<p>Not enough to worry about.</p>
<p>Most neutrinos we detect on Earth come from the sun.  The black hole at the centre of the galaxy is not so close to a star as the Earth is so you would expect the neutrino flux to be lower than on Earth,  let's assume it is similar: about <span class="math-container">$10^{11}$</span> neutrinos per <span class="math-container">$\text{cm}^2$</span> per second. But each neutrino is light, even including its kinetic energy it has only a few hundred KeV (again based off solar neutrinos for the sake of a rough estimate). Converting that to mass (<span class="math-container">$m= E/c^2$</span>) gives the mass of neutrinos per cm² per second.  It's about <span class="math-container">$5×10^{-20}$</span> kg/cm²/s</p>
<p>The black hole is relatively small in diameter, even if you account for the fact that the effecive radius is larger than the actual event horizon radius, since things will tend to be bent towards the black hole, the cross sectional area is about 10²⁵ cm², so roughly 500000 kg of neutrinos might fall into the black hole every second.  (Although this is very much in the spirit of a Fermi estimate, this is probably an overestimate)</p>
<p>But the mass of the black hole is about 8,000,000,000,000,000,000,000,000,000,000,000,000kg</p>
<p>So 500,000kg is negligible.</p>

CC BY-SA 4.0
2020-06-11T01:35:35.820Z
Sounds reasonable, even if we neglect the (very slow) neutrinos comprising the cosmic neutrino background.
PM 2Ring
CC BY-SA 4.0
2020-06-11T09:05:37.083Z
There are a few stars nearby that are way brighter than the Sun (and way hotter, meaning they are even brighter in neutrino energy flux). There may be even more Sun-size stars, but they are not bright enough to be visible from Earth. So the above estimate may be 1 or 2 orders of magnitude off because of that. And stil negligible compared to the light absorbed.
fraxinus
CC BY-SA 4.0
2020-06-11T09:19:42.643Z
Are you even sure that Neutrinos would interact ? After all even with the best detectors we "catch" only 3 per day from the sun (of those trillions "created" in the sun each and every second )
eagle275
CC BY-SA 4.0
2020-06-11T09:33:21.327Z
@eagle275 According to general relativity, absolutely. There is no interaction required - there is no (forward in time) _path_ to the outside of a black hole. But even if you think about gravity as an interaction (rather than a question of geometry), the answer is still yes - neutrinos aren't especially non-interactive, their elusiveness comes from their lack of interaction through the electro-magnetic force specifically; most of the captures we detect come from the weak interaction, which is very weak at distance - the interaction cross section between a nucleus and a neutrino is tiny.
Luaan
CC BY-SA 4.0
2020-06-11T09:55:37.310Z
well the highest conversion rate I heart was 43% (or was it 42% ) - so not every mass reaching the event horizon falls into the black hole .. most is heated up and sent away in plasma beams
eagle275
CC BY-SA 4.0
2020-06-11T10:41:39.953Z
thanks for the answer @JamesK !   Just to absolutely recap, you figure about 500 tonne of neutrinos, per second ?  Amazing.
Fattie
CC BY-SA 4.0
2020-06-16T08:48:52.103Z
The BH at the center of the MW sits in the middle of a very dense star cluster, which includes some very luminous stars, all of which will be putting out neutrinos. So your neutrino rate is probably a bit low. (But, yeah, the mass gain from neutrinos will still be negligibly small.)
Peter Erwin
CC BY-SA 4.0
2020-06-17T17:57:53.797Z
Perhaps, but the black hole is further from those stars than we are from the core of the sun.  After all, even in the middle of a dense star cluster, the stars are still points of light in a dark sky.  So the inverse square law applies.  But I don't know what kind of nuclear reactions might be happening in the business end of the accretion disc. But the temperature in the disc is "only" 1e6K, an order of magnitude less than the core of the sun (and we are only 1 AU from the sun!)
James K
CC BY-SA 4.0
2020-06-10T21:09:57.590Z
2020-11-20T04:30:58.963Z
2020-11-20T04:30:58.963Z
James K
<p>If we are immersed in a sea of neutrino, then a black hole should be constantly capturing the ones crossing its event horizon, and gaining mass.</p>

<p>How much neutrino mass falls inside a black hole per year? For example, the one at the center of our galaxy.</p>

<black-hole><neutrinos>
How much mass per year may the black hole at the center of the Milky Way gain by capturing neutrinos?
tutizeri
# How much mass per year may the black hole at the center of the Milky Way gain by capturing neutrinos?

(Asked by: username_0 on 2020-06-10)

If we are immersed in a sea of neutrino, then a black hole should be constantly capturing the ones crossing its event horizon, and gaining mass.

How much neutrino mass falls inside a black hole per year? For example, the one at the center of our galaxy.

**Tags:** black-hole, neutrinos

---

## 1 answers

### Answer by username_1 on 2020-06-10. Score: 12 | Accepted answer

Not enough to worry about.

Most neutrinos we detect on Earth come from the sun. The black hole at the centre of the galaxy is not so close to a star as the Earth is so you would expect the neutrino flux to be lower than on Earth, let's assume it is similar: about $10^{11}$ neutrinos per $\text{cm}^2$ per second. But each neutrino is light, even including its kinetic energy it has only a few hundred KeV (again based off solar neutrinos for the sake of a rough estimate). Converting that to mass ($m= E/c^2$) gives the mass of neutrinos per cm² per second. It's about $5×10^{-20}$ kg/cm²/s

The black hole is relatively small in diameter, even if you account for the fact that the effecive radius is larger than the actual event horizon radius, since things will tend to be bent towards the black hole, the cross sectional area is about 10²⁵ cm², so roughly 500000 kg of neutrinos might fall into the black hole every second. (Although this is very much in the spirit of a Fermi estimate, this is probably an overestimate)

But the mass of the black hole is about 8,000,000,000,000,000,000,000,000,000,000,000,000kg

So 500,000kg is negligible.

> **Comments:**
> **username_2**: Sounds reasonable, even if we neglect the (very slow) neutrinos comprising the cosmic neutrino background.  
> **username_3**: There are a few stars nearby that are way brighter than the Sun (and way hotter, meaning they are even brighter in neutrino energy flux). There may be even more Sun-size stars, but they are not bright enough to be visible from Earth. So the above estimate may be 1 or 2 orders of magnitude off because of that. And stil negligible compared to the light absorbed.  
> **username_4**: Are you even sure that Neutrinos would interact ? After all even with the best detectors we "catch" only 3 per day from the sun (of those trillions "created" in the sun each and every second )  
> **username_5**: @eagle275 According to general relativity, absolutely. There is no interaction required - there is no (forward in time) _path_ to the outside of a black hole. But even if you think about gravity as an interaction (rather than a question of geometry), the answer is still yes - neutrinos aren't especially non-interactive, their elusiveness comes from their lack of interaction through the electro-magnetic force specifically; most of the captures we detect come from the weak interaction, which is very weak at distance - the interaction cross section between a nucleus and a neutrino is tiny.  
> **username_4**: well the highest conversion rate I heart was 43% (or was it 42% ) - so not every mass reaching the event horizon falls into the black hole .. most is heated up and sent away in plasma beams  
> **username_6**: thanks for the answer @JamesK !   Just to absolutely recap, you figure about 500 tonne of neutrinos, per second ?  Amazing.  
> **username_7**: The BH at the center of the MW sits in the middle of a very dense star cluster, which includes some very luminous stars, all of which will be putting out neutrinos. So your neutrino rate is probably a bit low. (But, yeah, the mass gain from neutrinos will still be negligibly small.)  
> **username_1**: Perhaps, but the black hole is further from those stars than we are from the core of the sun.  After all, even in the middle of a dense star cluster, the stars are still points of light in a dark sky.  So the inverse square law applies.  But I don't know what kind of nuclear reactions might be happening in the business end of the accretion disc. But the temperature in the disc is "only" 1e6K, an order of magnitude less than the core of the sun (and we are only 1 AU from the sun!)  

---

