“Standing beside railroad tracks, we are suddenly startled by a relativistic boxcar traveling past us as shown in the figure. Inside, a well-equipped hobo fires a laser pulse from the front of the boxcar to its rear. (a) Is our measurement of the speed of the pulse greater than, less than, or the same as that measured by the hobo? (b) Is his measurement of the flight time of the pulse a proper time? (c) Are his measurement and our measurement of the flight time related by Eq. 37-9?” - Fundamentals of Physics 10E by Halliday, Resnick

I am a homeschooled high school student who has been trying to learn about special relativity, but I was stumped by the answer to part (b). For part (b), the answer is “no (the start and end of the flight are spatially separated)”, however, isn’t the frame of reference for the hobo inside the boxcar? Furthermore, the term “proper time” is the “time interval between two events that occur at the same location in an inertial reference frame.”

If the above definition is true, and the hobo’s reference frame is inside the boxcar, why isn’t the hobo’s measurement the proper time? I thought the laser pulse was still considered to be inside the same reference frame (the boxcar) even when the boxcar itself was moving.

  • $\begingroup$ Proper time is the time measured by an observer at rest with the concerned reference frame ... $\endgroup$ Mar 13, 2018 at 6:51

5 Answers 5


It is useful to draw a spacetime diagram (a position-vs-time graph, with time running upward by relativity conventions). I've drawn it on rotated graph paper to make it easy to show the tickmarks along various segments.

Observers are represented by [thin]* curves ("worldlines"), which advance upward inside the lightcone at each event. The events on a worldline are "timelike-related" to each other. (By thin*, we mean no thickness or extent. To model an "extended object" [like a boxcar], we need a family of worldlines.)

The RED vertical worldline represents the observer at the station.
The GREEN worldline represents the rear of the boxcar.
The BLUE worldline represents the front of the boxcar and the Hobo that rides there.

Note that "proper time" is the "wristwatch time", the time elapsed along an observer's worldline recorded by her wristwatch. Along that observer's worldline, the event at the start and the event at the finish (and in fact, all events along that worldline) are "at the same position [for that observer]".
For example,

  • Along OQ, the proper time is 4, as measured by an observer at the rear of the boxcar.
    [The station observer would say that the elapsed-time between O and Q is 5.]
  • Along PY, the proper time is 6, as measured by the Hobo, the observer at the front of the boxcar. [The station observer would say that the elapsed-time between P and Y is 7.5.]
  • However, OY is not along an observer's worldline [O and Y are "spacelike-related".] However, OY is simultaneous according to the rear of the boxcar and to the front of the boxcar. OY=10 is the proper-length of the boxcar.

The item of interest is the segment PQ,
where P the emission-of-light event at the front of the boxcar
and Q is the reception-of-light event at the back of the boxcar.

  • The segment of interest PQ is not along an observer's worldline [P and Q are "lightlike-related"]. Therefore the "interval" PQ is not a proper-time-interval. In fact, since a light-ray connects events P and Q, no observer worldline can visit both events P and Q.

Proper Time Boxcar - rotated graph paper

In my opinion, the answer "(the start and end of the flight are spatially separated)" is rather weak. Events O and Q are spatially separated according the station-observer, but not to the rear-boxcar observer... so OQ is a proper-time-interval for the rear-boxcar observer.

What it should say say is that "(the start and end of the flight are spatially separated for every observer)"... so PQ cannot be a proper-time-interval.


I think the problem comes from ambiguity of the question itself. Of course hobo can make a clock out of traveling photon, and claims that every passage of photon from front to rear and vice versa is for example 1 second. This hypothetical clock will measure the proper time of hobo himself though. What question is asking here imo is that does hobo measure the proper time of moving photon? Of course not! He is not at the rest with respect to the photon after all.


The requirement for proper time fails in this case because the front and rear of the boxcar are in different locations. The hobo cannot observe the time the beam reaches the rear of the boxcar simultaneously with the actual event because he is not there, and the information of that event must travel back to him.

  • $\begingroup$ Suppose he has a clock there that is synchronized with the clock at his location such that its time is recorded when the beam reaches it. Wouldn't that be equivalent to him being there, as far as the measurement of time is concerned? $\endgroup$ Aug 4, 2020 at 1:06
  • $\begingroup$ @Not_Einstein can you move a clock and keep it synchronized with a stationary clock? $\endgroup$
    – Bill Watts
    Aug 6, 2020 at 7:34
  • $\begingroup$ A frame of reference is assumed in principle to have an array of clocks that have been synchronized after they have been moved into position. Otherwise, the time of an event in a given frame of reference would have no meaning if it depended on the location of the event in the frame. So the time recorded by the hobo's clock at the rear of the boxcar should agree with the time on the clock next to him at the front of the boxcar. $\endgroup$ Aug 6, 2020 at 12:46

A good rule of thumb is that proper time is the time ticked off by a single clock. In the case of the hobo firing the laser, you would need two clocks to measure the interval, one next to the hobo to record the time the laser was fired, and the other at the rear of the boxcar to record the time the light arrived there. That means the duration of the interval depends on the synchronisation of the clocks, and in reference frames moving relative to the boxcar the clocks will appear out of synch owing to the relativity of simultaneity.

If the hobo had thrown a clock the length of the boxcar, then the clock would record the proper time of its flight. However, since the hobo had fired a beam of light, and it is not possible for any clock to travel at the speed of light, it's not possible for a single clock to be present at the start and end of the light's journey along the boxcar.


When the Hobo says "one second of proper time has passed, because the clocks hanging on the walls of the boxcar say so", then that is the truth, in all frames, because the clocks on the boxcar walls have proceeded one second in all frames, I mean the second hands of the clocks have moved 1/60th of a full circle, in all frames.

Let's say the Hobo fires the laser from the rear towards the front, because I just prefer that direction. And let's say the speed of the boxcar is 0.99 c relative to the track.

When the Hobo says "the laser pulse has now at this moment reached the front", then those people that have the opinion that the front wall is moving rapidly away from the point from where the laser pulse was fired, those people disagree with the Hobo's statement, because those people have the opinion that the front wall is moving at speed 99% of c, and the laser pulse is moving at speed 100% of c to the same direction as the wall, which means that the pulse is approaching the wall at closing speed 1% of c, so it takes quite a long time for the pulse to reach the front wall.


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