Kinetic energy of an object falling into an event horizon

[Andrew Wright]

TL;DR Summary

Objects falling into an event horizon seem to gain infinite kinetic energy.

Hi,

When objects fall in a gravitational field, they convert gravitational potential energy into kinetic energy. Because energy is always conserved:

amount of kinetic energy gained = amount of gravitational potential energy lost.

Now the gravitational energy lost should be equal to the amount of energy gained by doing the journey in reverse.

This is where I have a problem. An object that is near the event horizon would require a near infinite amount of energy to make the journey in reverse out of the black hole. So this means a near infinite amount of gravitational potential energy is lost. In turn, that the kinetic energy gained by falling into an event horizon is infinite. So, this feels wrong since black holes are real objects that exist out there and infinite energy sounds like nonsense.

Have I made a mistake? Why/Why not?

 

[Ibix]

Kinetic energy as measured by who? It only tends to infinity for a sequence of hovering observers instantaneously at the same altitude as the infaller as they approach the horizon (edit: that's probably a little strong, but certainly there are a lot of descending observers for whom the kinetic energy of the infaller does not approach infinity). But it's not possible to have a hovering observer at or below the event horizon so that value is finite where defined. An observer above the horizon can't see the infaller cross the horizon in order to measure their energy anyway, and I don't think they would see the value they measure (to the extent there's a unique value to measure) tend to infinity, basically due to gravitational redshift.

Another way to look at it is to note that your approach only works where gravitational potential can be defined. That turns out to be only outside the horizon, and hence the potential is finite everywhere it is defined.

 

[PeterDonis]

this feels wrong since black holes are real objects that exist out there and infinite energy sounds like nonsense

@Ibix has given the correct answer to the "infinite kinetic energy" problem. However, here is an additional thing to consider that might help with this.

Consider an object at rest very, very far from the black hole. We now let it free-fall into the hole. How much energy does it add to the hole?

The answer is not "infinity" or "close to infinity". It is "the rest energy of the object".

In terms of "converting potential energy to kinetic energy", this is true because the potential energy of the object is negative, and gets more negative as the object falls and the kinetic energy gets more positive. So the total energy of the object, which is the energy that will get added to the hole, is constant--it is just the original rest energy of the object. (Another way of saying this is that the total energy of the object is a constant of free-fall motion--"converting potential energy into kinetic energy" is just another way of expressing that.)

 

[Andrew Wright]

Thanks. Appreciated :)

 

[jartsa]

An object that is near the event horizon would require a near infinite amount of energy to make the journey in reverse out of the black hole.

As we now know more about falling in, what happens to that assumption above?

 

[Ibix]

As we now know more about falling in, what happens to that assumption above?

It's perfectly true. The notion of "escape velocity" carries across well from Newtonian gravity, as long as you don't touch the event horizon. In Peter's terms, the large positive KE and large negative GPE become progressively nearer zero, both reaching zero at infinity in the case of exact escape velocity.

This "energy at infinity" (KE+GPE) is perfectly well defined. However, it's worth noting that what I'm blithely calling "KE" is specifically the KE measured by hovering observers. Other observers are available, but would typically need to monkey around with the KE they actually measure if they want to get the KE I'm referring to.

 

[pervect]

Summary:: Objects falling into an event horizon seem to gain infinite kinetic energy.

Hi,

When objects fall in a gravitational field, they convert gravitational potential energy into kinetic energy. Because energy is always conserved:

That's true in Newtonian theory. It's not quite so simple in General Relativity.

There's a good FAQ on the general topic of energy conservation in GR at http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

But this general FAQ may not answer your specific question. "Is energy conserved in general relativity? Well, it depends on what you mean by energy, and what you mean by conserved" is a totally true statement, but it's so general it may not be helpful. But it does indicate that the topic of energy conservation in GR is a bit tricky.

If we take the specific question of an object falling into a black hole, there is a conserved "energy-at-infinity" for the infalling object, but it's not usual to split this energy into a "kinetic" part and a "potential" part.

https://www.fourmilab.ch/gravitation/orbits/ has some of the formula, which are taken from the textbook "Gravitation" by Misner, Thorne, and WHeeler.

For the case where there is no angular momentum (which is also a conserved quantity), we can set L=0 from the webpage, then we can write, in geometric units where the gravitational constant G and the speed of light c are both set to 1

$$\left( \frac{dr}{d\tau} \right) ^2 + 1-\frac{2M}{r} = \tilde{E}^2$$

Here $\tilde{E}$ is the energy/ unit rest mass. It's probably best to consider the whole equation as normalized for a "unit infalling mass". $\tau$ is proper time, which makes $dr/d\tau$ a sort of velocity, but it's not quite the same as the sort of velocity you're used to, both because dr doesn't measure distance, and because $d\tau$ measures a change in proper time $\tau$ rather than coordinate time t.

Note that $\tilde{E}$ and in fact all of the terms above all of the terms are normalized for a unit mass. If we were doing a Newtonian analysis (which we are not), the kinetic energy would be $mv^2/2$, and the normalized kinetic energy would be just $v^2/2$.

In Newtonian terms, we might write

$$\frac{v^2}{2} + \tilde{U}(r) = \tilde{E}$$

The GR formula

$$\left( \frac{dr}{d\tau} \right) ^2 + 1-\frac{2M}{r} = \tilde{E}^2$$

has a similar formal structure, but the detials are all different and would take some study to master.

 

[jartsa]

It's perfectly true. The notion of "escape velocity" carries across well from Newtonian gravity, as long as you don't touch the event horizon. In Peter's terms, the large positive KE and large negative GPE become progressively nearer zero, both reaching zero at infinity in the case of exact escape velocity.

This "energy at infinity" (KE+GPE) is perfectly well defined. However, it's worth noting that what I'm blithely calling "KE" is specifically the KE measured by hovering observers. Other observers are available, but would typically need to monkey around with the KE they actually measure if they want to get the KE I'm referring to.

Okay. But OP's almost infinite escape energy led to OP's almost infinite kinetic energy after falling. So that escape energy in OP's mind should change somehow, I think ... It should change to almost infinite escape energy according to observers hovering nearby - which energy according to observers at infinity is almost rest mass of the escaping object times c squared.

 

[Ibix]

So that escape energy in OP's mind should change somehow, I think ...

I'm not sure what you mean. An object that's going to freefall upwards to infinity does have a high kinetic energy as measured by a local hovering observer. A distant hovering observer needs to parallel transport the four momentum to them before measuring it (that's the formal mathematical version of "observing the object's four momentum from a distance") and that will, I think, redshift the resulting kinetic energy down to the same value as they'd measure if they just waited for the object to get to them. If the observer is very far from the black hole, this will be the quantity pervect called $\tilde E$.

 

[timmdeeg]

In turn, that the kinetic energy gained by falling into an event horizon is infinite. So, this feels wrong since black holes are real objects that exist out there and infinite energy sounds like nonsense.

You can't say the kinetic energy "is infinite", it isn't. Kinetic energy depends on relative velocity. The radial velocity of an infalling object as measured by a shell observer (a hovering observer) at $R = r$ is $dr/dt = -(2M/r)^{1/2}$, whereby the shell observer is at rest relativ to an accelerated frame. But there exists no such frame at $r = 2M$ relativ to which an observer could be at rest. So the prediction of this formula must be understood as a limiting case.

 

[jartsa]

I'm not sure what you mean.

Correct understanding of escape energy would not lead to the incorrect infinite energy after a long fall. So the Original Poster's (OP's) idea about escape energy must be wrong.

Winching a potato up from very close to the event horizon of a black hole requires only the energy of one potato at that position to which the potato is winched. Those two potatoes are identical.

 

[Ibix]

Correct understanding of escape energy would not lead to the incorrect infinite energy after a long fall. So the Original Poster's (OP's) idea about escape energy must be wrong.

But the kinetic energy of a free-falling body, as measured by local hovering observers, does tend to infinity as you approach the event horizon. It just doesn't reach infinity, because the point at which a naive reading of the maths says it would is when the argument underlying the maths breaks down - there are no hovering observers at the event horizon.

Winching a potato up from very close to the event horizon of a black hole requires only the energy of one potato at that position to which the potato is winched.

This is a concrete way of making a distant measurement of the gravitational potential energy of the stationary potato, which is equal to the escape kinetic energy. As I said in my last post, this is redshifted compared to the local measurement. It will, indeed, not exceed $mc^2$ in magnitude.