Recently, I've read an article on scientific regarding the possibility of a stairwell into a blackhole (unsurprisingly, it isn't possible).

I've found the following question more interesting:

Consider a model of the universe with an isolated black hole (distant from other matter and light sources) and a certain thermal background temperature (in the case of the Cosmic Microwave Background, 2.7K).

For an observer sufficiently close to the black hole, he will experience the external radiation blue shifted; but the temperature of the black hole itself is the Hawking temperature, which (at infinity) is only:

$$ T_H = {1 \over 8 \pi M}. $$

So, can he exploit this temperature difference to do work? How much work can be done?


2 Answers 2


Whether you see the incoming background radiation as blueshifted depends on your relative motion compared to the large scale Hubble flow. Even in the absence of a black hole you can accelerate in a direction and see blue shift in the forward direction.

And that's fundamentally what is happening in the black hole, to stay at a fixed distance from the black hole you have to accelerate away from it, and that is why you see the CMBR blueshifted in the direction away from the black hole.

Now, that's just about how and why you see blueshifted radiation. Let's bring up some difficulties with using a black hole. A black hole has a negative heat capacity, so if you throw energy or heat into it, it gets colder (bigger).

That is possible. And even regular stars can have a negative heat capacity, it is fairly common for gravitational bound systems to have a negative heat capacity and just get larger when energy is supplied.

This is why the outer part of our sun will get much larger (and colder) when the core heats up as it switches to burning heavier nuclei. The flux of heat balances out even though it decreases temperature because it increases the outward facing area.

So basically your black hole is going to get larger (and colder), so your temperature gradient is only going to get better as time goes on (until your external radiation finally cools to less than the black hole, which will happen).

So you might be able to assume your black hole is very very very cold. However, lots of practical issues come up. One problem with the black hole is not falling in. If you built a strong shell a ways away, then that will be fine for a while, but as that black hole gets colder and larger it would eventually envelope you. So you need to move away as time goes on. That will require work, did you want to face the consequences, or do you need to save that energy to move outwards, and only want the excess? This is like wanting to live off the thermal gradient for a while versus wanting to mine it for distant future use.

If you are going to build a shell around the black hole to cheaply avoid falling in, then you can simply be very absorbing and try to soak up all the radiation coming in. But then you start to get close to the same temperature as the incoming radiation, at which point you stop getting work. This will depend on your heat capacity and you yourself need to start out very very very very very cold yourself, or else the radiation from far away isn't much of a source.

Edited to respond to comments.

In the toy universe, does the black hole would absorb most of the energy of the system until the background temperature equates the black hole's temperature?

The black hole has a negative heat capacity, as it absorbs energy it actually gets colder. And it gets larger. But since it is colder than the radiation getting larger just makes it get worse since it has more area to collect more radiation. So it is going to grow and grow and grow, getting colder and colder. Thermal energy is exchanging, but this is not causing equilibrium to occur. Black holes have a negative heat capacity and so this idea that absorbing energy brings things closer to equilibrium needs to be reexamined and is actually false in this case.

Equilibrium can occur, but only if the radiation itself makes itself colder. So only if the universe expands enough to redshift that radiation all the way down in temperature until it is teh same temperature as the black hole. If the radiation gets to a lower temeprature, then the black hole sends net energy into space and gets smaller (and hotter) and so again there is no equilibrium until the black evaporates completely.

But even in thermal equilibrium, wouldn't the observer near the black hole still experience a blue shifted background light?

Equilibrium? Are you talking about your engine approaching the temperature of the radiation? At that point you've probably gotten all the thermal work you are going to get. But frankly, how did you get colder than the radiation to begin with? In many situations, work was done to get you colder and now you are just getting that back, so I don't think you are extracting energy from the black hole at all. And since the black hole is just gettign more massive as time goes on, I think it is safe to say that you are not "extracting energy" from the black hole, at best you make it grow less quickly so are steal some of its expected growth.

People can extract energy from black holes, but they do it by stealing rotational energy.

Then should I conclude this external radiation is "breaking" somehow the object so that the work required to keep in orbit is greater or equal to the work acquired by the thermal machine?

Work is frame dependant quantity, so its hard for to figure out what you are talking about. If you coupled together a 2d mesh of springs then it can compress as it contracts around the hole until it reaches a fixed distance from the hole. Then yes the 2d mesh of springs it will see the radiation as blueshifted. But in the frame of the 2d mesh of springs you also have some energy from the compression of all those springs, its just that you need that to stay in place and not fall in. You got that spring compression energy in some sense from the black hole. But you would lose that climbing out to back where you started. But if your springs were at 2.7K originally (which is reasonable), and then they compressed down as you contracted it into the gravity well, and then you saw blueshifted radiation, then you can let yourself warm up until you get close to the blueshifted temperature and you can exploit that thermal flux to store some of the thermal energy flux as "useful" energy.

However, energy is itself a source of gravitation, and those springs have to fight the gravity of the black hole as well as their own gravity, so you have two options, shoot the energy you get from work out into space (in which case it has to fight the gravity well and thus arrives at destination with less energy) or else have more energy on site, and contract and get closer and closer to the black hole. The answer about what energy you can ship home now, energy you can ship home later, or energy you can use up on site are all different answers. And they are different questions.

  • $\begingroup$ Thank you for the answer. So in our toy universe the black hole would absorb most of the energy of the system until the background temperature equates the black hole's temperature. But even in thermal equilibrium, wouldn't the observer near the black hole still experience a blue shifted background light? Then should I conclude this external radiation is "breaking" somehow the object so that the work required to keep in orbit is greater or equal to the work acquired by the thermal machine? $\endgroup$
    – Real
    Apr 12, 2015 at 8:24
  • $\begingroup$ I wish I could upvote more than once :) So let me clarify the equilibrium condition: if the energy in the system is finite and the black hole starts off cool enough, then it will grow, but there should be a point where the energy density of the system is so low the black hole emits more radiation than it absorbs, right? I mean, under some assumptions about the system, it will probably always absorb strictly more than it emits, but it should tend to a limit, if we assume it doesn't oscillate somehow. $\endgroup$
    – Real
    Apr 12, 2015 at 19:13
  • $\begingroup$ So in this case the thermal machine orbiting the black hole, in my understanding, is observing a large temperature anisotropy around him, which should be usable to extract work, correct? (Work in the sense of the frame of the machine only) Even if the machine at one time reaches an equilibrium temperature, this anisotropy seems exploitable (by for example, having a radiator pointing in the coolest direction). So in this condition where everything is seemingly in equilibrium there is still usable work. $\endgroup$
    – Real
    Apr 12, 2015 at 19:20
  • $\begingroup$ So the 2nd law of thermodynamics should imply this thermal machine in fact needs to do work to maintain this orbit. Does that make sense? $\endgroup$
    – Real
    Apr 12, 2015 at 19:20

The answer to the question is simply the Carnot efficiency:

$$ Efficiency = 1-{T_U \over T_B} $$

Where $T_U$ means the temperature of the universe and $T_B$ means the temperature of the black hole, and Efficiency means the maximum efficiency that is possible.

It does not matter if you place the heat engine close to the black hole or far away from it, the ratio $T_U / T_B$ stays the same.

(If the temperature of cosmic microwave background is doubled by blue shift, then the temperature of Hawking radiation is doubled by the halving of the redshift of the Hawking radiation)


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