![]() 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.Īs 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. This integral is the temporal component of four-momentum $p_t$. 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. In other words, you could understand the mass more as total gravitating energy as felt by observers at infinity. 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. 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. It is also these observers through which we define the notion of mass. 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. 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. The black hole mass must be, in fact, defined by some coordinated measurements of privileged observers.įurthermore, 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! 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. Of course, the full relativistic problem has to be considered more carefully. 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. ![]() The quantity which is conserved during the infall is the total energy $E = T + V + E_ = m_0 c^2$, where $m_0$ is the object's rests mass. 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?Ĭonsider the non-relativistic problem of a particle falling into a potential well and releasing all its energy in there. Can someone clarify what is the total gravitational energy released by a mass m falling from infinity to the singularity of a black hole?Ī 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 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. Therefore it would seem that the mass of the black hole after consuming this object would increase more than just by m. ![]() 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. ![]()
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