We were doing a flyby of a black hole at a radius of 20M, which is relatively safe conseidering our velocity of 0.993% the speed of light. However, just as we were passing by our engine failed, and now we're falling, I sure hope we didn't speed too much.
We'll we're in deep now, the best we can do is bide our time, so why not find out if we actually fall into the black hole, ey? How to go about this you ask, well to find out that we're going to have to figure out a couple of things;
does the black hole spin?
how much is our total energy?
and as we'll se we need to find the magnitude of the aformentioned spin of the black hole in order to find this.
Let's take it one step at a time then, why does the black hole need to spin? Well, that has to do with the garavitational potential around the black hole. You can see a depiction to the left, and we'll touch a bit on the mathmatics as to why it is this way, but what we need to know for now is that when the black hole has a spin, you require energy in order to enter into it.
How much energy you ask? Well to determine this we'll first look at a more mundane neutonic example; the potential energy of a slope with no friction. We'll still be able to use the graph, but imagine it's a slope for now. In order to cross the peak of it, also known as the critical radius, you need to enter with a high enough velocity. In other words you need the total energy to be higher than the potential energy. The total energy in a system can be expressed as \(A = B\vec{\dot x} + V(x)\) (1), where V(x) is the potential energy we're after, I go into more detail of this in the calculations and therefre won't touch on the process too much here. Once we found an expression for the total energy: \(\frac E m\), we can use it to fin an expression for \(\frac{\Delta r}{\Delta \tau}\), which we can then form into the shape above (1). Giving us the formula for the potential energy: \(V_{eff} = \sqrt{(1-\frac{2M}r)[1+(\frac{L/m}r)^2]}\). Great, now we know how the gravitational potential varies with our distance (r) to the black hole, now we wish to find the critical radius, which is found by deriving the potential and setting it equal to 0: \(r_{extrem}=\frac{(L/m)^2}{2M}(1 \pm\sqrt{1-\frac{12M^2}{(L/m)^2}})\). We can tell from the graph, that of the two values we get from this, the smallest one (the one farthest to the left / when we subtracet) needs to be the maximum. We also find an expression for L/m, using a few cross products and relativistic momentum to arrive at \(L/m = R\gamma_{sh}v_{sh}sin\theta\). Now we're going to use the tiny bit of power remaining to compare our total energy, which we can find from our total energy by the shell, which is the area 20M away from the mass centre where we lost our thrust. You can think of the shell as a shpere which dictates what frame of reference within the accelerated field you will have, and within any given shell you'll have a special relativistic frame. In this 'shell' it's a lot easier to find various components needed for the total energy, so let's look at how our simulation turned out shall we?
By the good wobbly one who watches over us from above, How much time do we have left? Well about that, we're swiftly approaching the event horizon so you might want to find that out kinda sharpish!
To find the time it takes for us astronauts to traverse from the event horizon into the singularity, we firstly assume the spin is now zero. We're gonna be falling straight down baby. We have an expression for \(\Delta r\) with \(\Delta \tau\) in it which we used to find the potential earlier.
If we now move \(\Delta \tau\) on its own and move the rest of the expression over to the opposite side we get \(\tau = \int_\limits{0}^{2M}\frac 1 {\sqrt{(\frac E m)^2 - (1-\frac{2M} r)}} dr = \int_\limits{0}^{2M}\frac 1 {\sqrt{(1 - \frac{2M} r) \gamma^2 - (1-\frac{2M} r)}}dr\) which with some fansy numerical integrals gives us: 0.242M = 4.7592s.
In other words, we don't got long left chief.
Well I guess all that is left now is to describe in excruciating detail what is going to happen to us as we pass over the event horizon?
At this point we can only travel towards the centre, and as we do the gravitational pull will become increasingly stronger the closer to the singularity you get. At a certain point the difference in gravitational pull from your head to your feet will become so great that your body is stretched out faster and faster like a strand of spaghetty, which has the charming name of spaghettification. Put simply we'll enjoy our final seconds.
I will see you folks on the other side