This might be a dumb question but I'm still going to ask it.
So basically I've been trying to learn Special Relativity through this lecture by Brian Greene, and the way he justifies time dilation is through a Light Clock. I think I understand this, but I just don't know how I could picture this with a normal clock, or without a clock at all. If someone could explain this to me, that would be great.
Any help is much appreciated. Thank you!
Forget about it being called a light 'clock'.
Just imagine you flash a light at a mirror on the Moon, and using an ordinary stop watch you time how long the light takes to be reflected back to you.
Now imagine a friend and I are in a pair of cars travelling in convoy. My friend goes past you in her car just as you flash the light, and makes a note of the time on her ordinary clock. I happen to be going past you in my car just as the light returns to you from the distant mirror, and I make a note of the time on my ordinary clock.
When my friend and I get together and compare notes later, we can calculate the duration of the light's trip by subtracting her reading from mine. In the frame of our convoy, the light's path was longer than in your stationary frame. So my friend and I will conclude that the duration of the light's journey was longer, which means the time elapsed according to your stop-watch will be less than the time elapsed according to our clocks.
That shows that even using ordinary clocks you can work out the effects of time dilation.
Of course, the thought experiment I set out above would require much more accurate timing than you could easily manage in practice, but in principle it can be done. Indeed there have been hundreds of experimental tests of time dilation.
The light clock is used in these examples because it makes things really easier to see. I don't know of a simple and intuitive explanation of why a quartz clock or an atomic clock will also undergo the same effect, but I can assure you that they do (see, e.g., Wikipedia for experimental evidence).
The key point in the argument is then that a light clock measures time just as well as any other clock (for example, you could compare the ticks on your watch with those of the light clock and see that they match). Hence, if the light clock undergoes time dilation, every other clock undergoes as well. The effect is not a consequence of the way the clock works, but rather of how time itself works. It will affect every other clock not because it changes the mechanism of the clock or something similar, but because time behaves in a non-intuitive way.
I just don't know how I could picture this with a normal clock
This would only be possible if you knew the relativistic law that governed the operation of a given clock. The light clock is used because, by the second postulate, the relativistic law that governs it is easy to analyze. Then, by the first postulate we know that all other clocks must show the same effect also.
The next most simple law that I can imagine for a clock would be a clock based on radioactive decay. The non-relativistic law is $$N=N_0 e^{-\lambda t}$$ where $N_0$ is the initial number of radioactive particles and $\lambda$ is the decay constant. The relativistic law is very similar $$N=N_0 e^{-\lambda \tau}$$ where $d\tau^2=dt^2-(dx^2+dy^2+dz^2)/c^2$ is the proper time. A simple bit of algebra gives $$N=N_0 e^{-\lambda \int d\tau}$$ $$\int d\tau= \int \frac{d\tau}{dt} dt$$ $$ =\int \sqrt{1-\frac{v^2}{c^2}} dt$$ $$=\frac{1}{\gamma}\int dt$$ $$N=N_0 e^{-\lambda t/\gamma}$$ So indeed, this type of clock also displays the same relativistic time dilation as any other clock.
But what’s the point? To do the analysis of a radioactive decay clock I had to know the relativistic law for radioactive decay. That law produces the same time dilation as a light clock, but without the clear benefit of a simple relativistic law. It is far clearer to use a light clock and then invoke the first postulate in extending the result to other types of clock.
I suggest you look into Lorentz transformations in detail. Let's assume the reference frame of a still observer with the 2 axes: the x-axis and time axis t. Let's say there's another observer within that frame moving with a constant velocity v with respect to the first observer (who's still). The two transformed coordinates for this moving observer are given by x' and t' in a way (by Lorentz transformation):
$x' = \frac{x-vt}{\sqrt{1-v^2}}$ ... (1.1)
$t' = \frac{t-vx}{\sqrt{1-v^2}}$ ... (1.2)
assuming relativistic units.
This comes from rather simple algebra once graphed out assuming you're familiar with the concepts of invariants else it may look like the hypotenuse of a triangle is not the longest side ;)
Anyway, in equation 1.2, we can clearly see the relation between the time frame of the still observer and the time frame of the moving observer.
Since in eq 1.2, the denominator is (always) less than 1, t' > t which means that time passes slower for our moving observer.
That's time dilation.
