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If I have a black hole with a mass that's exactly the same mass as a star, why does the black hole warp spacetime so much more (light can’t escape) than a star (light can escape) with the exact same mass?

Is it due to the black hole having a singularity, or it being more dense than the star, or something else?

Peter Mortensen
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Rory Shaughnessy
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    It doesn't. This question can be answered in spirit using Newtonian gravity only. A point mass has gravitational potential proportional to $1/r$. A perfectly spherical mass distribution also has this potential but only outside its radius. A star is usually taken to have a finite radius $r_0$, so the $1/r$ formula (which goes to $\infty$ as $r\rightarrow 0$ is only valid for $r>r_0$. By contrast, in this context we can take a black hole to be a point mass, so the potential can be increased indefinitely by approaching its center. – Bence Racskó Jul 31 '18 at 16:00
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    @Uldreth, shouldn't that be an answer rather than a comment? – Hal Hollis Jul 31 '18 at 16:29
  • Why do you claim that black holes warp spacetime so much more? What do you mean by this? – sammy gerbil Aug 01 '18 at 13:24

2 Answers2

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They don't. The gravitational field is the same outside all spherical objects with a given mass. But a black hole is much much smaller than a star with the same mass, so you have access to regions much closer to the center, where the gravitational field is stronger. You can certainly try going inside a star to get close to its center, but then the field stops increasing because most of the mass is now outside you.

To illustrate, a black hole with the mass of the Sun would have a radius of around 3 km, while the Sun's radius is 700000 km. You have the same mass concentrated in a ball which is 1/200000 of the size, leading to a gravitational field at the black hole's surface 200000² times stronger than that at the Sun's surface.

Javier
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    ...Or, if you were to orbit that black hole at a radius of 700 000 km, then the strength of gravity at that distance would be the same as the strength of gravity at the surface of the Sun. – Solomon Slow Jul 31 '18 at 16:18
  • Thanks, I get it using Newtonian Gravity, but is there anyway to describe it in terms of general relativity (like how it warps spacetime), or can you just use Newtonian gravity to describe the gravity that black holes have? – Rory Shaughnessy Jul 31 '18 at 16:56
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    @Murphy you can't use Newtonian gravity to describe a black hole, but the idea works the same in general relativity. The field no longer goes exactly as $1/r^2$, but that's a technical detail. – Javier Jul 31 '18 at 17:27
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    Dumb question, but how do you figure a black hole with the Sun's mass has radius 3 km? Aren't black holes point masses? A 3 km object with the Sun's mass would have only an order of magnitude higher density than a neutron star. – Reinstate Monica Jul 31 '18 at 18:59
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    @Solomonoff'sSecret Conventionally, the radius of a black hole is the radius of its event horizon (the Schwarzchild radius). That's where the 3km number comes from. – zwol Jul 31 '18 at 19:10
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    @Solomonoff'sSecret That concept of point masses exist inside black holes is a mathematical idea. We (by definition) have no way get information about what happens inside the event horizon. – JimmyJames Jul 31 '18 at 20:43
  • @JimmyJames Well, we do know what happens: Time and space switch roles (things get weird inside the event horizon). You are correct that we can't test those mathematical representations, but we can model them and be pretty confident that the model is accurate. – Draco18s no longer trusts SE Jul 31 '18 at 23:08
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    We can have no such confidence. The model assumes there's nothing fundamentally different about the universe inside a black hole, which is just that - an assumption. – Turksarama Aug 01 '18 at 02:28
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    Nitpick: No. If you orbit the star at the distances where the black hole's gravity is extreme you're deep inside the star and all the mass outside your orbit contributes nothing to the gravity you experience. Thus the gravity never gets nearly as extreme in the star as it does near the black hole. – Loren Pechtel Aug 01 '18 at 04:01
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    @LorenPechtel I don't think that's what Javier is saying; but on second reading, it certainly is a bit confusing. Maybe it needs saying explicitly that once you get into the mass of the star, the gravity "flattens out" and the gradient no longer increases . – Luaan Aug 01 '18 at 06:34
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    Is it really correct to say that the gravitational field is stronger? AFAIK, the sum of gravity is the same, it's just that the "center field" is much more smaller and spread over a smaller volume. – phresnel Aug 01 '18 at 07:14
  • @phresnel I think you're right. In fact, the gravitational field density is higher (same field, less space), but I'm not sure if "field density" is commonly used with gravitation. – JimmyB Aug 01 '18 at 10:46
  • Your last sentence needs the words "at the surface" added to it; and maybe size replaced with radius (as size for a 3 dimensional object is ambiguous; for a black hole, maybe 50% more ambiguous...) – Yakk Aug 01 '18 at 13:12
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    @phresnel There's no such thing as "sum of gravity" or "field density". The field strength is just the field strength, proportional to the gravitational force. It is definitely stronger at the surface of a black hole than at the surface of a star. – Javier Aug 01 '18 at 13:14
  • @Javier: It was just a model for thinking (should have been "sum of 'gravity packets'"), like when you approximately integrate the constant \pi in a Monte Carlo experiment. But thanks for editing your answer (at the black hole's surface), which was the crucial piece missing. – phresnel Aug 01 '18 at 15:23
  • @phresnel That isn't a thing either. Can you provide some reference for the idea that the field can have a density instead of being stronger? – Javier Aug 01 '18 at 15:26
  • @Javier: I know that this isn't a thing either; this is why I said "thought model". Let's not escalate this discussion and just agree that your sentence made no sense without your edit, because outside the original body's circumference (before collapsing into a black hole), there is no difference in gravity. – phresnel Aug 01 '18 at 15:28
  • @Luaan Look at Stilez's answer for a much more detailed exploration of what I am saying. – Loren Pechtel Aug 01 '18 at 23:54
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Note: this is a simplified answer, aimed at the level of the question. It isn't technically precise, but concepts like "below" are probably much easier to visualise than more exact terminology.

The answer by Javier is correct, but some elaboration might help.

When we discuss ordinary isolated astronomical objects like planets, stars, neutron stars, and black holes, the intensity of the gravitational force is governed by two things: how much mass there is, "below" you, and how far from the "centre" you are.

How much mass "below" you

Using fairly ordinary calculus, we can show that if we have a spherically symmetrical mass (which roughly describes any planet, star or black hole), then the only gravity you experience from it, comes from the mass "below" you.

Example: the earth is a sphere about 8000 miles radius.

  • If you stand on the surface, you feel the force of gravity from the entire earth's mass, at a distance of 8000 miles from its centre.
  • But if we could go 2000 miles down, you would experience the force of gravity for a sphere with the mass of only the inner 6000 miles of the earth, not the whole 8000 miles, which would be weaker. The outer 2000 miles wouldn't have any gravitational effect overall, either plus or minus.
  • But.... at 2000 miles down, you would also be much closer to the centre of the earth as well, and this would also make the force of gravity stronger, somewhat counteracting the reduced mass.

How close to the "centre" you are:

The closer you get to a mass, the more intense the force of gravity from it.

One exception is the example above - if being closer also means being inside it, then in effect, there will be less mass to act on you.

Your question: star vs. black hole:

Imagine the sun, compared to a black hole with the sun's mass. In this case you aren't "inside", so the only things affecting how intense gravity is, are the mass - which is the same - and the distance from the "centre".

The sun has a radius of 700,000 km. The black hole has a radius of 3km. They both have the same mass "below" them. The intensity of the gravitational field is proportional to the square of the distance.

Because the mass "below" is the same, but the distance is 233,000 x less, gravity is 233,000^2 = 55 billion times stronger at the 3km "boundary" of the black hole. (3km is the closest anything can get to the black hole without being "lost" to us within it)

That 55 billion times stronger gravity is why light is much easier to see bending round the edge of a black hole than the edge of the sun. It's also why any closer and light can't escape a black hole, while it can escape the sun.

But the effect exists for both of them. Using very careful measurement, we can see the mass of the sun also bending light, and warping spacetime. It's just that the effect is so small it would be very hard to detect with the naked eye.

Stilez
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  • All assuming no rotation? Rotation and the resulting oblateness will affect the orbits of anything orbiting. The effect is very significant for artificial satellites in low earth orbit (say, 4 ° per day). – Peter Mortensen Jul 31 '18 at 21:32
  • But in broad terms, to understand the core principles in the OP, rotation is almost totally irrelevant, or at best a minor footnote. In practical non-esoteric cases, it doesn't have significant relevance to the OP. – Stilez Jul 31 '18 at 22:03
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    you would experience the force of gravity for a sphere with the mass of only the inner 6000 miles of the earth, not the whole 8000 miles, which would be weaker. Strictly speaking, the mass of the shell above you actually has zero effect on you (the mass above you pulls in one direction, the mass at the same elevation on the other side of the planet pulls in the other; there's more of it but it's also farther away). https://en.m.wikipedia.org/wiki/Shell_theorem – Draco18s no longer trusts SE Jul 31 '18 at 23:12
  • Yes. Although I think at this simple level, this is mostly semantics. If gravity inside a shell has a net zero effect, then you would "experience" only the effect of gravity from the non-shell portion. At this level, a net zero element/effect plus a positive element/effect is reasonable to describe as only * experiencing * the positive element. – Stilez Aug 01 '18 at 08:14