The mass of a bound system is the mass of its part minus the binding energy. For instance if you put an electron and a proton together to form a hydrogen atom you get a $13.6 \text{ eV}$ photon and the mass of the atom is lower than the masses of its parts by the same amount (divided by $c^2$). Another example: you put two protons and two neutrons together to form an alpha particle, you get $28 \text{ MeV}$ out and the mass of the alpha particle is $28 \text{ MeV}/c^2$ less than the sum of the masses of two protons and two neutrons. The same should be true regardless of the nature of the force exerted between the objects (electromagnetic, nuclear, gravitational...).

Now I propose the following thought experiment: consider the system of an object of mass m at some far distance from a black hole of mass $M$. The mass of the whole system is $m+M$. Now let $m$ fall into the black hole. It will gain kinetic energy, and one can compute that when it reaches the event horizon (viewed from outside, otherwise that never happens) the kinetic energy is exactly equal to $\frac{mc^2}{2}$. So one may surmise that the total energy of the object is $mc^2+\frac{mc^2}{2}=\frac{3mc^2}{2}$ when it merges with the black hole, and thus the object increases the black hole mass by $\frac{3}{2}m$. Now the total mass of the system is $\frac{3}{2}m+M$. Where was the extra $m/2$ in the initial situation? Was it somehow stored in the gravitational field (knowing that the notion of energy density of a gravitaional field is ill-defined in general relativity, see this post)? Does any object really possess in fact 1.5 times its own mass when considering that it is in an excited state since it may fall into a black hole? That makes no sense to me, so I will be grateful if someone could point out the mistake in my argument.

  • $\begingroup$ black holes are relativistic, you cannot use a classical approximation for the kinetic energy $\endgroup$ – user83548 Oct 30 '15 at 15:29
  • $\begingroup$ The subject line is a good question. The first paragraph is correct. The second paragraph is error ridden. :) The second paragraph should agree with the first paragraph. To answer the good question: We always get the same number when we measure the mass of a proton, it does not matter on what planet our laboratory is located. $\endgroup$ – stuffu Oct 30 '15 at 16:16
  • 1
    $\begingroup$ Short answer: we do, see this answer. $\endgroup$ – DilithiumMatrix Oct 30 '15 at 17:06
  • $\begingroup$ As far as the hydrogen atom goes, have you computed the fractional change? $\endgroup$ – dmckee Oct 30 '15 at 18:58
  • $\begingroup$ Actually, the near inevitability of Hawking radiation (if for no other reason than the third law of thermodynamics) demonstrates almost beyond doubt that gravitating systems are not even bound. They are merely metastable. $\endgroup$ – CuriousOne Oct 30 '15 at 20:54

Now I propose the following thought experiment: consider the system of an object of mass $m$ at some far distance from a black hole of mass $M$. The mass of the whole system is $m+M$.

It's only because they are so far away from each other that the mass from even farther away looks so close to $m+M.$ So that's not entirely a trivial thing.

Now let $m$ fall into the black hole. It will gain kinetic energy,

The two objects will move towards each other and not just by getting closer but also by picking up relative velocity. Correct.

and one can compute that when it reaches the event horizon (viewed from outside, otherwise that never happens)

Now you've got that backwards. Its exactly from the outside it never happens.

the kinetic energy is exactly equal to $\frac{mc^2}{2}$.

And that's completely and utterly wrong, even not withstanding that from the outside you never see the mass reach the event horizon.

So one may surmise that the total energy of the object is $mc^2+\frac{mc^2}{2}=\frac{3mc^2}{2}$

No one ever said mass is additive. You made that point right at the beginning. An alpha particle has less system mass than the sum of the masses of its parts. So if no one ever said mass is additive, why are you now a ting like it is. I'd love to tell you what you are doing wrong, but it seems like you are just assuming something.

So instead I'll cover the basics of how you feel a mass from a system in general relativity by covering a mathematically simpler example. A spherical collapsing star. Its like your example except not as extreme (just a star, not a black hole) and more symmetric (lots of small masses arranged in a spherical shell, all falling inwards).

The first thing to note is that you can measure the mass by being far away and orbiting it. When you are far away from a mass M in Newtonian mechanics then you feel an acceleration $GM/r^2$ and you need an acceleration of $v^2/r$ to go in a circle. And that circle has a circumference of $C=2\pi r$ and you travel it in time $T=C/v.$

These last two things, $C$ and $T$ can be measured right there while staying far from the star where things act similar to Newtonian mechanics. And in Newtonian mechanics we have $GM/r^2=v^2/r$ so $M=v^2r/G.$ And $v=C/T$ and $r=C/2\pi$ so you get $M=C^3/(GT^22\pi).$ And that's what the $M$ means. It means that way far out there are nearly circular orbits that have $C^3/(GT^22\pi)$ be approximately the same for all those concentric circles no matter how far out they are, as long as they stay far out.

So that's what the $M$ is. It is not the sum of the masses of the things that went it. Or is it?

See, spacetime naturally curves even far from any source and that's actually what mass does is it allows different vacuum type curvatures to be sewn together. So when the outside all had the same time the same $C^3/(GT^22\pi)$ then everything is fine. But inside that outer shell of matter, since it is regular ordinary positive energy density matter you end up with a smaller $C^3/(GT^22\pi)$ on the inside than the outside.

And that's fine. Positive matter allows a curvature of type $M+m=C^3/(GT^22\pi)$ on the outside of the spherical shell to be sewn up to a curvature of type $M=C^3/(GT^22\pi)$ on the inside of that spherical shell.

It's like if you made a shallow funnel and a steep funnel and you cut them each in two and put the deeper one on the outside of the shallower one, right where they are the same size.

So now that outer shell can contract. And you are correct that it picks up speed and its energy density increases because each part goes faster. And the energy density increases because the spherical shell reduces its surface area.

But spacetime can only curve with a constant (vacuum) type when there isn't matter. So as the matter contracts it has to create more of the vacuum type curvature of type $M+m=C^3/(GT^22\pi)$ on the outside of the spherical shell and some of the old type of spacetime curvature of type $M=C^3/(GT^22\pi)$ gets destroyed.

And its spookier than that. Since the two types of curvature are of a different type, you literally see more space get of the new type get created outside the spherical shell than you see get destroyed from inside. Your shell gets farther away from the outside things than it gets closer to the inside things. That inevitable when you have two different types of curvature, a type $M+m=C^3/(GT^22\pi)$ on the outside and a type $M=C^3/(GT^22\pi)$ on the inside.

The only way this is possible is for time inside to start ticking slower as seen by people in the outside.

So people are disagreeing about a lot, how fast time is ticking, how much closer further farther away things are, who's moving, etcetera. And we have to check all our assumptions and baggage and expectations and loom at the math of what is happening in the model to learn what happens in reality.

Now what is interesting is that even though the energy density of all the stiff increased and even though the total energy on the entire shell increased, that larger energy on that smaller shell turns out to remain exactly the amount of energy you need to sew those two types of curvature together.

And it has to be for thibgs to be consistent. You had some curvature of each type on either side of the shell and they could only stay what they are on their side. But the boundary can move. And it can move more from the outside than from the outside because it is possible to create space, and it turns out it is common to do so.

That's how we get strongly curved spacetime. You take a type of spacetime and make more of that type, but make in the direction where that type gets more curved. So the type with $M+m=C^3/(GT^22\pi)$ gets more of itself farther in where it naturally is more curved (stronger gravity).

And that's how spacetime gets curved in the first place. So you had some mass and spacetime was curved. As the mass contracts the spacetime changes. It makes more spacetime but makes it of the same type as was seen far outside, the type with parameter $M+m=C^3/(GT^22\pi).$

And that new spacetime (new as in more, but of that same old type) is stronger curved, but still of the same type.

There is not some other definition of mass. We say there is some mass $M$ in some region over there solely as a shorthand to say there is a certain curvature, a curvature of type $M$ over here. And the curvature gets frozen in to one of these static vacuum type curvatures.

If your system wasn't super symmetric it might have a dynamically changing spacetime for a while as gravitational waves are emitted (which would make the final mass less) and the mass might rub against each other and give up some energy as light (which means it has less than the right energy needed so it has to make some slightly less curved type spacetime outside it so you see type $M+m=C^3/(GT^22\pi)$ out beyond the light and see type $M+fm=C^3/(GT^22\pi)$ between the light and the shell of hot matter and you you still see type $M=C^3/(GT^22\pi)$ inside of the shell.

That's real too. Pluto sees a more massive sun than the earth does because the light reaching us bow hasn't reached Pluto yet so we start seeing $M+fm=C^3/(GT^22\pi)$ (with $0\lt f\lt 1$) when Pluto still sees $M+m=C^3/(GT^22\pi).$

And it could settle down to a different type such as a type for a rotating source.

But since gravity is about curvature we have to learn how gravity allows spacetime to curve far from matter then we have to learn how matter allows these different types of vacuum curvature of spacetime to sew up.

And none of it is as obvious as just adding up masses. But its also not totally unexpected if you think it through.


Let me try to answer my own question, helped by this post on the binding energy of neutron stars (many thanks DilithiumMatrix for pointing that out). Indeed the example of the neutron star, though not conceptually different from the example of the black hole which I used in my original question, is easier because we do not get distracted by the problems linked with black holes.

So, just as in the case of the electron of an atom or a parton in a nucleus, if we lift a neutron from the surface of a neutron star we can surmise that we give it gravitational potential energy and in effect increase its mass -- the potential energy is stored as an extra excitation mass, leaving us with the usual neutron mass we know about and can measure here on Earth. So the answer to the question would be: we do already account for it, it's part of the mass of the free object. Then in the case of an object falling into a black hole, the gained kinetic energy is obtained from the potential energy which itself would be stored as part of the mass of the object -- so the falling object gains speed and at the same time it loses most (perhaps all???) of its rest mass.

While that answer seems satisfactory to me at fist glance, it makes me uneasy. I am used to thinking about the mass of neutrons as arising from the strong field (QCD interactions between quarks). I can understand that the mass of a neutron gets lower when the neutron is bound to a nucleus -- we are still talking about the strong force (the nuclear force is a manifestation the strong force), so I can see that the neutron finds itself in a more energetically favourable configuration in the nucleus, no problem. Now if you tell me that a significant part of the neutron mass is lost when the neutron binds to a neutron star -- a loss which has nothing to do with the strong force, but rather the gravitational force -- it leaves me with more questions. How can the gravitational field influence the way the strong field generates the neutron mass? Same thing when considering any particle (eg an electron) falling into a black hole: we think that we understand the mechanism which generates the mass, but in the end the mass is basically just gravitational -- how can that be? (EDIT: thsi is now posted as a separate question).

  • $\begingroup$ Potential energy doesn't randomly pick one of the objects at random and increase that object's mass. And when an object falls, its rest mass doesn't decrease. Everything about your answer is wrong. And it seems to be built on fundamental misunderstandings of more basic results, like thinking that in Newtonian gravity that potential energy belongs to a random object instead of to the system of multiple objects. $\endgroup$ – Timaeus Nov 14 '15 at 18:12
  • $\begingroup$ This answer is the only satisfactory one I could find because otherwise, where would the object get its energy from when falling and gaining speed? Does it somehow instantaneously "suck" it from the mass of the whole system? I am very unfamiliar with the concept of energy being non-local on a large scale. $\endgroup$ – Philippe Mermod Nov 14 '15 at 18:56
  • $\begingroup$ Being wrong shouldn't be satisfying, and if you need additional explanations about other answers you should ask. Newtonian gravity has non local energy. If you want a theory with local energy, consider classical electrodynamics instead. In that theory there is energy in the fields and it can travel from here to there and be located in between too. For infalling matter it isn't more massive in its new comoving frame, and there's nothing instantaneous. The 4d spacetime it goes through is curved, so it might seem like is speeding up, but it's going straight in a curved spacetime. $\endgroup$ – Timaeus Nov 15 '15 at 15:53
  • $\begingroup$ Please rethink "So, just as in the case of the electron of an atom or a parton in a nucleus," physics.stackexchange.com/questions/149744/… . It is not true so your analogy is not true $\endgroup$ – anna v Dec 10 '15 at 4:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.