Also dont understand this question i found on a past exam paper (31% got it correct): A robot is heading radially towards the surface of a planet in the Hoth system at a constant speed of 0.85c. Observers on the surface of the planet observe it at a time when it is a distance x above the surface in their reference frame. The observers calculate the time that the robot will take to reach the surface of the planet as 784 microseconds.

Which one of the following best describes the time of the robot’s descent to the planet surface as measured by the robot, and the time as measured by the observers on the surface of the planet?

A. They are both measurements of proper time in their own reference frames.

B. Neither are measures of proper time.

C. Only the observers measure the proper time.

D. Only the robot measures the proper time.

The answer is D, but I thought it would be A if anything, since all reference frames are equal and the times observed on reference frames moving relative to observers' own stationary frames will be slower, and the times as measured in their own frames can both be considered proper times, no?

  • $\begingroup$ Isn't proper time as a technical term defined as the time shown by a clock that is attached to the moving object? $\endgroup$ – Steeven Apr 17 at 12:38
  • $\begingroup$ idk, i was just taught that proper time was the time of an event measured in a stationary frame. though if that is the technical definition, couldnt the planet be considered the "object" too? meaning again, theres 2 proper times? my brain is srsly starting to hurt :I $\endgroup$ – seb aye Apr 17 at 12:45
  • $\begingroup$ @sebaye " i was just taught that proper time was the time of an event measured in a stationary frame " This is inaccurate. Proper time is the time of an event measured only in that frame which is either at both the events or in which frame, both events happen at the same place. $\endgroup$ – silverrahul Apr 17 at 13:11

From Britannica.com

The proper time would be measured by any clock moving along the straight world line between the two events.

In other words, the proper time is the time measured by that clock which is moving with a uniform velocity between the 2 events.

Here, the 2 events are the event of the robot being at a certain point above the planet's surface and the event of the robot reaching the earth's surface.

Out of the given options , only one observer's clock satisfies the condition of moving from one event to the other with uniform velocity and that is the robot observer's clock.

Hence, only the robot measures proper time


Edit: I noticed one of your other questions so I'll preface this answer with an additional point, in special relativity it is better to think of reality as if everything that ever happened or ever will is fixed in stone. It turns out all the weirdness that arises from special relativity basically comes from looking at this from different angles, e.g. if you draw a square on a sheet of paper and look at it from above, it looks like a square, look at it from a different angle and it looks like a rectangle, look at it from a different angle again and you have a rhombus but all these shapes came from the one thing you drew (the one reality), and if you think hard you can work out the original shape even if you are looking at one of the deformed shapes.

In special relativity you are drawing the world lines of observers in space time. If you take a piece of paper you can say one direction is time, and one direction is space and the world lines are now just lines that describe the path sometime takes from time a to time b. Single points on your page aren't locations like we normally think of them but rather are events, that is a combination of a world line, and a time.

Original answer:

The proper time is the time measured by an observer that travels between two different events (points in space, as well as time). For this question the two events are 1) robot at distance x above planet surface, and 2) the robot lands on the surface.

If you draw the space-time diagram and include the robot and observers then you can see that the path of the robot (its worldline) passes through these two events, so the time it measures is a proper time between the two events. The observer just stands off to the side and doesn't pass through either of the events. Instead they partition all of space time into different time slices and work out the time between the two relevant slices. But this second process could be different for different people.

If the robot sent a messenger at near the speed of light to the observer when it was at distance x and then lands next to the observer you can say the time measured by the messenger plus the time spent waiting by the observer is also a proper time (because you have now drawn a line between the two events). But the observers are only measuring the time between 1) we see the robot at distance x (not when the robot is at distance x) and 2) the robot lands next to us. This is different to the previous example.

Note that these proper times can be different since they are different paths, however this makes sense since the amount of time that passes for someone depends on the path they take (e.g. the twin paradox).

  • $\begingroup$ Oooh this all makes sense now! thanks so much $\endgroup$ – seb aye Apr 17 at 13:18

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