We all know that what goes up must come down, unless escape velocity has been achieved. We also know that whatever escapes the tug of gravity loses some speed in its effort to resist the pull of gravity. But, what happens when the thing can't slow down? What if the thing that was launched into space was a beam of light?
As you would expect, even though the beam of light cannot lose any of its speed, it does lose energy. It loses energy by slowing its frequency. This loss of frequency can be viewed as a slowing of the flow of time. Let's hold on to that thought, we'll come back to it later.
So, how do we calculate just how much energy is lost?
In order to do that we have to switch to considering what happens to a particle with mass. The amount of energy lost leaving a gravity well is just the negative of the kinetic energy gained by a particle falling to the surface from infinity. That energy is equal to the work it took for gravity to bring the particle in from infinity to the surface of the planet (or star, or whatever else).
Let's let
$$\begin{align} m & = \text{mass of particle} \\ M & = \text{mass of planet} \\ R & = \text{radius of planet} \\ G & = \text{Gravitational constant} \\ c &= \text{speed of light} \end{align}$$
Then, Newton gives us
$$\begin{align} F & = ma \\ F & = \frac{GMm}{x^2} \end{align}$$
where \(x\) is the particle's distance from the center of the planet.
At this point, you might be wondering why we're using Newton's equations when we're after gravitational time dilation. We can use Newtonian gravity to calculate the escape velocity, which isn't affected by relativity at all. The reason why it's left untouched by relativity is that when an object falls into a planet along a straight line headed directly for the planet's center, the acceleration it experiences is only affected by its position. Any relativistic time dilation that that particle experiences (relative to the planet) amounts to nothing more than a change of time variable within an integral. It doesn't affect the value of the integral.
If we let \( t_R = \text{time the particle impacts the planet} \) and \( x(s) = \text{position of the particle at time s} \), then we have
$$ \begin{align} v(t_R) & = \int_{\infty}^{t_R} \dot v(s) ds \\ &= \int_{\infty}^{t_R} a(s)ds \\ &= \int_{\infty}^{t_R} \frac{GM}{x(s)^2} ds \end{align}$$
We won't evaluate this integral directly, but it does show us that changing the time variable doesn't alter the velocity at impact. So, we're free to use Newton to calculate the escape velocity. We'll do that the only way I know how, by calculating the (non-relativistic) kinetic energy at impact after falling from infinity (with initial velocity of 0).
As mentioned earlier, the kinetic energy is equal to the work done by gravity to pull the object from infinity to the planet. It is given by the integral
$$ \begin{align}Work & = \int_{\infty}^{R} \frac{GMm}{x^2}dx \\ &= - \frac{GMm}{R} \\ &= \pm E_{kinetic} \end{align}$$
We don't care about the sign. Under Newtonian physics,
$$ E_{kinetic} = \frac{1}{2}mv^2 \\ \Rightarrow v = \sqrt \frac{2GM}{R}$$
That gives us \( v \), the escape velocity, which is (after a change in sign) the velocity a particle would have hitting the surface of a planet (with no atmosphere) after falling to its surface from infinity.
Now for the relativity part. We calculated non-relativistic kinetic energy above because all we cared about was calculating the escape velocity, but it wasn't the actual kinetic energy. That is given by the relativistic formula
$$ E_{kinetic} = \frac{mc^2}{\sqrt {1 - \frac{v^2}{c^2}}} - mc^2$$
The total energy of a particle is then given by adding the relativistic kinetic energy to the rest energy (the famous \( mc^2\))
$$ E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} $$
We can now calculate the ratio of the energy of the particle at rest on the surface of the planet to the energy after it has escaped the pull of gravity and is at rest at infinity.
$$ \begin{align} \frac{E_{ground}}{E_{\infty}} &= \frac{\left \{ \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} \right \}_{| v = 0} }{ \left \{ \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} \right \}_{|v=\text{escape velocity}}} \\ &= \sqrt{1 - \frac{v^2}{c^2}} |_{v = \text{escape velocity}} \\ &= \sqrt{1 - \frac{2GM}{Rc^2}} \\ &= \frac{t_{ground}}{t_{\infty}} \end{align}$$
Here \( E_{ground}\) and \( E_{\infty} \) refer to the energy of a particle on the surface of the planet and the energy of a particle at infinity, respectively.
This ratio holds for everything that escapes the planet's gravity (maybe because everything can be viewed as a particle, maybe for some other reason. I'm not sure). So, it also holds for photons. This takes us back to what we were thinking about in the second paragraph. The loss of energy for photons is the same as a decrease in their frequency. If we hold the frequency the same, however, it means that time has slowed.
I hope this post makes some sense to you, and that gravitational time dilation no longer seems as mysterious as it once may have. If you actually understand the math behind General Relativity, I'm sure you'll object to some of the steps I made above. I'll try to keep working on this and hopefully it will become more comprehensible with time.
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