As I understand it, the proper time, $\tau$, between to events in spacetime is defined in terms of the spacetime interval $ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$, such that $$d\tau =\sqrt{-ds^{2}}$$ (where we are using the "mostly +" signature with $c=1$).

Now, for time-like intervals, for which $ds^{2}<0$, it is clear that proper time is well-defined since the quantity $\sqrt{-ds^{2}}$ is positive, and furthermore, one can always find a frame in which the two events occur at the same point in space, such that one can construct a worldline connecting the two events, along which an observer can travel, at rest with respect to both events, such that $d\tau =\sqrt{-ds^{2}}=dt$.

However, why is it the case that for space-like, $ds^{2}>0$, and light-like intervals, $ds^{2}=0$, the notion of proper time is undefined (or perhaps ill-defined)?

For the space-like case, I get that heuristically, one cannot construct a path between the two events along which an observer can travel and so in this sense proper time is meaningless, since a worldline connecting the events does not exist and so no clock can pass through both events. However, can this be seen purely by examining the definition of proper time in terms of the spacetime interval? Is it simply that the quantity $\sqrt{-ds^{2}}$ will become imaginary and so clearly cannot be used to represent any physical time interval?

Likewise, for a light-like interval, only a beam of light can pass between both events and since there is no rest frame for light one cannot construct a frame in which a clock is at rest with respect to the beam and passes through both events. However, purely in terms of the spacetime interval, is it simply because the quantity $\sqrt{-ds^{2}}$ equals $0$, and so the notion of proper time is ill-defined since there is no invertible map between reference frames (here I'm thinking in terms of time dilation, $t =\gamma\tau$ and so for a light-like interval, $\gamma\rightarrow\infty$ meaning that the inverse relation $\tau =\frac{t}{\gamma}$ is ill-defined)?!

  • $\begingroup$ I think this is essentially correct. However to be precise you need to talk in terms of geodesics, or at least smooth curves which are everywhere (space|time)-like I think, because you can construct everywhere timelike curves which connect two spacelike-separated events if you are willing for them to be not everywhere smooth (just go forwards to some event in the causal future of both events, and then backwards to the second event), or not everywhere timelike (round the corners of the previous curve). I'm not putting this as an answer because I don't think it is a good one. $\endgroup$
    – user107153
    Nov 1, 2016 at 11:26
  • $\begingroup$ Although I like the question, I would suggest abandoning the search for the "deeper physical meaning". The math is really all there is to it. You can interpret $ds^2 < 0$ as you wish, but it won't change any physical prediction of the theory. $\endgroup$ Nov 1, 2016 at 15:20
  • $\begingroup$ @SolenodonParadoxus My main issue is why proper time is not defined for space-like and light-like intervals? $\endgroup$
    – user35305
    Nov 1, 2016 at 15:34
  • $\begingroup$ @user35305 That's what I'm saying - it just isn't. Why do you think there's ought to be a reason? $\endgroup$ Nov 2, 2016 at 8:53
  • $\begingroup$ @SolenodonParadoxus I guess I'm just trying to over complicate things for myself, trying to extrapolate some deep meaning from a definition. $\endgroup$
    – user35305
    Nov 2, 2016 at 9:47

2 Answers 2


I'm not sure what kind of answer you're looking for; you seem to have said everything in your question. The causal structure of spacetime is such that there are light cones, and the square of vectors may be positive or negative. When the vector is timelike, we call its length proper time. When it is spacelike, we call the length proper distance. Proper time is not defined for spatially separated events precisely because they're spatially sperated. Why would you expect the notion of time to make sense in such a context?

Still, you might appreciate a more physical reason. Most people, Einstein included, would say that the definition of proper time along a world line (that is, a curve in spacetime) is the time measured by a clock carried by an observer moving along that world line. Well, observers can't move faster than light. And relative to some (arbitrary) fixed observer, the closer you go to the speed of light the smaller your proper time, so lightlike intervals have null proper time.

You'll notice that I'm basically saying the same thing twice: proper time is not defined for spacelike intervals by definition. When the interval is spacelike we don't put the minus sign and we call it proper distance. You might as well ask why proper distance is undefined for timelike intervals, and you should realize that you would get the same answer again but with the words exchanged.

  • $\begingroup$ I think I've been overthinking the situation and trying to gain some deep physical meaning from something that's just a definition. I was hoping to understand if there was a physical motivation for defining proper time as $d\tau=\sqrt{-ds^{2}}$?! It seems to me that if one defines the proper time of an object as the time measured in the rest frame of that object, then in this frame we have that $ds^{2}=-d\tau^{2}$ and then, since $ds^{2}$ is Lorentz invariant, this automatically guarantees that proper time is a Lorentz invariant quantity?! $\endgroup$
    – user35305
    Nov 2, 2016 at 9:53
  • $\begingroup$ @user35305: I would say that's it; proper time, like rest mass, is defined in a particular frame and so is invariant by definition. It then turns out that it is equal to the square of some four vector. $\endgroup$
    – Javier
    Nov 2, 2016 at 10:10
  • $\begingroup$ Ah ok. So is the reason why it does not make sense to discuss the notion of a photon having proper time because there is no rest frame for a photon? Similarly, for space-like separations, there is no worldline between the two events and so again a notion of proper time is not possible?! $\endgroup$
    – user35305
    Nov 2, 2016 at 11:45
  • $\begingroup$ @user35305: Yes, indeed. $\endgroup$
    – Javier
    Nov 5, 2016 at 0:23

You are right, except regarding lightlike intervals.

One important element which you are perhaps missing and which may be the reason for your incomprehension is that space and time are not on the same footing, their symmetry is strictly limited to Lorentz symmetry. This is why there is no use to look for any similarity between space and time beyond Lorentz symmetry, and in the same way there is no law of symmetry between spacelike and timelike intervals beyond Lorentz symmetry.

As you wrote, you can distinguish timelike, lightlike and spacelike intervals. Timelike worldlines are generating proper time. The proper time of spacelike worldlines is meaningless because no particle and no other phenomenon can travel on spacelike worldlines, and for this reason, spacelike worldlines cannot generate any proper time.

In contrast, lightlike worldlines are different from spacelike worldlines because the zero interval is within the domain of definition of the parabolic root function, while the negative intervals of spacelike worldlines are not. Particles and other phenomena (such as fields) moving at v = c are traveling on lightlike worldlines, and their proper time is 0, it is not undefined!

Also, the proper time of lightlike worldlines is not "ill-defined": You are right when you say that the Lorentz factor is going towards infinity. But the inverse Lorentz factor in the equation $τ = dt/γ$ is going to zero, that means that any observed coordinate time of lightlike phenomena corresponds to zero proper time. However, the fact that the proper time of lightlike phenomena is zero prohibits Lorentz transformation which would lead to problems of divisions by zero: Lightlike phenomena have no reference frame which could be transformed into other reference frames. But the equation $τ = dt/γ$ does not depend on Lorentz transformation because it can be derived directly from the two postulates of special relativity.

  • $\begingroup$ So, in terms of the spacetime interval, is proper time not defined for space-like intervals because the definition $d\tau=\sqrt{-ds^{2}}$ implies that proper time would become imaginary and therefore unphysical?! Also, I would've thought that proper time for a photon would be undefined, if not purely because there is no rest frame for a photon and hence it is not possible for a "clock" to travel along the worldline of a photon?! $\endgroup$
    – user35305
    Nov 1, 2016 at 14:06
  • $\begingroup$ Spacelike intervals: Yes the physical situation reflects exactly to the mathematical equation. $\endgroup$
    – Moonraker
    Nov 1, 2016 at 14:14
  • $\begingroup$ Lightlike intervals: Their constellation is very particular, take for example a massless particle (photon). As the spacetime interval is zero that means that the point of emission and the point of absorption are adjacent in spacetime, they are at the same place. Your clock will show zero because its traveled distance (in spacetime) is zero. $\endgroup$
    – Moonraker
    Nov 1, 2016 at 14:21
  • $\begingroup$ My point though is that you can't be at rest with respect to a photon and so you can't define a proper time for it, since coordinate time is only equal to proper time in the rest frame of the object. An observer who sees a photon being emitted by a source and then subsequently by a detector will certainly measure a non-zero (coordinate) time interval between the two events. $\endgroup$
    – user35305
    Nov 1, 2016 at 14:26
  • $\begingroup$ You argument is not pertinent because you are mixing up the time interval (which is not zero) and the spacetime interval (which is zero). $\endgroup$
    – Moonraker
    Nov 1, 2016 at 14:29

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