Two Questions on Black Holes

curiousasker

Hey all! Two things came to my mind about black holes, and I hope some of you can help me out :)

1. Say you have a non rotating, chargeless black hole of a certain size. Its radius will be R.

If you go through the black hole, from your perspective then time will be fine, right? Then that generalizes to everything going into a black hole will experience time normally, from its perspective.

So say you have a gigantic planet of radius 10R going into the black hole, moving very very fast. What happens to the planet?

2. I've read about quantum entanglement not being able to transfer information when across the event horizon because it doesnt transmit information, only noise, since you can't control the result of your measurement.

But what if I took 52 particles and entangled them into 26 pairs. Then I take 26 of those particles with me inside the event horizon of a supermassive black hole.

I have set up a system (that the other person with the other 26 particles knows) wherein if I measure the spin of the first particle, that's letter A. And if I measure the spin of the second, that's B. And so on.

So, within the event horizon, I could theoretically describe what it feels like inside the black hole. By measuring the spins of the particles in a specific order, it won't matter to me if the result is up or down. The other person just has to know what letter each particle represents and in which order I measured them (which, in terms of the two particle spins being forced into one spin and another anti spin, seems to be simultaneous, defying light speed). I could spell out "its dark" and that is definitely information being transfered from inside the event horizon to outside it.

What are your takes on this procedure?

Nugatory

Entanglement doesn't work that way. You cannot tell whether the other guy has actually measured the spin, you just know that when does (whether before or after you've measured), he'll get the opposite result from you. So.... no message transmission.

bcrowell

What do you mean by "go through?" What do you mean by experiencing time normally?

Why do you refer to a "gigantic" planet? Even a small planet is much, much bigger than a the event horizon of a typical black hole.

The result of the planet's encounter with the black hole would depend on the relative speed. The planet would be ripped apart, but some material might escape.

DaleSpam

Time is fine, but what happens to the planet depends on the tidal forces. Smaller black holes have larger tidal forces at the horizon, and you appear to be describing a reasonably small black hole. So the planet would probably be torn apart due to the tidal forces.


If it doesn't transmit information then it cannot transmit information. Throwing it down a black hole doesn't make something that didn't work suddenly start working.

[QUOTE=curiousasker;4231296]

curiousasker

Hi all. Thanks for the answers. I'd just like to clarify my questions :)

Nugatory - Maybe, but I have measured the spin, if and only if you will get the opposite. Which means, if you get the opposite, I must have measured the spin.

bcrowell - Go through means, if the black hole is an egg and the planet is a basketball, I will throw the basketball at the egg very quickly.

Experiencing time normally means the proper time from the perspective of someone who is on the planet.

Also, the part about relative speed - that's why I asked what if the planet went into the black hole very quickly. Would a sufficiently large speed cause the black hole to burrow into the planet while the rest of the planet escapes?

DaleSpam - I realize that if a planet comes close to a black hole, the black hole will eat the planet. But that's why I asked what if the planet comes hurtling into the black hole at a great speed. What would happen then? And from the perspective of someone on the surface of the planet, what would it look like when the planet collides with the black hole?

Also, I do believe your assertion that using entanglement to transmit not working, doesn't provide an easily understandlable counterpoint to my set up. I was asking why it will or will not work. Imagine this graphical example:

O = the particle has not been measured yet
U = the particle has an up spin
D = the particle has a down spin

I brought with me 5 x 26 = 130 entangled particles, segragated into 5 groups of 26. Each group will represent a letter, and what letter that is will be determined by which place they have on some order I predetermined with my partner.

So when I bring the particles with me inside, it will look like this.

OOOOO OOOOO OOOOO OOOOO OOOOO O
OOOOO OOOOO OOOOO OOOOO OOOOO O
OOOOO OOOOO OOOOO OOOOO OOOOO O
OOOOO OOOOO OOOOO OOOOO OOOOO O
OOOOO OOOOO OOOOO OOOOO OOOOO O

Now, from inside, I measure specific particles, but only one from each group.

OOOOO OOUOO OOOOO OOOOO OOOOO O
OOOOD OOOOO OOOOO OOOOO OOOOO O
OOOOO OOOOO ODOOO OOOOO OOOOO O
OOOOO OOOOO ODOOO OOOOO OOOOO O
OOOOO OOOOO OOOOU OOOOO OOOOO O

When the partner decodes it, it will look like this:

ABCDE FGHIJ KLMNO PQRST UVWXY Z

OOOOO OOUOO OOOOO OOOOO OOOOO O
OOOOD OOOOO OOOOO OOOOO OOOOO O
OOOOO OOOOO ODOOO OOOOO OOOOO O
OOOOO OOOOO ODOOO OOOOO OOOOO O
OOOOO OOOOO OOOOU OOOOO OOOOO O

Translated into: Hello.

Why won't that set up work? Why specifically? Is there some good, solid argument for or against it?

DaleSpam

Regarding the entanglement communication idea, this has nothing to do with black holes, it doesn't work in everyday flat spacetime either. If you and I have a bunch of entangled particles and if you measure them first then you get a bunch of random values. If I measure them first and then you measure them you get a bunch of random values. If I measure half first and then you measure you get a bunch of random values. Regardless of what I do when you measure you get a bunch of random values.

DennisN

DaleSpam said it nicely above, but I'd like to make an addition.
Entanglement does not work like that. All you (A) can do with your entangled particle pA is measure it; you can not force it into taking a particular value. And you can not tell whether the other particle pB has been measured or not, unless you communicate with the other guy B in a classical way. The same goes for guy B (in this case outside the black hole). This means there is no way for A and B to communicate via entangled particles.

I'll try to describe the issue with a very simple analogy (although I'm usually not fond of this particular analogy when discussing entanglement, but in this case I think it works);

Imagine you have two persons A and B at different locations in space, lightyears apart, and you have told them you will be sending them notes in a particular manner; you will take two pieces of paper and on one paper you will write "Up" and on the other paper you will write "Down". You put the papers in two envelopes and sends them off with two rockets to the two persons. When person A receives his envelope, he opens it and reads either "Up" or "Down". He then instantly knows that when and if person B reads his note, he will read the opposite (and the same goes for B, of course). Now the questions are: (1) has any information been transmitted between A and B? and (2) can this setup be used for communication betweeen A and B? .....

The answers are no, and no, of course not. It doesn't matter how many envelopes or notes you send, this setup will simply not work. Another interesting way to look at it is that you can "lie"; if you send two notes both saying "Up", none of the recipients would be able to know something was wrong until they compared their notes with eachother in a classical way .

Nugatory

This part of the discussion might be better off in the quantum mechanics section, but the reason why it won't work is that when you "look" at the particle you will never observe the O state, only U or D. The only way to know whether the particle's spin is in the O state is to measure it... And if you measure it, it's not in the O state any more.

For example, if I look at one member of the entangled pair and I see a D, I have no way of knowing whether:
1) both particles were in the O state, and now that I've seen a D we know that the other guy will see a U if he looks at his particle; OR
2) both particles had been in the O state, but the other guy has already looked at his particle and seen a U, so I had to see a D when I looked at my particle.

curiousasker

Hey all. Thanks for the response. I understand what you're saying. Nugatory's comment, in particular, helped me realize you could never observe the O state, which was where I was stumbling on.

So now that question 2 has been answered, anybody have any ideas on question 1?

If you were standing on the edge of a planet of radius 10R, hurtling very very fast towards a black hole of radius R, what would you see? As you enter the event horizon and the rest of the universe is lensed into a circle behind you, would the entire planet go inside with you, or will only chunks of it get in?

If it helps, imagine that we have a really, really big black hole (so that the event horizon is a few hours away from the singularity), but a really really big planet as well (how it formed isn't a concern - maybe we took a large amount of planets and merged them together to make a sufficiently large planet).

It seems absurd to think of a planet that is as big as that, but I think it will help us think in "slow motion" better. Small black holes will, as DaleSpam said, rip the planet apart fairly quickly. It happens too fast to appreciate what's going on inside. But a huge black hole with an even bigger super planet could survive just long enough for some speculation.

pervect

An actual planet won't be strong enough to hold together - tidal forces will rip it apart.

Depending on the size of the black hole, the planet might not be terribly large. A solar mass black hole would have a Schwarzschild radius (which isn't quite the same thing as a diameter) of 3km.

So, we can make the problem more definite by considering a 200 solar mass black hole (Schwarzxschld radius of 600km) and a 6000km radius planet (about the size of the Earth).

Tidal forces from the black hole are given by 2GM/r^3 (very simlar to the newtonian formula) , so at the event horizon (r = 600km) I get tidal forces on the order of 10,000 g's (10^5 m/s^2) per meter.

One lesson you should learn from special relativity that also applies to general relativity is that there is no such thing as a rigid object.

I'm not sure about the details of what happens after the planet is torn apart by tidal forces - it'd be a messy calculation. You can certainly calculate the trajectory of any piece of the planet you want, using the formulae from http://www.fourmilab.ch/gravitation/orbits/, but that wouldn't account for deviations that would occur due to the interactions of the pieces.

So, I'll stick with "torn apart by tidal forces" as a sufficient answer. Some of the pieces might orbit in an accretion disk for a while. I believe the accretion disk spirals in slowly as it radiates energy, but I haven't seen any detailed calculations or papers on the topic of exactly how they behave. (I'm not sure it's even known - I think there are some observed "beaming" phenomenon that people don't quite understand yet.)

SJBauer

My take on you procedure is that it is fun science fiction - great imagination.

Per Question 1 assumption, ” If you go through the black hole, from your perspective then time will be fine, right?” You cannot go through a black hole. You can only go into a black hole. Perhaps you are confusing it with the wormhole theory. As for time, time is increasingly dilated toward increasing gravitational acceleration. As for Question 1 itself, “What happens to the planet?” The planet will be torn apart and the matter will be reorganized with the black hole (whether the black hole is rotating or not).

Per Question two assumption, “So, within the event horizon, I could theoretically describe what it feels like inside the black hole. By measuring the spins of the particles in a specific order, it won't matter to me if the result is up or down.” The inertial frame of reference is drastically different from one in which quantum entanglement has been observed. Any mass travel across the Event Horizon of a black hole is reorganized at an atomic level and maybe even at a quantum level.

PeterDonis

A couple of notes:

With reference to objects being torn apart by tidal forces, that depends on the relative sizes of the hole and the object. Pervect's answer was correct for the parameters given in the OP, but if the hole were large enough, an Earth-sized planet could fall into it without being torn apart.

As far as matter being "reorganized" at a quantum level, with our best current understanding of how quantum mechanics interacts with GR, that's speculative. One line of thought says that quantum effects provide large corrections to classical behavior at the horizon even for very large holes; another line of thought says that a large enough hole has curvature at the horizon that is small enough that the classical model is a good approximation. This is one of the main unresolved questions that awaits an actual theory of quantum gravity.

SJBauer

PeterDonis - Since we are all talking theory anyway - While I appreciate the math behind the explanation of tidal forces within a black hole, there is more than just tidal forces to deal with. My statement was directed toward the reorganization of mass/matter within the accretion disk of a black hole.

Incidentally, the concept of any object of positive mass density surviving at light speeds is very controversial at best. It is well known that photons travel at light speed because they are close to massless particles. Ergo, to travel at the speed of light requires that the positive mass object shed its mass or transition to a super dense state that would be able to withstand disintegration within the accretion disk of a black hole. Both are highly unlikely without the reorganization of mass at an atomic level.

pervect

That's a good point. If you have a sufficiently big black hole, an Earth-sized planet can fall into it.

The OP was asking, however, about a planet that was 10x as big as the black hole. I'm not quite sure of his motivation for asking this, but I thought it would be best to take one specific scenario with a reasonable sized planet and analyze it in detail.

I think that tidal forces will probably always be significant given the constraints of the problem as originally stated (a planet 10x as big as the hole), but I haven't worked this out in detail, my intuition could be playing me wrong.

I think I should also point out to the OP that one cannot "walk" across an event horizon. I suspect this could be one of the motivations of the original question (or maybe not).

Anyway, (and I'm sure Peter already knows this, it's for the benefit of the OP), an event horizion is a lightlike surface, so from the point of view of someone falling into a black hole, they are standing still and the event horizon approaches them at c.

(Actually, this is only exactly true for the Killing horizion, I think, but it should be a good enough apprxoimation for most of the various sorts of horizon).

DaleSpam

I have the same intuition. Certainly, the size of the object is important for tidal forces ( http://en.wikipedia.org/wiki/Tidal_f...ical_treatment ). So I would think that constraining the planet to be large wrt the horizon would constrain the tidal forces to be large wrt the planet.

PeterDonis

I agree; I think that for tidal forces to be negligible at the horizon for an infalling object, the size of the object must be much less than the horizon radius (or the horizon circumference, if we want to avoid any issues with the definition of the radial coordinate). I think this may be an exercise in MTW; when I get a chance I'll dig out my copy and see if I can find it.

PeterDonis

Ah, ok; that is outside the horizon, so it's different from what I was talking about.

Only massless objects can travel *at* the speed of light. But objects with nonzero invariant mass can travel very, very close to the speed of light; particle accelerators routinely accelerate particles to speeds within 1 part in a million or better of the speed of light.

Also, you use the term "mass density" when you should be using the term "invariant mass". They are not the same thing. Invariant mass is what has to be zero for objects moving at the speed of light. A system composed of such objects can still have a positive mass density, however.

I'm not sure, but it seems like you believe that an object must travel at the speed of light to fall through the event horizon of a black hole. That's not correct; objects with nonzero invariant mass can fall into a black hole just fine, and they don't have to reach the speed of light to do so. If there is an accretion disk, of course they have to make it through the accretion disk, but even if a macroscopic object gets "reorganized" by this process, the individual particles making it up, which will have nonzero invariant mass, can survive just fine and end up falling through the event horizon of the hole.

An infalling object will see the horizon moving past it at the speed of light, but that is because the horizon itself is an outgoing null surface: the *horizon* is moving outward at the speed of light.