Can we communicate with people outside a black hole?

[cragar]

My teacher today was saying that you can't communicate with people outside of a black hole.
But what if you went in on a spaceship and then shot off rockets inside like 1 every two seconds to represent like a letter. Or one every 3 seconds. And the person outside could measure the change in the Gravitational field outside. And let's say that the black hole is pretty big so that I have a decent amount of time and that tidal forces don't rip me apart.
Or could I do something with entangled particles? Or am i crazy.

 

[Drakkith]

What would rockets do to the gravity of a black hole? No matter or energy escapes other than possibly Hawking Radiation.

 

[cragar]

it would change its position inside the black hole, and when it was spiraling around the field would be stronger and then get weaker when it spirals around to the other side.

 

[DrGreg]

Changes in "gravity fields" are propagated by gravitational waves that move at the same speed as light. As light cannot pass upwards through the event horizon, gravitational waves cannot either.

 

[cragar]

are you implying that if mass goes into a black hole that i won't feel the added pull from the now increased mass of the black hole. And are you saying that Gravity interacts with itself.

 

[Khursed]

are you implying that if mass goes into a black hole that i won't feel the added pull from the now increased mass of the black hole. And are you saying that Gravity interacts with itself.

Lets be honest, your question posits a black hole so big you're not destroyed instantly by the spaghettification, so that means a galaxy nucleus black hole, weighting at between several hundred of thousands, and possibly billions of star mass.

So given such a disproportionate massive black hole, and your puny rockets, we're taking something on the scale of 35 order of magnetude or so.

Take a star the sun's mass, 1,98E30kg, times say a 15 million sun mass black hole, roughly 3E37 kg of black hole, and say you're using 900 rockets to do your "message" do you know many ways of measuring gravity field disturbance of those magnetudes? You could be using rockets in the million ton range, and that would still be over 30 order of magnetude too small to be noticed in any meaningful way.

I mean just the error range of any machine would be well over too large to give any realistic results.

I also assume a non rotating black hole, otherwise I know the math goes crazy. Not to mention who knows what else the black hole is eating when you conduct your experiment.

Final nail in the coffin, who the heck would volunteer for such an experiment? :)

 

[physiqueper4]

My english is trrible and it seems to me I did not understand the idea as you are desplying it.
I only have one question: what everyou choos to do inside the hole how could the outsider observer know about it. No communication can be done through the surface of the hole. thx

 

[Nabeshin]

are you implying that if mass goes into a black hole that i won't feel the added pull from the now increased mass of the black hole. And are you saying that Gravity interacts with itself.

1) This is not what should be implied. A full description is a bit complicated in GR so I'll just give this qualitative description: As mass falls towards a black hole, the event horizon expands smoothly. So an outside observer measures the mass of the hole to slowly increase from its previous value, M, to a final value M + dM. The reason you cannot wiggle your arms around and generate gravitational waves to signal your buddies outside of a black hole is that gravitational waves follow timelike geodesics just like normal waves, and thus are bound to remain inside the horizon.

2) Gravity does indeed interact with itself. Einstein's GR is a nonlinear theory, which effectively means the gravitational field can interact with itself.

 

[Antiphon]

I voulenteer.

 

[skeptic2]

1) As mass falls towards a black hole, the event horizon expands smoothly. So an outside observer measures the mass of the hole to slowly increase from its previous value, M, to a final value M + dM. The reason you cannot wiggle your arms around and generate gravitational waves to signal your buddies outside of a black hole is that gravitational waves follow timelike geodesics just like normal waves, and thus are bound to remain inside the horizon.

You seem to be saying the event horizon increases in size BEFORE the mass crosses the event horizon. Is that true? Why?

Wouldn't the gravity of an infalling mass oppose the gravity of the BH such that directly below the infalling mass, (but still outside the EH) the escape velocity and thus the EH radius would be slightly less? I'm not suggesting the area of the EH would decrease, just that a slight dimple in the EH would form under the infalling mass.

 

[cragar]

1) This is not what should be implied. A full description is a bit complicated in GR so I'll just give this qualitative description: As mass falls towards a black hole, the event horizon expands smoothly. So an outside observer measures the mass of the hole to slowly increase from its previous value, M, to a final value M + dM. The reason you cannot wiggle your arms around and generate gravitational waves to signal your buddies outside of a black hole is that gravitational waves follow timelike geodesics just like normal waves, and thus are bound to remain inside the horizon.

2) Gravity does indeed interact with itself. Einstein's GR is a nonlinear theory, which effectively means the gravitational field can interact with itself.

Ok i see that a G wave can't escape the black hole. But what about the G field . What is the difference between the wave and the field. And I can obvisouly expirence the field outside of the black hole So something is being exchanged from the center of the black hole to me, So what is going on .
Thanks for your response by the way.

 

[Nabeshin]

You seem to be saying the event horizon increases in size BEFORE the mass crosses the event horizon. Is that true? Why?

This is correct :). I've given a description of it in many other threads, and Kip Thorne has a great picture illustrating this in Black Holes and Time Warps (I can't seem to find the image online though...). Here's what I wrote in a similar thread (ignore the maths at the top): https://www.physicsforums.com/showpost.php?p=3141245&postcount=4

Wouldn't the gravity of an infalling mass oppose the gravity of the BH such that directly below the infalling mass, (but still outside the EH) the escape velocity and thus the EH radius would be slightly less? I'm not suggesting the area of the EH would decrease, just that a slight dimple in the EH would form under the infalling mass.

You're thinking in terms of Newtonian gravity, where forces would add vectorally and linearly. General relativity doesn't work like that. If you want to think of it along such lines at all, the infalling mass adds to the potential well of the black hole, thus increasing the overall potential (since mass is only positive, it can only increase, not decrease). To make use of the much overused bowling ball on a sheet analogy, if I add a marble, the dip only gets deeper, not shallower.

 

[Nabeshin]

Ok i see that a G wave can't escape the black hole. But what about the G field . What is the difference between the wave and the field. And I can obvisouly expirence the field outside of the black hole So something is being exchanged from the center of the black hole to me, So what is going on .
Thanks for your response by the way.

Well the black hole solution is stationary in that it does not evolve in time. Therefore, the geometry of the hole is the geometry of the hole -- it does not need to be communicated to outside observers. At least in the classical relativity picture of the situation, there is curvature due to the black hole which exists everywhere in the entire universe, so all objects feel it -- there is no need for anything to propagate out in order for distant objects to move along geodesics (i.e. feel a force).

Of course, when you try to talk about quanta of gravity, i.e. gravitons, the situation is more complicated. I don't know too too much about this situation, but I suspect it has something to do with the fact that virtual gravitons would mediate a static gravitational force, and these odd particles have some strange properties... Someone more knowledgeable about QFT can probably answer though, so take this part with a healthy heap of salt.

 

[skeptic2]

This is correct :). I've given a description of it in many other threads, and Kip Thorne has a great picture illustrating this in Black Holes and Time Warps (I can't seem to find the image online though...). Here's what I wrote in a similar thread (ignore the maths at the top): https://www.physicsforums.com/showpost.php?p=3141245&postcount=4

You're thinking in terms of Newtonian gravity, where forces would add vectorally and linearly. General relativity doesn't work like that. If you want to think of it along such lines at all, the infalling mass adds to the potential well of the black hole, thus increasing the overall potential (since mass is only positive, it can only increase, not decrease). To make use of the much overused bowling ball on a sheet analogy, if I add a marble, the dip only gets deeper, not shallower.

Thanks. I've had Kip Thorne's book for many years and read your link which was helpful. I think the difference between my understanding and yours is best illustrated by the bowling ball analogy. I was thinking that escape velocity is not represented by the depth the bowling ball sinks into the rubber sheet but by the slope of the rubber sheet. As another mass approaches the bowling ball, it is true the overall depth increases but there is a small area between the bowling ball and the infalling mass where the local slope decreases slightly. Why do you feel escape velocity is better represented by the depth rather than by the slope of the rubber sheet?

 

[Nabeshin]

Why do you feel escape velocity is better represented by the depth rather than by the slope of the rubber sheet?

The slope is a very local phenomenon. Imagine an incredibly deep potential well, and we can agree that light can never propagate out of it. Now if you just put, somewhere deep in the well, an area of shallow slope, does that suddenly mean the light will propagate out? Of course not. It means that, perhaps the light will temporarily be able to move outwards, it will still eventually be captured. This is precisely the definition of an event horizon -- the surface beyond which light will never propagate out to null infinity. So the event horizon is a global phenomenon, and isn't well represented by small local behaviour of the potential.

 

[skeptic2]

It seems to me that the slope of the rubber sheet is a better representation of escape velocity than is the depth of the impression because the slope represents a rate of change just as escape velocity also represents a rate of change.

Consider this thought experiment.

A bowling ball is placed on a rubber sheet and a contour is drawn on the rubber sheet around the bowling ball that we will arbitrarily call the event horizon. For the sake of the experiment let's say the slope of the rubber sheet at the contour is one. Then a second bowling ball is added equidistant from the contour such that the contour is halfway between the two bowling balls. The slope of the rubber sheet at the contour and in the direction perpendicular to the contour is now zero.

If we start at the point exactly halfway between the two bowling balls and move perpendicular to a line connecting the centers of mass (tangent to the contour) the slope of the rubber sheet in the direction we are moving is always less than the original slope of one, where we drew the contour of the event horizon.

I concede that bowling balls and rubber sheets rarely demonstrate all the intricacies of GR and there may be effects I have not taken into consideration.

 

[Nabeshin]

I concede that bowling balls and rubber sheets rarely demonstrate all the intricacies of GR and there may be effects I have not taken into consideration.

Yeah I think this is just a failing of the rubber sheet analogy. I could try to fanangle some argument for what I had said earlier, but I'm not sure it would do much good. At some point you have to take off the training wheels and go on to the real thing.

The moral of the story is that it is really the potential which matters, not the "gravitational force" in the Newtonian analogue.