If Light Can't Escape a Black Hole, Why Not Stars of the Same Mass?

February 19, 2016 6:54 AM

I was stumbling around trying to explain to someone that sometimes a black hole is created at the end of a star's life cycle. And they said....

..."Well, if light can't escape a black hole because its mass creates enough gravity to prevent that, why did light escape the original star, which presumably had at least as much mass as the black hole it spawned?"

Ummm...?

Is concentration of mass a factor?

posted by justcorbly to Science & Nature (13 answers total)

Yes. The key concept here is the Schwarzschild radius. As an object gets denser, its escape velocity increases. When the escape velocity reaches the speed of light, light cannot escape.
posted by ubiquity at 7:00 AM on February 19, 2016 [11 favorites]

By the way, stars are constantly converting their mass into energy, much of which radiates away. So the original star had much MORE mass than it did before it collapsed. But it was a LOT bigger.
posted by ubiquity at 7:06 AM on February 19, 2016 [1 favorite]

Yes, density matters. If you shrunk the Earth down so that it's whole mass fit in the size of a pea, it would act as a black hole (on non-preview, that's the Earth's Schwarzchild radius). The moon would go on orbiting Earth just the same, and the Earth-black-hole would keep on orbiting the sun, because the mass is the same--it's just compressed into a smaller space. This is one consequence of Newton's Shell theorem: if a mass is contained in a spherical shell, when you're outside of the shell, you can treat the mass as if it's all at one point in the center.

So why isn't the Earth a black hole? Because the other consequence of the theorem is that, if you have a hollow, symmetric shell, and you're inside the hollow, you don't feel any gravitational forces from the shell itself--they cancel out.

So, let's say you're at a point midway between the Earth's surface and the core. You wouldn't feel any gravity from the earth's mass above you, only from what's beneath you (where "above" is anything further than you from the core, including on the other side of the planet; same thing for below). So the net gravity you'd feel as you moved towards the center of the earth would eventually decline to zero. If you could, somehow, emit light from the center of the earth and have it bypass all that mass, it would have no trouble escaping.
posted by thecaddy at 7:10 AM on February 19, 2016 [2 favorites]

To boil down what everyone above said to layman's terms, the reason that concentration of mass is a factor is because gravity gets much stronger over shorter distances. That's why we stick to the earth, but don't ourselves feel the gravity of the much larger sun.

In the case of a black hole versus a star of the same mass: Say the star is 1,000,000 kilometers wide. Light that is escaping its surface is 500,000 km from its center of mass, so gravity is weak. If the star becomes a black hole with the same mass but only 1 inch wide, the light at the surface is now just half an inch from the center of mass, so the gravity is millions of times stronger, and could be too strong for the light to escape.
posted by ejs at 7:27 AM on February 19, 2016 [6 favorites]

Yes to all previous answers. I'll add that in your original example, why did light escape the original star, light was escaping from the surface of the star, at a radius R from the center. Now collapse the start into a black hole, same mass but concentrated at the center, and light can still escape from that radius R. That former surface distance is not considered to be "inside the black hole", because R is much larger than the Schwartzchild radius. We say "no light escapes from the black hole" because we're generally talking about light generated by the massive object, and we've established that the mass is concentrated at the center point, inside the Schwartzchild radius.
posted by aimedwander at 8:07 AM on February 19, 2016 [1 favorite]

Note that Schwarzschild radius depends only on the mass of an object (plus constants). So, every object has a Schwarzschild radius that's relatively easy to figure out (fun exercise: compute Donald Trump's Schwarzschild radius).

The thing that makes a black hole special is that its physical radius (i.e. size) is contained to within its Schwarzschild radius. So, if you completed the exercise from earlier, you now know how small we need to compress Trump in order to never hear from him again.
posted by Betelgeuse at 9:10 AM on February 19, 2016 [2 favorites]

The location of the event horizon is indeed given by the Schwarzschild radius, if the black hole is not spinning. However if it is spinning, you should instead study the Kerr solution, for rotating black holes. A black hole spinning as fast as it can has an event horizon (at the closest point) that is only half of the Schwarzschild radius.

So, if you think there might be spin involved in thinking about Donald Trump, you have to compress him twice as much.
posted by nat at 9:44 AM on February 19, 2016 [1 favorite]

"The thing that makes a black hole special is that its physical radius (i.e. size) is contained to within its Schwarzschild radius. So, if you completed the exercise from earlier, you now know how small we need to compress Trump in order to never hear from him again."
--posted by Betelgeuse at 12:10 PM on February 19

I'm not a politician or an astrophysicist, but since sound waves travel much slower than light waves, wouldn't the required compression of Trump be of a different degree (larger radius?) if all we aimed to do was not hear him?
posted by baseballpajamas at 10:22 AM on February 19, 2016

To supplement the other good answers above, it's worth emphasizing that concentration of mass isn't just a factor -- it's (more or less) the only factor. In theory, you can have black holes at any mass at all, so long as the density is sufficiently high (ignoring quantum mechanics). But in general, the only way to get densities high enough (that is, to cram the entire mass of an object to within its Schwartzchild radius) is to rely on gravitational collapse, so you need the mass of a big star to create one, which is why we associate black holes with being really massive, in addition to being really dense.
posted by biogeo at 12:18 PM on February 19, 2016 [1 favorite]

To add on to the excellent, this is a case where the simple math might help understand how big the difference is.

The sun is 700,000 kilometers across. The Schwarzschild radius is 3 kilometers. Or, you are 200,000 times further away from the all that mass.

Since gravity decreases as the square of the distance, this means that if the sun did shrink so you were within just a few kilometers of that same mass, you would experience gravity that 200,000 * 200,000, or a mind bending 40 billion times, stronger.
posted by mark k at 12:01 AM on February 20, 2016

If you fancy a visual metaphor, take the perennial rubber-sheet-as-space idea, as beloved by pop science TV (square grid optional). As you'll have seen countless times, if you have a strong enough rubber sheet and you put a bowling ball on it, it'll create a depression in that sheet, which is a bit like how a mass distorts the space it's in. A marble rolling across the sheet that came close to the bowling ball would appear to be attracted towards it. If the marble's going fast enough, it'll be deflected on its path but will carry on; if slower, it'll run around the ball in a pleasingly orbital fashion.

But if you take a piece of some very dense material that weighs the same as a bowling ball but is only the size of a grain of rice, it'll make a much deeper impression on the rubber and perhaps even make a hole - much as you can push a needle through a balloon but not your finger, even though you use the same force. If the marble gets too close to the hole, no matter how fast it's going, it falls in and cannot come out; further away, it'll be deflected as before.

This is a bad metaphor that should not be used near a physicist (if they do get grumpy, however, and they're old enough that you can outrun them, ask how that quantum gravity thing's working out). However, as a first-order thought experiment that gets the -ahem - point across about gravity, mass and density, it's been useful for me in the past.
posted by Devonian at 7:19 AM on February 20, 2016

That's why we stick to the earth, but don't ourselves feel the gravity of the much larger sun.
It's certainly true that the gravity from the Sun is tiny in comparison to the gravity from Earth at the Earth's surface, but we're also in freefall around the Sun so you wouldn't feel it even if it were a lot stronger.
posted by edd at 3:01 AM on February 23, 2016

There's something to be said for it not quite simply being that gravity is too strong for light to escape. I mean, I can't reach escape velocity by jumping, but if you built me a big enough stair case I could walk up an awful lot further and only time, patience, and the essentials of staying alive would present any limit.

In other words, a Newtonian 'black hole' would be quite escapable, even if light wouldn't in practice do a good job of it. General relativity and all the spacetime bending makes such objects completely inescapable though, once you've got too close. It's not simply that the pull is really strong - it's that the pull is so strong that quite literally all your possible paths going forward in time take you closer in.
posted by edd at 5:57 AM on February 23, 2016