This question came to my mind after reading this.
I think it is safe to assume that the supposedly thrown camera would not reach the event horizon in one piece. So my question is :
If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon, and if so, why?
If the black hole has no surrounding matter, so that there is no violent radiation generated by accretion or the like, then it still depends on the mass of the black hole. If it is in the thousands of solar masses or more, it is possible for a realistic camera to survive free-fall to the horizon. This is essentially a non-issue for supermassive black holes. On the other hand, smaller black holes are quite a bit more punishing.
For Newtonian gravity with potential $\Phi$, in a free-falling frame, a particle at $x^k$ near the origin of the frame will be accelerated at $$\frac{\mathrm{d}^2x^j}{\mathrm{d}t^2} = -\frac{\partial\Phi}{\partial x^j} = -\frac{\partial^2\Phi}{\partial x^j\partial x^k}x^k\text{,}$$ where the the second derivatives of the potential, $\Phi_{,jk}$, form the so-called tidal gravitational field. Since $\Phi = -GM/r$ for a point-source, you should expect the tidal forces on a free-falling object to be proportional to $GM/r^3$ times the size of the object. Thus, at the Schwarzschild radius, this is on the order of $c^6/(GM)^2$.
Of course, black holes are not Newtonian. However, it turns out that for the non-rotating, uncharged (Schwarzschild) black hole, radial free-fall of a test particle has the same form as in Newtonian theory, except in Schwarzschild radial coordinate (not radial distance) and proper time of the particle (not universal time), so the above is essentially correct even for Schwarzschild black holes.
To be relativistically correct, the tidal forces on an free-falling object are described by the equation of geodesic deviation, in which the gravitoelectric part of the Riemann curvature provides forms the tidal tensor: $$\frac{\mathrm{D}^2x^\alpha}{\mathrm{d}\tau^2} = -R^\alpha{}_{\mu\beta\nu} u^\mu u^\nu x^\beta\text{.}$$ In Schwarzschild spacetime, this turns out to be $+2GM/r^3$ in the radial direction, stretching the free-falling object, and $-GM/r^3$ in the orthogonal directions, squeezing it. This stretching-and-squeezing due to gravitational tidal forces is sometimes called spaghettification.
Some example numbers: say the camera size is on the order of $0.1\,\mathrm{m}$. The the following are the approximate tidal accelerations near the horizon for black holes of different multiples of solar masses:
The curvature around a rotating black hole is more complicated, but the moral of the story is basically the same.
If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon?
I think so. I also think it will be destroyed before tidal forces have any effect.
If so, why?
Because it would otherwise end up falling faster than the speed of light.
Sounds strange I know, but take a look at the Shapiro delay: "the speed of a light wave depends on the strength of the gravitational potential along its path". Or see Professor Ned Wright's Deflection and Delay of Light: "in a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light":
Alternatively see this PhysicsFAQ article by Don Koks: "this difference in speeds is precisely that referred to above by ceiling and floor observers". He's referring to the way Einstein said light curves because the speed of light varies with position. See the Einstein digital papers for examples of that. Here's one from 1920, see the second paragraph:
Light doesn't curve because spacetime is curved. Einstein never actually said that. It curves because the speed of light is lower at a lower altitude, rather like the way sonar waves curve:
As to why this isn't common knowledge, I don't know. There's this myth that Einstein gave up on a varying speed of light in 1911, but he didn't, see this Wikipedia article and this example from 1914. I don't know why the reason matter falls down isn't common knowledge either. You know about pair production and electron diffraction and the wave nature of matter, just think of an electron as a wave going round a closed path, then simplify it to a square path, like this:
The horizontals bend downwards, so the electron falls down. Could it be any simpler? Anyway, your TV camera falls down because the speed of light is reducing with altitude. We tend to call this the "coordinate" speed of light nowadays, even though Einstein just called it the speed of light. But regardless of what we call it, you don't have to be the Brain of Britain, or the Brain of France, to work out that half down, there's some kind of problem. At this point the camera will be falling as fast as the "coordinate" speed of light at that location. And it isn't going to slow down. Things always fall faster, not slower. But matter can't go faster than light, because of the wave nature of matter. When matter is made of waves, there is no way that matter can move faster than the speed of those waves.
So what's going to happen? I can see no other option: that wave has to break. Again it sounds strange, but take a look at an old version of the firewall article on Wikipedia. Follow the reference 7 link to Friedwardt Winterberg's Gamma-Ray Bursters and Lorentzian Relativity: "If the balance of forces holding together elementary particles is destroyed near the event horizon, all matter would be converted into zero rest mass particles which could explain the large energy release of gamma ray bursters". I think this is what happens to your TV camera. Flash! It turns into a gamma-ray burst. It would be like an atom bomb, but a lot more efficient. So make sure you drop it from a safe distance.
