Reciprocal time dilation

This diagram is a continuation of the special relativity diagrams.

Time dilation is one of the most popular consequences of special relativity, and the most common example to explain this effect is the light clock. It clearly shows how an observer notices that all clocks in movement run more slowly than his. But it poses a paradox, movement is relative, so all observers notice the same thing: observer A notices that B's clock slows, and Observer B also notices that A's clock slows. How's that possible?

To understand it, we must take into account that time dilation is a consequence of the space-time continuum (Diagram 10) and the relativity of simultaneity (Diagram 9):

Space-time is a continuum with all past, present and future events.
A "world" is the set of simultaneous events for a given observer at a given instant T. It's the equivalent of a film frame. It's a SUBSET, a three-dimensional (x, y, z) "slice" of four-dimensional space-time events (ct, x, y, z).
The "worlds" of observer A are different from the "worlds" of observer B due to the relative motion between the two observers.
The "worlds" of observers A and B include events where the other observer's clock lags behind his. Let's see it:

Diagram 11: reciprocal time dilation

We start from the origin event (0, 0) where the clocks of the inertial observers A (grey color) and B (blue color) are synchronized.
When observer A's clock reads T = 5, observer A's "world" is world 1, which includes the event where observer B's clock reads T= 4. Therefore, A observes that B's clock slows.
When observer B's clock reads T = 5, observer B's "world" is world 2, which includes the event where observer A's clock reads T = 4. Therefore, B observes that A's clock slows.
Let's keep in mind that clocks don't "appear to slow", clocks really slow down. World 1 is real for A, and world 2 is real for B, there is no absolute or privileged frame of reference. This phenomenon is real and is due to the existence of the space-time continuum, it cannot be explained without it.

Diagram 12: axis calibration (mathematically)

From the invariance of the interval it follows that the time dilation measured by an observer for moving clocks corresponds to the Lorentz factor: $Δt' = Δt\gamma$.
Therefore, when observer A reads a second according to his clock since the origin event (0, 0), he measures that for observer B has elapsed: $Δt' = 1\gamma$.

Mathematical derivation of time dilation

We're going to derive the time dilation measured by an inertial observer for clocks in relative motion. To do this, given an observer and a clock, we'll consider the invariance of the interval between two events: the origin event (0, 0) where the clock ticks T = 0 seconds, and event E1 where the clock ticks T = 1 second.

We'll consider two cases:

The clock remains at rest with respect to the observer. In the diagram, event E1 (clock ticks T = 1 second) would correspond to E1_A.
The clock moves with relative motion with respect to the observer. In the diagram, event E1 (clock ticks T = 1 second) would correspond to E1_B.

Due to the invariance of the interval, we have:

$Δs^2 = -c^2 Δt^2 + Δx^2 = -(c'Δt')^2 + (Δx')^2$

In the first case, the clock is at rest, therefore $Δx = 0$
In the second case, the clock moves at a speed v, therefore $Δx' = vΔt'$

$-c^2 Δt^2 + 0 = -c^2Δt'^2 + v^2 Δt'^2$
$-c^2 Δt^2 = -c^2Δt'^2 (1 - v^2/c^2)$
$Δt = Δt' \sqrt{1 - v^2/c^2} \implies \boxed{Δt' = Δt\gamma}$

Diagram 13: axis calibration (graphically)

We have that in both reference frames A and B the interval is invariant:
$Δs^2 = -c^2 Δt^2 + Δx^2 = -(c'Δt')^2 + (Δx')^2$
Tracing on the Minkowsky diagram the hyperbolae $-c^2 Δt^2 + Δx^2 = ±1$ we obtain on the axes the tick marks corresponding to the unit values.
For example, if we draw the curve $-c^2 Δt^2 + Δx^2 = -1$:
- On the ct axis, for $Δx = 0$, we have $cΔt = ±1$.
- On the ct' axis, for $Δx' = 0$, we have in the interval equation: $-(c'Δt')^2 = -c^2 Δt^2 + Δx^2 = -1$. Therefore, we have that $cΔt'= ±1$.
The 2, 3, 4... tick marks can be obtained drawing the curves $-c^2 Δt^2 + Δx^2 = ±2, ±3, ±4, etc$.