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This question has confused me for quite some time now. I have searched it up online, and the basic answer is: 'Mass is a form of energy. When black holes die they release the amount of energy that they should. Mass is conserved.' But there's a problem with that answer. Energy is conserved, but mass isn't; it's turned into another energy store. From what I've learned in school, the amount of mass in the universe is always the same, and that's conservation of mass. But if mass is just another type of energy and can be transferred into other types of energy, mass is certainly not conserved, and the conservation of mass doesn't exist. It's like saying conservation of kinetic energy. Kinetic energy can be transferred into the electrical or thermal or other energy stores, and although the overall energy in the universe is the same, the amount of kinetic energy has changed? Am I being stupid or is my life a lie?

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  • $\begingroup$ the amount of mass in the universe is always the same, the observable universe (which appears to be shrinking due to accelerating expansion), or the 'whole' universe (the mass of which might be ∞)? And can the interior of a black hole be considered part of either? $\endgroup$ – ᆼᆺᆼ Mar 10 at 3:29
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Conservation of mass is not a fundamental law of nature so the violation of it is not a problem. Mass is just a form of energy, and energy is the quantity that is supposed to be conserved. So since black holes do not violate the conservation of energy the problem you are worried about does not exist.

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    $\begingroup$ This is incorrect, mass is conserved, as it must be since energy and momentum are conserved and $m^2 c^2=E^2/c^2-p^2$ $\endgroup$ – Dale Mar 9 at 16:26
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    $\begingroup$ @Dale Mathematically this is right, but for the example you mentioned in your answer it seems weird to define the "mass of the system" when the system is not a bound state. $\endgroup$ – proton Mar 9 at 16:43
  • $\begingroup$ There is no restriction that all systems must be bound, and the conservation law works regardless. It can be applied to bound systems also with the same results $\endgroup$ – Dale Mar 9 at 17:55
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    $\begingroup$ @Dale OP seems to talk about the sum of rest masses of constituent particles / subsystems rather than about the rest mass of the system as a whole. $\endgroup$ – John Dvorak Mar 9 at 18:20
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    $\begingroup$ This answer is really oversimplifying the situation. It's actually nontrivial to define energy in general relativity. In fact, the most direct definition (via Noether's theorem) gives a quantity that is not conserved. $\endgroup$ – knzhou Mar 10 at 11:48
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Energy is conserved, but mass isn't; it's turned into another energy store.

Contrary to some other comments and answers, mass is in fact also locally conserved in modern physics, but it is important to understand what that means since there are some subtleties that cause confusion.

With relativity time and space were unified into a single construct called spacetime. Similarly, energy and momentum were unified into a single construct called the four-momentum, where energy is the timelike part of the four-momentum and momentum is the spacelike part of the four-momentum.

Mass is the norm of the four-momentum: $m^2 c^2 = E^2/c^2-p^2$. Since the relativistic four-momentum is conserved we automatically get the classical laws of conservation of energy, momentum, and mass. Note also, for $p=0$ we recover the famous $E=mc^2$.

So why do people claim that mass is not conserved? It is conserved, but it is not additive. The mass of a system of particles is not equal to the sum of the masses of the system.

For example, in the anhilation of a positron and an electron the four-momentum of the initial particles is $(0.5,0,0,0)+(0.5,0,0,0)=(1,0,0,0)$ where the first component is the energy and the last three components are the x, y, and z components of the momentum, and the units are MeV. After the anhilation if the x axis is aligned with the resulting photons then their four-momentum would be $(0.5,0.5,0,0)+(0.5,-0.5,0,0)=(1,0,0,0)$. I leave it as an exercise to the reader that the mass of both systems are 1 MeV, and that the mass of each photon is 0 MeV, showing that mass is conserved but that the mass of a system is not the sum of the masses of its constituents.

So, back to your question, the statement “Mass is a form of energy. When black holes die they release the amount of energy that they should. Mass is conserved” is correct. They are using the modern understanding of mass as the norm of the four momentum. The conservation of the four momentum implies the conservation of energy, momentum, and mass.

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    $\begingroup$ I can't disagree with that. $\endgroup$ – PM 2Ring Mar 9 at 16:31
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    $\begingroup$ I think, your answer is as valuable as proton's answer: You give the in-detail, modern, relativistic understanding answer, and do a pretty good job at making it clear. Proton's answer aligns with the classical understanding of mass as rest-mass, which is indeed converted into energy as the BH evaporates. Just like rest-mass is converted into usable energy in nuclear fusion/fission. As such, proton's answer is better for novices to get a first, rough understanding, while your answer is better for physics students. Both answers are valuable, and both have my vote. $\endgroup$ – cmaster Mar 9 at 19:31
  • $\begingroup$ I think I understand what you're saying; that if energy and momentum are conserved, mass must also be. Correct me if I'm wrong. But what actually happens to the mass consumed by a black hole? At the end of its life when it releases a load of energy, where does that energy come from? $\endgroup$ – Kieron Mar 10 at 9:45
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    $\begingroup$ This answer is somewhat oversimplifying the situation. It's actually nontrivial to define energy and momentum in general relativity. In fact, the most direct definition (via Noether's theorem) gives a quantity that is not conserved. $\endgroup$ – knzhou Mar 10 at 13:16
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    $\begingroup$ @knzhou yes, I was simplifying since the answer was already long as it is. The key word is “locally” conserved. However, since the geometry under consideration is asymptotically flat there is a conserved mass, but it is not as simple as just the four momentum. $\endgroup$ – Dale Mar 10 at 13:29

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