# Meaning of simultaneity in special relativity

In Special Relativity, time is no longer a parameter, but a variable. As in, two events simultaneous in one frame of reference need not be simultaneous in another frame of reference that is inertial with respect to the first. So, does that mean that if one observer sees two rocket ships collide and explode at one instant (the two rockets are simultaneously in the same location), we can find an inertial frame of reference where this doesn't happen? In that case, how does one explain the debris of the exploded spaceships in the frame where the two rockets didn't collide?

• The collision would be a single event. Dec 14, 2020 at 5:52

The relativity of simultaneity refers to a pair of spatially-separated events.

The collision event of two objects meeting is not spatially-separated. All frames of reference will mark that collision with a single event.

UPDATE (to address the comment by the OP):

A spacetime diagram is useful to clarify the ideas.
Furthermore, with my diagram, you can see that [from a geometric point of view],
"absolute simultaneity" (that if Alice says $$t_P=t_Q$$, then all observers say $$t'_P=t'_Q$$) is really an exceptional case (rather then the rule).

Using my "spacetime diagrammer" https://www.desmos.com/calculator/dha8izuxz1 ,
with event $$P(x,t)=(0.5,1)$$,

• Alice (in RED) says that distant event P is simultaneous with tick-1 on her clock... (because Alice assigned the same t-coordinate to those events),
• however, Bob (in GREEN) says that distant event P happened before "tick-1 on Alice's clock".... (because Bob assigned a smaller t-coordinate to P than to tick-1-on-Alice's-clock). The "lines of simultaneity (i.e, equal-assignments-of-t)" are determined by the tangents to the "circles" [in the respective geometries], which is a hyperbola in Special Relativity (arising from the invariance of the speed of light).

The "circle" is determined by the set of all observer "tick-1"s.
The "circle" defines perpendicularity: the tangent is perpendicular to the radius...
"the Space[line] is Perpendicular to the Time[line] (worldline)", as Minkowski described in his "Space and Time".

• For the Galilean case, open the "Metric" folder and slide E to 0. Note how "absolute simultaneity" is obtained since the tangent lines through Alice's-tick-1 and Bob's-tick-1 coincide.
• Proceed further to E= -1 to obtain Euclidean geometry. Note how the tangent lines do not coincide, as in Minkowskian geometry (special relativity).
• Thus, the Galilean case is the exception. But since this is our "everyday intuition" (since we don't rely on super-accurate-clocks [except for GPS] or regularly interact with particles at high-relative-speeds), it's a layer of common sense we have to let go in order to understand relativity.  • I think I see your point. Are you suggesting that the collision itself will be a single event? If that is the case, right upto the moment of collision, the two objects would be spatially separated, right? And if the observers perceive them differently, how do they all manage to concur on the collision? Is it just that the velocities are different? Then in that case, what is so profound about simultaneity in SR?
– user280877
Dec 14, 2020 at 6:00
• @Manimatics What do you mean by it being profound? Dec 14, 2020 at 7:31
• @Manimatics But you know what is different in SR about simultaneity. It's that two events can be simultaneous in one frame but not in another frame, which is completely counterintuitive for anyone who hasn't seen it before. In the world of Galilean transformations this makes no sense. Dec 14, 2020 at 13:16
• @Manimatics There is a maximum time difference between the events which depends on the spatial distance between the two. When the distance (as observed) shrinks, so does that (observed) difference. I not sure the distance is something all observers agree (my gut-feeeling says probably not) but they all agree it gets smaller. Imagine two spaceships going in a straight line towards each other from some space station. They shine a light such that an observer in the middle sees that event as simultaneous. People on station A see the light from ship A first, B from B... Dec 14, 2020 at 14:43
• @Manimatics ... but both agree the difference shrinks as the two approach. (I think) There is no observer that observes the difference greater than the people on the stations. Dec 14, 2020 at 14:44

The answer by robphy is complete, but I thought that I would anyway add some illustrations (Minkowski spacetime diagrams) that, at least in my opinion, possibly show better the relativity of simultaneity.

I assume that you are somewhat familiar with special relativity and/or spacetime diagrams. If not, for the latter you might be interested in my post here.

The coordinate grids represent simulaneity on the $$x$$ and $$x'$$ axis and the same location on the $$y$$ or rather $$ct$$ and $$ct'$$ axes (not only the axes, but all grid lines).

It can easily be seen that if two events happen at the same point at the same time in $$K$$ (which is the green, "stationary" frame), they are also simultaneous and at the same position in $$K'$$ (the red frame). This is because such two events are rather one event that we can represent by a point in our diagram.

• Events that happen at the same time but at different locations for one observer might happen at different times for another observer.
• Events that happen at the same time and at the same place are simultaneous for all observers.

we can find an inertial frame of reference where this doesn't happen?

No. If the two spaceships are at the same point at the same time for one observer (they collide), all observers will see them collide.

• Additionally, events can also happen at different locations for other observers Dec 14, 2020 at 20:16
• Also, since almost any two macroscopic events happening at the same time and place are effectively identical for practical purposes (i.e. two massive macroscopic objects cannot occupy the same space at the same time), one could say that there are basically no two events that are simultaneous for all observers. Dec 15, 2020 at 2:11
• @Obie2.0 This is true, but saying it that way might lead to exactly the misunderstanding evidenced in the original question. Dec 15, 2020 at 16:28