I'm asked to draw the worldline of an observer that would see two events as simultaneous. My thought process was to draw the light cones of the two events and see where they intersect, then connect these points to obtain the worldline of the observer. Below is a very simple illustrating diagram made by desmos. At the first point of intersection, (0,0) the observer emits two light pulses, Pulse 1 to event A (the intersection of red lines) and pulse 2 to event B (the intersection of the green lines). The light pulses then follow the light cones (the red and green lines) of the two events respectively. When they reflect from the two events, they also follow the light cones and meet again at the second point of intersection between the light cones (5,15). That way, the two light pulses were emitted at the same time and received at the same time, making the events simultaneous, at least according to my understanding of simultaneity. The line connecting these points (the black line) is then the worldline of the observer who sees the events to be simultaneous. I'm not entirely sure about my answer/reasoning, especially that the book offers a different solution which I don't quite understand. I'd explain the book's solution, but the question has gone on for too long already. It is problem 2.6 in D'inverno's relativity book. I edited the question to include the book's solution (2nd figure) which is constructed as follows. Draw a circle centered at event B (green intersection) with radius equal to the distance between B and A (green intersection and red intersection). Then it connects the points where the light cone of A intersects the circle and the resulting line (the orange line) is the worldline of the observer seeing A and B to be simultaneous. I don't understand this solution at all. enter image description here

enter image description here

  • $\begingroup$ Perhaps include your attempt as a figure in the question? That would make it easier to follow your thought process. $\endgroup$
    – pela
    Jul 5, 2022 at 10:06
  • $\begingroup$ @pela I did now. Does that clear things up a bit? $\endgroup$
    – EM_1
    Jul 5, 2022 at 12:41
  • $\begingroup$ Can you share what the book's solution is? And how your answer differs from it? You have enough information in your plot to compute the coordinates of the Red and Green events in the frame of the black worldline. Alternatively, can you construct the line of simultaneity for the black worldline? It might be good to construct the midpoint $\endgroup$
    – robphy
    Jul 5, 2022 at 13:53
  • $\begingroup$ @robphy The books solution draws a circle centered at the red event, with radius equaling the distance between the red and green events. It then draws the light cone of the green event, the one on the circle, and connects the two points where the light cone intersects the circle, and this is the worldline of the observer. $\endgroup$
    – EM_1
    Jul 5, 2022 at 14:29
  • 1
    $\begingroup$ @robphy I edited the question and added the book's solution. $\endgroup$
    – EM_1
    Jul 5, 2022 at 16:05

2 Answers 2


Congratulations, your solution and method and reasoning looks correct to me, and I prefer it because it is in the spirit of relativity (by a construction using the tools of spacetime geometry--in particular, the light-cone structure and the affine structure (parallel lines)).

(As a check, the product of the slopes of the observer-axes should equal 1.)

Since the textbook solutions uses a circle, it is (in my opinion) not in the spirit of spacetime geometry.

The reason the textbook solution works is that the light-cone on the circle forms a [Euclidean*] right-angle (which implicitly uses the unit-choice so that light-rays are at 45-degrees). Since that resulting triangle is a [Euclidean*] right-triangle, that orange chord is really a diameter. The resulting triangle is a radar measurement so that the event at the corner on the circle is simultaneous with the event at the center (which is the midpoint of the diameter). [Maybe my logic can be tightened-up... but I think you can get the idea.] Although it "works", it's not a spacetime-geometric construction because of the use of the circle.

*Note that the two light rays are NOT Minkowski-perpendicular. The Minkowski dot-product of those lightlike vectors is not zero. So, describing that as a "right-angle" is not appropriate in spacetime geometry.

UPDATE (to address the question about parallelism in the comments)

This spacetime diagram might help.



Consider the spacelike segment AB (the bold dotted violet line),
then form the "causal diamond" associated with it, which is marked by the intersection of the light-cones of A and of B.

The spacelike diagonal AB is simultaneous according to worldline along the timelike diagonal (the light dotted violet line). (These diagonals are Minkowski-perpendicular [the Minkowski-dot-product of 4-vectors along these diagonals is zero].)

Now consider a family of worldlines parallel to the timelike diagonal .

  • Consider the segments bounded by the light-cones of A and of B. These are associated with radar-measurements of events A and B. (See d'Inverno eq. (2.2) in Sec. 2.7.)
  • Note that the events along AB bisect these segments.
  • A and B are simultaneous according to this family of parallel worldlines.
  • $\begingroup$ My problem with the book's solution is that I don't see the motivation behind the construction of the circle. But I do have a question, regarding the parallelism of worldlines. If there's observer (2) that has a worldline parallel to that of observer (1) who sees A and B to be simultaneous, does that mean that observer (2) also sees A and B simultaneous but occurring at a different time than observer (1)? Or does he not see the simultaneous at all? Does the line of simultaneity define simultaneity in an absolute sense or differently with respect to different observers? $\endgroup$
    – EM_1
    Jul 6, 2022 at 9:57
  • $\begingroup$ Clarification: I meant absolute or different among different observers that are at rest with respect to each other. $\endgroup$
    – EM_1
    Jul 6, 2022 at 10:05
  • $\begingroup$ @user626542 I added a diagram to address the parallel-worldlines question. $\endgroup$
    – robphy
    Jul 6, 2022 at 14:46


"[...] problem 2.6 in D'Inverno's relativity book" "[...] to draw the worldline of an observer that would see two events as simultaneous."

D'Inverno's statement of problem 2.6 is rather asking for the worldline of an inertial observer

  • who took part in one given event (event $O$, in D'Inverno's Fig. 2.14), and

  • who reached a certain conclusion (involving event $O$ and event $G$, which are given as spacelike separated from each other).

"the resulting line (the orange line), OP 2nd figure [...] I don't understand this solution at all."

In a nutshell: the solution you sketched above is surely correct (as D'Inverno intended), and it obtains as an application of (a variant of) Thales's theorem: If two chords through a given circle meet in a right angle in one point on the circle circumference, then the remaining two endpoints of these two chords define a straight line through the center of the circle; a.k.a. a diameter of this circle.

Now, to justify and to discuss this in more detail I prefer to assign individual names to relevant identifiable participants, as well as to relevant events in which they took part; and I consider the naming scheme in the following figure closer to Einstein's (and also to Comstock's) descriptions concerning simultaneity than that in D'Inverno's Fig. 2.14: Simultaneity configuration: <span class=$\,$ Simultaneity configuration: $A$'s indication $A_J$ and $B$'s indication $B_Q$ are simultaneous. ($M$ is identified as "middle between" $A$ and $B$.)" />

Shown are two given participants ("material points") $A$ and $B$ who are at rest wrt. each other (in a flat spacetime region); a third participant $M$ who is at rest wrt. $A$ as well as wrt. $B$ (thus all three being members of the same inertial system), and who is explicitly identified as "middle between" $A$ and $B$ through the shown lightcone structure.

Also shown are some auxiliary participants (who are not at rest wrt. $A$, $B$, and $M$; but not necessarily at rest wrt. each other) so that each relevant event is uniquely identifiable as coincidence event by the participants who took part it in together.

Important features of the lightcone structure involving these participants are that

  • after having taken part in event $\varepsilon_{MK\_\text{sign}}$ (which involved an observable recognizable signal indication of $M$), $M$ observed the corresponding reflection indications of $A$ and of $B$ togther (a.k.a. "in coincidence"),

  • after having taken part in event $\varepsilon_{AH\_\text{sign}}$ (which involved an observable recognizable signal indication of $A$), $A$ observed the corresponding reflection indication of $B$ in coincidence with observing $M$'s reflection indication of $A$'s reflection indication of $M$'s corresponding reflection indication, and

  • after having taken part in event $\varepsilon_{BP\_\text{sign}}$ (which involved an observable recognizable signal indication of $B$), $B$ observed the corresponding reflection indication of $A$ in coincidence with observing $M$'s reflection indication of $B$'s reflection indication of $M$'s corresponding reflection indication.

As a consequence, by Einstein's (and, supplementing, by Comstock's) definition, $A$'s indication $A_J$ (i.e. $A$'s part of event $\varepsilon_{AJ\_\text{refl}}$) and $B$'s indication $B_Q$ (i.e. $B$'s part of event $\varepsilon_{BQ\_\text{refl}}$) are simultaneous.

Further, arguably also as a consequence of having verified the described lightcone structure, it is thereby established that

$s^2[ \, \varepsilon_{AH\_\text{sign}}, \varepsilon_{AJ\_\text{refl}} \, ] = s^2[ \, \varepsilon_{AJ\_\text{refl}}, \varepsilon_{AK\_\text{rec}} \, ]$, and

$s^2[ \, \varepsilon_{BP\_\text{sign}}, \varepsilon_{BQ\_\text{refl}} \, ] = s^2[ \, \varepsilon_{BQ\_\text{refl}}, \varepsilon_{BR\_\text{rec}} \, ]$.

Further, since $A$ is a member of an inertial system,

$s^2[ \, \varepsilon_{AH\_\text{sign}}, \varepsilon_{AK\_\text{rec}} \, ] = 4 \, s^2[ \, \varepsilon_{AH\_\text{sign}}, \varepsilon_{AJ\_\text{refl}} \, ] = 4 \, s^2[ \, \varepsilon_{AJ\_\text{refl}}, \varepsilon_{AK\_\text{rec}} \, ]$, and

$s^2[ \, \varepsilon_{BP\_\text{sign}}, \varepsilon_{BR\_\text{rec}} \, ] = 4 \, s^2[ \, \varepsilon_{BP\_\text{sign}}, \varepsilon_{BQ\_\text{refl}} \, ] = 4 \, s^2[ \, \varepsilon_{BQ\_\text{refl}}, \varepsilon_{BR\_\text{rec}} \, ]$.

By the conventions on how to represent these geometric and lightcone relations as 1+1-dimensional Minkowski diagramm with

  • each member of an inertial system as straight line,

  • equal timelike spacetime intervals of one member of an inertial system as straight line segments of equal length,

  • lightcone structure as grid of sraight parallel lines intersecting at right angles,

therefore (by Thales's theorem) event $\varepsilon_{AJ\_\text{refl}}$ is drawn as center point of the circle through events $\varepsilon_{AH\_\text{sign}}$, $\varepsilon_{AK\_\text{rec}}$ and $\varepsilon_{BQ\_\text{refl}}$; corresponding to the solution to D'Inverno's problem which you know already.

Likewise, event $\varepsilon_{BQ\_\text{refl}}$ is drawn as center point of the circle through events $\varepsilon_{BP\_\text{sign}}$, $\varepsilon_{BR\_\text{rec}}$ and $\varepsilon_{AJ\_\text{refl}}$.


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