Is it possible to deduce the phenomena of time dilation from a carefully constructed Minkowski diagram?

For example, consider the image below.

minkowski diagram


  1. Let us say the Euclidean distance between the blue and pink dots is 1 unit of time in the unprimed frame of reference.
  2. The line marked ct'=1.25 is a line of simultaneity in the primed frame of reference - all points on that line are simultaneous in the primed frame.
  3. Therefore, the point of intersection of the ct'(x'=0) line and the ct'=1.25 line (let us call that point P) should be simultaneous with the pink dot, in the primed frame of reference.

Given this, does the fact that the Euclidean distance between the blue dot and P is greater than the Euclidean distance between the blue and pink dots, represent time dilation? I'm guessing it doesn't, since we should be measuring Minkowski distances? Also, what is the physical significance of the time component of P in the unprimed frame?

To re-iterate my question: is it possible to deduce the phenomena of time dilation from a carefully constructed Minkowski diagram?

  • 2
    $\begingroup$ See Spacetime Physics by Taylor and Wheeler or A Traveler's Guide to Spacetime by Moore. They do it very well. $\endgroup$
    – Bill N
    Sep 29, 2015 at 20:30
  • $\begingroup$ Or Takeuchi's little book. $\endgroup$ Sep 29, 2015 at 21:37
  • $\begingroup$ The euclidean distance is not relevant. What matters is the lorentzian distance $\sqrt{ t^2-x^2 }$ (which in this case is 1.25 from the blue dot to $P$). $\endgroup$
    – WillO
    Sep 30, 2015 at 0:36
  • $\begingroup$ Atriya: "does [...] represent time dil.?" -- As a construction of/in Euclidean geometry, the given diagram is indeed suggestive of$$\begin{align}~&\frac{\text{d[ blue dot, pink dot ]}}{\text{d[ blue dot, intersection of green lines ]}}=\cr~&\sqrt{1 - \left( \frac{\text{d[ blue dot, pink dot ]}}{\text{d[ blue dot, intersection of green line and red line ]}}\right)^2}.\end{align}$$It doesn't show, however, how to decide which participants belonged together to one "(inertial) reference frame", how to compare durations, how to compare (chronometric) distances etc. for deducing SR time dil. $\endgroup$
    – user12262
    Oct 1, 2015 at 10:13

2 Answers 2


All of special relativity is captured by spacetime diagrams like the one you've drawn. The lorentzian (or, if you prefer, minkoskian) distance from the blue point to $P$ is 1.25, meaning that a clock traveling along the green worldline will record 1.25 ticks between those points. The lorentzian distance from the blue point to the pink point is 1, meaning that a "stationary" clock (i.e. one with the red worldline) will record 1 tick between those points. The traveler will therefore say that the stationary clock ticked only once in 1.25 minutes, i.e. it is running slow.

So, yes, you can see the time dilation in the diagram.

The unprimed coordinates of point $P$ show the time and location that the stationary traveler assigns to point $P$. He therefore says that the $P$-event occurs after the pink event, while the traveler says that the $P$ event is simultaneous with the pink event.

  • 1
    $\begingroup$ I think that this is a good explanation of time dilation as viewed from an observer in the moving (or 'primed') reference frame. As an exercise, readers who understand this explanation may also want to think about how they would demonstrate time dilation using the same Minkowski plots but by an observer in the stationary (or 'unprimed') frame of reference. $\endgroup$
    – user93237
    Sep 30, 2015 at 1:32
  • $\begingroup$ Thanks. Why exactly is the Minkowski distance, and not the Euclidean distance, relevant here? You state that as a fact without explanation. My question then is - is there a way to deduce by means of some construction within the diagram itself, that the Minkowski distance must be used? A naive user of the diagram might simply calculate Euclidean distances. Surely it can be deduced from features of the diagram itself (perhaps constructed explicitly for this purpose) that this is wrong? $\endgroup$
    – Atriya
    Sep 30, 2015 at 19:24

The Minkowski diagram captures time dilation even without invoking the Minkowski distance, provided we account for the different units/scales along the unprimed and the primed axes, see for example "Minkowski diagram in special relativity" (Wikipedia link). The calculation is exactly equivalent to that using the Minkowski distance as explained by WillO, but uses the Euclidian distance and usual trig/geometry.

Let the unit along the $x$ and $ct$ axes be $U$, and that along the $x'$ and $ct'$ axes be $U'$. They are related as $$ U' = U \sqrt{\frac{1+\beta^2}{1-\beta^2}} $$ for $\beta = v/c$ as usual.

In the case of the "pink" and "blue" events on your diagram, let the time difference between them in the unprimed frame be $ct$. This is just the Euclidian distance from "blue" to "pink" along the $ct$ axis in units of $U$. The time difference between these events as perceived in the primed frame is the Euclidian distance between the two intersections of green lines in units of $U'$. Let us first calculate this Euclidian distance according to usual geometry, in units of U.

In the triangle formed by "blue", "pink", and the 2nd intersection of green lines (first one being "blue" itself), denote $\overline{ct'}$ the desired Euclidian distance along the ct' axis and $\theta$ the angle between the $ct$ and $ct'$ axes. Calculate the other angles in terms of $\theta$ and apply the sine theorem to obtain $$ \frac{ct}{\sin\left( \frac{\pi}{2}-2\theta \right)} = \frac{\overline{ct'}}{\sin\left(\frac{\pi}{2}+\theta \right)} \;\; \text{or} \;\; \frac{ct}{\cos 2\theta} = \frac{\overline{ct'}}{\cos \theta} $$ Since $\tan \theta = \beta$, we have $\cos \theta = \frac{1}{\sqrt{1+\tan^2 \theta}} = \frac{1}{\sqrt{1+\beta^2}}$ and $\cos 2\theta = 2\cos^2 \theta -1 = \frac{1-\beta^2}{1+\beta^2}$, and we find $$ \overline{ct'} = \frac{1}{\sqrt{1+\beta^2}} \frac{1+\beta^2}{1-\beta^2} ct = \frac{\sqrt{1+\beta^2}}{1-\beta^2} ct $$ Now express $\overline{ct'}$ in units of $U'$ to obtain the correct time observed in the primed frame as $$ ct' = \frac{\overline{ct'}\cdot U}{U'} = \frac{\sqrt{1+\beta^2}}{1-\beta^2} ct \sqrt{\frac{1-\beta^2}{1+\beta^2}} = \frac{ct}{\sqrt{1-\beta^2}} $$ So, if the time difference between "blue" and "pink" in the unprimed frame is $ct$, the corresponding time difference $ct'$ in the primed frame is dilated by a factor $\gamma = \frac{1}{\sqrt{1-\beta^2}}$, which is the time dilation we were looking for.

In a similar way we can calculate the time dilation perceived in the unprimed frame relative to events in the primed frame.

  • $\begingroup$ Thank you very much for this detailed answer! I was not looking to actually calculate ct' as you have shown, but simply deduce by reasoning (no mathematics allowed) that ct' > ct, and do this by reasoning (simple inequalities only - no mathematics) with Euclidean and not Minkowski distances. I actually think I've devised a way to do this now, but am not sure if it's correct. $\endgroup$
    – Atriya
    Oct 2, 2015 at 15:40

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