Black Holes & Sound

Reader Abby Normal(!) writes in with an excellent question:

Something in your post about Physics in Star Trek, May 18, 2009, has been bouncing around my brain. You stated that a black hole has the same mass, and therefore gravitational pull, as it's parent body. That makes perfect sense. But as I understand it a typical black hole is formed by a collapsing star. (Ignoring supermassive black holes, which I know form differently.) So why then does a star emit light prior to it's collapse but afterward, assuming it becomes a black hole, can light no longer escape? Does it have to do with the size of the star "spreading" the gravity over a larger area? If so, does that mean that the diameter of the event horizon is always smaller than that of the parent body shortly after formation?

In fact Abby has pretty much answered her own question quite accurately. All we have to do is fill in some detail, and we'll start by doing a few thought experiments.

Imagine yourself standing outside in a vast and open lanscape, perhaps the High Plains of Colorado. Somewhere in the middle of this wide space, you have a standard orchestra and choir spread out uniformly over a few acres. They're singing something suitably dramatic. I would choose the St. Matthew Passion, but you may feel free to substitute as desired in your own thought experiment.

Standing perhaps a half-mile from the choir, you're much farther away from the choir than the size of the choir itself. You can hear them, though it's probably a gentle and somewhat faint sound. Now give a prearranged signal and the company rearranges itself as it keeps playing, collapsing toward the center and forming itself into a more usual configuration that fits neatly into an area the size of an orchestra pit and stage. From your faraway vantage point the loudness of the choir will not change. It's still the same number of musicians putting out the same amount of sound.

But repeat the experiment, except this time you start off standing on the fringes of the widely scattered choir. Now when you order the choir to collapse down into its compact form the the volume at your location still won't change if you stay put. But if you follow along with the choir so that now you're standing right at the edge of the compactly arrayed choir then the song will be quite loud indeed. The loudness isn't determined by the physical dispersion of the choir, so long as the number of musicians doesn't change. Instead the loudness is determined by how far you are from the center of the choir.

Stars collapsing are the exact same, except the singers represent mass and the sound represents the gravitational field.

Well I'm sure that sounds a little sketchy. To the extent that it is, it's that sound is an imperfect metaphor for a gravitational field and the musicians are an imperfect metaphor for each little atom of mass in the star. And there are some provisos you have to attach to the geometry both of the choir and the star - mostly, that the density is uniform (or at least only a function of distance from the center) and they're arranged in a circle (for the choir) or a sphere (for the star).

But if so then the analogy is reasonably close*. A black hole's gravity is not any stronger as seen from a distance, it's that now instead of coming from a wide area it's all coming from one point. As you get close to that point the field strength will be very strong.

Correspondingly and in answer to the last part of the question, the event horizon is always smaller than the parent body. Usually a lot smaller. An event horizon would correspond to a particular loudness, and that loudness won't exist anywhere until the choir reaches its appropriate critical density. For a star, once that critical gravitational field strength is reached during the collapse a black hole is formed.

Running through the mathematics of all this to work the bugs out of the analogy will be a project for another time. If you'd like a head start, I'd read up a bit on the shell theorem and the hoop conjecture. If you are very very brave - braver than me - you might try Gravitation by Misner, Thorne, and Wheeler. Until then, happy choir listening.

*So long as you never walk into the choir itself. Gravity decreases as you go into the interior of a massive object as the mass above you cancels the mass below. Sound can't approximate that in any reasonable way.

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Hey look, I'm famous! Thanks Matt, for answering my questions. I'm glad to know that I was conceptually on the right track. The orchestra analogy helped. But interestingly it was the caveat about gravity diminishing as one enters the interior of an object that really filled in the missing piece of the puzzle for me, as well as a forehead smacking "Duh!" moment. I love those moments.

By Abby Normal (not verified) on 04 Jun 2009 #permalink

And here I expected a post on dumb holes (which, in my opinion, are one of the best ways of getting a handle on black holes).

By Physicalist (not verified) on 04 Jun 2009 #permalink

"Gravity decreases as you go into the interior of a massive object as the mass above you cancels the mass below"

I'm having a picky moment. This is not quite accurate, and the actual relationship I found quite surprising - and simple - when the problem was posed in High School Physics:

As you descend towards the center of a sphere, consider at any point that you are standing on the surface of a sphere below your feet, with a concentric spherical shell above you. The gravity you experience when standing on the surface of a sphere (or anywhere in space higher than that) is the familiar inverse-square law (GM1M2/r^2) relating your mass, the sphere's mass, and the distance from you to the center of the sphere. The surprising bit was the effect of the shell above you when you are closer to the center than the complete sphere's radius. Turns out that inside a hollow sphere the gravitational force is everywhere constant and zero. It's too long ago now for me to remember the form of the integral that proves this. Its a similar result to the one that gives a constant time to traverse a linear chord between two arbitrary points on the surface of a solid sphere assuming only gravity is at work, the tube / conveyance contact is frictionless, and you start from rest at one surface end of the tube. (IIRC, about 40 minutes for the Earth).

So the gravity you experience is that of the sphere below your feet. That mass is not cancelled by anything. However the shell above your head cancels itself out completely. Therefore, as you approach the center, the mass below you decreases and consequently so does the gravitational force you experience. Interesting also is the rate of decrease. The mass is proportional to r^3, and the g force is proportional to 1/r^2. So (I'm guessing in elderly informed way here) the g force ends up being proportional to r. A linear relationship, once you get inside.


The next obvious question to answer is, what/why an Event Horizon? Not so simple as this little bit of solid geometry. And I do not have the math for it.

By GrayGaffer (not verified) on 04 Jun 2009 #permalink

Funny, I just explained this very thing to some movie folks in a meeting last week. Indeed, the trick to a black hole's strength is the simple fact that you can get much closer to it than when it was a physically large star. More or less the same mass is there, just highly compacted into a smaller object and with the inverse square attenuation of gravitational strength in effect you eventually reach a point around the object where the escape velocity exceeds the speed of light (ie the event horizon).

Along this line, you might enjoy the paper by Hamilton and Lisle, The River Model of Black Holes. This does rotating black holes as well as non rotating.

By Carl Brannen (not verified) on 05 Jun 2009 #permalink

so bad & irritating