Does the mass of a star change as it collapses into a black hole?

I know (I think!) that when a really big star collapses on itself it creates a black hole.

My question: When a star collapses, is the mass equal to the mass of the star when it's not a black hole? Or does it change while collapsing?

This question came to me and my friend while studying Newton's law: $$F=G \frac{m_1 \cdot m_2}{r^2}$$ If the mass of the star doesn't change, then it can't have enough force to "eat" light (unless it has that force in the first place). Does the force change because of the density?

• but if the law contains r, when the star collapses r decreases, so F increases – adhara99 Apr 15 '14 at 19:46
• I've edited your question to improve the wording. Could you verify that it still accurately reflects what you wanted to ask? – David Z Apr 15 '14 at 19:52
• Yeah thanks a lot! As I said English is not my mother tongue... – Peterix Apr 15 '14 at 19:53
• – Qmechanic Aug 14 '14 at 20:52

The formula $F=G \frac{m_1 \cdot m_2}{r^2}$ is valid only for point masses. However, it can be applied to non-point masses if its spherically symmetric. Enter Shell Theorem:

1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.

So, when a spherically symmetric massive star attracts an object at its surface, its like its actually attracting that object from a distance equal to its radius.

2.If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

So, if you put something near center of that star, it can still escape because it is experiencing force due to only mass below it.

But, when that star collapses to a smaller volume, force due to whole mass on surface increases (because its inversely proportional to $r$) which makes escaping tougher (required escape velocity increases). When this radius is decreased to Schwarzschild radius, the escape velocity exceeeds $c$.

During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly.

Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) For a black hole, however, the escape velocity at the event horizon (the "edge," in some sense) of the black hole is the speed of light, $c = 300,000 \text{ km/s}$. For anything within the event horizon to escape the pull of a black hole, it must exceed the speed of light, a physical impossibility. There's a certain mass-dependent radius - the Schwarzschild radius - to which an object must shrink in order to become a black hole.

Newton's Law of Universal Gravitation, which you stated, doesn't apply in its standard form to light. Rather, you need to use Einstein's general relativity, which considers gravitational forces in a much different light than Newton did. However, Newton's Law of Gravitation can intuitively apply here: when a star collapses, $m$ does decreases. However, $r$ becomes much smaller, so the net effect of these changes is the creating of a stronger gravitational force on the surface of the remaining object.

• Your first paragraph is very misleading. A 15 solar mass star will blow off something like 13.5 solar masses, leaving a 1.5 solar mass neutron star. Larger stellar remnants (>3 solar masses) will create that size black hole. – Kyle Kanos Apr 15 '14 at 21:26
• @KyleKanos I've made an edit; thanks for the catch. – Shivam Sarodia Apr 15 '14 at 22:15
• Is the escape velocity for Earth 11.2 km/s on the surface, at the edge of space, or everywhere? (And yes, there is an edge of space). – trysis Apr 16 '14 at 0:49
• @trysis: On the surface. And no, it doesn't account for the presence of the atmosphere, so it's not actually sufficient to escape the Earth's gravity if you're starting from the surface. (In fact, an object moving at 11.2 km/s or faster near the Earth's surface would most likely shortly turn into a ball of incandescent gas due to aerodynamic heating. To actually get into space from the Earth's surface, you need to start relatively slowly until you've cleared most of the atmosphere.) – Ilmari Karonen Apr 16 '14 at 1:29
• I don't think this really answers the question. For a start, some stars (probably) collapse directly to a black hole without a supernova. Secondly, the question is about whether a collapsed black hole has the same mass as the material from which it formed. I think the answer needs careful discussion considering all contributors to the stress-energy tensor. – Rob Jeffries Apr 11 '16 at 16:58

I was considering this question as well. Neither matter/energy can be created or destroyed, only converted from one form to the other. Consumption of star stuff by a singularity would convert that star stuff into star energy, but with no other way of expressing that energy there seems no choice but for nature to convert with 100% efficiency that matter into gravitational energy. When I model this in my mind I can envision a neutron star that collides with an object that tips the scales, the straw that breaks the camels back, and the neutrons give up their integrity and the whole thing drops into a singularity, and as this happens an immense gravity wave, a gravity tsunami, propels itself into the universe, leaving an object that has more mass than it did before it collapsed. Exactly how much more is beyond my math skills.

• welcome to Physics.SE. There are numerous inaccuracies with this answer. Answers on this site should be constrained to factual, scientific content instead of opinions and imaginations. – DilithiumMatrix Apr 11 '16 at 15:47
• A spherically symmetric collapse produces no gravitational waves (NB: not "gravity waves"). – Rob Jeffries Apr 11 '16 at 16:59

protected by Qmechanic♦Apr 11 '16 at 15:22

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