2
$\begingroup$

According to Wikipedia,

proper time along a timelike world line is defined as the time as measured by a clock following that line.

This makes sense to me, but my book defines proper time in the following way:

Proper time is the time between two events measured in a frame in which the events happen at the same position.

I don't quite understand why those two definitions would be equivalent.

$\endgroup$
  • 1
    $\begingroup$ Did you notice the subtle difference between "proper time along a worldline" and "proper time between two events"? How can the two be reconciled? Does that help with your confusion? $\endgroup$ – dmckee Oct 18 at 18:24
  • $\begingroup$ hmm, what I get from the second definition is the sense that the proper time is the time evaluated in the rest frame of an object moving from one event to the other. $\endgroup$ – daljit97 Oct 18 at 18:51
  • $\begingroup$ What book is this second definition from? $\endgroup$ – Greg.Paul Oct 18 at 23:17
  • $\begingroup$ @Greg.Paul The definition is from "Relativity, Gravitation and Cosmology" by Robert J.A. Lambourne (page 25) $\endgroup$ – daljit97 Oct 19 at 11:30
  • $\begingroup$ @daljit97 Did you get a clear idea of ​​the concept of "proper time" from the answers and literature? This link claims, that it is the observer, who was both at start and end. physics.bu.edu/~duffy/ns547_spring10_notes03/time_dilation.html ? But the both observers are present at both events; an observer at "rest" measures by means of two synchronized clocks; if he was at start, he had second synchronized clock at finish, the moving uses one clock. Astronauts clock measures shorter time interval, than two synchronized clocks. I think that explanation in the book is confusing. $\endgroup$ – Albert Oct 21 at 8:40
2
$\begingroup$

I prefer to think of proper time as the 'distance' between the two events in spacetime. Consider some sort of a world-line in a four dimensional spacetime

$x^\mu = \left(ct, \mathbf{r}\right)^\mu$

Where $c$ is the speed of light, $t$ is time and $\mathbf{r}$ is position. Lets define some point (event) on this curve as the 'start': $x_0^\mu =\left(ct_0, \mathbf{r}_0\right)$

Consider now an event on the same world line that is close to the 'start': $x_\delta^\mu =\left(ct_0+c\delta t, \mathbf{r}_0+\delta \mathbf{r}\right)$

What is the four-distance ($\delta s$) between these two events? The square of the distance is:

$\delta s^2 = c^2\delta t^2-\delta r^2$

Provided we deal with time-like world-lines (i.e. $\delta s^2 > 0 $), one can, by going in small steps and adding small distances, find the full four-distance between any two events on the world-line. It is therefore convenient to parametrize the world-line by this distance (also known as arc-length):

$x^\mu=x^\mu\left(s\right)=\left(ct\left(s\right),\mathbf{r}\left(s\right)\right)^\mu,\quad x^\mu_0=x^\mu\left(0\right)$

We may now choose to measure distance in seconds by introducing proper time $\tau=s/c$. That's it - no clocks involved at all. It is all about arc-length. And since this arc-length is Lorentz-invariant, all observers will agree on it.

Now if you do want clocks back, think of the world-line in the rest-frame ($\bar{S}$) of the observer moving along this world-line. For that observer the world-line will be straight and 'vertical' (at least locally), i.e. along the temporal axis only:

$\bar{x}^\mu\left(s\right)=\left(c\bar{t}\left(s\right),\mathbf{0}\right)^\mu$

So the distance between two closeby events on this world-line is, by definition:

$\delta s^2=c^2\delta \tau^2 = c^2 \delta \bar{t}^2 - 0$

Hence, $\delta \tau = \delta \bar{t}$ i.e. the clock carried by the rest-frame observer is measuring the proper time.


As a bonus, from here it is easy to get to Lorentz factor. Consider the derivative of the world line with respect to its own arc-length:

$\frac{dx^\mu\left(s\right)}{ds}=c^{-1}\frac{d}{d\tau}\left(ct,\mathbf{r}\right)^\mu$

We can define $\frac{dt}{d\tau}=\gamma$ as the Lorentz factor

Then:

$\frac{dx^\mu\left(s\right)}{ds}=c^{-1}\gamma\cdot\left(c,\frac{d\mathbf{r}}{dt}\right)^\mu=c^{-1}\gamma\cdot\left(c,\mathbf{v}\right)^\mu$

Where $\mathbf{v}$ is the velocity. The step between the two nearby events on the world line is:

$x^\mu\left(s+\delta s\right)^\mu-x^\mu\left(s\right)^\mu=\delta s\, c^{-1}\gamma\cdot\left(c,\mathbf{v}\right)^\mu$

Clearly, the distance between these two event is $\delta s$, so:

$\left|x^\mu\left(s+\delta s\right)^\mu-x^\mu\left(s\right)^\mu\right|^2=\delta s^2 = \delta s ^2 \, c^{-2}\gamma^2 \cdot\left(c^2-v^2\right)$

Therefore:

$\gamma^2=\frac{c^2}{\left(c^2-v^2\right)}$

irrespective of whether the world-line is straight, or curved.

$\endgroup$
1
$\begingroup$

Consider two timelike related events.
Now draw a worldline for an observer that visits [was present at] both events.
The proper time for that worldline is the time elapsed on that observer's wristwatch.
In the frame of that observer, those events are in the same position (here at the origin).

For another worldline visiting those two events, one gets a proper time for that worldline, which is generally different from the first. (This is the Clock Effect.)

By contrast, for a worldline that doesn't visit one of the events, the two events are not in the same position in that frame. Since this worldline did not visit both events, the elapsed time measured by this observer would generally not [strictly speaking] be "a proper time for a worldline between those two events".

You might have heard of the "proper time interval between two [nearby] events".
That would correspond to the proper time for the inertial worldline that visits both events.


UPDATE

A spacetime diagram might help.
I've drawn it on rotated graph paper to help us see and count the ticks.
Unlike lengths of segments on this diagram, the areas of these light-clock diamonds are Lorentz-invariant. These light-clock diamonds are traced-out by the light-signals in an standard-issue light-clock traveling with each observer.

As described in the Wikipedia link on Proper Time (above in the OP),
"proper time" is associated with a worldline--not just the endpoint-events.
The adjective "proper" is a reference to "ownership" or "property" [not "correct" or "the opposite of improper"].
(Minkowski used "eigenzeit", which Google Translate translates to "own time".)

Wristwatch-time (in Taylor/Wheeler's "Spacetime Physics") or private-time (in Bondi's "Relativity and Common Sense) may be better terms.


Below are multiple worldlines from event O to event Z.
For the observer along each worldline,

  • their clocks following along these worldlines visited both events O and Z.
  • O and Z are at the same position in their frame: "here at (x,y,z)=(0,0,0)".

RRGP-multiple-worldline-properTimes-robphy

I'll leave you to count the light-clock diamonds to determine the proper-time along each worldline. That these proper-times don't all agree is called the "Clock Effect" in relativity. The worldline with the longest elapsed-proper-time from O to Z is the inertial one.

[The "Twin Paradox" is the misuse of the principle of relativity to
mistakenly-equate "being allowed to consider one's self at rest" (which they all can)
with "being allowed to consider one's self as inertial" (only one of the five above is entirely inertial from O to Z).]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.