I have a situation similar to that underlying the twin 'paradox' - the observer m1 below remains stationary while the observer m2 goes off on a space-time trajectory, and then returns to m1's position. Here is the Minkowski diagram:

minkowski diagram

My reasoning is as follows.


The event at A is observed by m2 at B. m1 observes m2 observing A, at C. At C, m1 reasons: "m2 is observing the event A now, whereas this event is already in the past for me. Therefore, time must be running slower for m2 than for me."


The event at X is observed by m2 at Y. m1 observes m2 observing X, and Z. At Z, m2 reasons: "m2 is observing the event X now, whereas this event is in the future for me. Therefore time must be running faster for m2 than for me."

Is my reasoning for none/one/both of these parts correct? If not, why?

Part 2 in particular seems very suspect to me, given the apparent ability of m2 to see m1's future, and the ability of m1 to note that m2 can see its future. After all, isn't m1 always supposed to conclude that m2's time is running slower?

  • 1
    $\begingroup$ I'm voting to close this question as off-topic because check-my-work questions are off-topic. $\endgroup$
    – ACuriousMind
    Oct 2, 2015 at 18:50
  • $\begingroup$ The post you linked to refers to homework questions. This isn't a homework question. I am not currently taking a special relativity course, and did not receive this as a homework problem. In fact, if I had just filled in Part 1 above and left Part 2 empty, asking for help, this wouldn't be a check-my-work question anymore. Is there any reason why that shouldn't be closed, but this should? My Part 2 is probably wrong. It's just that I actually made an attempt. Surely the question shouldn't be closed for that? $\endgroup$
    – Atriya
    Oct 2, 2015 at 19:12
  • $\begingroup$ @Atriya I think your question would be safer if you edit it to drop the opening line. No info would be lost and it (hopefully) could avoid the "check-my-work" pitfall. $\endgroup$
    – udrv
    Oct 2, 2015 at 20:18
  • 2
    $\begingroup$ As a point of reference rather than a judgement about which category this post belongs in, the [homework-and-exercises] tag (and the rules that go with it) applies to any question that has primarily pedagogical value no matter what context it came up in. So "this isn't an assignment" has no weight in deciding if the tag belongs or not. $\endgroup$ Oct 2, 2015 at 22:45
  • $\begingroup$ Now, with that said, somewhere on the site is a post referencing Einstein's phrase "moment of acceleration" which applies to the particular approach to analyzing the twin paradox you appear to be attempting. I personally found musing on the phrase to be very profitable, though I still approach the twin paradox from a invariant interval point of view. $\endgroup$ Oct 2, 2015 at 22:47

2 Answers 2


Your confusion arises from an incorrect reading of event coordinates. It is a frequent beginner's mistake, so it's worth it to trace its roots step by step. Here's how:

1) Representing events: Every point in the Minkovski diagram represents an event. The same point represents the event both in the unprimed and in the primed frames. That is, we do not associate different points to a given event in order to represent it relative to different frames. This means your event A is always represented by point A. Point B is an entirely different event, relative to both the primed and the unprimed frames.

2) Event coordinates: The coordinates of a given event in one frame or another must be read off the corresponding coordinate axes. For the unprimed frame, the coordinates are read by orthogonal projections onto the $x$ and $ct$ axes, while for the primed frame we must read them by means of skew projections parallel to the $x'$ and $ct'$ axes respectively. When you read the coordinates of event A in the primed frame, the AB line you drew is meant to show that the $ct'$ coordinate of A is read along the $ct'$ axis at point B, not that A is observed at B. To see why, consider the $x'$ coordinate for A and B: A will have a negative $x'_A < 0$, while B occurs at $x'_B = 0$. On the other hand, it is absolutely true that in the primed frame events A and B occur simultaneously, at time $ct'_0 = ct'_A = ct'_B$. So the primed frame sees event A at $(x'_A < 0, ct'_0)$ and event B at $(0, ct'_0)$.

3) Answer to PART 1: if m1 observes event A at unprimed coordinates, then m2 also observes event A, albeit at primed coordinates, but definitely not at event B. Likewise, if m2 observes event B, m1 also observes event B, but at the corresponding unprimed coordinates read off the $x$ and $ct$ axes. Event C is completely different from events A and B. To m2, A and B are simultaneous, but occur at different positions along $x'$. To m1, B and C are simultaneous, but occur at different positions on $x$, while A and C occur at the same position $x=0$, but at different times. You are correct though when you say that to m1, B lies in the future of A, since $ct_B > ct_A$, while C lies in the future of B, since $ct_C > ct_B$.

4) PART 2 gets a similar treatment.

5) About one frame being able to see into the other's past and future: this is formally correct and amounts to relativity of simultaneity. Take event O at the common origin. Events at different positions along $x$ that appear to m1 simultaneous with O actually are observed at different times by m2. Some may be in m2's past relative to O, some may be in his future. This doesn't mean that m1 can inform O about his future or vice-versa: the speed of light limit prevents it, but that is another story.

I think you were looking for an answer to your previous question on representing time dilation in the Minkowski diagram via a geometric construction. If so, you may want to use 2 diagrams, one as above with the unprimed frame as stationary, the other with the primed frame as stationary. If you insist on showing mutual time dilation on the same diagram, then you might want to consider a variant called the Loedel diagram (it is said Max Born used it too). It has the advantage that the coordinate axes for the 2 frames are symmetric, although both become skew.


Here's the right way to say this (and to highlight the symmetry between the outbound and inbound journeys).

Label the origin $O$. Label the point where the voyager returns to earth $P$.

At time $C$, the earthbound observer says "$C$ minutes have passed since my friend departed at time $O$. But right now, his clock shows time $A$. Therefore his clock is slow."

At time $Z$, the earthbound observer says "In $P-Z$ minutes, my friend will return to earth. But his clock shows that the journey, starting right now, will take only $P-X$ minutes. Therefore his clock is slow."

  • $\begingroup$ Since both observers started out at the same place and time, so they would both agree the outbound observer started at time O. Unless you are saying O=Z, the statement "the journey, starting right now" is inaccurate. $\endgroup$
    – user854
    Mar 17, 2016 at 2:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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