I’m not sure if this should be in GQ or here, but i thought i’d try here first…
In this column Cecil says that the more massive the black hole, the less dense it is. A black hole with the same mass as our galaxy would have the same density as a thin gas.
Is there an easy way to intuitively explain why this is so? I can handle some formulae, but nothing too complicated.
Also, if this means that the universe is a black hole wouldn’t that mean that the universe is “closed” and therefore will end in a big crunch? Does it make any sense to ask what this would look like to someone in the “higher” universe in which our universe is a black hole?
IANACosmologist but as the mass increases, the “radius” of the black hole expands. Now we know that the escape velocity of the BH is c so …
R = 2*GM/c^2
The density is M/(4/3piR^3) so we see that the density is porportional to the inverse square of its mass (1/M^2).
I think maybe he meant “ten billion miles across”. IIRC the Universe itself is only judged to be 10-20 billion years old. He’s asserting that our galaxy would collapse into a black hole the size of the observable universe.
Cecil’s point about the black hole not being dense is true, when you consider how large the event horizon is. The area of no return can encompass a huge amount of space, which is mostly empty. All the matter/energy is in the singularity.
After all, the milky way itself isn’t dense, it is mostly empty space. When you compress it into a singularity, the singularity is infinitely dense, not because it has infinite mass, but because it is infinitely small (supposedly). There is a definite quantity of mass, and when you divide that amount of mass by the huge amount of space defined by the black hole’s event horizon, you get something that is not dense, over all.
However, Cecil does make some other errors in his article which we’ll chalk up to the state of the science in the early 80’s. Black holes don’t have to have an accretion disk. Time does not slow down for the one falling into a black hole, it only appears that way to an outside observer. You need a star with forty times the mass of our Sun to create a black hole. And tidal gravity forces won’t necessarily tear you apart immediately upon entering the event horizon.
Hhmmm. If I plug in for a Vesc of c and use 100 billion suns I get roughly 1000 AUs in “radius”. Maybe Chronos will be by to correct us. Or maybe this was answered years ago.
For stellar-mass BHs (black holes formed when a large star exploads), I’m pretty sure that tidal forces would indeed tear you apart as you approach the event horizon. For a large-enough BH, perhaps like those believed to be in the centers of galaxies, the tidal forces might not be so great near the event horizon. But I’m not sure how massive a BH would have to be so that the tidal forces would not tear you apart near the event horizon.
As for accretion disks, I guess that it’s technically true that BHs don’t have to have one, but given the abundance of interstellar dust and gas, I think that almost all of them are sure to have one, even if it is not very dense or active. In fact, if I remember my astronomy, I believe that most stars are part of a multiple-star system. In many multiple-star systems, you have two stars orbiting each other at fairly close range, such that if one of them becomes a BH, the other feeds material into it.
Some day, many trillions of years from now, all the BHs will have eaten up their accretion disks and will only grow larger from chance encounters with other objects. But for now, I think it’s a safe bet that any BH you are likely to find will have an accretion disk.
I assume the calculation of the radius of the black hole takes into account relativistic effects from the intense gravity?
Also, is his proposition that we could be living in a massive black hole correct? Wouldn’t all the matter in the universe contract to form a black hole if this is correct?
When it comes to calculating the radius of a black hole, amateur physicists are lucky. Doing it the old-fashioned way makes two mistakes which cancel out, giving you the same answer you’d get if you did it the relativistic way.
A black hole the mass of our galaxy would have a radius equal to…
… 3×10[sup]16[/sup] cm
… 184 billion miles
… 0.31 light years
… 1974 AUs.
However, a black hole the mass of the observable universe would have a radius of about 10 billion light-years. Maybe that’s what Cecil meant.
The first is supposing that the kinetic energy of a photon is equal to 1/2 mv[sup]2[/sup], and the second is that the gravity from a black hole can accurately be described with the formula F = GMm/r[sup]2[/sup]. Neither of these things is true, of course. I think they’re both off by a factor of 2 (in opposite directions), but I’m not positive.
This is according to The Physical Universe, by Frank Shu. I’m not qualified to explain the subtle differences between a classical treatment and a relativistic one.
Noone said anything about spaghettification, which is a cool thing about black holes.
The gravitational pull increases exponentially the closer you get to the middle of the black hole. So when you’re pretty close, assuming your flying towards it, say, head first, the pull on your legs is a minute fraction of the pull on your head, because your head is closer to the centre so it’s further down the gravity well. With the massive gravitational pulls involved, this means that you get stretched out like spaghetti. Spectacular way to die. If it made a noise, the noise would be “skloompf”.
It should be mentioned that the “size” of a BH always refers the size of its event horizon. The BH itself is a point singularity.
Nothing important happens physically at the event horizon. If the BH is big enough that tidal forces are small, you wouldn’t even notice when you crossed over. Once you cross, though, there’s no going back.
Of course, calculating the size of the event horizon takes relativity into account: without relativity, there is no event horizon.
On your second question, if we are living in the type of universe that ends in a “Big Crunch”, that would be similar to being inside a BH and ending up at the singularity. Recent observations make this scenario unlikely.
What Achernar means is that if you set the Newtonian escape speed of an object to c, you’ll end up with the correct formula for the Schwartszschild radius, but it’s really just a coincidence that it’s right. The calculation for escape speed depends on the formula for the energy of an object.
As for spaghettification (which, incidentally, is the same thing as “being torn apart by the tidal forces”, so it was mentioned earlier), does anyone know the ultimate tensile strength of the human body? I’ll work from the assumption that it’s about ten times a person’s (Earthly) weight, or 10 kN in round numbers (I’m using a 100 kg person, 2 m long). Tidal force is equal to [symbol]D[/symbol]F = 2GMmL/r[sup]3[/sup], where G is Newton’s constant, M is the mass of the gravitating object, m is the test mass, L is the length of the test mass, and r is the distance between them. In other words, this is the difference between the force on your head and your feet. We want to know how big a hole has to be, so that you don’t get noodled when crossing the event horizon, so r = 2GM/c[sup]2[/sup]. So now we have 10000 N = 100 kg * 2 m * c[sup]6[/sup] / (2GM)[sup]2[/sup]. Solving this, I get that M = 2.86*10[sup]34[/sup] kg, or about 14000 solar masses. A typical stellar black hole is on the order of 3-10 solar masses, so it’s quite capable of turning you into pasta, but a typical galactic black hole is a million solar masses or more, so you’re quite capable of surviving past the event horizon.
As for time till you reach the center: In a nice, safe reference frame outside of the hole, it’ll take forever for you to reach the horizon. But if you’re falling in, that’s not what you care about: You want to know how long it’ll be in a free-falling reference frame. There, the news isn’t so good: In your reference frame, you’ll reach the center in a very finite time indeed. The time it takes you to reach the center is of the order of the mass of the hole (with appropriate factors of G and C thrown in to make the units right), which works out to about ten microseconds per solar mass. So even if you’re falling into one of the billion-solar-mass leviathans, it’ll still only take you a few hours to reach the center.
I always wondered why Captain Kirk and his Enterprise always had so much trouble leaving our galaxy. I chalked that up to the small-minded 60’s approach to sci-fi. HOWEVER … this galaxy-as-black-hole makes that episode make a lot more sense. Can’t quite figure out though, why we would have no trouble travelling TO the event horizon and then find it impossible to pass throught it. Shouldn’t it get progressively harder to travel outwards from the singularity of a BH?
The Enterprise couldn’t leave the galaxy (or even explore a very large portion of it) due to the huge distances involved – it had nothing to do with gravity. The speculation above is that the universe could be a black hole, not the galaxy. Two very different concepts – look them up if you don’t know the difference. No one is suggesting that gravitational attraction is keeping you from leaving the galaxy (assuming you can achieve escape velocity), just that it would take a very long time.
And no one said that one would be able to travel back to the event horizon from the inside, but be unable to pass through it. Once past the event horizon, you can’t travel backwards (away from the singularity) at all. All you could conceivably do is slow yourself down somewhat. The event horizon marks the boundary between where you (at least theoretically) could apply enough acceleration to stop and back off, and where you must continue forward to the singularity, regardless of your acceleration.
bryanmcc, that’s assuming you’re talking about a Schwarzschild black hole with a singularity of mass M at the center. However, if the Universe is something like a black hole then it’s certainly not that. I admit I don’t know how black holes with non-singular mass distributions behave.
Galaxy-as-black-hole would make a lot of sense, I think, keeping someone in the galaxy, but unfortunately, Cecil’s column seems to have an editing error on this part, and no such thing is the case.
