Serious Scientific Answers to Absurd Hypothetical questions

[Delay]

How can a black hole even rotate? If I understood correctly, the "black" of the black hole is just the area where light can't reach your eye (i.e. event horizon?), right?

It isn't a physical border, just a point at which the lack of photons starts. So only the singularity itself could rotate?

[WinkAllKerb'']

28 minutes ago, Delay said:

How can a black hole even rotate? If I understood correctly, the "black" of the black hole is just the area where light can't reach your eye (i.e. event horizon?), right?

It isn't a physical border, just a point at which the lack of photons starts. So only the singularity itself could rotate?

you know the fun about black hole, you ll never know accurately until you travel to it's exact center core, wich it's seem you can't , for now, there is too much for you to handle in a single location ; )

this may change at some point, who know ^^

Edited November 13, 2017 by WinkAllKerb''

[wumpus]

On 11/11/2017 at 6:56 PM, Delay said:

If you could orbit around a black hole with your eyes exactly at the event horizon (or better yet; one inside, the other outside), what would you see?

In addition to Slashy's point that you would be dead, orbital speed at the even horizon is greater than c.  And there is a bigger zone where orbiting a black hole means you can't reach escape velocity.  Scott Manley had a video on just this, although at the end he had to admit that everything changes if the black hole is rotating (although it wouldn't change the fact that orbital velocity at the event horizon >c).

[p1t1o]

2 hours ago, Delay said:

How can a black hole even rotate? If I understood correctly, the "black" of the black hole is just the area where light can't reach your eye (i.e. event horizon?), right?

It isn't a physical border, just a point at which the lack of photons starts. So only the singularity itself could rotate?

Exactly right, a "rotating event horizon" is meaningless - like a rotating shadow - its the mass itself that can have a rotation.

It is hypothesised that if it spins fast enough, the singularity itself could form a ring rather than a point, and further, that if it spins even-faster-enough, that the singularity itself could extend past the event horizon forming a "naked singularity".

If that sounds somewhat noodle-baking, it is, and it is not clear whether or not it represents reality (clearly a singularity not covered with an event horizon has a lot to answer for) but thats what some hypotheses's math says is possible at this point.

There is still a lot of scientific mystery surrounding BH's.

[GoSlash27]

1 hour ago, wumpus said:

orbital speed at the even horizon is greater than c.  And there is a bigger zone where orbiting a black hole means you can't reach escape velocity.

As I understand it, the "event horizon" is the radius where Vesc>C. But if that's the case, then it *is* possible to cross back out, it's just impossible to leave the SoI. Since Vesc=Vorb*sqrt(2), then it'd stand to reason that the actual "abandon hope all ye who enter here" is at half the Schwarzschild radius, where Vorb >/= C.

 I may be missing something, tho'...

[p1t1o]

3 minutes ago, GoSlash27 said:

As I understand it, the "event horizon" is the radius where Vesc>C. But if that's the case, then it *is* possible to cross back out, it's just impossible to leave the SoI. Since Vesc=Vorb*sqrt(2), then it'd stand to reason that the actual "abandon hope all ye who enter here" is at half the Schwarzschild radius, where Vorb >/= C.

 I may be missing something, tho'...

I think there is some complicated space-warping where beyond the EH, no matter what direction you travel in, you get closer to the centre. So it may be more involved that a simple matter of speed. I would not expect Newtonian mechanics to be dominant.

[mikegarrison]

46 minutes ago, GoSlash27 said:

As I understand it, the "event horizon" is the radius where Vesc>C. But if that's the case, then it *is* possible to cross back out, it's just impossible to leave the SoI. Since Vesc=Vorb*sqrt(2), then it'd stand to reason that the actual "abandon hope all ye who enter here" is at half the Schwarzschild radius, where Vorb >/= C.

 I may be missing something, tho'...

No, by definition it is not possible to cross back out. The one known method for anything to "leave" a black hole is when pair production occurs and one particle falls into the hole while the other does not ("Hawking radiation"). The definition of the event horizon is not "Vesc>C" in a Newtonian sense. The definition is that "all lightlike paths" must fall further into the hole.

Also, the time dilation stuff is supposedly only seen from the viewpoint of an outside observer. From far away, it appears that time slows down and it takes an infinite amount of time to actually cross the event horizon. But in the time stream of the particle itself, it experiences no strange effects and simply approaches and then crosses the event horizon.

"Event horizon" is actually a more general feature of relativity and not specific to black holes. As I type this message, there is an event horizon expanding away from me at the speed of light. Anything on the other side of that event horizon can not be influenced by my message, because the information from my message can not possibly have reached it yet. The analog to this is the Mach cone, where the air outside the cone can not be disturbed by anything inside the cone because the pressure wave can not have reached it (because it would have had to travel faster than sound).

Edited November 13, 2017 by mikegarrison

[GoSlash27]

p1t1o,
 I've always found this stuff confusing since it's often counterintuitive, so bear with me...

I hear "Schwarzchild radius" and "event horizon" used interchangeably, and the sources I've picked up this stuff from insist that they are one and the same. Perhaps this is incorrect?

All of the proofs I've seen re. the Schwarzchild radius are straight- up Newtonian; the radius at which Vesc=c.

That's where my puzzlement stems from, because I wouldn't *expect* the Schwarzchild radius to be a true event horizon. It would just appear as such from outside the SoI since nothing can reach Ve beyond that point, but within the SoI light and even matter would be able to cross that radius and come back out.
 From within the SoI, the true event horizon would be at half that, where space gets so bent that escape is truly impossible and all directions point in.

 I'd love to have someone clear this discrepancy up for me

Best,
-Slashy
 

[Green Baron]

6 minutes ago, GoSlash27 said:

p1t1o,
 I've always found this stuff confusing since it's often counterintuitive, so bear with me...

No, it is not intuitive. I am not a physicist but may be i can take away some confusion and replace it by a new one.

6 minutes ago, GoSlash27 said:

I hear "Schwarzchild radius" and "event horizon" used interchangeably, and the sources I've picked up this stuff from insist that they are one and the same. Perhaps this is incorrect?

Schwarzschild radius is a subset of event horizon. The latter being a name for everything we cannot look behind. Like our Hubble sphere, the observable universe. It's border is an event horizon, but no Schwarzschild r., which is a property of a mass. I have one (which is very small and inside), the sun has one (it is 3km or so), and a black hole has one, which is bigger than the mass concentration at its center, the "singularity", witch nobody can describe because our models have no solution.

6 minutes ago, GoSlash27 said:

All of the proofs I've seen re. the Schwarzchild radius are straight- up Newtonian; the radius at which Vesc=c.

Yep, that's my understanding as well. The Lorentz term gets undefined as something approaches the horizon, as seen from outside.

6 minutes ago, GoSlash27 said:

That's where my puzzlement stems from, because I wouldn't *expect* the Schwarzchild radius to be a true event horizon. It would just appear as such from outside the SoI since nothing can reach Ve beyond that point, but within the SoI light and even matter would be able to cross that radius and come back out.

It is a true event horizon for an observer outside of it, not for someone crossing it.Crossing it is no problem in the reference frame of the traveler, probably not even recognized if it is a huge BH with a large horizon and relatively small gravitational gradient. I assume with SoI you mean the horizon ? "out" is a concept from the frame of an outside observer, and no, for him nothing returns from the event horizon. In fact, nothing can reach it, see the Lorentz term as it approaches and apply it to the time passing, it converges to infinity.

6 minutes ago, GoSlash27 said:

 From within the SoI, the true event horizon would be at half that, where space gets so bent that escape is truly impossible and all directions point in.

I don't understand what you mean with SoI and "true". SoI is nothing natural, it is a game concept or a simplification for slow traveling spaceships. Gravity of a body extends to infinity.

6 minutes ago, GoSlash27 said:

 I'd love to have someone clear this discrepancy up for me

Best,
-Slashy
 

That probably wasn't much help ?

The discrepancy might arise from the attempt to lump the two relative reference frames together ...

 

[GoSlash27]

Green Baron,
 By SoI, I mean "Sphere of Influence", and yeah, it is a bit of an oversimplification.
 What I mean is that a particle *should* be able to cross the threshold of the Schwarzchild radius and exit it again (down to a depth of half that), but it just won't be able to reach hyperbolic escape velocity afterwards. It will be doomed to be stuck inside the black hole's gravity well forever in a highly eccentric orbit. A periapsis of *half* the Schwarzchild radius should create an absolute event horizon, since nothing inside it can ever interact with the outside universe again. Orbital velocity is equal to or higher than c at that point, so all directions point in.
 From far away, the Schwarzchild radius would *appear* to be an event horizon since nothing that enters can be visible to you. Stuff goes in, but nothing comes out as far as you can tell. I would expect that the event horizon would appear to shrink and you would see more and more trapped particles and radiation as you get closer that wasn't previously visible. It's size would appear to be Rs at an infinite distance, but approach Rs/2 as you approach Rs/2 yourself.

 This is the way it would seem to be to me, but apparently it's not and I don't grasp why it isn't.

Hope this explains my confusion,
-Slashy

 

 

Edited November 14, 2017 by GoSlash27

[mikegarrison]

1 hour ago, GoSlash27 said:

Green Baron,
 By SoI, I mean "Sphere of Influence", and yeah, it is a bit of an oversimplification.
 What I mean is that a particle *should* be able to cross the threshold of the Schwarzchild radius and exit it again (down to a depth of half that), but it just won't be able to reach hyperbolic escape velocity afterwards. It will be doomed to be stuck inside the black hole's gravity well forever in a highly eccentric orbit. A periapsis of *half* the Schwarzchild radius should create an absolute event horizon, since nothing inside it can ever interact with the outside universe again. Orbital velocity is equal to or higher than c at that point, so all directions point in.
 From far away, the Schwarzchild radius would *appear* to be an event horizon since nothing that enters can be visible to you. Stuff goes in, but nothing comes out as far as you can tell. I would expect that the event horizon would appear to shrink and you would see more and more trapped particles and radiation as you get closer that wasn't previously visible. It's size would appear to be Rs at an infinite distance, but approach Rs/2 as you approach Rs/2 yourself.

 This is the way it would seem to be to me, but apparently it's not and I don't grasp why it isn't.

Hope this explains my confusion,
-Slashy

 

 

I think the problem is that the orbital mechanics you are trying to understand this with are Newtonian.

[magnemoe]

1 minute ago, mikegarrison said:

I think the problem is that the orbital mechanics you are trying to understand this with are Newtonian.

Yes, if the gravity field or gradient is large you get an significant braking effect because of relativity, this is that causes the inward spiraling disc around a black hole, around an normal body it would be an stable ring. 

 

[GoSlash27]

15 minutes ago, mikegarrison said:

I think the problem is that the orbital mechanics you are trying to understand this with are Newtonian.

Right... But again, the definition of the Schwarzchild radius *is* Newtonian. What @magnemoe and I are saying is the same thing (as far as I comprehend it). The Schwarzchild  radius is Newtonian, while the event horizon is relativistic. They should be 2 different radii, not one and the same...

Best,
-Slashy

Edited November 14, 2017 by GoSlash27

[GoSlash27]

Or phrased another way...

 By my way of thinking, the event horizon shouldn't be where all light and particles are trapped hopelessly in orbit about the singularity, it should be where anything that enters can never leave. These are two entirely different concepts.

HTHs,
-Slashy

 

[WinkAllKerb'']

, )

- 1 - let's travel to the center of the sun

- 2 - let's travel to the center of a black hole 

one thing at time, don't burn the steps peops ^^

Edited November 14, 2017 by WinkAllKerb''
(oops i had too) sometime biped have some weird idea if you ask me ^^

[DAL59]

The questions here are supposed to be kind of absurd...so what if the Earth's mantle was magically replaced with an equal volume of water?  What exotic ices would form?  

[Steel]

4 hours ago, GoSlash27 said:

Right... But again, the definition of the Schwarzchild radius *is* Newtonian. What @magnemoe and I are saying is the same thing (as far as I comprehend it). The Schwarzchild  radius is Newtonian, while the event horizon is relativistic. They should be 2 different radii, not one and the same...

Best,
-Slashy

This is where your problem is. The definition of the Schwarzschild radius is NOT Newtonian, you have to do GR and solve the field equations to get a general solution.

Luckily for high school teachers all over the world, it just happens that if you do the basic assumption V_esc = c and then solve with Newtonian gravity you get the same result as if you'd done a full solution of the Schwarzschild metric for the black hole. 

This is a great way to make the topic more accessible and easier to visualise, but is not a true reflection of the physics going on.

Don't make the mistake of thinking that this means Newtonian gravity is going to help you much where a black hole is concerned.

Edited November 14, 2017 by Steel

[DDE]

17 hours ago, ModZero said:

By "idiots" you mean engineers? Because they're overrepresented in right-wing extremist groups. Perhaps luckily (unless you're trying to build a new nuclear power plant or something) nuclear engineers are rather uncommon amongst engineers to begin with, though.

Was it ISIS or Al-Qaida that ran a campaign showcasing that their suicide bombers were not basement-dwelling losers?

[Bill Phil]

4 hours ago, GoSlash27 said:

Right... But again, the definition of the Schwarzchild radius *is* Newtonian. What @magnemoe and I are saying is the same thing (as far as I comprehend it). The Schwarzchild  radius is Newtonian, while the event horizon is relativistic. They should be 2 different radii, not one and the same...

Best,
-Slashy

The definition of the Schwarzschild radius is not Newtonian. It is also not necessarily the event horizon. Other factors may come into play, while finding the Schwarzschild radius is only based on mass and the speed of light. It's an upper bound for the event horizon, if I understand it correctly. The horizon itself could be smaller.

[Zeiss Ikon]

6 hours ago, DAL59 said:

The questions here are supposed to be kind of absurd...so what if the Earth's mantle was magically replaced with an equal volume of water?  What exotic ices would form?  

Well, none, any time soon.

Most of the internal heat inside the Earth is due to radioactive decay of heavy elements in the core ("impurities" in the nickel-iron core material).  Since your replacement doesn't affect the core, that heat source would still be present.

What you would have is far more vigorous convection (millions of times the rate with a partially molten rock mantle).  So, now you have the equivalent of a planet-sized pressure cooker, carrying off heat from the core under conditions where the water cannot boil (pressure is far beyond critical, so there's no phase change between room temperature and "hot enough to dissolve quartz like sugar").  The convection would bring hot water into contact with the underside of the crustal rocks, and in pretty short order dissolve away most of the minerals (did I mention quartz is soluble in water at something like 1000 C and 500 atmospheres?).

Time frame for this "dissolve the entire crust" I don't know.  I've gotten the impression that dissolution of quartz (which is what deposits the crystals inside a geode, among other less obvious effects) that this is a slowish process, so I'm going to say years to centuries.  That would mean there'd be a period when the crust is precariously balanced, floating/not-floating on the less dense water beneath.  Weak places in the crust (current spreading and subduction zones, at a minimum) would lead to blocks tilting, breaking loose, and sinking into the vast depths, events that would tend to destabilize the nearby crust.  I'd guess the crust would break up and sink, plunging every human and animal into not-yet-boiling water, long before the core can heat the shallower water enough to even start dissolving the crust (water, after all, is a rather poor conductor of heat and has an immense specific heat capacity).

Longer term, the mass of water in the new mantle would nearly equal the mass of the core.  The core, high in iron and nickel, is a far better conductor of heat than water is, so it would cool through as the water heated.  Over a period of a few centuries, then, the core would solidify as it cooled.  Eventually, an equilibrium would be reached -- most likely one in which the Earth would have a small molten inner core, a much larger solid outer core, a deep ocean formed from the water mantle, and a water surface (presuming there's still enough atmosphere to keep the water from boil/freezing in vaccuum).

All this upheaval might still not kill all oceanic life -- plankton and things that feed on it, all the way up to large sharks and whales, might survive if the water temperature at the surface doesn't get too hot.  I'm inclined to think it would boil vigorously for a while, but it's time to go to work, so I'm not going to try to do the thermodynamics.

[p1t1o]

6 hours ago, DAL59 said:

The questions here are supposed to be kind of absurd...so what if the Earth's mantle was magically replaced with an equal volume of water?  What exotic ices would form?  

Seems like a question fairly easily answered. The behaviour of water under various degrees of pressure/temperature is well characterised. Calculating pressure at a chosen depth is straightforward enough, but I have no idea what the temperature profile would be though. The chart goes all the way to 10megabar, which represents a [water] depth of approx 100,000km.

It should be noted that if you replace the mantle with water, the mass of the planet will be significantly changed, as will its gravity. Now that I think about it, this does complicate estimating pressure at depth.

Pressure at the Earth's core in its current configuration is apparently approx. 330-360GPa. Convert mantle to water and it will be somewhat less, but you get an idea of the ballpark.

Temperature at the Earth's core in its current configuration is apparently approx. 5700K. I dont know how that would change if you made your changes, and as you can see from the chart, the hotter you are, the more likely you are to be liquid or gas - to the right of the critical point liquid and gas can coexist and to the right and above the critical point you have a supercritical fluid.

So the answer is - a complex, layered combination of ice VII, VIII, X and XI and at some depth, depending on the temperature profile, there could be some liquid water.

 

 

[ARS]

If we somehow able to fire current day weapons in space, which one that's most effective in terms of damage, accuracy and range between lasers, bullets and missiles?

[YNM]

23 hours ago, Delay said:

How can a black hole even rotate? If I understood correctly, the "black" of the black hole is just the area where light can't reach your eye (i.e. event horizon?), right?

It isn't a physical border, just a point at which the lack of photons starts. So only the singularity itself could rotate?

It's... as fuzzy as like how electrons have spin. Or how they have magnetic moment. Things do sometimes go crazy.

6 hours ago, Bill Phil said:

The definition of the Schwarzschild radius is not Newtonian. It is also not necessarily the event horizon. Other factors may come into play, while finding the Schwarzschild radius is only based on mass and the speed of light. It's an upper bound for the event horizon, if I understand it correctly. The horizon itself could be smaller.

Schwarzchild radius is newtonian, but you can derrive so from GR as well, and you can only correctly derrive from GR when it comes to trajectory details (metrics etc).

As the term "schwarzchild radius" refers to the location of the event horizon (I presume it should result in a null or such) for a non-spinning, uncharged black holes, the end result of where the event horizon is from GR derrivation are exactly the same as with the assumption using newtonian gravitation. I mean, black holes existence were presumed long before general relativity (was it Eddington or such ?) so it doesn't differ much indeed. Only when you start to add different energies (electromagnetic, rotational etc.) does it massively differ.

[DDE]

1 hour ago, ARS said:

If we somehow able to fire current day weapons in space, which one that's most effective in terms of damage, accuracy and range between lasers, bullets and missiles?

Missiles, full stop. Effectively unlimited range, higher impact velocity than regular guns. Lasers still suffer from target coupling - it’s more efficient to punch through targets than to melt them - and they don’t actually have perfect accuracy.

Thus far, lasers haven’t been tested in orbit. There has been a 23 mm cannon trial (R-23M in Kartech-1 gun pod on Almaz stations) and de facto missile strikes in the form of air-to-space missiles (fired by F-15s and possibly MiG-31Ds) and coorbiting kamikadze satellites (various versions of IS, operational since 1967, over two dozen intercepts).

[p1t1o]

1 minute ago, DDE said:

Missiles, full stop. Effectively unlimited range, higher impact velocity than regular guns. Lasers still suffer from target coupling - it’s more efficient to punch through targets than to melt them - and they don’t actually have perfect accuracy.

Thus far, lasers haven’t been tested in orbit. There has been a 23 mm cannon trial (R-23M in Kartech-1 gun pod on Almaz stations) and de facto missile strikes in the form of air-to-space missiles (fired by F-15s and possibly MiG-31Ds) and coorbiting kamikadze satellites (various versions of IS, operational since 1967, over two dozen intercepts).

 

2 hours ago, ARS said:

If we somehow able to fire current day weapons in space, which one that's most effective in terms of damage, accuracy and range between lasers, bullets and missiles?

 

Raw power? An ICBM, absolute maximum on all criteria (range, accuracy, power etc.). Quite well suited for space operations.

I wouldnt discount 21st century tank cannon and their kinetic projectiles. Impact energy can easily be of the same order as an average missile warhead (though there are many types of missile - I automatically think of an AMRAAM or similar when I hear the word "missile"). 

The present day ASAT is almost perfect though, since it is actually designed to operate in space, has a manouvering system, space-optimised IR seeker and a kinetic impact equivalent to being hit with a modest sized artillery round. More even, since we wont be firing it "upwards" so much, or through an atmosphere.

Honourable mention: Recoil-less guns also have obvious advantages, even if they arent the most powerful things around. Definitely useful if you are talking man-portable.

 

Need more information for a better answer.

Can we use ICBMs or does it need to be able to fit into a briefcase? Or a small vehicle? Space Shuttle?

What is the target? Is it armoured/hardened? Is it a swarm of small targets? What are its maneuvering capabilities.

 

Unsurprisingly the best weapon depends on the context. Even a knife would be useful in a hand-to-hand between two astronauts. Although now that I think about it I think a mace would be more effective (spacesuits ought to be highly resistant to blades, due to their durable construction).

 

Lasers have certain disadvantages, DDE describes one, but they also produce copious amounts of heat which is largely unavoidable, demanding large and complex cooling equipment. 

There are hypothetical nuclear one-shot devices on paper, but their effectiveness at range is less than one would expect. No, the dreaded Casaba howitzer is not the ultimate weapon. 

Railguns consume vast amounts of power, likely necessitating a nuclear reactor.

Theres a lot to be said for tried&tested technologies - High-velocity, rocket-boosted kinetic projectiles are my gut-feeling tip for a future space war. Essentially advanced, space-based versions of the ASAT.