c^2\Delta t/\Delta x=(9\times 10^{16})(2\times 10^{-6})/900=2.0\times 10^8\text{m/s}=\frac{2}{3}c$.

Condition: $\Delta t^{\prime }=0\Rightarrow \gamma (\Delta t-v\Delta x/c^2)=0$. Since $\gamma \ne 0$, we need $\Delta t=\frac{v\Delta x}{c^2}\Rightarrow v=\frac{c^2\Delta t}{\Delta x}=\frac{(3\times 10^8{)}^2(2\times 10^{-6})}{900}$ $v=\frac{9\times 10^{16}\times 2\times 10^{-6}}{900}=\frac{1.8\times 10^{11}}{900}=2.0\times 10^8\text{m/s}$ As a fraction of $c$: $v/c=2.0\times 10^8/3\times 10^8=0.667$, i.e. $v=\frac{2}{3}c<c$. In Figure s02 this is exactly the tilt that lays the $S^{\prime }$ line of simultaneity through both A and B. A valid sub-light frame exists precisely because the pair is spacelike ($|\Delta x|>c\Delta t$). If they'd been timelike, this formula would demand $v>c$ — impossible — which is why causal order can never be flipped.

Step 1 — $\gamma$ (why: ultra-relativistic, expect large $\gamma$). $\gamma =\frac{1}{\sqrt{1-0.995^2}}=\frac{1}{\sqrt{1-0.990025}}=\frac{1}{\sqrt{0.009975}}=\frac{1}{0.099875}\approx 10.01$ Step 2 — signed gap formula with event 2 $=$ the detector the muon flies toward (larger $x$), $\Delta x=+600\text{m}$ in the ground frame $S$ where they're simultaneous: $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(10.01)(0.995c)(600)}{c^2}=-\frac{(10.01)(0.995)(600)}{3\times 10^8}$ $\Delta t^{\prime }\approx -1.99\times 10^{-5}\text{s},|\Delta t^{\prime }|\approx 19.9\mu \text{s}$ (b) Sign: negative → the leading detector (larger $x$, the one the muon flies toward) fires first in the muon frame — the amber clock in Figure s01 once more. A modest 600 m gap in space blows up into a $\sim 20\mu \text{s}$ time gap because of the enormous $\gamma$ and near-$c$ speed. Same physics as the Time Dilation muon-lifetime result, seen from the simultaneity side.

Why this matters: the sign of $\Delta x$ is just a labelling choice; the physics (which detector fires first) must not change. This is the missing edge case. $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(10.01)(0.995c)(-600)}{c^2}=+\frac{(10.01)(0.995)(600)}{3\times 10^8}\approx +1.99\times 10^{-5}\text{s}$ Now $\Delta t^{\prime }=t_2^{\prime }-t_1^{\prime }>0$, meaning $t_2^{\prime }>t_1^{\prime }$: event 2 (the trailing detector) fires later, i.e. the front/leading detector still fires first. Identical physics to L5.2 — only the bookkeeping label flipped. Rule: a positive $\Delta x$ (event 2 leads) gives a negative $\Delta t^{\prime }$ (event 2 first); a negative $\Delta x$ (event 2 trails) gives a positive $\Delta t^{\prime }$ (event 1 first). Either way, the leading event happens first — exactly the amber clock in Figure s01.

The trigger for disagreement ::: nonzero spatial separation $\Delta x$ (in the simultaneous frame). The formula's hidden assumption ::: events are simultaneous in $S$, i.e. $\Delta t=0$. Sign of $\Delta t^{\prime }$ vs order ::: $\Delta t^{\prime }=t_2^{\prime }-t_1^{\prime }<0$ means event 2 (larger $x$) happens first. If you flip the labels so $\Delta x<0$ ::: $\Delta t^{\prime }$ flips sign but the leading event still fires first. When order can flip ::: only for spacelike pairs ($|\Delta x|>c|\Delta t|$). Which clock reads behind ::: the leading one (front, in the direction of motion) — amber clock, Figure s01.