Consider a massive particle falling into a black hole with a very high kinetic energy, for instance, an a particle with travelling at 99.9999999999% of the speed of light towards it.

The mass-energy equivalence implies that the total energy of the electron $E$ is the sum of its rest energy plus its kinetic energy, so the electron has much more energy than its rest mass.

The black hole thus must increase its mass by $E/c^2$. According to John Duffield answer in question: "Proper mass" and "gravitational binding energy", the black hole does not increase in mass from to the kinetic energy of the electron, but only from its rest mass.

Why is this the case? Where has the kinetic energy of the electron gone?

This does not make sense, since a photon with no rest mass but with energy must contribute to the black hole's mass when falling into it.

  • 2
    $\begingroup$ I think one cannot go wrong if one thinks in four vectors being added to the total four vector of the black hole. $\endgroup$
    – anna v
    Aug 31, 2023 at 4:41
  • $\begingroup$ Using a charged particle make things more complicated: physics.stackexchange.com/q/21830 $\endgroup$
    – PM 2Ring
    Aug 31, 2023 at 8:41

2 Answers 2


John Duffield is talking about dropping an electron into a black hole. i.e. You start it from rest and the only energy it has is its rest mass energy when it is far from the black hole.

In this case, the black hole only gains the rest mass of the electron. The total energy/mass of the system cannot change as measured by a distant observer. In Newtonian mechanics it would be the equivalent of saying that the increasing kinetic energy of the electron as it falls is exactly cancelled by its increasingly negative potential energy.

However, if you were to fire the electron inwards with total energy $E$ then the black hole would gain a mass $E/c^2$ (where $E$ is measured far from the black hole). The same would be true if you were to add photons of energy $E$ (measured far from the black hole).

  • $\begingroup$ But why does it only gain the rest mass of the electron? $\endgroup$
    – Manuel
    Aug 31, 2023 at 12:10
  • $\begingroup$ @Manuel why would it have more? The total energy of the system is conserved as the electron falls into the black hole. $\endgroup$
    – ProfRob
    Aug 31, 2023 at 13:12
  • $\begingroup$ The electron has rest mass and kinetic energy from its speed right before being absorbed by the black hole. Where does this kinetic energy go? $\endgroup$
    – Manuel
    Aug 31, 2023 at 15:51
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    $\begingroup$ It also has negative gravitational potential energy, and, in the case where you dropped it from infinity, this cancels the positive kinetic energy. $\endgroup$
    – Ghoster
    Aug 31, 2023 at 17:36
  • $\begingroup$ @Ghoster Your comment above applies only to the Newtonian gravity, but not to General Relativity where the potential energy does not exist. The correct answer is that the mass of the falling particles (measured far from the black hole) decreases proportionally to the increase of the kinetic energy, so the total energy remains the same. The mass of any falling object (measured from afar) becomes zero at the horizon due to the infinite time dilation. Mass measured locally is always the same, but measured remotely directly depends on the time dilation. $\endgroup$
    – safesphere
    Sep 2, 2023 at 16:09

Manuel wrote: "According to John Duffield the black hole does not increase in mass from to the kinetic energy of the electron, but only from its rest mass."

You can also have a black hole made only of photons who have only kinetic and no rest energy, so that can't be correct in general, rather the opposite.

You have conservation of energy and momentum. If you shoot two particles from opposite sides into a Schwarzschild black hole so its net momentum gain is zero, the particles add the energy

$$\rm E=-p_t=mc^2 \sqrt{\frac{1-r_s/r}{1-v^2/c^2}}$$

to the black hole, which is a conserved quantity along the path if we neglect the small losses of kinetic energy due to gravitational waves. If you arrange your infalling particles as concentric shells you can avoid any gravitational radiation since we have the Birkhoff theorem in that case and the energy is truly conserved while the momentums average out.

If the free fall is with the negative escape velocity $\rm v=\pm c \sqrt{r_s/r}$, they add exactly $\rm E=mc^2$ to the black hole. If they fall faster the energy is higher, if they fall slower lower.

In the limit that the the particles hover stationary at the horizon their contribution to the black hole's energy is $0$, so if you decelerate an infalling particle to extract some kinetic energy, for example by attaching it to a string powering a generator with its pull, it will contribute less mass energy to the black hole by that amount.

  • $\begingroup$ What is r, rs, and v in your equation? What do you answer the question with a two particle example? $\endgroup$
    – Manuel
    Aug 31, 2023 at 1:34
  • $\begingroup$ If you gattle your bullets one sided into the black hole it will not only absorb the mass but also the momentum of the bullets, but you can transform that away by switching to the new rest frame of the black hole after getting its kick. Your other questions about the Schwarzschild radius and the velocity in the Lorentz factor can be resolved by Google very easily. The velocity is relative to local stationary observers. $\endgroup$
    – Yukterez
    Aug 31, 2023 at 5:47
  • $\begingroup$ What “Einstein” has downvoted a correct answer? +1 $\endgroup$
    – safesphere
    Sep 2, 2023 at 15:50
  • $\begingroup$ In the limit that the the particles hover stationary at the horizon their contribution to the black hole's energy is 0” - This is exactly correct no matter how big the rest mass of these particles is. Not everyone understands this. $\endgroup$
    – safesphere
    Sep 2, 2023 at 15:57

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