Maybe some spacetime diagrams will help.
Here's one showing a photon (thin blue line) being emitted by an observer (thick blue line) and hitting a reflecting surface (thick red line) and traveling back (thin red line) to the observer. This is for the rest frame of the observer and the reflecting surface:
As you can see, the separation between the observer and the reflecting surface is four feet and it takes 8 nsecs for the photon to make the round trip. It takes an equal amount of time for the photon to get to the reflecting surface as it takes for the reflection to get back to the observer.
Now we will transform this diagram using the Lorentz Transformation process to see what it looks like in a frame traveling at 60%c to the right:
At 60%c, gamma is 1.25 which means the distance between the thick blue and red lines gets contracted to 4/1.25=3.2 feet as you can see in the diagram at the Coordinate Time 0 for example. Also the photon takes longer to go through its cycle by the factor of gamma so now it takes 8 times 1.25 or 10 nsecs. But note that the photon gets to the reflecting surface in just 2 nsecs and it takes another 8 nsecs for the reflection to get back to the observer.
Does this make perfect sense to you?
That's not true.
Let's assume we do the same experiment except with an electron traveling at 80%c. Here is a diagram depicting an electron as the thick black line:
As you can see, it takes 10 nsec for the electron to make the round trip since it is going slower than the photon was. At 80%c, it takes 5 nsecs to go 4 feet (4/5=0.8) and another 5 nesecs to get back.
And as before, we will transform this to a frame moving at 60%c to the right:
Again, the separation between the observer and the reflecting surface is contracted to 3.2 feet but this time the electron leaving the observer is traveling slower than before, it looks like it's taking 2.7 nsecs to go one foot which is a speed of 37%c. But on the way back, the electron is taking 8.6 nsecs to go 8 feet for a speed of 93%c. The net result is that it takes a total of 12.5 nsecs which is the same Time Dilation factor that we got for the photon (12.5/10=1.25).
You can use the
velocity addition formula to calculate the exact speeds of the electron and the photon in these two experiments. For the case where the speed of light equals 1 the formula is:
s = (v+u)/(1+vu)
For the case of the photon, v=-0.6 and u=±1.
First, let's calculate it for u=+1:
s = (v+u)/(1+vu) = (-0.6+1)/(1+(-0.6)(1)) = (0.4)/(1-0.6) = (0.4)/(0.4) = 1
The photon is traveling at c away from the observer.
Now we do it for u=-1:
s = (v+u)/(1+vu) = (-0.6-1)/(1+(-0.6)(-1)) = (-1.6)/(1+0.6) = (-1.6)/(1.6) = -1
The reflected photon is traveling at -c (towards the observer).
For the case of the electron, v=-0.6 and u=±0.8.
First, let's calculate it for u=+0.8:
s = (v+u)/(1+vu) = (-0.6+0.8)/(1+(-0.6)(0.8)) = (0.2)/(1-0.48) = (0.2)/(0.52) = 0.385
The electron is traveling at 38.5%c away from the observer.
Now we do it for u=-0.8:
s = (v+u)/(1+vu) = (-0.6-0.8)/(1+(-0.6)(-0.8)) = (-1.4)/(1+0.48) = (-1.4)/(1.48) = -0.946
The electron is traveling at -94.6%c (towards the observer)
To summarize, these two cases, for the photon and the electron, both show Time Dilation and they both use the same equations, either the Lorentz Transformation or the Velocity Addition Formula.
Does this all make perfect sense to you? Any questions?