The formula for simultaneity

We have already seen, in the story of the Good, the Ugly and the Bad, that two events simultaneous in one reference frame (SR) are no longer so in another SR in relative motion. Now let us take the quantitative step: we derive the exact formula for the time difference those two events show in the other frame. It is one of the three fundamental formulae of relativity, alongside time dilation and length contraction.

Setup. Consider two events $E_1$ and $E_2$ that in SRA occur simultaneously and are separated by the vector $\vv{L}_{12(A)}$ (i.e. $E_2$ is “ahead” of $E_1$ along the direction of SRB’s motion). Call $\vv{v}$ the velocity of SRB relative to SRA and $\vv{\beta} = \vv{v}/c$. We ask what $\Delta t_{12(B)} = t_{2(B)} - t_{1(B)}$ is in SRB.

Derivation via events and equations of motion. Let us take the western duel as a concrete example. The Good, in the middle, fires two simultaneous laser beams in SRA. Event $E_1$ is the arrival of the first beam at the Bad (on the left, against the direction of the vulture’s motion); $E_2$ is the arrival of the second beam at the Ugly (on the right, along the direction of motion). In SRA, by symmetry, $E_1$ and $E_2$ are simultaneous. In SRB, instead, we see the Ugly and the Bad as a single contracted rod of length $L_{(B)} = L_{(A)}\sqrt{1 - v^2/c^2}$: the Ugly moves towards the vulture at speed $v$ and the Bad moves away at the same speed.

The Good fires in SRB at event $E_0$, placed for convenience at $t_{0(B)} = 0$ and $s_{0(B)} = 0$. The beam towards the Bad travels at speed $-c$ (law $s = -ct$) and must reach the Bad, who moves in the positive direction with law $s = vt + L_{(B)}/2$; the beam towards the Ugly travels at $+c$ (law $s = ct$) and chases the Ugly, with law $s = vt - L_{(B)}/2$. Solving the two linear systems gives the arrival times:

$t_{1(B)} = \frac{L_{(B)}/2}{c-v}, \qquad t_{2(B)} = \frac{L_{(B)}/2}{c+v}$

The difference is a single algebraic expression:

$\Delta t_{12(B)} = t_{2(B)} - t_{1(B)} = \frac{L_{(B)}}{2}\left(\frac{1}{c+v} - \frac{1}{c-v}\right) = -\frac{v\,L_{(B)}}{c^2 - v^2}$

Substituting $L_{(B)} = L_{(A)}\sqrt{1-v^2/c^2}$ and simplifying, everything collapses into a clean result:

$\ev{\Delta t_{12(B)} = -\,\beta\,\gamma\,\frac{L_{(A)}}{c}}$

The minus sign is the most instructive part: it recalls that in SRB the event “downstream” of the motion (the one in the direction of the velocity) happens first. Simultaneity is not absolute, and whoever is moving sees what lies ahead of them, in the direction of motion, happen first.

In vector form, valid for a separation $\vv{L}_{12(A)}$ oriented in any way:

$\Delta t_{12(B)} = -\,\gamma\,\frac{\vv{\beta}\cdot\vv{L}_{12(A)}}{c}$

Only the component of $\vv{L}_{12(A)}$ along the motion matters: events separated transversally to the motion remain simultaneous even in the other frame.

Key formula

$\Delta t_{12(B)} = -\gamma\,\frac{\vv{\beta}\cdot\vv{L}_{12(A)}}{c}$ Holds for two events simultaneous in SRA, with $\vv{L}_{12(A)}$ directed from $E_1$ to $E_2$. The event downstream of the motion happens first in the other frame.

Links

Topics: Special relativity Concepts: Relativity of simultaneity Skills: Changing reference frame Methods: Minkowski diagram

Related exercises: Space-type or time-type · Problem — Tunnel at 0.8c · Problem — Simultaneity (true or false)