# Correct interpretation of time dilation using spacetime interval and proper times

I grasped my head around this topic a lot and I want to ask if my interpretation of the problem is now correct in the most clear way.

I'm trying to measure time dilation using the spacetime interval but adding the ingredient of the proper time, the measurable time, being compared between two observers (this part is usually overlooked).

Suppose two observers 1 and 2 in Minkowski spacetime ($$c=1$$). From the perspective of 1, 2 moves in a given direction with a velocity v. Assume 1 and 2 are colinear with the velocity vector. Two events happen: 2 passes through a point A and then through a point B. The interval of the events in the reference system of 2 is

$$\Delta S^2$$=-$$\Delta t^2$$

As 2 is at rest for the events ($$\Delta x=0$$), we say that the time measured by him is the proper time as he is the proper system of these two events ($$\Delta \tau$$=$$\Delta t$$). So in this case,

$$\Delta S^2$$=-$$\Delta \tau^2$$

Now, from the perspective of 1, 2 is moving with a velocity v in x'-direction. The interval of the events in the reference system of 1 (coordinates are x' and t') is

$$\Delta S^2=-\Delta t'^2+\Delta x'^2 = -\Delta t'^2+v^2 \Delta t'^2= -(1-v^2)\Delta t'^2$$

Common approaches stop here and say that, since $$\Delta S^2$$ is an invariant then

-$$\Delta \tau^2=-(1-v^2)\Delta t'^2$$

$$\Delta t'=\gamma(v) \Delta \tau$$

so there we have it. I believe we are missing something because $$t'$$ is a coordinate and not measurable time. Now I want to do a seemingly obvious step further that I think clarifies this result. I believe that, since we already used a proper time for this event, we have to take into consideration another event that happens and is related to this one that can lead us to use the time measured by 1. The perfect example is the one thing why I believe some people have a hard time understanding measurement in relativity: We are not taking into explicit consideration what is measuring time by an observer outside the proper system. That example is a series of events: the light reaching the observer's light cone. Since 1 and 2 are colinear with the movement of 2 (more simple), then we can talk about the second pair of events: 1 receives a light ray that is sent from 2 when he passes through A and another one that is sent when 2 passes through B.

Now the special part: Since 1 is at rest with these two events, he is now the proper system and measures proper time $$\Delta \tau'$$. The first light ray travels a certain distance and the second one travels an extra $$c\Delta t'$$ ($$\Delta t'$$ since $$c=1$$). So the difference between the light rays (the interval between the events) measured by 1 is $$\Delta t'$$, which corresponds to the total time measured $$\Delta \tau'=\Delta t'$$. Now we have the result

$$\Delta \tau'=\gamma(v) \Delta \tau$$

We now compare two measurable times and now we can say that 1 measure that 2 takes a longer time reaching B from A that the time it takes for 2 to pass through those points. It is the same result, but I believe this completes the picture.

• To complete the picture, it might help to draw a spacetime diagram. Jan 11 at 4:16

I believe we are missing something, because 𝑡′ is a coordinate and not measurable time.

$$t'$$ corresponds to the time on observer $$1$$'s wristwatch, and is therefore perfectly measurable. More precisely, in the reference frame of observer $$1$$, the two events "my stopwatch reads $$t'$$" and "observer 2 passes through point $$B$$" are simultaneous.

We are not taking into explicit consideration what is measuring an interval of events by an observer outside the proper system.

The reason we don't take that into account is because the actual mechanism by which observer $$1$$ concludes that the two aforemented events are simultaneous is irrelevant. There is no speed-of-light delay which needs to be accounted for here. To put things in the rod-and-clock language, we imagine that observer $$1$$ has placed synchronized clocks at every point in space, and $$t'$$ is the time on the clock at point $$B$$ when observer $$2$$ passes it.

Now we have the result that $$\Delta \tau'=\gamma(v) \Delta \tau$$

I don't exactly follow what you're doing. If observer 2 sends light signals to observer 1 when they pass points A and B respectively, then the time between observer 1 receiving those two light signals is equal to $$\gamma(v) \Delta \tau(1+ v/c) = \Delta \tau \sqrt{\frac{1+v/c}{1-v/c}}$$. This calculation arises e.g. when computing the relativistic Doppler effect.

• Thanks, it clarified a lot, although some teacher always said that coordinate time is not measured by anyone (and is repeated on the internet a lot). The example with the light rays was to make some argument on why proper time of 1 with this new pair of events coincides with t' measured before, but i believe i oversimplified and the calculations are wrong, as you point out, although i wasn't trying to calculate the doppler effect, i was merely talking about relations of time intervals. Jan 11 at 6:01
• @ManuelMizrahi I suspect what your instructors were trying to say was that coordinates are not physical in and of themselves - they are merely labels which we apply to events in spacetime, and we may devise whatever alternative labeling scheme we like as long as it smooth and consistent. The key point is that coordinates need to be interpreted, and (especially in GR) it can be very subtle work indeed to tie a particular coordinate system to the observations that would be made by some experimenter. Jan 11 at 6:46
• @ManuelMizrahi In SR, the familiar Cartesian coordinates in which the metric components take the form $\eta_{\mu\nu} = \mathrm{diag}(-1,1,1,1)$ are to be interpreted via a carefully constructed (imaginary) orthogonal grid of measuring rods and clocks which have been synchronized via the Einstein synchronization convention. This coordinate system provides the canonical interpretation of events and coordinates in SR, and you can frame any observations in those terms. Jan 11 at 6:50
• Crystal clear now, thank you very much. Jan 11 at 16:15

[This] example is the one thing why i believe some people has a hard time understanding measurement in relativity: We are not taking into explicit consideration what is measuring an interval of events by an observer outside the proper system.

You are simultaneously wrong and correct.

There is a lot of bad pedagogy out there, and indeed there is a lot of confusion in students when learning about relativity, particularly on the measurement parts. However, the problem is not what you think it is. It is that good teachers will emphasise the importance of measurement at the beginning of teaching SR, and yet students will not realise that it is important at that stage.

i.e. It is not that "we are not taking into explicit consideration what is measuring", but rather that you forgot that this was part of the preliminary work.

In particular, the following is wrong:

I believe we are missing something, because $$t^\prime$$ is a coordinate and not measurable time.

and it is obviously wrong because you correctly presented a way for it to become directly measured later on.

Of course, there will be plenty of bad pedagogy that fails to state beforehand that you need to explicitly re-evaluate how to set up coördinate systems $$(t,\vec x)$$ under the new assumptions of SR, but it is necessary to understand how they are working before we can make the predictions and claims of SR.

The first step to understanding SR is that naïve measurements tends to be relative to certain frames of reference, and we have to construct objective measurement data from them. This implies that measurements are first considered. It is not possible to teach SR properly without taking measurements into explicit consideration.

• I thought the same, although i don't recall a lot of time invested in measurement. In fact, they used to say that coordinate time was not measured by anyone's clock. And i believe the example i gave to make it reconcile with some proper time is badly calculated. Jan 11 at 6:03