Deduce time dilation from a Minkowski diagram?

Question

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

For example, consider the image below.

Here,

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?

Answer 1

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.

Answer 2

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.