Imagine the following:

A->  []  <-B

where [] is a black hole, and A and B are two objects of same mass, traveling with the same speed right at the black hole.

A and B have immense kinetic energy. When they crash into the black hole, the black hole does not feel any net force; thus it will not accelerate to bring away the kinetic energy. Nor would it heat up, due to the no-hair theorem dictating that black holes don't have a "temperature". Neither would angular momentum change since we assume A and B go absolutely straight towards the black hole.

Where does the energy go? Does it get converted to mass and added into the black hole mass? If so, what is the exact mechanism? Obviously "conservation of energy", but in each case there is a more low-level description (say, when A and B crash into a solid object, the heating comes from the kinetic energy of the objects agitating the atoms making up the object).


2 Answers 2


Does it get converted to mass and added into the black hole mass?

Yes, exactly. But asking this question in the way you have exposes a subtlety in the definition of mass: the amount of energy that counts as a system's mass doesn't all have to be "intrinsic mass." Different kinds of energy can contribute to mass as well.

The mass of any system is really just the minimum energy the system has in any inertial reference frame. Consider two photons traveling in opposite directions, for example. If each photon has an energy $E$, the system of the two has mass $2E/c^2$ because there is no reference frame in which the total energy of the two photons is any less than $2E$. This is an example of how what we call mass can be produced out of nothing but kinetic energy, simply by defining our system such that the motion is "hidden" inside.

For a more traditional example, consider a proton. A proton has a mass of about $1\text{ GeV}/c^2$, in the sense that you can't reduce its energy any lower by boosting to a different reference frame, but the intrinsic masses of the quarks that make it up contribute only a few percent of that. About half of the $1\text{ GeV}$ comes from kinetic energy of the gluons, and another almost half from kinetic energy of the quarks, but outside of particle physics, we don't make a distinction between the energy contributed by the intrinsic mass of the quarks and the kinetic energy. It all just gets called mass.

In your situation with the two objects, once the objects get inside the event horizon, they're part of the black hole. In the reference frame in which you formulated the problem, the total energy of the black hole (now including the objects) is

$$M_\text{BH(original)}c^2 + 2m_\text{object}c^2 + 2K_\text{object}$$

($K$ is kinetic energy), and there's no other inertial reference frame in which the energy is any less than that. Therefore, this total amount of energy divided by $c^2$ gives the (new) mass of the black hole. The objects may still be inside the black hole, moving toward the center, or they might have been ripped apart, or they might have reacted with antimatter and converted into photons, but it doesn't matter. As long as energy is conserved, the mass of the black hole after it absorbs the objects is given by the above expression. (I'm ignoring all the issues surrounding whether energy conservation actually applies in or near a black hole; someone else would have to address that.)

  • $\begingroup$ Hmm. So since kinetic/whatever energy is not separable from the mass energy from the outside of the black hole, thus all that energy would by definition be part of the "mass" of the black hole? $\endgroup$
    – ithisa
    Apr 26, 2014 at 9:15
  • $\begingroup$ Yeah, pretty much. $\endgroup$
    – David Z
    Apr 26, 2014 at 16:36

The source for the spacetime curvature is not simply mass, but rather the stress-energy tensor. When written in coordinates in which the system is time independant only the $T_{00}$ component is non-zero, and this corresponds to mass (or energy using $E = mc^2$).

However your objects are not stationary in the coordinates you're using, so when you write down the contribution they make to the stress-energy tensor you'll find that other components of the tensor are non-zero. The Wikipedia article I linked describes the stress-energy tensor for a moving particle here.

So, as David says, the kinetic energy of the particles does contribute to the gravity. You can give an approximate idea of the contribution by equating the kinetic energy to a mass increase, but to really understand what is going on you need to look at the stress-energy tensor.


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