I am a bit confused regarding a commonly cited definition of proper time. For example, in Spacetime and Geometry, Sean Carroll writes:

the proper time between two events measured the time elapsed as seen by an observer moving on a straight path between the events

Similarly, in Gravity, James Hartle writes:

proper time - the time that would be measured by a clock carried along the world line [between two events]

Let me explain my confusion: I will try and be as precise as possible so this post might be a bit long.

Defintion of proper time

The definition of proper time that I am working with is:

$$ d\tau^2 = -ds^2/c^2 $$ where $ds^2$ is the space time interval between two points.

Defintion of a clock

I am assuming a model of a light emitter/detector and a mirror separated by a distance $L$ in the $y$ direction. The clock emits light, it travels to the mirror, bounces off, and is detected by the detector. The clock ticks each time a pulse of light is detected by the detector. Thus each tick is separated by an interval of $\Delta t = 2 L /c$. Therefore, the clock measures the coordinate time interval between an emission and detection of a pulse*.

Definition of an intertial frame

We set up spatial coordinates via a grid of rigid rods, in a manner commonly found in most textbooks. We can define our time coordinate by using a clock that is stationary with respect to our grid. Thus one tick of the clock (defined above) is considered as one unit of coordinate time.

With the above definitions out of the way, let us measure the proper time between two events: the emission (A) and detection (B) of a light pulse of our clock. We will consider it in two frames, namely the rest frame of the clock, and a frame in which the clock is moving. Now, the world line between our two events is a trajectory in spacetime and thus should be independent of the coordinate system we use to describe it; that is, it's length should be the same in both frames we consider.

Rest frame of the clock

In the rest frame, the clock emits a pulse of light and some coordinate time later $\Delta t$ detects it. How far (in spacetime) has the clock moved meanwhile? Since this is the rest frame of the clock, $\Delta x^i =0$, so $$ \Delta s^2 = -(c\Delta t)^2 \rightarrow \Delta \tau^2 = \Delta t^2 $$

Thus in this case, the coordinate time interval between and emission and detection of the pulse (for a clock carried on a worldline between A and B) matches precisely the proper time. This makes complete sense given the definition.

Moving frame (where my confusion lies)

Consider a frame boosted with a velocity $-v$ relative to the clock's rest frame in the $x$ direction. In this frame, the clock is moving with velocity $v$ in the $x$ direction. The spacetime trajectory of the clock is still the same (it should be invariant; after all, the path the clock takes shouldn't change based on how we think of it), but it looks different under the new coordinates. The clock carried along this trajectory still ticks every time it detects a pulse. In these coordinates, the say the clock detects a pulse after $\Delta t'$. Then the spacetime interval between $A$ and $B$ is $$ \Delta s^2 = -(c\Delta t')^2 + (\Delta x')^2 $$ which reduces to (noting $\Delta x' = v\Delta t'$) $$ \Delta s^2 = -(c\Delta t')^2(1 - v^2/c^2) \rightarrow \Delta \tau = \Delta t' \sqrt{1 - v^2/c^2} $$

which is the standard result that a moving clock runs slower.

My confusion is that, in this case, the time measured in this frame by the clock carried from A to B is $\Delta t'$, NOT $\Delta \tau$. Even though the clock is carried on the same spacetime trajectory (described in different coordinates), it runs slower ($\Delta t' > \Delta t = \Delta \tau$). Thus the time measured by this clock is not the proper time, even though it is on the world line between two points.

I know I am missing something here with the definitions but I can't tell what. If a clock on the world line between two events measures the proper time, and proper time is invariant, then clocks moving between A and B should run at the same interval regardless of if our coordinate system has A and B at different spatial positions. What is causing my confusion?


1 Answer 1


The proper time along a worldline between two events is like the distance along a path between two points. Proper-time is worldline-dependent--this is the Clock Effect. Different world lines from chronologically related events A to B have different proper times.

(Given a particular worldline from A to B, all observers agree on the proper time elapsed on that worldline. For that worldline, all observers agree on the readings of the clock on that worldline. )

The Spacetime [square-]interval is like the squared-distance between two points in the plane (taken along a straight line path joining the points). In Minkowski spacetime, the square-interval from chronologically related events A to B is the square of the proper time along the inertial worldline from A to B.

  • $\begingroup$ I believe I understand these points. My confusion comes in regarding what it means to “carry a clock along a world line”. A moving clock will tick slower than a stationary clock, and the coordinate time recorded by such a clock will not be the proper time. In what sense then does a clock carried along a world line measure the proper time? I understand how this is the case if we are in the clock’s stationary frame. If we are not though, the coordinate time (measured by the clock) will not match up with the proper time. $\endgroup$
    – gabe
    Aug 15, 2020 at 17:45
  • $\begingroup$ By carrying a clock along a worldline, I mean that the clock [or better wristwatch] is worn by the observer. So, the wristwatch is at rest with respect to the observer and the watch and observer trace out the same worldline on a spacetime diagram. So, this watch measures the proper time of the observer. Geometrically, this watch measures the Minkowski-arclength of the worldline on a spacetime diagram, just like an odometer of a car measures the arclength of the path of a car on the road. $\endgroup$
    – robphy
    Aug 15, 2020 at 19:23

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