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I understand - I think! - that the gravity field of a black hole is so great that light photons from within the event horizon (or Schwarzschild limit)  cannot leave it - hence the name. However I recently read that the mass of a black hole - which is after all just the crushed core of its parent star - is the same mass that the core had before the star's collapse. That seems to make sense. No additional mass has been acquired in the meantime, I assume. It would follow, it seems to me, that its gravity cannot be any greater than the gravity the core had beforehand. Yet we never hear about the star, before its death, having a gravitational field so strong that light could not escape from it. Why and how is the black hole's gravity so much stronger after the core collapse, that even light cannot escape, when its mass is no greater than it was before when light could escape it perfectly easily?

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I have never looked at it that way, good question, i have a sneaking suspicion a gentlemen will respond soon, (Vlaiv) with the answer you seek.

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This puzzles me too for example if our sun suddenly transformed into a black hole of the same mass would our orbit change? 

Alan

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The sun is not massive enough to form a black hole.

It is the energy density that matters

 In the star it is too low by when it contracts enough the density gets high enough for an event horizon to form.

Far enough away the "pull" is the same before and after (assuming no mass energy is lost).

Regards Andrew

 

 

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This is a gross oversimplification but...

Remember that the mass if he star has collapsed into a point of infinite density and this causes the curvature of spacetime at the point of singularity to become infinite as well.  It's this that prevents anything from escaping.   Think of it this way.  A photon entering the region outside the black hole's influence can travel in any direction.  As it approaches the black hole, the directions of travel available to it begin to disappear until it can only head into the singularity itself.  That point is defined by the event horizon.

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18 minutes ago, Alien 13 said:

This puzzles me too for example if our sun suddenly transformed into a black hole of the same mass would our orbit change? 

Alan

No.  The Earth would experience the same gravitational pull as it did before the Sun collapsed (which it actually couldn't - too little mass available).  It would continue to orbit normally.  

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Even if we don't account for curvature of space time and energy density - there is explanation for this. Even in Newtonian gravity, highest gravity that one can "feel" on a spherical solid body is at surface. When you start descending into a shaft and reach some depth - gravity pull on you is equal to what it would be if you took all the mass that is located at shorter distance to center than you and you put it at single point at the center. All the mass that is "above" you - simply has no gravitational effect on you because it cancels out.

Two important points:

1. only stuff that is closer to center than a given point will act as mass exerting gravity at that point

2. gravity acts as if all that mass is concentrated in center (for spherical objects)

This means from number two that anyone sitting in orbit around the star when it collapses into a black hole - will be ok and continue to orbit - gravity pull is equal to all the mass that is closer to center (again whole star) and as if all that mass was in a single point (after it collapses it effectively is in a single point - but it also acted like that before collapse).

And it also means that density needs to change. If earth had only half of radius it now has - Moon would still orbit it regularly, but we would feel much larger gravity force on the surface. It is this reduction in size and increase in density that causes bending of space time and ultimately causes singularity - like Andrew said above - it is energy density that governs shape of space time.

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lightcone-bh.png.78f2e87af77d281a8076d75c7d3957fc.png

This image shows how the light cones of objects are effected as a black hole forms. Far enough away nothing happens and the world line continue as before, but as you get closer you see the light cones tip towards the singularity. At the event horizon the light ray touch the world line of the event horizon. Further in toward the singularity all world lines tilt more and more and terminate on the singulariry.

Strictly speaking the singularity is not a point in space time.

Regards Andrew

Edited by andrew s

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I'm not sure that graph is entirely correct - light cones of object past event horizon can't turn more than 45 degrees, can they? In that case - light is traveling backward in time?

 

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8 minutes ago, vlaiv said:

I'm not sure that graph is entirely correct - light cones of object past event horizon can't turn more than 45 degrees, can they? In that case - light is traveling backward in time?

 

It is diagrammatic. The small light cones are drawn as for a flat space time and should be valid only at an infinitesimal point.  A better diagram would have the cones narrowing and eventually being time like so the light lines don't tip back in time as you point out.

Regards Andrew 

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16 hours ago, piff said:

 Why and how is the black hole's gravity so much stronger after the core collapse, that even light cannot escape, when its mass is no greater than it was before when light could escape it perfectly easily?

For a collapsed star the diameter of the event horizon is much smaller than the diameter of the original star so any photon emitted at the same location as the surface of the star was would have no difficulty escaping even after the collapse.  After collapse, as you get closer to the black hole  the force of gravity increases as the inverse square of the distance to the point where at the event horizon the photon cannot escape

Edited by robin_astro

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My understanding of the original question and the framing of the answer is that I am not very smart. I was asked this very same question by a phd in Chemistry a month ago and tried to explain but realised I didn’t have the scientific tools to answer correctly.

In landscape gardeners terms which can be shortened to layman’s terms, take all the material of the object (red super giant) crush it down under its own gravity to a point where it becomes so dense (super nova) that the fabric of space time can no longer support it, so it warps the fabric of space so massively it ‘is’ a black hole.

Before the supernova of the RSG it’s mass is the same but not condensed into a tiny space so warps space time much less. Thank you Professor Brian Cox.

Now to get shot down by loads of people who are not Landscape gardeners😂

Marv

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You don't need a lot of mass to form a black hole, you just need a lot of density. If the Earth was squeezed into a sphere 17.4mm in diameter, it would be a black hole. If you spread out the same matter over an Earth sized sphere the gravity at the surface is only 1g or course. By the way, as you travel into a uniform sphere the gravitational force experienced is equal to the matter closer to the center of the sphere than you. So if you went to the center of the Earth you would be floating about in free fall!

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HI Marvin - thanks for your input. But doesn't the degree of "warping" depend purely on the mass involved, rather than the volume? This is where I got confused and why I posed the original question - the volume is the only thing that has changed. The mass remains the same. 

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It's the density that matters and the density that causes the curvature of spacetime to increase.  When the star runs out of fuel, there is no longer sufficient pressure of fusion to hold the outer layers up against gravity, the star is no longer in hydrostatic equilibrium and the outer layers rapidly collapse into the core which then further collapses until the density is infinite.  

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9 hours ago, piff said:

HI Marvin - thanks for your input. But doesn't the degree of "warping" depend purely on the mass involved, rather than the volume? This is where I got confused and why I posed the original question - the volume is the only thing that has changed. The mass remains the same. 

In Newtonian gravity - force depends on distance, right? Imagine you are standing on the surface of the earth - you will experience certain force (we experience right now). Now imagine all that mass of the earth is condensed into sphere half the radius of the earth - do the math and you will see that force at the surface of such object has gone up by factor of 4 (depends on square of the distance). How come? It is the same mass, you are standing on the surface of it - it is the density that changed.

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To add to that, even if Earth is compressed into a black hole, the gravitational strength experienced at 6371km from the black hole would still be 1g. 

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12 hours ago, piff said:

the volume is the only thing that has changed. The mass remains the same. 

You are confusing the strength of the gravitational field (which  depends only on the mass and is therefore unchanged by the collapse) with the force felt by an object in that field which is proportional to the mass and inversely proportional to the square of the distance. After the collapse the mass is concentrated into a single point so you can venture much closer to the centre of mass, compared to when it was a star,  where the force is much higher, until eventually at the event horizon it is high enough to prevent escape of the photon

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29 minutes ago, robin_astro said:

You are confusing the strength of the gravitational field (which  depends only on the mass and is therefore unchanged by the collapse) with the force felt by an object in that field which is proportional to the mass and inversely proportional to the square of the distance. After the collapse the mass is concentrated into a single point so you can venture much closer to the centre of mass, compared to when it was a star,  where the force is much higher, until eventually at the event horizon it is high enough to prevent escape of the photon

Or if you want to think of it in terms of the warping of spacetime, the warping is only sufficient to prevent the photon escaping when at a distance from the singularity much smaller than the radius of the original star

Edited by robin_astro

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@vlaiv here is a more accurate diagram for a non rotating black hole in Kerr metric coordinates. The radius of the event (Schwarzchild) horizon is r = 2GM/c^2 or about 3 km for a 1 solar mass object.

1954890201_bhlightcones.gif.74befb4a17365fd8df08a1493f1350b1.gif

See here www.astro.ucla.edu/~wright/bh-st.html for the explanation.

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Thanks, everyone who has contributed. I think I have grasped the idea now.

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