Black Hole Event Horizon: Atom Effects & Wave Detection

[accdd]

Are the atoms of objects ripped off when they cross the event horizon?
Does a metal rod that partially crosses the event horizon maintain its lattice structure of atoms?
If I put a sound wave generator on the end of the bar that has crossed the event horizon how far can the waves be detected?

 

[PeroK]

Are the atoms of objects ripped off when they cross the event horizon?

Why would they?

 

[Nugatory]

Are the atoms of objects ripped off when they cross the event horizon?
Does a metal rod that partially crosses the event horizon maintain its lattice structure of atoms?

It depends.
If we're considering a free-falling rod what matters is the tidal force, the difference between the force on bottom end and the top end (you can get an intuition for how this works by considering tidal forces in ordinary Newtonian gravity). The more massive the black hole the less the tidal force (the forces on the two ends will be greater but less different); the shorter the rod the less the tidal forces will be. Thus you could approach and fall through the event horizon of a sufficiently massive black hole without even noticing, and a bacterium might be able to survive passage through the event horizon when a person would be torn apart well before they reach the horizon.

If we're trying to hold the rod stationary instead of letting it fall, the force required to hold any given atom in place against gravity increases as we get nearer to the horizon and is infinite at the horizon. No matter how strong the rod and how powerful the force we use to hold onto the upper end, if we try lowering the rod towards and through the horizon the bottom end will be torn away.

If I put a sound wave generator on the end of the bar that has crossed the event horizon how far can the waves be detected?

Assuming that the bar remains intact (the free-fall, short enough bar, massive enough black hole case above) the sound wave will always reach the upper end of the bar, but not before the entire bar has passed through the horizon.

 

[Dale]

Does a metal rod that partially crosses the event horizon maintain its lattice structure of atoms?

You will have to be more specific. Is the outer end being held fixed? Is the rod free falling? Is the rod long or short relative to the Schwarzschild radius?

 

[Grinkle]

not before the entire bar has passed through the horizon.

I know the EH is not anything detectable, please bear with my phrasing to the contrary - I hope my question makes sense.

Is it the case that the bar in free-fall sees the EH approaching it at c as it is crossing the EH? In other words, the EH traverses the free-falling bar locally at c - so no signal, sound or otherwise which originated from the end of the bar that has crossed the EH, can leave the observable end of the bar before that observable end has crossed the EH. Is that correct?

 

[jbriggs444]

Is it the case that the bar in free-fall sees the EH approaching it at c as it is crossing the EH? In other words, the EH traverses the free-falling bar locally at c - so no signal, sound or otherwise which originated from the end of the bar that has crossed the EH, can leave the observable end of the bar before that observable end has crossed the EH. Is that correct?

Yup.

 

[cianfa72]

Assuming that the bar remains intact (the free-fall, short enough bar, massive enough black hole case above) the sound wave will always reach the upper end of the bar, but not before the entire bar has passed through the horizon.

That's because no whatever signal/physical process can cross from the EH's inside to the EH's outside, I believe.

 

[Demystifier]

Are the atoms of objects ripped off when they cross the event horizon?

Short answer - no.

Does a metal rod that partially crosses the event horizon maintain its lattice structure of atoms?

Short answer - yes.

If I put a sound wave generator on the end of the bar that has crossed the event horizon how far can the waves be detected?

All the way to the second end.

 

[Vanadium 50]

As @Demystifier says, the short answer is "no". If I have a large enough hole, I can make the tidal forces at the horizon as small as I would like.

If you are thinking "yeabut..."

To ionize an atom, you need a high tidal force between a neutron and a proton. That means you need a small BH. Roughly the mass of a small planet or large moon. It us not certain that these even exist.

It is not clear to me what it means quantum mechanically to say "the electron is closer to this pioint than the proton". (uding a hydrogen atom as an example. I am certain it is possible to formulate this sensibly, but don't think tthis is it. It probably involves solving the Scroedinger Equation in parabolic coordinates. This is not something that can be done at B-level and probably not I-level.

If an electron falls into a neutral BH, it is no longer neutral. That means it is no longer a Schwatzchild BH, but` a Kerr BH. The event horizon will move. Enough to capture the proton too? Perhaps. This is definitely A level.

 

[Ibix]

If an electron falls into a neutral BH, it is no longer neutral. That means it is no longer a Schwatzchild BH, but` a Kerr BH

Reissner-Nordstrom or Kerr-Newman, to be precise. Kerr is uncharged but rotating; Reissner-Nordstrom is charged but unrotating; Kerr-Newman is charged and rotating. I expect the hole would pick up angular momentum (however tiny) in this exchange too.

(Strictly all of these are eternal black holes so it's actually none of the above, but I suspect a black hole absorbing a single electron would settle down to indistinguishable from an eternal hole in a very short timescale.)

I am unclear on how reliable we believe our models of black holes swallowing quantum mechanical objects to be. The behaviour of quantum fields near event horizons is the basis of Hawking radiation, but I know little more than that. And that's definitely not B level, I agree.

 

[PeterDonis]

Are the atoms of objects ripped off when they cross the event horizon?
Does a metal rod that partially crosses the event horizon maintain its lattice structure of atoms?
If I put a sound wave generator on the end of the bar that has crossed the event horizon how far can the waves be detected?

All of these questions appear to me to have a hidden assumption that is wrong. The hidden assumption is that the event horizon is a place in space. That's wrong; it's not. It's a surface composed of radially outgoing light rays. That means that, from the viewpoint of an object falling into the black hole, the horizon moves outward across the object at the speed of light. (See post #5.)

I suggest rethinking all of your questions based on the above.

 

[pervect]

One thing should be clarified. The event horizon of a Schwarzschild black hole is a null surface. Another example of a null surface is the surface of a flash of electromagnetic radiation. Or any other surface generated by something traveling at the speed of light.

It is somewhere between a little and a lot misleading to think of any null surface as a "place" when it is really a null surface. A null surface is different than a spatial surface.

I'm reasonably sure the original poster here is trying to think of the event horizon as a place, i.e. as it it were a surface in space, which is generating some confusion. I'm not sure that my explanation of the problem will make sense to them, but I don't know how to say it better or more simply.

 

[Demystifier]

event horizon is ... a surface composed of radially outgoing light rays.

By "light rays", do you mean mathematical geometrical objects or actual physical light carrying energy?

 

[Ibix]

By "light rays", do you mean mathematical geometrical objects or actual physical light carrying energy?

@PeterDonis is referring to just the geometry, but in B-level-friendly terms. Technically, it's a null surface, and all of these models are vacuum spacetimes so there's nothing actually there.

 

[cianfa72]

As the OP I've some difficulty to grasp the intuitive concept of null surface. For $r > 2M$ the locus $r=\text {const}$ is a 3D hypersurface parametrized by $(t,\theta, \phi)$ and it is the set of timelike worldlines of hovering observers at fixed $(\theta, \phi)$. What about $r=2M$ 3D hypersurface? At fixed $(\theta, \phi)$ it is an outgoing light ray, so it should move towards increasing values of coordinate radius $r$, do you ?

 

[DrGreg]

As the OP I've some difficulty to grasp the intuitive concept of null surface. For $r > 2M$ the locus $r=\text {const}$ is a 3D hypersurface parametrized by $(t,\theta, \phi)$ and it is the set of timelike worldlines of hovering observers at fixed $(\theta, \phi)$. What about $r=2M$ 3D hypersurface? At fixed $(\theta, \phi)$ it is an outgoing light ray, so it should move towards increasing values of coordinate radius $r$, do you ?[/i]

All motion is relative to something. "Outgoing" is relative to a local inertial object (i.e. freefalling into the hole).

You may find it useful to look up Rindler horizons, which are null surfaces in flat spacetime (i.e. in the absence of gravity) and the associated Rindler coordinates (which are analogous to Schwarzschild coordinates -- a Rindler horizon is a surface in which one of the Rindler coordinates is constant).

 

[Nugatory]

What about $r=2M$ 3D hypersurface? At fixed $(\theta, \phi)$ it is an outgoing light ray, so it should move towards increasing values of coordinate radius $r$, do you ?

This is where the “tipping light cones” model really helps. It’s not right (any attempt to draw in Euclidean space a two- or three-dimensional picture of curved four-dimensional space is a lost cause) but it does help a lot with intuition about outgoing light at the horizon.

 

[pervect]

The following expansion on my original comment may be of some help. To recap my original point briefly, the event horizon of a black hole is a null surface, and not a place. The key difference between a place and a null surface is that you can stop at a place. But you can't stop in the same manner at a null surface - the null surface is light-like, and no matter can move at the speed of light. Thus, it is impossible to have a material object stationary at the event horizon. Any object at the event horizon will have a velocity of "c", the speed of light, relative to the horizon, because the horizon is a null surface, not a location in space.

One of the interesting, though mildly obscure, features of special relativity is that while no object can move as fast as light, it is possible for an accelerating object, given a sufficiently high acceleration, and a sufficiently large head start, to stay "ahead" of a light beam indefinitely, as long as it keeps accelerating.

This is essentially the way that an object can avoid falling into a black hole. For a sufficiently large black hole, the effects of space-time curvature can be ignored as a first approximation, and avoiding the event horizon is just like outrunning a light beam. This takes a very high acceleration - the magnitude of the acceleration required is c^2/d, c being the speed of light, and d being the distance one wants to stay ahead of the event horizon. Thus it takes an acceleration of about 9*10^16 meters/second^2 to "hover" at one meter above the event horizon of a very large black hole, which is the same acceleration it would require to stay ahead of a light beam in the flat space-time of special relativity given a 1 meter head start.

An acceleration of 10 m/s^2, in contrast, requires approximately a 1 light year head start. More exactly, given a 1 light year head start, one needs to accelerate at 9.5 meters/second^2 to stay ahead of a light beam. Equivalently, to "hover" one light year away from the event horizon of a sufficiently large black hole, one needs to accelerate at 9.5 meters/second^2, approximately 1 Earth gravity.

Note that for smaller black holes one does need to account for the additional effects of space-time curvature, but the simplest case to understand the problem is to imagine a very large black hole so that one can ignore the curvature issues.

 

[cianfa72]

Sorry, on the $\Sigma$ null hypersurface $r=2M$ the Schwarzschild metric even if singular becomes: $$ ds^2 = 0dt^2 + 4M^2 (d\theta^2 + sin^2\theta d\phi^2) $$ That means the vector field $\partial_t$ is a null vector (its length is zero). Now $\partial_t$ belongs to the tangent space $T_p\Sigma$ and at the same time is orthogonal to itself and to the vector fields $\partial_{\theta}$ and $\partial_{\phi}$. So $\Sigma$ is an hypersurface such that at each $T_p\Sigma$ there is a basis of a null + 2 spacelike vectors.

In other words a null hypersurface is defined as an hypersurface such that the normal vector to each $T_p\Sigma$ is a null vector (note that the normal null vector belongs itself to each $T_p\Sigma$).

 

[martinbn]

Sorry, on the $\Sigma$ null hypersurface $r=2M$ the Schwarzschild metric even if singular becomes: $$ ds^2 = 0dt^2 + 4M^2 (d\theta^2 + sin^2\theta d\phi^2) $$ That means the vector field $\partial_t$ is a null vector (its length is zero). Now $\partial_t$ belongs to the tangent space $T_p\Sigma$ and at the same time is orthogonal to itself and to the vector fields $\partial_{\theta}$ and $\partial_{\phi}$. So $\Sigma$ is an hypersurface such that at each $T_p\Sigma$ there is a basis of a null + 2 spacelike vectors.

In other words a null hypersurface is defined as an hypersurface such that the normal vector to each $T_p\Sigma$ is a null vector (note that the normal null vector belongs itself to each $T_p\Sigma$).

Why are you saying this? Are these questions or statements? Don't get me wrong, what you say is correct but it is well known. The definition of a null hypersurface (or the causal character of a hypersurface) is in any GR book. It even has a wiki article https://en.wikipedia.org/wiki/Null_hypersurface

 

[cianfa72]

The definition of a null hypersurface (or the causal character of a hypersurface) is in any GR book. It even has a wiki article https://en.wikipedia.org/wiki/Null_hypersurface

Yes, my doubt was about spacelike vectors belonging to the tangent space $T_p \Sigma$ at each point on it. From your wiki link all vectors in such a tangent space are spacelike except in one direction, in which they are null.