Imagine you live on a planet orbiting a black hole. What would your life be like? Black holes are incredibly interesting phenomena, even light can't escape their immense gravity.
You certainly couldn't (provided you got close enough). It's time we explored how black holes would effect you if you lived around one, so you'll be prepared just in case that ever happens. And if it doesn't, you'll have learned something along the way, which isn't a bad thing! Promise!
So let's imagine you're living on a planet that orbits a black hole. Your work has sent you to a satellite that orbits the planet, not moving with respect to the planet. A spaceship is falling freely past you, radially inwards towards the black hole with a velocity \(v\) at the moment when it's passing you. You are both emitting light signals, blue from the spaceship and red from the satellite, at fixed time intervals. We assume that you both see the light immediately when it's turned on, meaning that the speed of light is technically infinitly large.
First, what sort of observers do we have? The planet is in orbit around the black hole, and as you are not moving relative to the planet, you are a shell observer. The free falling spaceship will be the freely falling observer. We don't have a far away observer at the moment.
Now we need to know the energy of the falling spaceship. We know that \(\gamma=E/m\), and as \(\gamma\) a conserved size, so is \(E/m\). So it makes sense to use 
the definition of \(E/m\):
\(\frac{E}{m}=(1-\frac{2M}{r})\frac{dt}{d\tau}\)
where M is the mass of the black hole, r is the distance to the Schwartzchild radius, dt is an infinitly small timestep for a potential far away observer and \(d\tau\) is the proper time of the spaceship. All units are natural, meaning that mass is given in meters, \(M_m=GM_{kg}/c^2\).
As we don't have a far away observer, we should transform this general expression for our shell observer instead. We use the Schwarzshild line element:
\(\Delta s^2 = (1-\frac{2M}{r})\Delta t^2-\frac{\Delta r^2}{(1-2M/r)}-r^2\Delta\phi^2\)
where we have that the shell observer never changes their distance from the black hole, \(r\), which leaves us with: \(\Delta s^2 = (1-\frac{2M}{r})\Delta t^2\). As we've said before, spacetime and proper time are the same, \(\Delta s^2 = \Delta \tau^2\). In other words (letters?): \(\Delta s^2 = (1-\frac{2M}{r})\Delta t^2 = \Delta \tau^2=\Delta t^2_{shell}\). By transforming this, we find our expression for the energy, where we can use the time the shell observer sees, rather than the far away observer:
\(\frac{E}{m}=(1-\frac{2M}{r})\frac{dt}{d\tau}=\sqrt{1-\frac{2M}{r}}\frac{dt_{shell}}{d\tau}\)
The distance between the shell observer and the black hole, r, becomes the radius of the shell. This distance is pretty much impossible to measure, because once you pass the event-horizon of the black hole, the gravitation will be so strong that light cannot escape. How can you, the shell observer, then measure the distance to the center? Not with a ruler, that's for sure. Even if you could see past the event horizon, length contraction would come into play, and you still wouldn't be able to push the ruler to the center. You can, however, find the circumference of the black hole!
Now, we should be familiar with the expression \(dt/d\tau=\gamma=\frac{1}{\sqrt{1 -v^2}}\). In our case, both t and v are locally observed by the shell observer, and so we insert this expression for \(dt_{shell}/d\tau=\frac{1}{\sqrt{1-v_{shell}^2}}\), ultimately giving us
\(\frac{E}{m}=\sqrt{1-\frac{2M}{r}} \gamma_{shell}\)
for a shell observer at position r. In other words, a shell observer will remain at rest and see an object move past it with a constant velocity, as we have that \(\gamma\) is constant.
Using the expression above, with the locally observed shell velocity, we can find the energy per mass of the falling spaceship. We use the first expression for energy, \(\frac{E}{m}=(1-\frac{2M}{r})\frac{dt}{d\tau}\), and find the relation: , \(\Delta \tau = \frac{1-\frac{2M}{r}}{E/m}\Delta t\). If we insert our expression for the relation between the time of a shell observer and the time of a far away observer, \(\Delta t^2_{shell} = (1-\frac{2M}{r})\Delta t^2 \), for \(\Delta t\) we will see that the relation between the proper time and the time of the shell observer can be expressed as
\(\Delta \tau = \frac{1-2M/r}{E/m}\Delta t\)
We now want to use the light signals you and the spaceship send out, in order to find a relation between a time interval \(\Delta\tau\) in the falling spaceship and a time interval \(\Delta t_{shell}\) in the planet frame.
Imagine that you are in the top of a steep hill, with the black hole at the bottom, and the spaceship in the middle. Light has to do a bit of work if it wants to get from the spaceship to you, rather than fall into the black hole. Light has no mass, but it does have energy, as we established when we talked about the momenergy of photons. The blue light will use this energy, and therefore become redshifted as it approaches you, while the red light from you will become blueshifted as its energy increases while it's 'falling downwards'.
We have the general description of a wavelength \(\lambda = c\Delta t\), which we use in the last expression we found, and we see that the relation between the two wavelengths \(\lambda\) and \(\lambda_{shell}\) is
\(\lambda = \frac{\sqrt{1-2M/r}}{E/m}\lambda_{shell}\)
where \(\lambda\) is the wavelength that you send out from the sattelite and \(\lambda_{shell}\) is the wavelength observed from the spaceship.
We know that the number under the root will be very small, as the distance to the black hole is a lot smaller than the mass of the black hole. Let's look at expressions both for the light you send out and for the light your friend sends out:
\(\lambda = \frac{\sqrt{1-2M/r}}{E/m}\lambda_{shell}\longleftrightarrow \lambda_{shell} = \frac{E/m}{\sqrt{1-2M/r}}\lambda\)
From this we can see that the value you send out from the sattelite will be a greater value (as it is divided by a small number), and so the light is blueshifted. The opposite happens to the light sent out from the spaceship, it will get redshifted as predicted.
In order to do these calculations we've said that light is observed instantaneous, meaning that it must have infinite velocity, but it does not. What would change if the speed of light is actually the speed of light, as we normally know it?
In special relativity we assumed that the speed of light was invariant, but that is not the case in general relativity! We can actually measure the speed of light in Schwarzchild-coordinates, \((r,t)\):
\(v_r=\frac{dr}{dt}=-(1-\frac{2M}{r})\)
This is the equation for the radial light speed, and from it we can judge that as the speed of light gets closer to 2M, in other words as it approaches the black hole, the velocity will get closer and closer to 0- the light will appear to stands still. So, while in free fall towards the black hole, will the signals sent from the spaceship slow down in the satellite, the closer the spaceship gets to the black hole?
It turns out that this is not the case for you, in your shell. This is only true for afar away observer!
We can check if the time between each signal sent increases. If we want to do this, though, we need to keep in mind that the light will take a longer time to reach you as the distance between the spaceship and the satellite increases.
It turns out that the distance \(\Delta r_\gamma\) that a photon approaching the spaceship travels during a time interval \(\Delta\tau\) on the spaceship clock is given by \(\Delta r_\gamma=-\frac{1-\frac{2M}{r_\gamma}}{1-\frac{2M}{r}}\frac{E}{m}\Delta\tau\), where \(r_\gamma\) is the position of the photon and r is the position of the spaceship.
Judging from this, the light does not seem to slow down as it approaches the black hole (it is not past the event horizon at this point). The light will only appear slower to a far away observer, while a shell observer will always measure the speed of light as 1.
We can plot the light signal number and the time when you (in the satellite) receive the signal as a function:
We see that the time difference between when the light hits the satellite grows the further away the spaceship is, which makes sense as the distance is irrelevant when the speed of light is infinite.
We can do the same for a different perspective, rather looking at the signal from the satellite to the spaceship.
These results again show us the time when you receive the signal with infinite velocity (blue) and with finite velocity, c, (orange).
There is a bit of a difference between the lights sent from the spaceship and the light sent from the satellite. Seen from the satellite, the time intervals between each beam from the spaceship grows exponentially, while the time intervals between each beam you send out from the satellite grow more linearly. The reason for this is that the great gravity of the black hole means that time dilation will cause the time to move slower for the spaceship than for the satellite.
This can be seen from the fact that we know the signals are sent out with exact time intervals, and yet the time between each signal sent out from the spaceship increases. Similarly, the time interval between when you send out a signal and the spaceship receives it is shorter, as time moves faster for you.
We mentioned that the speed of light isn't constant, which is still true, only for a far away observer. So, we've seen how time dilation gets greater the larger the gravity is (the closer we are to the black hole), and know now that a great gravity greatly affects time and light!