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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