<p>You got some very good answers already. I just want to point out this:</p>

<p>The "temperature" of a black hole is more like just "a way of speaking". It's not temperature as normally understood.</p>

<p>There is this process called the Hawking radiation where vacuum near a black hole produces a stream of particles, borrowing energy from the black hole's gravity to create those particles - and so it appears like the black hole "emits" radiation. Since this is radiation, you could in theory measure its temperature. But that's just the temperature of the Hawking radiation.</p>

<p>Obviously you couldn't stick a thermometer into a black hole.</p>

CC BY-SA 4.0
2018-05-29T17:54:12.673Z
2018-05-29T17:54:12.673Z
Florin Andrei
<blockquote>
  <p>Black holes absorb every form of energy, even light. Absorption of energy should raise its temperature but still it is extremely cold, why?</p>
</blockquote>

<p>Because of the infinite gravitational time dilation. The thing to understand is that temperature is a measure of motion. A hot gas is one where the molecules are, on average, moving faster than in a cold gas. See the Wikipedia <a href="https://en.wikipedia.org/w/index.php?title=Temperature&amp;oldid=843448637" rel="nofollow noreferrer">temperature</a> article and note this: <em>"The coldest theoretical temperature is absolute zero, at which the thermal motion of all fundamental particles in matter reaches a minimum"</em>. Gravitational time dilation means things are moving slower. When the gravitational time dilation is infinite, things aren't moving at all. This is why the black hole was originally known as the frozen star. </p>

<p>Stephen Hawking wrote a <a href="https://projecteuclid.org/euclid.cmp/1103858973" rel="nofollow noreferrer">paper</a> in 1972 with Brandon Carter and Jim Bardeen where they said <em>"It should however be emphasized that κ/8π and A are distinct from the temperature and entropy of the black hole. In fact the effective temperature of a black hole is absolute zero”</em>. </p>

<p>Robert Wald said much the same in <a href="https://books.google.co.uk/books?id=Ry7pCAAAQBAJ&amp;pg=PA54&amp;lpg=PA54#v=onepage&amp;q&amp;f=false" rel="nofollow noreferrer">black hole physics</a>. On page 69 he said in classical black hole physics <em>“κ has nothing whatever to do with the physical temperature of a black hole, which is absolute zero by any reasonable criteria”</em>. </p>

<p>The black hole is said to have an effective temperature by virtue of Hawking radiation, but as Fraser Cain said, it's very low. And as Mark said, <em>"under GR alone, BHs are utterly black and don't have a temperature at all"</em>. More importantly, Hawking radiation is said to be emitted from outside the event horizon. So it isn't actually the temperature of the black hole. Just as <em>"the accretion disk that surrounds a BH can be very hot indeed"</em> but it isn't actually the temperature of the black hole.  </p>

CC BY-SA 4.0
2018-05-29T11:14:12.713Z
2018-05-29T11:30:27.023Z
2018-05-29T11:30:27.023Z
John Duffield
<p>Black holes do radiate, see Hawking radiation. And the more matter they absorb, the colder they get</p>

<p>For a black hole to evaporate, energy has to completely escape from its potential well. To make a rather crude analogy, if we fire a rocket from the surface of the Earth then below the escape velocity the rocket will eventually fall back. The rocket has to have a velocity greater than the escape velocity to completely escape the Earth.</p>

<p>When we are considering a black hole, rather than the escape velocity we consider the gravitational red shift. The red shift reduces the energy of any outgoing radiation, so it reduces the energy of any radiation emitted by the hotter vacuum state near the event horizon. If the red shift is infinite then the emitted radiation gets red shifted away to nothing and in this case there will be no Hawking radiation. If the red shift remains finite then the emitted radiation still has a non-zero energy as it approaches spatial infinity. In this case some energy does escape from the black hole, and this is what we call the Hawking radiation. This energy comes ultimately from the mass energy of the black hole, so the mass/energy of the black hole is decreased by the amount or radiation that has escaped.
One can arrange a heat-producing reaction to take place inside the event horizon of a black hole. For example, I can drop two cold blocks of matter on trajectories so that they collide inside the horizon, producing heat. There is nothing special about the spacetime inside the horizon in this respect except that the heat from the collision will not be seen by outside observers due to the horizon. What is unusual about this region is that in a short time (as experienced by the objects) they - and the heat emissions - will meet the singularity and at this point we have no theory describing what happens.
Since the topology of the region is such that the singularity is more like a point in time than a place in space there is also no lingering heat in the interior space nor any sense of temperature of the singularity.</p>

<p>Event horizons do not care if things crossing them are energy or matter. The reasons for the accretion disks and jets are different: non-black hole object like stars being formed and neutron stars also have disks and jets. Basically disks happen because matter is interacting and slowly shedding angular momentum and potential energy through turbulent interactions, and jets happen because the resulting plasma produces strong magnetic fields and blocks radiation in the equatorial direction.</p>

<p>I have taken reference from some answer like. John rennie.
 <a href="https://physics.stackexchange.com/questions/251385/an-explanation-of-hawking-radiation/252236#252236">https://physics.stackexchange.com/questions/251385/an-explanation-of-hawking-radiation/252236#252236</a></p>

<p>And Anders Sandberg's </p>

<p><a href="https://physics.stackexchange.com/questions/476882/can-there-be-temperature-inside-of-a-massive-black-hole/476896#476896">https://physics.stackexchange.com/questions/476882/can-there-be-temperature-inside-of-a-massive-black-hole/476896#476896</a></p>

CC BY-SA 4.0
2019-10-31T11:28:59.543Z
2019-10-31T12:27:52.787Z
2019-10-31T12:27:52.787Z
आर्यभट्ट
<p>Under General Relativity (GR) alone, a Black Hole's (BH's) <a href="https://en.wikipedia.org/wiki/Event_horizon" rel="noreferrer">event horizon</a> is a point of no return -- anything that passes through the event horizon is lost and gone forever, and <em>nothing</em> comes out. Hence, under GR alone, BHs are utterly black and don't have a temperature at all.</p>

<p>This is why the absorption of radiation (or anything else) by a BH doesn't raise its temperature -- it just gets swallowed up and lost. (It's mass, angular momentum and charge do remain, but that's all -- see the <a href="https://en.wikipedia.org/wiki/No-hair_theorem" rel="noreferrer">No Hair Theorem</a>.)</p>

<p>(Note: The <a href="https://en.wikipedia.org/wiki/Accretion_disk" rel="noreferrer">accretion disk</a> that <em>surrounds</em> a BH can be very hot indeed, but that's another thing entirely.)</p>

<p>Stephen Hawking discovered that applying quantum mechanics to BHs showed that BHs <em>would</em> emit a random spray of radiation, and that that radiation was precisely what a black body would emit -- <a href="https://en.wikipedia.org/wiki/Black-body_radiation" rel="noreferrer">black body radiation</a>.  This is called <a href="https://en.wikipedia.org/wiki/Hawking_radiation" rel="noreferrer">Hawking radiation</a>.</p>

<p>Black body radiation is simply the thermal emission of a perfect absorber of radiation, and leads to the inescapable conclusion that a BH <em>does</em> have a non-zero temperature.  Interestingly, Hawking's analysis showed that the effective temperature of the BH is inversely proportional to its mass and that solar-mass BHs (which are the smallest for which we have actual evidence) would have a temperature of about 0.00000006 K. Kinda cold, but still not zero.</p>

<p>Note that, unintuitively, a solar mass BH get <em>colder</em> as it absorbs radiation. Because any radiation (or anything else) it absorbs increases its mass, and since higher mass BHs are colder, the more energy you dump into one, the colder it gets!</p>

CC BY-SA 4.0
2018-05-28T18:35:21.480Z
2018-05-28T18:35:21.480Z
Mark Olson
<blockquote>
<p>&quot;The most massive black holes in the Universe, the supermassive black
holes with millions of times the math [sic] of the Sun will have a
temperature of 1.4 x <span class="math-container">$10^{-14}$</span> Kelvin. That's low. Almost absolute zero,
but not quite. A solar mass black hole might have a temperature of
only 0.00000006 Kelvin.&quot;</p>
<p><a href="https://phys.org/news/2016-09-cold-black-holes.html" rel="noreferrer">September 5, 2016 by Fraser Cain, Universe Today</a></p>
</blockquote>
<p>Black holes absorb every form of energy, even light. Absorption of energy should raise its temperature but still it is extremely cold, why?</p>

CC BY-SA 4.0
2018-05-29T02:03:44.137Z
I've found the source of the quote. Interestingly, the typos ("math" instead of "mass", and an errant decimal point) are in the linked doc on phys.org, and are repeated in many other places. The article itself is not very accurate, and fails to mention Hawking's fundamental proposition that a BH's temperature is inversely proportional to its mass. Supermassive = super-cold. Absorb more mass/energy, get even colder.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2018-05-29T03:47:27.620Z
Closely related to [What is the temperature inside a Black Hole?](https://astronomy.stackexchange.com/questions/13081/what-is-the-temperature-inside-a-black-hole) (asked Jan 5 '16).
Rob
<black-hole><universe><supermassive-black-hole><heat>
Why black holes are extremely cold?
Paran
# Why black holes are extremely cold?

(Asked by: username_0 on 2018-05-28)

> "The most massive black holes in the Universe, the supermassive black
> holes with millions of times the math [sic] of the Sun will have a
> temperature of 1.4 x $10^{-14}$ Kelvin. That's low. Almost absolute zero,
> but not quite. A solar mass black hole might have a temperature of
> only 0.00000006 Kelvin."
>
> [September 5, 2016 by Fraser Cain, Universe Today](https://phys.org/news/2016-09-cold-black-holes.html)

Black holes absorb every form of energy, even light. Absorption of energy should raise its temperature but still it is extremely cold, why?

**Tags:** black-hole, universe, supermassive-black-hole, heat
> **Comments:**
> **username_1**: I've found the source of the quote. Interestingly, the typos ("math" instead of "mass", and an errant decimal point) are in the linked doc on phys.org, and are repeated in many other places. The article itself is not very accurate, and fails to mention Hawking's fundamental proposition that a BH's temperature is inversely proportional to its mass. Supermassive = super-cold. Absorb more mass/energy, get even colder.  
> **username_2**: Closely related to [What is the temperature inside a Black Hole?](https://astronomy.stackexchange.com/questions/13081/what-is-the-temperature-inside-a-black-hole) (asked Jan 5 '16).  

---

## 4 answers

### Answer by username_6 on 2018-05-28. Score: 20 | Accepted answer

Under General Relativity (GR) alone, a Black Hole's (BH's) [event horizon](https://en.wikipedia.org/wiki/Event_horizon) is a point of no return -- anything that passes through the event horizon is lost and gone forever, and *nothing* comes out. Hence, under GR alone, BHs are utterly black and don't have a temperature at all.

This is why the absorption of radiation (or anything else) by a BH doesn't raise its temperature -- it just gets swallowed up and lost. (It's mass, angular momentum and charge do remain, but that's all -- see the [No Hair Theorem](https://en.wikipedia.org/wiki/No-hair_theorem).)

(Note: The [accretion disk](https://en.wikipedia.org/wiki/Accretion_disk) that *surrounds* a BH can be very hot indeed, but that's another thing entirely.)

Stephen Hawking discovered that applying quantum mechanics to BHs showed that BHs *would* emit a random spray of radiation, and that that radiation was precisely what a black body would emit -- [black body radiation](https://en.wikipedia.org/wiki/Black-body_radiation). This is called [Hawking radiation](https://en.wikipedia.org/wiki/Hawking_radiation).

Black body radiation is simply the thermal emission of a perfect absorber of radiation, and leads to the inescapable conclusion that a BH *does* have a non-zero temperature. Interestingly, Hawking's analysis showed that the effective temperature of the BH is inversely proportional to its mass and that solar-mass BHs (which are the smallest for which we have actual evidence) would have a temperature of about 0.00000006 K. Kinda cold, but still not zero.

Note that, unintuitively, a solar mass BH get *colder* as it absorbs radiation. Because any radiation (or anything else) it absorbs increases its mass, and since higher mass BHs are colder, the more energy you dump into one, the colder it gets!


---

### Answer by username_3 on 2018-05-29. Score: 3

You got some very good answers already. I just want to point out this:

The "temperature" of a black hole is more like just "a way of speaking". It's not temperature as normally understood.

There is this process called the Hawking radiation where vacuum near a black hole produces a stream of particles, borrowing energy from the black hole's gravity to create those particles - and so it appears like the black hole "emits" radiation. Since this is radiation, you could in theory measure its temperature. But that's just the temperature of the Hawking radiation.

Obviously you couldn't stick a thermometer into a black hole.


---

### Answer by username_4 on 2018-05-29. Score: 3

> Black holes absorb every form of energy, even light. Absorption of energy should raise its temperature but still it is extremely cold, why?

Because of the infinite gravitational time dilation. The thing to understand is that temperature is a measure of motion. A hot gas is one where the molecules are, on average, moving faster than in a cold gas. See the Wikipedia [temperature](https://en.wikipedia.org/w/index.php?title=Temperature&oldid=843448637) article and note this: *"The coldest theoretical temperature is absolute zero, at which the thermal motion of all fundamental particles in matter reaches a minimum"*. Gravitational time dilation means things are moving slower. When the gravitational time dilation is infinite, things aren't moving at all. This is why the black hole was originally known as the frozen star.

Stephen Hawking wrote a [paper](https://projecteuclid.org/euclid.cmp/1103858973) in 1972 with Brandon Carter and Jim Bardeen where they said *"It should however be emphasized that κ/8π and A are distinct from the temperature and entropy of the black hole. In fact the effective temperature of a black hole is absolute zero”*.

Robert Wald said much the same in [black hole physics](https://books.google.co.uk/books?id=Ry7pCAAAQBAJ&pg=PA54&lpg=PA54#v=onepage&q&f=false). On page 69 he said in classical black hole physics *“κ has nothing whatever to do with the physical temperature of a black hole, which is absolute zero by any reasonable criteria”*.

The black hole is said to have an effective temperature by virtue of Hawking radiation, but as Fraser Cain said, it's very low. And as Mark said, *"under GR alone, BHs are utterly black and don't have a temperature at all"*. More importantly, Hawking radiation is said to be emitted from outside the event horizon. So it isn't actually the temperature of the black hole. Just as *"the accretion disk that surrounds a BH can be very hot indeed"* but it isn't actually the temperature of the black hole.


---

### Answer by username_5 on 2019-10-31. Score: 0

Black holes do radiate, see Hawking radiation. And the more matter they absorb, the colder they get

For a black hole to evaporate, energy has to completely escape from its potential well. To make a rather crude analogy, if we fire a rocket from the surface of the Earth then below the escape velocity the rocket will eventually fall back. The rocket has to have a velocity greater than the escape velocity to completely escape the Earth.

When we are considering a black hole, rather than the escape velocity we consider the gravitational red shift. The red shift reduces the energy of any outgoing radiation, so it reduces the energy of any radiation emitted by the hotter vacuum state near the event horizon. If the red shift is infinite then the emitted radiation gets red shifted away to nothing and in this case there will be no Hawking radiation. If the red shift remains finite then the emitted radiation still has a non-zero energy as it approaches spatial infinity. In this case some energy does escape from the black hole, and this is what we call the Hawking radiation. This energy comes ultimately from the mass energy of the black hole, so the mass/energy of the black hole is decreased by the amount or radiation that has escaped.
One can arrange a heat-producing reaction to take place inside the event horizon of a black hole. For example, I can drop two cold blocks of matter on trajectories so that they collide inside the horizon, producing heat. There is nothing special about the spacetime inside the horizon in this respect except that the heat from the collision will not be seen by outside observers due to the horizon. What is unusual about this region is that in a short time (as experienced by the objects) they - and the heat emissions - will meet the singularity and at this point we have no theory describing what happens.
Since the topology of the region is such that the singularity is more like a point in time than a place in space there is also no lingering heat in the interior space nor any sense of temperature of the singularity.

Event horizons do not care if things crossing them are energy or matter. The reasons for the accretion disks and jets are different: non-black hole object like stars being formed and neutron stars also have disks and jets. Basically disks happen because matter is interacting and slowly shedding angular momentum and potential energy through turbulent interactions, and jets happen because the resulting plasma produces strong magnetic fields and blocks radiation in the equatorial direction.

I have taken reference from some answer like. John rennie.
<https://physics.stackexchange.com/questions/251385/an-explanation-of-hawking-radiation/252236#252236>

And Anders Sandberg's

<https://physics.stackexchange.com/questions/476882/can-there-be-temperature-inside-of-a-massive-black-hole/476896#476896>


---

