# Energy of falling into a black hole

When an object of the mass m (small in comparison) falls into a black hole from infinity, the object gains a certain speed and therefore kinetic energy. Therefore it would seem that the mass of the black hole after consuming this object would increase more than just by m. In fact, in a classical (not applicable) way of thinking, the energy released in a fall to a singularity would be infinite, but surely it is not infinite in General Relativity. Can someone clarify what is the total gravitational energy released by a mass m falling from infinity to the singularity of a black hole?

A different interpretation of this may be the Frozen Star where the object never actually crosses the event horizon in the frame of a remote observer. In this case the speed actually is reduced to zero at the event horizon, but what happens with the kinetic energy? What is the extra mass added by the object in this case, as easily measured by a remote observer based on the change in his speed and the size of his distant orbit around the black hole?

• Note the difference between kinetic energy and potential energy and total energy. I think you've neglected potential energy in your thinking. And the singularity is a logical mathematical extreme of a theory that does not include quantum theory or anything like it which is relevant to small size and high energy density. – StephenG Sep 7 '17 at 22:47
• I did not neglect anything, because I am asking a question, not answweing it. Whoever answers it is free to account for the potential energy or anything else. The question is well defined. From the parameters of my orbit I know the mass of the black hole. An object of the rest mass m falls in. How does it change the mass of the black hole that defines my orbit? M+m+? – safesphere Sep 8 '17 at 0:07
• i think the object(m) might already becomes a black hole along the way and then quickly explodes in forms of energy, lol – user6760 Sep 8 '17 at 2:26
• The mass will reach the event horizon with finite velocity, after that, nothing else will be released to the outside. – user167013 Sep 8 '17 at 3:00
• @safesphere: My question though is not about energy conservation, but how much an object of mass m falling from the infinity increases the mass of a black hole of mass M. Mass and energy are equivalent in relativity. You can refer to either the "ADM mass" or the "ADM energy." It means the same thing. – user4552 Sep 18 '17 at 15:54

Consider the non-relativistic problem of a particle falling into a potential well and releasing all its energy in there. The quantity which is conserved during the infall is the total energy $E = T + V + E_{internal}$, where $T$ is kinetic energy, $V$ is the potential energy, and $E_{internal}$ is some internal "chemical" energy of the particle.

Now we assume that the particle starts at rest at infinity where the potential is zero so that $E = 0 + 0 + E_{internal} = E_{internal}$. As the particle starts falling to the potential well, nothing happens to internal energy, $V$ becomes negative and since energy is conserved, that must be countered with a positive $T$. When we arrive all the way in the potential well, no matter what happens with the kinetic and potential energy, we always have $E= E_{internal}$ and when the energy release in the well comes about, it is exactly $E= E_{internal}$ which is released.

You can think similarly about the black hole along with the realization that $E_{internal} = m_0 c^2$, where $m_0$ is the object's rests mass. In other words, if a particle of (rest) mass $m_0$ falls into a stationary black hole starting at rest at infinity, the black hole will receive exactly $m_0 c^2$ in terms of energy, no matter what happens to the kinematic or potential parts of the energy.

EDIT

Of course, the full relativistic problem has to be considered more carefully. First of all, what is exactly the mass of a black hole? One of the postulates of relativity is that a freely falling observer never feels gravity - so a freely falling observer will judge the black hole to be probably weightless and thus there cannot exist any local, frame-independent notion of the black hole mass. The black hole mass must be, in fact, defined by some coordinated measurements of privileged observers.

Furthermore, notice that no frame is privileged and thus there is no notion of big or small velocity! To define a notion of velocity, you also need a privileged frame with respect to which you are measuring it!

Sometimes it so happens that the black hole is in such a state that there exists a family of observers at infinity who collect measurements in which the black hole field appears as stationary. In such a case, we call the black hole stationary and the time in which these observers measure any physical process will be our privileged notion of time throughout the space-time.

It is also these observers through which we define the notion of mass. Since they define a notion of rest and the space-time around them is almost flat, they feel the gravity of the black hole in the weak-field, Newtonian limit. The mass of the black hole is defined exactly and only as the apparent Newtonian mass $M$ in the $\approx -M/r^2$ gravitational force these observers at infinity feel. In other words, you could understand the mass more as total gravitating energy as felt by observers at infinity.

Since the background is stationary with respect to this time, there will be a respective integral of motion for the evolution of test bodies moving in this space-time due to Noether's theorem. This integral is the temporal component of four-momentum $p_t$. Now, at infinity the space-time is asymptotically Minkowski space-time and if the test particle starts there at rest, we will just have $p_t = -m$ and other component of four-momentum equal to zero.

As the particle then falls into the black hole, $p_t = -m$ will never change and will play exactly the same role as $-E$ in the argument given above. Similarly, the growth of coordinate velocity (as measured by the observers at infinity) as a compensation of some kind of potential energy can also be traced to the conservation of $p_t$, at least in the Newtonian limit.

Of course, one could argue that this does not automatically make $-p_t$ the contribution to the black hole mass once the particle is absorbed by the central, mathematically ill-behaved singularity. It is an elegant and favorable candidate, because the sum of $-p_t$s from many particles should not be possible to annihilate by Noether's theorem and by some kind of assumption of adiabaticity of the black hole growth, so the $-p_t$ must go "somewhere". But this just tells us that the contribution to the black hole should be a linear function of $-p_t$.

The argument why it is exactly $-p_t$ to increase the black hole mass can come only from the consideration of the infalling body as a non-test object, that is an object which perturbs the black hole background with its own gravitational field. From such arguments we know that a part of the infalling energy almost always gets radiated away by gravitational radiation, but if the body gets sufficiently light, the contribution of mass to the resulting black hole indeed converges to $-p_t$.

• I need your help to understand the answer. Consider 2 cases. (1) A rock drops on a planet from afar with a great speed gained due to gravity. (2) We gently lower the same rock to the same planet on a rope from a spaceship. In the former case we have the planet, rock, plus the kinetic energy converted to heat, so the planet is heavier. If we gently join 2 planets of mass M each, would the total not be 2M? But if they fall onto each other from afar, the total would be higher, would it not? Your answer doesn't work without explaining the physical meaning of the "negative potential energy". (TBC) – safesphere Sep 17 '17 at 22:50
• I can also argue that there is no such thing as "potential gravitational energy". The energy conservation per the Noether theorem is a result of the time uniformity. Time is not uniform in gravity. My head moves faster in time than my feet. Therefore lifting a mass up consumes energy while dropping down releases energy. Energy is not conserved, unless we use a workaround by introducing an imaginary value we call "potential energy". This works, because the process is reversible and thus allows us to "renormalize" energy conservation as if time were uniform and "potential energy" really existed. – safesphere Sep 17 '17 at 23:04
• The logic in the comment above works fine up to the event horizon where time stops for a remote observer. However the fall down to the singularity intuitively would seem to release an infinite energy, if the potential is still reverse proportional to the radius. How does the gravitational field potential depend on the radius in the Schwarzschild solution? Using the counterintuitive concept of not exactly real "negative potential energy" is a fine mathematical method and workaround, but it doesn't answer my question, only shifts it from energy to even less tangible "potential energy". – safesphere Sep 17 '17 at 23:20
• Thanks for your help. I upvoted your answer, although it would take me more time to comprehend it. Can we look at a simpler example with no black holes, no event horizons, and no strong gravity? (1) Using my mighty powers, I gently put a rock of mass m on a planet of mass M and radius R. Question 1: Is the resulting total mass M+m? (2) Now I'm out of magic powers and I let a rock of mass m fall from infinity onto a planet of mass M and radius R. The kinetic energy of the impact is E, it is absorbed and added to the total per the mass-energy equivalence. Question 2: What is the total mass? – safesphere Sep 19 '17 at 4:00
• I added a paragraph to the answer which defined what do we actually mean by black hole mass - it is the gravitating mass as felt by observers at infinity. The funny thing is exactly that gravitating mass in relativity is sensitive to all kinds of forms of energy, including the potential or gravitational-binding energy. Then you get (1) $M' = M + m - m \Phi_{surface}$ (in Newtonian limit, where $\Phi_{surface}$ is the surface potential), and (2) $M' = M + m$ (minus possible radiation losses). – Void Sep 19 '17 at 7:36