In Minkowski spacetime time is subjective [or more precisely: time is different for every particle/ reference frame]. It is the coordinate time of an observer whose reference frame travels up the $ct$-axis. Before Minkowski, we thought that time was the same for everybody, but now we know that the slightest movement of an observer destroys synchronization with regard to the other observers. Each bus driver lives slower than a prisoner because he is moving, even if the difference is not perceivable. Time is an accumulation of a huge number of relative Minkowski diagrams.

Does that mean that there is no generally applying physical definition of time dimension left? Or is there an absolute dimension of time of the universe in which we are living, and which is independent from the observer?


2 Answers 2


You shouldn't use the "subjective/objective" distinction for a place where "relative/absolute" is much more appropriate, because they mean different things. For something to be subjective, it must be dependent on the knowledge or state of mind of an observer.

As an example, suppose we define "depth" as "length along the direction an observer is facing". This direction can be used to define a "depth axis", and measurements of depth would depend on projections to that axis. Then how deep an object is depends on the circumstances of the observers, and so is relative. But it is not subjective: it doesn't matter is the observer is delusional about they're facing, etc.

In special relativity, time is rather analogous to this. In spacetime, your four-velocity is kind of like 'direction you're facing' would be in ordinary space. It defines the temporal axis of your inertial frame, and time measurements would be projections to that axis, similarly to the previous case.


It means that time is no longer an absolute concept, yes. The time a specific observer experiences in a specific frame of reference, i.e. his proper time depends on the path (worldline) he takes through spacetime. In other words, it depends on his state of motion, the way he accelerates. This is the reason for the famous twin paradoxon: the resolution is that both twins move on different worldlines and hence end up having different proper times. This can even be easily understood geometrically: between two points in spacetime, the proper time interval of the shortest path connecting them is the greatest. With this knowledge, the reason for the twin-paradox should be evident from the following picture:

enter image description here

Clearly, the black lines corresponding to the travelling twin represent a longer part than that of the stationary one.

This does not mean, however, that time is not well-defined as a physical quantity. It just means that when you measure it, you may get different results, depending on how you move with respect to somebody else.

  • $\begingroup$ Thank you. My issue is not the twin phenomenon, for me it was interesting that there is no complementary absolute concept of a time dimension. $\endgroup$
    – Moonraker
    Commented Jul 3, 2014 at 21:14
  • $\begingroup$ @Moonraker: I know that you did not ask for it, but I thought that it might be a good example to illustrate the relativity of time. $\endgroup$ Commented Jul 3, 2014 at 21:15
  • $\begingroup$ In the true twin paradox, i.e. in the Einstein's "clock paradox", both clocks move in exactly the same way. Best example are two spaceships traveling in opposite directions and without accelerations - both frames are inertial throughout the whole experiment (and are both accelerated in exactly the same way before). And yet the equation shows time dilatation, although both frames are indistinguishable. Conclusions? Proper time is always the same for all frames (Feynman admitted this), and therefore can be considered absolute. $\endgroup$ Commented Jul 5, 2014 at 20:41
  • $\begingroup$ @brightmagus: the twin paradox is about asymmetric movement, i.e. one accelerates and the other does not. $\endgroup$ Commented Jul 6, 2014 at 0:29
  • $\begingroup$ @FredericBrünner: Originally it wasn't. Accelerations were added, because no-one could solve it for inertial frames. Still, accelerations can be symmetric, as I have shown. $\endgroup$ Commented Jul 6, 2014 at 6:01

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