<p>For simplicity, let's consider a Schwarzschild black hole, so that the spacetime is spherically symmetric and static. In particular, the Schwarzschild time $t$ coordinate gives a direction in which the geometry 'stays the same' (a Killing vector field), and its inner product with the orbital four-velocity $u$ is conserved:
$$\epsilon = -g(u,\partial_t) = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{,}$$
where $\tau$ is the proper time of the orbiting particle. One can think of this as the specific (per-mass) energy, including mass-energy: a particle escaping to infinity that becomes asymptotically at rest with respect to stationary observers would have $\epsilon = 1$.</p>

<blockquote>
  <p>If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite.</p>
</blockquote>

<p>The energy as measured by whom? Imagine a family of stationary observers everywhere surrounding the black hole, or at least along an infalling particle's trajectory. Those observers measure the speed and energy of the particle as it falls past them. As the particle nears the horizon, they will report speeds arbitrarily close to the speed of light and arbitrarily high energies.</p>

<p>But to a far-away stationary observer, those observers near the horizon are experiencing increasingly divergent gravitational time dilation, and the orbital energy of the particle stays constant. Flinging a particle of mass $m$ into the black hole will increase the mass of the black hole by $m\epsilon$.</p>

<blockquote>
  <p>What's the critical flaw in this argument?</p>
</blockquote>

<p>All energy gravitates, so I would say the main flaw is forgetting to add the gravitational potential energy of the particle to the mass of the black hole as well. Or better put, you should be concerned with the orbital energy, not just its mass+kinetic parts ('relativistic mass').</p>

<p>In this situation, you have a well-defined conserved orbital energy. If you insist on measuring the Lorentz factor according to close-by stationary observers, then yes, it diverges to $+\infty$, but then you would have to admit a gravitational potential energy term that diverges to $-\infty$, because their sum must resolve to be the orbital energy.</p>

CC BY-SA 3.0
2016-03-14T23:02:29.423Z
2016-03-14T23:02:29.423Z
Stan Liou
<p>Let's reformulate your argument as follows (using idealized assumptions for simplicity):</p>

<ol>
<li><p>An object some distance away from the black hole has potential energy equal to the kinetic energy it would have at the event horizon.</p></li>
<li><p>This is the same amount of energy that would be required to pull the object from the event horizon back to its original position (and velocity).</p></li>
<li><p>We would need to accelerate the object to the escape velocity to return it to its position.</p></li>
<li><p>Since the escape velocity is greater than <em>c</em>, it would take an infinite amount of energy to return it to its position.</p></li>
</ol>

<p>The incorrect point is <strong>3</strong>. We can pull an object out of a gravitational field slowly by exerting a force on it the whole way. We would only need to accelerate it to the escape velocity if we wanted it to escape "on its own", i.e. without us exerting a force to keep pushing it away from the black hole.</p>

<p>EDIT: I'm less sure of my answer after doing some back-of-the-envelope calculations. It would be helpful for a real expert to weigh in. I'll leave my answer here anyway.</p>

CC BY-SA 3.0
2016-03-14T20:35:32.913Z
2016-03-14T20:59:25.030Z
2016-03-14T20:59:25.030Z
Era
<p>I was musing today about black holes, and came across what seems to be a contradictory fact about black holes. If any matter were to actually fall into a black hole, it seems like it would need to gain infinite energy.</p>

<p>This is seen by considering an object close to the event horizon of a black hole. If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite. As we approach the event horizon, it seems like we should be able, then, to define an arbitrarily small distance across which the particle gains an arbitrary amount of energy. </p>

<p>As the object approaches arbitrarily close to the event horizon, it seems like its relativistic mass would therefore grow arbitrarily high, resulting in a corresponding increase in the gravity of the black hole, which seems impossible.</p>

<p>If we extrapolate back to the formation of the black hole, it then seems like it should be impossible for any particles to fall into the black hole, even as it forms. Instead, mass should be distributed in a probability cloud whose density asymptotically approaches zero at the event horizon of the black hole. New particles could then enter this cloud of incredibly dense (but not black hole dense) matter, where they would be captured rather than falling past the event horizon, within which there would be a vacuum.</p>

<p>If this were true, there would be no black holes, only hollow spheres of radiation and matter which externally look like black holes, since their gravity would be great enough that particles entering the sphere would exit at an incredibly slow rate. Interestingly, such spheres wouldn't suffer from the apparent information paradox that black holes do.</p>

<p>That or I'm missing an obvious explanation for why that's all wrong, which seems more likely because I'm neither a physicist nor an astronomer. What's the critical flaw in this argument?</p>

CC BY-SA 3.0
2016-03-14T19:11:03.917Z
You're assuming the object falling in actually reaches the speed of light.  It doesn't and obviously can't, as that would require an infinite amount of energy.  The in-falling velocity is quite a bit less than the speed of light, though you're correct on it taking an infinite amount of energy to pull a particle from the event horizon.   This question is related.  http://physics.stackexchange.com/questions/24319/will-free-fall-object-into-black-hole-exceed-speed-of-light-c-before-hitting-b
userLTK
CC BY-SA 3.0
2016-03-14T19:40:10.630Z
@userLTK Sort of. My conclusion was that reaching the event horizon would require infinite energy, so it must be impossible to cross the event horizon, and as such nothing can fall into the black hole and all accumulated matter would collect in a sphere outside the event horizon, which would increase in radius as the event horizon grows.
ckersch
CC BY-SA 3.0
2016-03-14T20:03:46.490Z
@ckersch But why would it require infinite energy? Accelerating matter to sub-lightspeed velocity requires only finite energy, and the black hole does not accelerate matter to light speed.
Era
CC BY-SA 3.0
2016-03-14T20:12:49.327Z
@ckersch So the matter outside the event horizon would move outwards, against gravity as the black hole grows?
userLTK
CC BY-SA 3.0
2016-03-14T20:13:34.147Z
@Era it seems like, as I get arbitrarily close to the event horizon, I can choose an arbitrarily small distance across which the change in potential energy of a particle increases by an arbitrarily large amount. As escape velocity from a specific depth approaches the speed of light, the kinetic energy of the particle would then approach infinity.
ckersch
CC BY-SA 3.0
2016-03-14T20:16:29.993Z
@ckersch I think you're mixing up the escape velocity with the velocity at which matter falls towards the black hole. The matter accelerating towards the black hole can increase in speed immensely, but it doesn't approach the speed of light as it approaches the event horizon.
Era
CC BY-SA 3.0
2016-03-14T20:22:31.817Z
@Era Why not? If a particle falls from arbitrarily far away, all of its potential energy should convert to kinetic energy. The kinetic energy of the particle should then be equivalent to the amount of energy required for it to return to its original position. Reaching the event horizon should require shedding an infinite amount of potential energy, which would accelerate a particle to the 'speed of light', which should be impossible.
ckersch
CC BY-SA 3.0
2016-03-14T20:23:52.817Z
Interestingly, I think if we assume the spherical black hole is a fermi liquid like a quark-gluon plasma with all possible states being filled, the exclusion principle should make it impossible for incoming fermions to enter  or cross the matter shell closest to the event horizon, assuming that shell is filled.
ckersch
CC BY-SA 3.0
2016-03-14T20:25:20.647Z
@userLTK Yes, the shell would increase in radius as more matter falls in, which makes sense if the innermost particles are gaining energy (and therefore velocity) from gravitational interaction with the infalling particles.
ckersch
CC BY-SA 3.0
2016-03-14T20:28:54.013Z
@ckersch I see what you're getting at now. Remember that it's not necessary to reach escape velocity in order to return to your original position after falling. Imagine that I'm some distance away from the earth and release a bowling ball on a rope: after it hits the ground, I can *slowly* pull it back up, putting in at least as much work (energy) as was gained in kinetic energy by its falling. Escape velocity does not take into account any power source that might be exerting a force on you: it's the velocity at which you would escape *if there were no other forces* acting on you.
Era
CC BY-SA 3.0
2016-03-14T20:36:43.050Z
@Era The question, though, is how much energy it gains by falling. Assuming there is no friction, the kinetic energy of your bowling ball will be exactly equal to the potential energy it loses during the fall. For something approaching the event horizon, where escape energy should be infinite, kinetic energy should asymptotically approach infinity. The asymptotic nature of that approach should mean we need to impart arbitrarily large amounts of energy over an arbitrarily small distance to escape.
ckersch
CC BY-SA 3.0
2016-03-14T20:40:27.263Z
@ckersch The *kinetic energy* of the *escape velocity* at the event horizon is infinite. The point is that you don't need to reach the kinetic energy of the escape velocity in order to escape, you just need as much energy as you gained by accelerating. The "escape energy" is just the potential energy, which is finite. (Edit: minor point- the kinetic energy also depends on the mass of the object, not just the velocity.)
Era
CC BY-SA 3.0
2016-03-14T20:45:40.850Z
@Era Why would the potential energy be finite? The potential energy differential between two points will always exactly equal the kinetic energy required to move between them, regardless of whether you apply that kinetic energy quickly or slowly.
ckersch
CC BY-SA 3.0
2016-03-14T20:57:53.883Z
@ckersch I don't know what it means to apply kinetic energy slowly. My point is that the energy differential is the **energy** required to move between the points. It doesn't matter whether that energy comes from kinetic energy or from somewhere else (as in the case with the bowling ball example: I put energy in by doing the work of pulling the rope, but the bowling ball's KE remains low because it is moving slowly).
Era
CC BY-SA 3.0
2016-03-14T21:04:37.103Z
Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/36980/discussion-between-ckersch-and-era).
ckersch
CC BY-SA 3.0
2016-03-14T21:41:12.570Z
Paper on gravitational collapse http://www.aei.mpg.de/~rezzolla/lnotes/mondragone/collapse.pdf Suffice to say the maths is hard.
James K
CC BY-SA 3.0
2016-03-15T07:13:27.960Z
To understand the energetics of black holes it *is* necessary to understand the rudiments of GR. They certainly *cannot* be understood in terms of Newtonian physics and "escape velocities".
ProfRob
<black-hole><astrophysics>
Does conservation of energy make black holes impossible?
ckersch
# Does conservation of energy make black holes impossible?

(Asked by: username_0 on 2016-03-14)

I was musing today about black holes, and came across what seems to be a contradictory fact about black holes. If any matter were to actually fall into a black hole, it seems like it would need to gain infinite energy.

This is seen by considering an object close to the event horizon of a black hole. If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite. As we approach the event horizon, it seems like we should be able, then, to define an arbitrarily small distance across which the particle gains an arbitrary amount of energy.

As the object approaches arbitrarily close to the event horizon, it seems like its relativistic mass would therefore grow arbitrarily high, resulting in a corresponding increase in the gravity of the black hole, which seems impossible.

If we extrapolate back to the formation of the black hole, it then seems like it should be impossible for any particles to fall into the black hole, even as it forms. Instead, mass should be distributed in a probability cloud whose density asymptotically approaches zero at the event horizon of the black hole. New particles could then enter this cloud of incredibly dense (but not black hole dense) matter, where they would be captured rather than falling past the event horizon, within which there would be a vacuum.

If this were true, there would be no black holes, only hollow spheres of radiation and matter which externally look like black holes, since their gravity would be great enough that particles entering the sphere would exit at an incredibly slow rate. Interestingly, such spheres wouldn't suffer from the apparent information paradox that black holes do.

That or I'm missing an obvious explanation for why that's all wrong, which seems more likely because I'm neither a physicist nor an astronomer. What's the critical flaw in this argument?

**Tags:** black-hole, astrophysics
> **Comments:**
> **username_1**: You're assuming the object falling in actually reaches the speed of light.  It doesn't and obviously can't, as that would require an infinite amount of energy.  The in-falling velocity is quite a bit less than the speed of light, though you're correct on it taking an infinite amount of energy to pull a particle from the event horizon.   This question is related.  http://physics.stackexchange.com/questions/24319/will-free-fall-object-into-black-hole-exceed-speed-of-light-c-before-hitting-b  
> **username_0**: @userLTK Sort of. My conclusion was that reaching the event horizon would require infinite energy, so it must be impossible to cross the event horizon, and as such nothing can fall into the black hole and all accumulated matter would collect in a sphere outside the event horizon, which would increase in radius as the event horizon grows.  
> **username_2**: @ckersch But why would it require infinite energy? Accelerating matter to sub-lightspeed velocity requires only finite energy, and the black hole does not accelerate matter to light speed.  
> **username_1**: @ckersch So the matter outside the event horizon would move outwards, against gravity as the black hole grows?  
> **username_0**: @Era it seems like, as I get arbitrarily close to the event horizon, I can choose an arbitrarily small distance across which the change in potential energy of a particle increases by an arbitrarily large amount. As escape velocity from a specific depth approaches the speed of light, the kinetic energy of the particle would then approach infinity.  
> **username_2**: @ckersch I think you're mixing up the escape velocity with the velocity at which matter falls towards the black hole. The matter accelerating towards the black hole can increase in speed immensely, but it doesn't approach the speed of light as it approaches the event horizon.  
> **username_0**: @Era Why not? If a particle falls from arbitrarily far away, all of its potential energy should convert to kinetic energy. The kinetic energy of the particle should then be equivalent to the amount of energy required for it to return to its original position. Reaching the event horizon should require shedding an infinite amount of potential energy, which would accelerate a particle to the 'speed of light', which should be impossible.  
> **username_0**: Interestingly, I think if we assume the spherical black hole is a fermi liquid like a quark-gluon plasma with all possible states being filled, the exclusion principle should make it impossible for incoming fermions to enter  or cross the matter shell closest to the event horizon, assuming that shell is filled.  
> **username_0**: @userLTK Yes, the shell would increase in radius as more matter falls in, which makes sense if the innermost particles are gaining energy (and therefore velocity) from gravitational interaction with the infalling particles.  
> **username_2**: @ckersch I see what you're getting at now. Remember that it's not necessary to reach escape velocity in order to return to your original position after falling. Imagine that I'm some distance away from the earth and release a bowling ball on a rope: after it hits the ground, I can *slowly* pull it back up, putting in at least as much work (energy) as was gained in kinetic energy by its falling. Escape velocity does not take into account any power source that might be exerting a force on you: it's the velocity at which you would escape *if there were no other forces* acting on you.  
> **username_0**: @Era The question, though, is how much energy it gains by falling. Assuming there is no friction, the kinetic energy of your bowling ball will be exactly equal to the potential energy it loses during the fall. For something approaching the event horizon, where escape energy should be infinite, kinetic energy should asymptotically approach infinity. The asymptotic nature of that approach should mean we need to impart arbitrarily large amounts of energy over an arbitrarily small distance to escape.  
> **username_2**: @ckersch The *kinetic energy* of the *escape velocity* at the event horizon is infinite. The point is that you don't need to reach the kinetic energy of the escape velocity in order to escape, you just need as much energy as you gained by accelerating. The "escape energy" is just the potential energy, which is finite. (Edit: minor point- the kinetic energy also depends on the mass of the object, not just the velocity.)  
> **username_0**: @Era Why would the potential energy be finite? The potential energy differential between two points will always exactly equal the kinetic energy required to move between them, regardless of whether you apply that kinetic energy quickly or slowly.  
> **username_2**: @ckersch I don't know what it means to apply kinetic energy slowly. My point is that the energy differential is the **energy** required to move between the points. It doesn't matter whether that energy comes from kinetic energy or from somewhere else (as in the case with the bowling ball example: I put energy in by doing the work of pulling the rope, but the bowling ball's KE remains low because it is moving slowly).  
> **username_0**: Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/36980/discussion-between-ckersch-and-era).  
> **username_3**: Paper on gravitational collapse http://www.aei.mpg.de/~rezzolla/lnotes/mondragone/collapse.pdf Suffice to say the maths is hard.  
> **username_4**: To understand the energetics of black holes it *is* necessary to understand the rudiments of GR. They certainly *cannot* be understood in terms of Newtonian physics and "escape velocities".  

---

## 2 answers

### Answer by username_5 on 2016-03-14. Score: 5

For simplicity, let's consider a Schwarzschild black hole, so that the spacetime is spherically symmetric and static. In particular, the Schwarzschild time $t$ coordinate gives a direction in which the geometry 'stays the same' (a Killing vector field), and its inner product with the orbital four-velocity $u$ is conserved:
$$\epsilon = -g(u,\partial\_t) = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{,}$$
where $\tau$ is the proper time of the orbiting particle. One can think of this as the specific (per-mass) energy, including mass-energy: a particle escaping to infinity that becomes asymptotically at rest with respect to stationary observers would have $\epsilon = 1$.

> If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite.

The energy as measured by whom? Imagine a family of stationary observers everywhere surrounding the black hole, or at least along an infalling particle's trajectory. Those observers measure the speed and energy of the particle as it falls past them. As the particle nears the horizon, they will report speeds arbitrarily close to the speed of light and arbitrarily high energies.

But to a far-away stationary observer, those observers near the horizon are experiencing increasingly divergent gravitational time dilation, and the orbital energy of the particle stays constant. Flinging a particle of mass $m$ into the black hole will increase the mass of the black hole by $m\epsilon$.

> What's the critical flaw in this argument?

All energy gravitates, so I would say the main flaw is forgetting to add the gravitational potential energy of the particle to the mass of the black hole as well. Or better put, you should be concerned with the orbital energy, not just its mass+kinetic parts ('relativistic mass').

In this situation, you have a well-defined conserved orbital energy. If you insist on measuring the Lorentz factor according to close-by stationary observers, then yes, it diverges to $+\infty$, but then you would have to admit a gravitational potential energy term that diverges to $-\infty$, because their sum must resolve to be the orbital energy.


---

### Answer by username_2 on 2016-03-14. Score: 0

Let's reformulate your argument as follows (using idealized assumptions for simplicity):

1. An object some distance away from the black hole has potential energy equal to the kinetic energy it would have at the event horizon.
2. This is the same amount of energy that would be required to pull the object from the event horizon back to its original position (and velocity).
3. We would need to accelerate the object to the escape velocity to return it to its position.
4. Since the escape velocity is greater than *c*, it would take an infinite amount of energy to return it to its position.

The incorrect point is **3**. We can pull an object out of a gravitational field slowly by exerting a force on it the whole way. We would only need to accelerate it to the escape velocity if we wanted it to escape "on its own", i.e. without us exerting a force to keep pushing it away from the black hole.

EDIT: I'm less sure of my answer after doing some back-of-the-envelope calculations. It would be helpful for a real expert to weigh in. I'll leave my answer here anyway.


---

