Henry Baker <hbaker1@pipeline.com> wrote:
A bigger black hole has a bigger Schwarzschild radius, and space in the vicinity of the Schwarzschild radius of a very large black hole is relatively flat.
Right. You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. Nor could you avoid the singularity.)
I.e., suppose the Sun were a black hole, whose Schwarzschild radius is quite small, so the Earth is very far from this radius. Now consider a Sun' whose mass is, e.g., twice as big as the Sun. Its Schwarzschild radius is bigger than before, but if we are still at 1 AU, would we be able to tell _just from the local curvature_ how much mass is in the center of the solar system?
Certainly, given that you know the distance to the sun. Just measure the local tidal force, i.e. how much less a test mass on Earth's surface weighs when the sun is overhead or underfoot than when it's on the horizon, after correcting for the moon's tidal force. This could perhaps be most sensitively measured by timing a pendulum clock. Getting back to black holes, I wonder if anyone has tried to measure the tidal force on Earth from the black hole in the center of our galaxy. I think it ought to be measurable. As an aside, since the tidal force is differential gravity, it falls off with the inverse cube of distance. Conversely, the volume of an object of given angular size increases with the cube of its distance. So the two effects cancel out. Hence the relative tidal forces from two objects of the same angular size is in proportion to their relative densitites. If the moon, which is about the same angular size as the sun, has about three times the tidal force as the sun, it's "because" the moon has about three times the density as the sun. This scaling law must be generally known, but I've never seen it. (I worked it out for myself years ago.) Another fun scaling law is that the ratio of the sun's absolute surface temperature to the Earth's ought to be the fourth root of the proportion of Earth's sky the sun's disk takes up. (This assumes the sun and Earth are both black bodies or both fall short of being black bodies by the same proportion, that the sun is Earth's only source of heat, that the sun-Earth distance is constant, and that everything is in equilibrium. By "Earth's sky" I include the half that's hidden by the ground.)
... curvature of space alone?
Nitpick: Curvature of spacetime, not of space alone. Fred Lunnon <fred.lunnon@gmail.com> wrote:
What about its charge? And (quantum-theoretically) temperature?
For a small black hole, any charge would rapidly dissipate by Hawking radiation. For any size black hole, its charge would disippate by preferential absorption of the opposite charge from surrounding material. But I suppose a large black hole in a very good vacuum could retain a charge for quite some time. The temperature depends entirely on the mass. Smaller is hotter. As a rough rule of thumb, the peak wavelength of the radiation is about equal to the apparent diameter of the event horizon.
Susskind (and many others) have missed a very interesting analogy between a black hole and going through a corner while driving a car. If you enter a corner too fast while driving a car, you may survive for most of the way through the corner, only to be catapulted into the air at the very end. If you "solve" a corner backwards, you can compute ahead of time the maximum speed at which you can enter the corner and still survive; that is your "critical velocity", or equivalently, your event horizon. You won't feel anything any different when entering the corner at too high a speed, but you won't be able to exit in one piece no matter how hard you brake and or attempt to steer out of the corner. There's an old joke about the difference between racing a front-engine vs a rear-engine car: you still go over the edge in the corner, only in one case you go over forwards, while in the other you go over backwards. At 10:36 AM 10/20/2013, Keith F. Lynch wrote:
Right. You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. Nor could you avoid the singularity.)
You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity.
I don't see why. If a black hole is massive enough, (my crude calculation, which those with better ALUs are welcome to correct, is 3.65x10^18 solar masses or about four million times the mass of the Miky Way galaxy) it has about the density of our galaxy as a whole and might be habitable.
Nor could you avoid the singularity.)
Why couldn't you be in orbit inside the event horizon. What this leads to is the question can the universe be distinguished from a black hole? Whit
I don't see why. If a black hole is massive enough, (my crude calculation, which those with better ALUs are welcome to correct, is 3.65x10^18 solar masses or about four million times the mass of the Miky Way galaxy) it has about the density of our galaxy as a whole and might be habitable.
This is why I've never understood the whole firewall debate (following the Almheiri-Marolf-Polchinski-Sully paper). As far as I know I might be living in a black hole, why would things be different when crossing the event horizon which would look nothing different to me? I assume there are good reasons to think this -- whether right or wrong -- but I don't get it. Charles Greathouse Analyst/Programmer Case Western Reserve University On Sun, Oct 20, 2013 at 3:17 PM, Whitfield Diffie < whitfield.diffie@gmail.com> wrote:
You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity.
I don't see why. If a black hole is massive enough, (my crude calculation, which those with better ALUs are welcome to correct, is 3.65x10^18 solar masses or about four million times the mass of the Miky Way galaxy) it has about the density of our galaxy as a whole and might be habitable.
Nor could you avoid the singularity.)
Why couldn't you be in orbit inside the event horizon.
What this leads to is the question can the universe be distinguished from a black hole?
Whit
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On 10/20/2013 12:17 PM, Whitfield Diffie wrote:
You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. I don't see why. If a black hole is massive enough, (my crude calculation, which those with better ALUs are welcome to correct, is 3.65x10^18 solar masses or about four million times the mass of the Miky Way galaxy) it has about the density of our galaxy as a whole and might be habitable.
It would take many lifetimes to fall from the event horizon to the singularity.
Nor could you avoid the singularity.) Why couldn't you be in orbit inside the event horizon.
You can't be in orbit around the singularity inside the event horizon because the singularity is in your future. It's time-like relative to you, not space-like. You could of course orbit some other body that was also waiting for the singularity to happen.
What this leads to is the question can the universe be distinguished from a black hole?
The universe seems to be expanding faster and doesn't have a singularity in its future. Brent Meeker
Whit
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So imagine a honking big black hole, where ordinary stuff can fall in without being tidally ripped to shreds. Let's make it large enough to take in a star cluster without disturbing the cluster too much. And let's suppose the star cluster has an ordinary garden variety big-star- end-of-life black hole. Do we expect anything special to happen when the big guy eats the little guy? An ordinary BH merger of equals is expected to be cataclysmic, emitting gravity waves every which way. What happens in the super-unequal case? Rich ------- Quoting meekerdb <meekerdb@verizon.net>:
On 10/20/2013 12:17 PM, Whitfield Diffie wrote:
You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. I don't see why. If a black hole is massive enough, (my crude calculation, which those with better ALUs are welcome to correct, is 3.65x10^18 solar masses or about four million times the mass of the Miky Way galaxy) it has about the density of our galaxy as a whole and might be habitable.
It would take many lifetimes to fall from the event horizon to the singularity.
Nor could you avoid the singularity.) Why couldn't you be in orbit inside the event horizon.
You can't be in orbit around the singularity inside the event horizon because the singularity is in your future. It's time-like relative to you, not space-like. You could of course orbit some other body that was also waiting for the singularity to happen.
What this leads to is the question can the universe be distinguished from a black hole?
The universe seems to be expanding faster and doesn't have a singularity in its future.
Brent Meeker
Whit
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Nothing very special. The gravitational radiation from merging BHs has a wavelength on the order of the BH size. So if you have a BH bigger than a galaxy then the wave length is going to be so long it will pass almost unnoticed, like a tsunami at sea. Brent Meeker On 10/20/2013 10:16 PM, rcs@xmission.com wrote:
So imagine a honking big black hole, where ordinary stuff can fall in without being tidally ripped to shreds. Let's make it large enough to take in a star cluster without disturbing the cluster too much. And let's suppose the star cluster has an ordinary garden variety big-star- end-of-life black hole. Do we expect anything special to happen when the big guy eats the little guy? An ordinary BH merger of equals is expected to be cataclysmic, emitting gravity waves every which way. What happens in the super-unequal case?
Rich
------- Quoting meekerdb <meekerdb@verizon.net>:
On 10/20/2013 12:17 PM, Whitfield Diffie wrote:
You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. I don't see why. If a black hole is massive enough, (my crude calculation, which those with better ALUs are welcome to correct, is 3.65x10^18 solar masses or about four million times the mass of the Miky Way galaxy) it has about the density of our galaxy as a whole and might be habitable.
It would take many lifetimes to fall from the event horizon to the singularity.
Nor could you avoid the singularity.) Why couldn't you be in orbit inside the event horizon.
You can't be in orbit around the singularity inside the event horizon because the singularity is in your future. It's time-like relative to you, not space-like. You could of course orbit some other body that was also waiting for the singularity to happen.
What this leads to is the question can the universe be distinguished from a black hole?
The universe seems to be expanding faster and doesn't have a singularity in its future.
Brent Meeker
Whit
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From: http://en.wikipedia.org/wiki/General_relativity http://en.wikipedia.org/wiki/Schwarzschild_radius http://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity http://en.wikipedia.org/wiki/Gravitational_wave Diffie black hole: 3.65*10^18 solar masses Meeker black hole: 21*10^9 solar masses Baker black hole: 5.066*10^7 solar masses BH = black hole SR = Schwarzschild Radius Baker BH SR = 1 AU = 8.3 light-minutes Meeker BH SR = 415 AU = 57.5 light-hours Diffie BH SR = 7.2*10^10 AU = 1,139,381 light-years Photon orbit radius = 3/2*SR Photon orbit period = 2*pi*(3/2*SR)/c = 3*pi*SR/c Baker BH photon orbit period = 1.3 hours Meeker BH photon orbit period = 22.6 days Diffie BH photon orbit period = 10,738,416 years Orbital lifetime from gravitational radiation Baker BH with Earth orbiting at 5*SR = 2.6*10^10 years Meeker BH with Earth orbiting at 5*SR = 4.5*10^15 years Diffie BH with Earth orbiting at 5*SR = 1.4*10^32 years Meeker BH with Baker BH orbiting at 5*SR = 265 years Diffie BH with Meeker BH orbiting at 5*SR = 2*10^16 years The actual lifetime is likely to be infinite, since the Earth will be torn apart by tidal forces and schmeared out like a ring of Saturn, at which point it will no longer generate a gravitational wave because it is rotationally symmetric. Gravitational wave radiated power Sun with Earth orbiting at 1 AU = 196 watts Sun with Earth orbiting at 5 SR = 2.1*10^37 watts Baker black hole with Earth orbiting at 5*SR = 8.2*10^21 watts Meeker black hole with Earth orbiting at 5*SR = 4.75*10^16 watts Diffie black hole with Earth orbiting at 5*SR = 1.6 watts
On 10/20/2013 10:36 AM, Keith F. Lynch wrote:
Henry Baker <hbaker1@pipeline.com> wrote:
A bigger black hole has a bigger Schwarzschild radius, and space in the vicinity of the Schwarzschild radius of a very large black hole is relatively flat. Right. You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. Nor could you avoid the singularity.)
For the largest know BH, 21e9 solar masses, the time from the event horizon to the singularity is about 3.5yrs. Brent Meeker
I.e., suppose the Sun were a black hole, whose Schwarzschild radius is quite small, so the Earth is very far from this radius. Now consider a Sun' whose mass is, e.g., twice as big as the Sun. Its Schwarzschild radius is bigger than before, but if we are still at 1 AU, would we be able to tell _just from the local curvature_ how much mass is in the center of the solar system? Certainly, given that you know the distance to the sun. Just measure the local tidal force, i.e. how much less a test mass on Earth's surface weighs when the sun is overhead or underfoot than when it's on the horizon, after correcting for the moon's tidal force. This could perhaps be most sensitively measured by timing a pendulum clock.
Getting back to black holes, I wonder if anyone has tried to measure the tidal force on Earth from the black hole in the center of our galaxy. I think it ought to be measurable.
As an aside, since the tidal force is differential gravity, it falls off with the inverse cube of distance. Conversely, the volume of an object of given angular size increases with the cube of its distance. So the two effects cancel out. Hence the relative tidal forces from two objects of the same angular size is in proportion to their relative densitites. If the moon, which is about the same angular size as the sun, has about three times the tidal force as the sun, it's "because" the moon has about three times the density as the sun. This scaling law must be generally known, but I've never seen it. (I worked it out for myself years ago.)
Another fun scaling law is that the ratio of the sun's absolute surface temperature to the Earth's ought to be the fourth root of the proportion of Earth's sky the sun's disk takes up. (This assumes the sun and Earth are both black bodies or both fall short of being black bodies by the same proportion, that the sun is Earth's only source of heat, that the sun-Earth distance is constant, and that everything is in equilibrium. By "Earth's sky" I include the half that's hidden by the ground.)
... curvature of space alone? Nitpick: Curvature of spacetime, not of space alone.
Fred Lunnon <fred.lunnon@gmail.com> wrote:
What about its charge? And (quantum-theoretically) temperature? For a small black hole, any charge would rapidly dissipate by Hawking radiation. For any size black hole, its charge would disippate by preferential absorption of the opposite charge from surrounding material. But I suppose a large black hole in a very good vacuum could retain a charge for quite some time.
The temperature depends entirely on the mass. Smaller is hotter. As a rough rule of thumb, the peak wavelength of the radiation is about equal to the apparent diameter of the event horizon.
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On the other hand, if the density of the universe is large enough so that it is closed (a 4-sphere), then we are living inside a black hole now! Cris On Oct 20, 2013, at 6:14 PM, meekerdb <meekerdb@verizon.net> wrote:
On 10/20/2013 10:36 AM, Keith F. Lynch wrote:
Henry Baker <hbaker1@pipeline.com> wrote:
A bigger black hole has a bigger Schwarzschild radius, and space in the vicinity of the Schwarzschild radius of a very large black hole is relatively flat. Right. You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. Nor could you avoid the singularity.)
For the largest know BH, 21e9 solar masses, the time from the event horizon to the singularity is about 3.5yrs.
Brent Meeker
I.e., suppose the Sun were a black hole, whose Schwarzschild radius is quite small, so the Earth is very far from this radius. Now consider a Sun' whose mass is, e.g., twice as big as the Sun. Its Schwarzschild radius is bigger than before, but if we are still at 1 AU, would we be able to tell _just from the local curvature_ how much mass is in the center of the solar system? Certainly, given that you know the distance to the sun. Just measure the local tidal force, i.e. how much less a test mass on Earth's surface weighs when the sun is overhead or underfoot than when it's on the horizon, after correcting for the moon's tidal force. This could perhaps be most sensitively measured by timing a pendulum clock.
Getting back to black holes, I wonder if anyone has tried to measure the tidal force on Earth from the black hole in the center of our galaxy. I think it ought to be measurable.
As an aside, since the tidal force is differential gravity, it falls off with the inverse cube of distance. Conversely, the volume of an object of given angular size increases with the cube of its distance. So the two effects cancel out. Hence the relative tidal forces from two objects of the same angular size is in proportion to their relative densitites. If the moon, which is about the same angular size as the sun, has about three times the tidal force as the sun, it's "because" the moon has about three times the density as the sun. This scaling law must be generally known, but I've never seen it. (I worked it out for myself years ago.)
Another fun scaling law is that the ratio of the sun's absolute surface temperature to the Earth's ought to be the fourth root of the proportion of Earth's sky the sun's disk takes up. (This assumes the sun and Earth are both black bodies or both fall short of being black bodies by the same proportion, that the sun is Earth's only source of heat, that the sun-Earth distance is constant, and that everything is in equilibrium. By "Earth's sky" I include the half that's hidden by the ground.)
... curvature of space alone? Nitpick: Curvature of spacetime, not of space alone.
Fred Lunnon <fred.lunnon@gmail.com> wrote:
What about its charge? And (quantum-theoretically) temperature? For a small black hole, any charge would rapidly dissipate by Hawking radiation. For any size black hole, its charge would disippate by preferential absorption of the opposite charge from surrounding material. But I suppose a large black hole in a very good vacuum could retain a charge for quite some time.
The temperature depends entirely on the mass. Smaller is hotter. As a rough rule of thumb, the peak wavelength of the radiation is about equal to the apparent diameter of the event horizon.
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participants (7)
-
Charles Greathouse -
Cris Moore -
Henry Baker -
Keith F. Lynch -
meekerdb -
rcs＠xmission.com -
Whitfield Diffie