More a general math question than a discussion of the overall article so I’m putting it here. Mods move if you must. Re Cecil’s recent articleI was quite surprised bordering on incredulous that a sphere the diameter of a proton could contain that much collapsed matter. Could someone detail for me how they came up with that mass calculation?
The principle of density and the resulting different gravity can be demonstrated at low densities.
There is no mathematics for the behaviour of matter inside the blackhole,
there is no observable behaviour.
They also can’t even get the radius of neutron stars accurately, in order to get a good idea of the surface gravity, but they are sure they are very very dense (but far less dense than a black hole )
here’s what they say about neutron stars density…
http://www.astro.umd.edu/~miller/poster1.html
(extracted sentences)
Estimating NS Masses and Radii
No easy task, this. Astronomical measurements are often challenging, because we can’t go to a star and experiment on it. Neutron stars are especially tough, because they are relatively small and far away:
One way to do this is to use Kepler’s laws. If we can figure out how far two stars in a binary are from each other, and the duration of their orbital period, we know something about their masses.
Estimating the radius is much more difficult than estimating the mass.
So, we need some kind of breakthrough in the evidence to allow us to further constrain the radii of neutron stars.
He calculated the mass using the Schwarzschild radius, the radius of the event horizon. The formula is
M = r x c^2
2G
where r is the radius, c is the speed of light, and G is the gravitational constant. A radius of 0.8768 fm results in 652 million tons.
One should also add that it’s even more counter-intuitive than collapsing hundreds of tons into the size of a proton, because in theory the mass of any black hole, no matter how massive, is actually collapsed into a dimensionless point. The “size” of a black hole is the size of the event horizon, not the singularity. Unless matter is being actively sucked in, the entire volume of a black hole is (in theory) empty space. But intuitive concepts beyond the event horizon are fairly meaningless.
Also note that, the smaller the black hole, the higher the density (by which is meant the density within the event horizon; since we can’t know what happens inside, we don’t care). A sufficiently-large black hole, like the ones in the cores of galaxies, can have a density comparable to water or even to air.
Incidentally, this does not depend in any way on our knowledge (or lack thereof) of neutron stars. Neutron stars are incredibly messy objects, composed of a wide variety of types of matter, interacting via multiple forces, including the Strong Force about which we have only approximate, kludged-together models. In fact, one of the most promising avenues for exploring the Strong Force is study of neutron stars: If we could manage to measure the masses and radii of a few of them (preferably both for the same object), that would go a long way to understanding the Strong Force. Black holes, though, are incredibly simple, and each of them can be completely described by three vectors and three scalars (and two of those scalars are almost never relevant, and two of the vectors and the direction of the third are basically trivial).
How big does a black hole have to be to avoid significant tidal forces as you cross the event horizon (say for a six foot astronaut or maybe for a 500 meter spaceship)?
What is the density of the galaxy? How big a black hole for that density?
“Death From the Skies” (by a illustrious former Doper- Phil Plait, the Bad Astronomer) goes into this in some detail. It’s a great read.
Based on what I read last night, essentially what the posters above are saying is that a black hole the mass of the Sun would physically be a dimensionless point (that’s a defining characteristic), but that since gravity is wholly dependent on mass, the gravitational effects at 1 AU from that black hole would be indistinguishable from those of the Sun, since the masses are identical.
The event horizon is at some distance from the actual black hole, and is basically the point at which the escape velocity to get away is equal to the speed of light.
This means that black holes have the dimensionless point and then dome distance outward to the event horizon that we’re assuming is empty space. For a sufficiently massive one, the density could be quite small.
Depends on what you define as “significant”. The relative acceleration between the top and bottom of an object of length d at the horizon of a black hole of mass M is
a = 2dc[sup]6[/sup]/(G[sup]2[/sup] M[sup]2[/sup]).
For a 2-meter-long astronaut falling headfirst into a black hole of mass 10[sup]6[/sup] solar masses, this works out to 0.16 m/s[sup]2[/sup] between one’s head and feet, or about 1.6% of one’s weight; not that noticeable. However, the tidal forces at the event horizon are inversely proportionally to the square of the black hole’s mass, so a black hole that’s ten times lighter would cause forces 100 times stronger — definitely noticeable and uncomfortable. A black hole 100 times lighter (10,000 times the mass of the Sun) would almost certainly not be survivable.
I know you didn’t ask this, but in terms of gravitational impact, a proton sized black hole would be able to pick you up off the ground, if you got within about 20 cm of it, but if you were a meter of two away, you probably wouldn’t notice it.
So there could be one sitting in the corner of your room, just waiting to nab you, and you wouldn’t even know it…
The gravitational pull is coming from…inside the house!
Just like a spider on the ceiling, it’s right above you, just waiting, waiting, … waiting. Don’t look now, it’s about to pounce!!
(aside)
I once had a dream that I found a black hole in the middle of my living room. Man, that was a cool dream.
It ain’t the gravity that does you in.
It’s the tides.
So… it’s possible in our 3D universe to have a dimensionless point with an event horizon the size of a proton that is generating the same gravitational field as an object weighing 652 million tons.
I’m sure the math involved says it’s true but objectively I find that difficult to believe.
You mean *subjectively *you find it difficult to believe! The math is pretty objective, though not necessarily intuitive. At the Big Bang, the entire universe was a dimensionless point, and space and time did not yet exist. We pretty much have to let go of our intuition about common experience at scales of the very small and very large.
652 million tons is a cube of granite about 250m on a side. Or about 1/4 mile on a side. The total amount of gravity that creates is damn near nil. If it was sitting in a field someplace you could certainly walk right along the edge of it and feel zero sensation of attraction.
What’s lacking is an intuitive sense of how much matter 652 million tons is, or more accurately isn’t, and also an intuitive sense of how extremely, extremely weak gravity is.
And the only reason you would feel the gravity from a black hole of that mass would be if you were much closer to it than you were to the center of the granite block. With a block of granite a 1/4 mile on a side, you can’t get any closer to the center than a furlong, but with a black hole the size of a proton, you could get as close as the radius of a proton.
A singularity is dimensionless by definition, isn’t it? That is, if a black hole wasn’t dimensionless, we wouldn’t call it a singularity, if I understand the meaning of the term correctly.
But it wouldn’t sit there, would it? Wouldn’t it suck itself down into the Earth and suck the Earth into it? I realize you’re just painting a picture here, but it raised the question in my mind. Is it possible to contain a black hole?
Yeah, no kidding. It’s interesting that the macro structure of the universe is dictated by the force that’s smaller than all the others, by many orders of magnitude, simply due to two factors (IIRC):
- it’s nonpolar (there’s no anti-gravity, no opposite force to cancel it out as there is for charge
- it diminishes with the square of the distance, unlike the strong and weak nuclear forces, which diminish faster with distance (cubed?)
The scale of the difference in force between gravity and even the next weakest (weak nuclear?) is staggering.
Forgive me for prating from memory and not checking my facts. As usual, I’ll appreciate any learned people fighting my ignorance!