# I Proper time?

1. Jan 11, 2019

### Pericles98

Hello everybody, I've just started studying special relativity and seem to be having a bit of a problem with understanding the concept of proper time.

In Modern Physics, by Serway, proper time is defined as the time interval between two events as measured by an observer that sees both events occur at the same point in space. Additionally, the book mentions that proper time will always be the shortest measurement of time for two given events.

I am having some difficulty understanding this last statement. Suppose that an observer, David, leaves Earth in a spaceship headed for the moon. John, a second observer, is left behind on Earth. In this particular example, if we are interested in calculating the time, as experienced by both observers, that it takes the spaceship to reach the moon, David would measure proper time (and therefore the shortest amount of time for the trip). This makes perfect sense when looking at the trip from John's frame of reference. John will see David's clock run slower and agree that David measures a shorter interval of time than his own. However, from David's point of view it will be John's clock that runs slower. David will claim that John measured a shorter amount of time for the trip than he did.

Then, what is the meaning of the statement: proper time will always be the shortest measurement of time for two given events? That seems only to be true from John's frame of reference. Why should we only consider his point of view? What am I missing?

2. Jan 11, 2019

### Matterwave

Don't get into the "John will see David's clock...and David will see John's clock..." kind of arguments - they will just serve to confuse the issue. The two events in this scenario are "David leaves Earth" and "David arrives on the Moon". For John, the two events happen at two different places since he's not moving with respect to the Earth. For David, the two events happen at the same place since he's moving. John will measure that the time interval between these two events is $t_J = \frac{D}{v_D}$ where $D$ is the distance between the Earth and Moon and $v_D$ is the speed of David. David, due to length contraction, will measure the time interval as $t_D = \frac{D}{\gamma v_D}<t_J$ - here $v_D$ is still the "speed of David" except in David's point of view it's more the "speed of the moon" (towards him). All that the second statement says is $t_D<t_J$ for any given $J$ observer who doesn't see the two events as happening in the same place (while observer $D$ does see the two events happen in the same place). There is no "but John see's David's clock as...and David sees John's clock as..." kind of shenanigans going on here.

3. Jan 11, 2019

### Staff: Mentor

You don't have two events in your thought experiment, you have three Remembering that an event is specified by a place and a time, and proper time between two events is the time measured by a clock present at both events, the three events are:
1) David and John set their wristwatches to zero, shake hands, and David departs for the moon.
2) David arrives at the moon.
3) John looks at his wristwatch and says "ha - David just landed". He knows this because he knows the distance to the moon and David's speed, so at this event his wristwatch reads $T=D/v$.

Note that #2 and 3 are different events because they happen in a different places.
David's wristwatch measures the proper time between #1 and #2; John's wristwatch measures the proper time between #1 and #3.

To go beyond this you'll want to understand the relativity of simultaneity as well; this often-overlooked concept is essential to understanding special relativity. For now, I'll just say that events #2 and #3 happen at the same time using a frame in which John is at rest, but they do not happen at the same time using a frame in which David is at rest.

Last edited: Jan 11, 2019
4. Jan 11, 2019

### PeroK

I'm not sure that is the most helpful description of proper time. It boils down to an application of the Lorentz Transfromation. Let's say that two events $A, B$ take place at the same point in space in a given inertial reference frame $S$. We may as well take the point as the origin and the times as $t_a = 0$ and $t_b = T$. And, in this case, $T$ is the length of the spacetime interval between the two events, and the time a clock stationary at the origin would record between the events. These are both ways to describe proper time, so $T$ is the proper time between the events.

In frame $S$, the time between the events is $T$, obviously.

If we look at events $A$ and $B$ in another inertial reference frame, $S'$ moving with some speed $v$ relative to S. We can take $v$ to be along the x-axis. Then the time of events $A$ and $B$ are:

$t_A' = 0$ and $t_B' = \gamma(t_b - vx_B/c^2) = \gamma T$

In frame $S'$ the time between the events is $\gamma T > T$.

In that sense, $T$ is the shortest time between the events.

But, wait a minute. Suppose a clock starts out at the origin at time $0$ in frame $S$, moves quickly at some speed $u$ and returns to the origin at time $T$, all as measured in frame $S$. As the clock has been moving relative to the clock at the origin, it will have recorded less (proper) time upon its return to the origin. E.g. if $u = 0.8c$, then $\gamma_u = \frac53$ and the "moving clock" will record only $\frac{3T}{5}$ when it returns to the origin.

So, we now have two proper times between events $A$ and $B$. The proper time of a clock stationary at the origin; and the proper time of a clock that moved away from and back to the origin. And, one is less than the other.

Moreover, as $u \rightarrow c, \ \ \gamma_u \rightarrow \infty, \$ and the time recorded by the moving clock tends to $0$. In that sense, there is no minimum time measured between the two events.

5. Jan 11, 2019

### Ibix

Longest, surely. Or am I more tired than I think I am?

6. Jan 11, 2019

### Matterwave

I think you are confusing the statement that a "geodesic between two events measures maximum proper time" with OP's statement about "shortest measurement of time for two given events". There are many ways one could travel from event A to event B - if one travels along a geodesic then the proper time is (generally - barring some funky causal structure) maximized for that path. Along other paths, though, the (still proper) time measured will be shorter - see e.g. the resolution of the Twin paradox. But OP is not talking about comparisons of different proper times between different paths between A and B. He's talking about comparing the actual proper time between A and B, and some coordinate time between A and B as measured by some other observer who doesn't travels from A to B.

7. Jan 11, 2019

### Staff: Mentor

No, he will not. The time dilation formula requires the time to be measured between two events which are at the same location in one frame and different locations in the other frame. Both frames agree which frame the events are co-located and which frame they are not. Both frames can correctly calculate the other frame’s measurement.

8. Jan 12, 2019

### Ibix

I was right - I was more tired than I thought.

Is that a helpful way to think about proper time? The usual statement, "proper time is the longest elapsed time along paths between two events", seems more fundamental than "proper time on the inertial path between events is less than or equal to coordinate time between them". In a Euclidean analogy it's like preferring "the distance between a point and a line through another point is longer if you don't travel perpendicular to the line" over (edit: the more familiar) "the shortest distance between two points is a straight line".

Last edited: Jan 12, 2019
9. Jan 12, 2019

### PeroK

I thought that was more or less the point I was making in post #4.

10. Jan 12, 2019

### robphy

For your boldfaced sentence, can you give a more specific... edition and page reference ?

11. Jan 12, 2019

### stevendaryl

Staff Emeritus
Well, it's an elementary observation.

$$\delta \tau = \sqrt{(\delta t)^2 - \frac{(\delta x)^2}{c^2}}$$

This is true in every inertial coordinate system. So:

$$\delta t = \sqrt{(\delta \tau)^2 + \frac{(\delta x)^2}{c^2}}$$

Again, that's true in every inertial coordinate system. So if there is an inertial coordinate system in which $\delta x = 0$, then $\delta t = \delta \tau$ in that coordinate system. For every other inertial coordinate system, $\delta t > \delta t$

12. Jan 12, 2019

### stevendaryl

Staff Emeritus
This argument is true for infinitesimal $\delta \tau$, only.

13. Jan 12, 2019

### PeroK

In my view, as I implied in post #4 "always" suggest more than just inertial reference frame to me. Especially with the twin paradox just round the corner.

In addition, proper time is really a function of a worldline and not just the end points.

14. Jan 12, 2019

### robphy

I presume you mean $\delta t > \delta \tau$.

My request for a specific reference is that I want to see the full context of this statement,
including its definitions of "proper time" and "interval".

As you suggest, this important aspect of proper time (that it depends on the worldline that meets the two events) is often left out,
and thus there is often a missing distinction between
"the proper time from event A to event B [along a specific worldline]"
and
"the interval [square-root of the square-interval] from event A to event B" ["the proper time from A to B along an inertial worldline from A to B"]

(For an inertial observer Zack, "the coordinate time from A to B" is defined regardless of whether or not Zack meets either event A or B.)

...which "time" is being referred to here, according to Serway.
(And why is this statement being made?)

So, I persist in my request from @Pericles98 for a specific reference: edition and page reference. Thanks in advance.

15. Jan 14, 2019

### Matterwave

I think in a more advanced treatment of SR/GR, that would not be a helpful way to think about proper time, but in elementary treatments (like Serway), this treatment is pretty common. I think it mostly stems from the Newtonian bias of treating coordinate time as a legitimate measure of time. There's a lot of talk in elementary textbooks about topics that I think aren't especially helpful...e.g. all the discussion about "John looks at Bob's clock and concludes...etc. etc."...but I'm not the one writing these textbooks so I don't think I get to complain too much.

16. Jan 14, 2019

### PeroK

There was a thread recently where someone was using Wikipedia as a reference, which shows an eye(!) on the y-axis representing an (or THE) observer in that reference frame. The poster would simply not accept, despite a long debate, that light travel time and what someone at one specific location actually sees are irrelevant - as they are in Newtonian physics. It is a real issue, as so much material on SR over-emphasises the role of a single observer.

The sooner one starts thinking in terms of reference frames (and a network of local observers) the better.

17. Jan 17, 2019 at 9:46 PM

### nitsuj

SR is kind of late in the "game" to be teaching relativity no? simply put it just carries over into geometry; that is everything physics lol

18. Jan 18, 2019 at 3:11 PM

### SiennaTheGr8

Yes, I've seen this confusion several times.

I think it's best to define proper time as something that "belongs" to a traveler (not to events). It's the aging or "wristwatch time" of any traveler, inertial or non-inertial.

Then we can say that the spacetime interval between a pair of timelike-separated events is equal to the proper time that elapses along an inertial traveler's journey from the earlier event to the later one.

A phrase like "the inertial proper time interval between the events" is fine, methinks, but "the proper time between two events" is bound to confuse beginners.

19. Jan 18, 2019 at 3:32 PM

### Staff: Mentor

Even that is only guaranteed unambiguous in flat spacetime; in curved spacetime there may be multiple inertial paths with different lengths between two events.