# When do I cross the event horizon?

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## Main Question or Discussion Point

@Grinkle recently asked a question about detecting crossing the event horizon, which got me thinking. I think that, at least in principle, I can deduce when I cross an event horizon with a "closed box" experiment, basically by measuring tidal forces. I'm planning to see if the maths works, but there's enough numerical work in doing so that I'd like to check I've not gone off the deep end before I try it.

To make things a little more tractable I'm going to assume a simple case - I was released from rest at some finite distance from a Schwarzschild black hole. My release radius was $R_u$ (u for unknown) and I was released at my proper time $-\tau_u$ (another unknown). It's messy, but I can actually write down a closed form expression for $\tau(r,\tau_u,R_u,R_S)$ and $t(r,R_u,R_S)$ (there's no $t_u$ because I can set the zero of $t$ at any time I like) in this case of purely radial motion. Sadly, they cannot be inverted to obtain $r$ as a function of $t$ or $\tau$, nor an explicit relationship between $t$ and $\tau$.

Inside my box, I have a free-floating radar set and a rocket. I release the radar set as near the center of mass of the box as I can, and the rocket a short distance away in the direction of the hole (which I can determine by releasing a cloud of test particles and watching for elongation of the cloud). Now I program the rocket to fire in order to maintain constant radar distance from the radar set, and measure the necessary thrust.

I think I can calculate (numerically) the worldline followed by the rocket. It'll be parameterised by the radar set's proper time $\tau$, using simultaneity established by radar. Given that, I can determine the proper acceleration of the rocket as a function of $\tau$, $\tau_u$, $R_u$ and $R_S$. So if I make three measurements of the proper acceleration at known proper times, I can get the black hole parameter and my "orbit" parameters. Barring degeneracy in the maths, I can then tell you the proper time corresponding to me crossing the horizon.

Am I nuts?

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## Answers and Replies

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PeterDonis
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I think that, at least in principle, I can deduce when I cross an event horizon with a "closed box" experiment, basically by measuring tidal forces.
If you can measure tidal forces, your closed box is large enough that it cannot be covered by a single local inertial frame. So even if your setup does allow you to tell when you cross the horizon, that doesn't contradict anything said in the other thread, because in the other thread, when people said nothing inside the spaceship could be used to tell when you crossed the horizon, that was assuming the spaceship was small enough that it could be covered by a single local inertial frame, so tidal forces would not be detectable inside it.

Note that tidal gravity at a black hole's horizon gets smaller as the mass gets bigger. So for a hole of one solar mass, for example, a local inertial frame would have to be much smaller than a kilometer in size for tidal gravity to be undetectable; a spaceship of ordinary size is probably large enough to detect the tidal gravity near the horizon of such a hole. But for the supermassive hole at the center of our galaxy, the scale on which tidal gravity is detectable would be a million times larger; an ordinary spaceship is small enough that tidal gravity would probably not be detectable inside it.

(These estimates are actually too generous, because for your scheme to work, you need to be able to measure tidal gravity not just at the horizon but well above it. That might be problematic for an ordinary sized spaceship even for a solar mass hole.)

I was released at my proper time $\tau_u$ (another unknown).
This isn't an unknown; you can measure it by looking at your clock. The simplest way is to just set your clock to zero when you are released; then $\tau = 0$ at release.

(there's no tutut_u because I can set the zero of $t$ at any time I like)
Sure, just set $t = 0$ when $\tau = 0$.

the direction of the hole (which I can determine by releasing a cloud of test particles and watching for elongation of the cloud)
This doesn't tell you which way along the direction of elongation the hole is. Also, the elongation is an effect of tidal gravity, so tidal gravity has to be detectable inside your closed box for this to work.

Fortunately, this doesn't matter because the rocket can be on either side of the center of mass (towards or away) and still give you a measure of tidal gravity. Or you could have one on each side, so each can serve as a check on the other.

I think I can calculate (numerically) the worldline followed by the rocket.
I assume you mean as a function of $R_u$ and $R_S$? Yes, of course you can; just solve the geodesic equation with those parameters left undetermined. Actually, I think what will go into the geodesic equation is the ratio $R_u / R_S$.

So even if your setup does allow you to tell when you cross the horizon, that doesn't contradict anything said in the other thread, because in the other thread, when people said nothing inside the spaceship could be used to tell when you crossed the horizon, that was assuming the spaceship was small enough that it could be covered by a single local inertial frame, so tidal forces would not be detectable inside it.
Indeed - I'm aware that I'm attempting to detect the failure of that assumption. I just wondered if there's an obvious reason it wouldn't work.
That might be problematic for an ordinary sized spaceship even for a solar mass hole.
You might be right - this old post quotes an "ouch radius" with a 1g difference over a 2m human which implies the ouch radius equals the Schwarzschild radius for a hole of mass around 1025kg, only slightly larger than the mass of the Earth, let alone a stellar black hole.

Edit: $c^3/G\sqrt g$≈1035kg, or about 50,000 solar masses, still a smallish black hole. I really must double check my mental arithmetic.
This isn't an unknown; you can measure it by looking at your clock. The simplest way is to just set your clock to zero when you are released; then $\tau = 0$ at release.
Sure - I was leaving open the possibility I didn't know the correct zeroing for my clock for some reason.
Actually, I think what will go into the geodesic equation is the ratio $R_u / R_S#d. I've left my notes at home but I think there were a few$R_S$not attached to an$R_u##. I'll have a look.

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stevendaryl
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