<p>I think this is really hard for the Milky Way because we are in it and have no accurate evaluation of how many stars it contains or its overall luminosity.</p>
<p>Some sort of answer would be to assume that the entire luminosity of the Milky Way is fuelled by the conversion of hydrogen to helium. This is likely to be far less uncertain than using an uncertain number of stars combined with an uncertain initial mass function and some sort of luminosity-weighting for that. An estimate for the Milky Way luminosity is around <span class="math-container">$2\times 10^{10} L_\odot$</span> (e.g., see <a href="https://people.ast.cam.ac.uk/%7Epettini/Intro%20Cosmology/Lecture16.pdf" rel="nofollow noreferrer">here</a>). Note that if there are 200 billion stars then the average luminosity is only <span class="math-container">$0.1 L_\odot$</span>. This is because the Sun is much more luminous than an average star (which would be an M-dwarf).</p>
<p>The creation of a helium nucleus from 4 hydrogen nuclei in either the <a href="https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain" rel="nofollow noreferrer">pp-chain</a> or <a href="https://en.wikipedia.org/wiki/CNO_cycle" rel="nofollow noreferrer">CNO cycle</a> releases about 26 MeV of energy that end up being radiated from stellar surfaces (a small amount escapes in neutrinos).</p>
<p><span class="math-container">$2\times 10^{10}L_\odot$</span> is the equivalent of <span class="math-container">$2\times 10^{48}$</span> <strong>helium nuclei being created every second.</strong></p>
<p>Assuming the Milky Way had a constant luminosity over its <span class="math-container">$\sim 10^{10}$</span> year lifetime, this corresponds to the creation of 2 billion solar masses of helium during that time. Note however, that 200 billion main sequence stars would contain about 50 billion solar masses in total, mostly in the form of hydrogen, so this is only increasing the helium inventory by a few per cent of the total, and less than that when you consider that much of the Milky Way (baryonic) mass is not inside stars. Another way of saying this is that only a very small fraction of the hydrogen fuel in our Galaxy has been &quot;used up&quot;.</p>
<p>The headline figure must be an upper limit because some of the luminosity of the Milky Way will be contributed by giant stars that are burning helium or heavier elements. But these stars, although luminous are still getting an appreciable fraction of their luminosity from hydrogen burning and are much rarer than main sequence stars, subgiant and first-ascent red giants, where the luminosity is due to hydrogen burning exclusively.</p>

CC BY-SA 4.0
2024-01-26T18:02:40.767Z
If you expand on idea of Anders Sanders in his comment you maybe should mention that.
Leos Ondra
CC BY-SA 4.0
2024-01-26T21:15:55.053Z
@LeosOndra I was busy writing my answer, so I didn't see the comment.
ProfRob
CC BY-SA 4.0
2024-01-26T21:37:58.120Z
Ok, I understand.
Leos Ondra
CC BY-SA 4.0
2024-01-27T10:52:45.267Z
$2\times10^{48}$ helium atoms weigh about the same as Pluto, or $\frac15$ as much as the Moon. At 1 atm pressure, they would occupy about the same volume as Uranus.
leftaroundabout
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2024-01-27T12:17:18.653Z
@leftaroundabout yes. Added-value paragraph included.
ProfRob
CC BY-SA 4.0
2024-01-26T16:50:49.290Z
2024-01-27T12:17:24.093Z
2024-01-27T12:17:24.093Z
ProfRob
<p>The Milky Way has 100-400 billion stars. Our sun is a fairly average example, and it converts 3.7× 10^38 protons to helium <a href="https://en.wikipedia.org/wiki/Solar_core#Energy_conversion" rel="noreferrer">each second</a>, through two different fusion pathways.</p>
<p>I would like to know the best estimate for how many new helium atoms are created across the entire Milky Way each second. I'm tempted to just multiply 3.7× 10^38 by 200 billion stars, but given the <a href="https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation" rel="noreferrer">mass-luminosity relation</a> this feels a bit naive. Is there a better approach?</p>

CC BY-SA 4.0
2024-01-26T16:33:01.390Z
What is "best" here? Do you need to get a very accurate number, or just an order of magnitude estimate? One way of estimating it is to assume the MW luminosity is all due to helium production, so 1.4e10*3.7e38 = 5.2e48 protons/s (a bit below the naive multiplication and reasonable due to the IMF and ML function)... but that likely just has one significant digit. Likely orders of magnitude is the only thing you can do.
Anders Sandberg
<milky-way><elemental-abundances><fusion>
Within the Milky Way, how much helium is made each second?
WaveInPlace
# Within the Milky Way, how much helium is made each second?

(Asked by: username_0 on 2024-01-26)

The Milky Way has 100-400 billion stars. Our sun is a fairly average example, and it converts 3.7× 10^38 protons to helium [each second](https://en.wikipedia.org/wiki/Solar_core#Energy_conversion), through two different fusion pathways.

I would like to know the best estimate for how many new helium atoms are created across the entire Milky Way each second. I'm tempted to just multiply 3.7× 10^38 by 200 billion stars, but given the [mass-luminosity relation](https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation) this feels a bit naive. Is there a better approach?

**Tags:** milky-way, elemental-abundances, fusion

> **Comments:**
> **username_1**: What is "best" here? Do you need to get a very accurate number, or just an order of magnitude estimate? One way of estimating it is to assume the MW luminosity is all due to helium production, so 1.4e10*3.7e38 = 5.2e48 protons/s (a bit below the naive multiplication and reasonable due to the IMF and ML function)... but that likely just has one significant digit. Likely orders of magnitude is the only thing you can do.  

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## 1 answers

### Answer by username_2 on 2024-01-26. Score: 12 | Accepted answer

I think this is really hard for the Milky Way because we are in it and have no accurate evaluation of how many stars it contains or its overall luminosity.

Some sort of answer would be to assume that the entire luminosity of the Milky Way is fuelled by the conversion of hydrogen to helium. This is likely to be far less uncertain than using an uncertain number of stars combined with an uncertain initial mass function and some sort of luminosity-weighting for that. An estimate for the Milky Way luminosity is around $2\times 10^{10} L\_\odot$ (e.g., see [here](https://people.ast.cam.ac.uk/%7Epettini/Intro%20Cosmology/Lecture16.pdf)). Note that if there are 200 billion stars then the average luminosity is only $0.1 L\_\odot$. This is because the Sun is much more luminous than an average star (which would be an M-dwarf).

The creation of a helium nucleus from 4 hydrogen nuclei in either the [pp-chain](https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain) or [CNO cycle](https://en.wikipedia.org/wiki/CNO_cycle) releases about 26 MeV of energy that end up being radiated from stellar surfaces (a small amount escapes in neutrinos).

$2\times 10^{10}L\_\odot$ is the equivalent of $2\times 10^{48}$ **helium nuclei being created every second.**

Assuming the Milky Way had a constant luminosity over its $\sim 10^{10}$ year lifetime, this corresponds to the creation of 2 billion solar masses of helium during that time. Note however, that 200 billion main sequence stars would contain about 50 billion solar masses in total, mostly in the form of hydrogen, so this is only increasing the helium inventory by a few per cent of the total, and less than that when you consider that much of the Milky Way (baryonic) mass is not inside stars. Another way of saying this is that only a very small fraction of the hydrogen fuel in our Galaxy has been "used up".

The headline figure must be an upper limit because some of the luminosity of the Milky Way will be contributed by giant stars that are burning helium or heavier elements. But these stars, although luminous are still getting an appreciable fraction of their luminosity from hydrogen burning and are much rarer than main sequence stars, subgiant and first-ascent red giants, where the luminosity is due to hydrogen burning exclusively.

> **Comments:**
> **username_3**: If you expand on idea of Anders Sanders in his comment you maybe should mention that.  
> **username_2**: @LeosOndra I was busy writing my answer, so I didn't see the comment.  
> **username_3**: Ok, I understand.  
> **username_4**: $2\times10^{48}$ helium atoms weigh about the same as Pluto, or $\frac15$ as much as the Moon. At 1 atm pressure, they would occupy about the same volume as Uranus.  
> **username_2**: @username_4 yes. Added-value paragraph included.  

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