The equation commonly taught to deal with Time Dilation is
$$t'=\gamma t=\frac{1}{\sqrt{1-v^2/c^2}}t$$
where $t$ is the time measured by the observer $A$ at rest (stationary frame of reference), and $t'$ is the time measured by the observer $B$ moving relative to the former observer with a velocity $v$ (moving frame of reference).
My first interpretation of the equation above was the following: given an event $E$, if $A$ measures the event $E$ to last $t_A$ seconds, then $B$ will measure $E$ to last $t_B=\gamma t_A$ seconds.
Question 1: is such an interpretation correct?
Question 2: if the answer to the former is "No", how are we to interpret -and when are we to use- the equation for Time Dilation? How can we, in general, given an event $E$ that $A$ measures to last $t_A$ seconds, determine how long that event will last from $B$'s perspective?
I explain in the rest of the post why I lean towards believing that the answer to the first question is in the negative. Consider the following;
Example 1: Suppose that $B$ is moving in the $x$-axis relative to $A$. If $B$ shines a ray of light in the $y$-axis towards a wall $1\ \text{meter}$ away stationary relative to $B$, then $B$ will measure the time the ray takes in reaching the wall to be $c^{-1}\ \text{seconds}$, while $A$ will measure the same event to take $\sqrt{c^2-v^2}^{-1}\ \text{seconds}$, as explained here. Thus we have $$t_B=\gamma ^{-1}t_A$$ as opposed to $$t_B = \gamma t_A.$$
Example 2: $A$ is positioned at the origin and measures that $1cs$ to its right there is a mirror. $B$ is moving in the $x$-axis with a velocity $v=+\frac{1}{2}c$ relative to $A$. When $B$ passes by $A$ they shine a ray of light towards the right. Consider the following two events:
$$E_1:= \text{the ray travels to the mirror and back to the origin.}$$
$$E_2:= B \ \text{travels to the mirror.}$$
It is easy to see that from $A$'s perspective both events will last $2s$. If the answer to my first question was in the positive there would exist a constant $\gamma$ such that, from $B$'s perspective, both events would last $\gamma \cdot 2s$. However, it can be shown that both events start but do not end simultaneusly to $B$ (so from $B$'s perspective the duration of $E_1$ will differ from the duration of $E_2$). In particular, $B$ will measure that the ray will return to the origin when $B$ has traveled only $2/3$ of its way to the mirror.
Best Answer
Events are not like "events", such as a gala or train ride. In SR, events are points in spacetime. As such, they occupy zero time (they are instants), and zero space (they are traditional Euclidean points).
The key to SR is that events are invariant, all observers see an event as the same thing, the only difference is the coordinates each assigns to it.
So consider to frame: $A$ and $A'$ (it's traditional in SR to have 'stationary' (not really) un-primed frame, and a moving (not really), primed frame). $A$ and $B$ will simply not do. An event, $E_0$, in $A$ will be at:
$$E_0 = (t_0, x_0,y_0,z_0)$$
while $A'$ will see it at:
$$E'_0 = (t_0', x_0',y_0',z_0')$$
Of course the coordinates are related by a Lorentz transform and its inverse.
If you have another event $E_1$, then you can talk about temporal and spatial separation. They key to relativity is that the interval between them, $\Delta s$, given by:
$$ \Delta s^2 = c^2(t_1-t_0)^2 - (x_1-x_0)^2 -(y_1-y_0)^2-(z_1-z_0)^2$$
is the same in all reference frames.
With that, we can look at the 1st part of Example 1. We immediately run into a problem. Everything is at rest to the moving $B$ frame, and the stationary $A$ frame is moving. That won't do. Let's put the mirror and light in $A$. Then there are 3 events:
Light is emitted, from the origin of $A$:
$$ E_0 = (0,0,0,0) $$
Light travels $L$ and hits the mirror:
$$ E_1 = (L/c, 0, L, 0) $$
and is reflected back to the origin:
$$ E_2 = (2L/c, 0,0,0) $$
To find out what $A'$ moving at $v$ along the $x$-axis sees: apply the Lorentz Transform:
$$ E'_0 = \Big(\gamma(t_0-\frac{vx_0}{c^2}),\gamma(x_0-vt_0),y_0,z_0\Big)=(0,0,0,0)$$
$$ E'_1 = \Big(\gamma(L/c-\frac{0}{c^2}),\gamma(0-vL/c),L,0\Big)=(\frac{\gamma L}c,-\frac v c \gamma L,L,0)$$
$$ E'_2 = \Big(\gamma(2L/c),\gamma(0-2vL/c),0,0\Big)=(2\frac{\gamma L}c,-2\frac v c \gamma L,0,0)$$
Then you can compare coordinates as needed. E.g., the invariant interval between emission and return is, in $A$:
$$\Delta s_{2,0}^2 = c^24t_2^2 = 4L^2 $$
while $A'$ has:
$$\Delta s_{2',0'}^2 = c^2(\gamma(2L/c))^2-(-2\frac v c \gamma L)^2=4L^2\gamma^2(1-\frac{v^2}{c^2})=4L^2$$
The two events occur at the same $(x,y,z)$ coordinate in $A$, so they are together "stationary" in $A$ (a single event can't move, but a pair can define linear motion). All of $\Delta s =2L$ comes from the time coordinate, hence, in a moving frame they occur at different points and the time interval must be larger (by $\gamma$)...moving clocks run slow.