<p>The standard treatment can be found in (<a href="https://press.princeton.edu/books/paperback/9780691130279/galactic-dynamics" rel="noreferrer">Binney &amp; Tremaine 2008</a>), but see also (<a href="https://arxiv.org/abs/astro-ph/9701131" rel="noreferrer">Adams &amp; Laughlin 1997</a>) for a good treatment. </p>

<p>The overall timescale for galactic evaporation is <span class="math-container">$$\tau_{evap}= 100\tau_{relax}\sim 10^{19}$$</span> years.</p>

<p>The relaxation timescale <span class="math-container">$$\tau_{relax}=\frac{R}{v}\frac{N}{12 \ln(N/2)},$$</span> where <span class="math-container">$R$</span> is the size of the system, <span class="math-container">$v$</span> is the typical random velocity, and <span class="math-container">$N$</span> is the total number of stars. This corresponds to the time it takes to completely randomize the velocity of a star by interactions with other stars.</p>

CC BY-SA 4.0
2020-02-23T05:06:47.470Z
This matches very closely the estimated time given here: https://en.wikipedia.org/wiki/Timeline_of_the_far_future
Michael
CC BY-SA 4.0
2020-02-23T09:30:23.397Z
Does this assume that \$N\$ is constant (i.e. the same number of stars are born as die)?
Graipher
CC BY-SA 4.0
2020-02-23T16:20:24.143Z
@Graipher - Yes, it assumes $N$ stays constant. In practice this does not matter much since star formation declines from the early stelliferous era, so the error is at most a small multiplicative factor totally swamped by other uncertainties like the velocity dispersion.
Anders Sandberg
CC BY-SA 4.0
2020-02-23T18:19:06.710Z
Funny thing is, the evaporation of the galaxy is surprisingly low. We have $\approx 10^{11}$ stars in our galaxy, leading to an evaporation rate of 1 star per 100million years. WIth other words, $\approx$ 140 stars ejected our galaxy until now.
peterh
CC BY-SA 4.0
2020-02-23T18:26:56.807Z
@peterh-ReinstateMonica - ~140 ejected by this kind of random event. There are a few more hypervelocity stars due to close encounters with the central black hole.
Anders Sandberg
CC BY-SA 4.0
2020-02-23T18:48:06.850Z
@AndersSandberg It looks the central black hole is a far more effective "evaporator". (Sorry for the unaccept, it looks this question is popular and I would like to attract attention. I will again accept in some days.)
peterh
CC BY-SA 4.0
2020-02-23T20:32:06.033Z
@peterh-ReinstateMonica - There are several processes discussed in Binney & Tremaine, with different efficiencies over time (could bring them into the answer, but it would not be very concise). The eventual dissolution is more due to the slow statistical effects than "hard encounters", but there is a fair bit of calculation involved.
Anders Sandberg
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2020-02-25T00:59:53.347Z
@AndersSandberg $10^{19}$ years is about a billion times the current age of the Universe. Does that evaporation timescale take into account the accelerating expansion of space due to dark energy?
Chappo Hasn't Forgotten
CC BY-SA 4.0
2020-02-25T02:14:22.450Z
@ChappoSaysSEDuddedMonica - Accelerated expansion does not strongly affect dynamics inside galactic systems (lots of other answers about that here). Long before the evaporation happens each galaxy will be in its own little island universe, and ejected stars soon (after a few billion years) will be alone in intergalactic space with nothing visible.
Anders Sandberg
CC BY-SA 4.0
2020-02-25T03:41:58.570Z
@AndersSandberg (after much more reading) yes, it's a dark and lonely future! In fact all the stars in our merged Milkomeda galaxy are expected to have exhausted their fuels after ~$10^{14}$ years, leaving just stellar remnants and brown dwarfs. If that's the case, our galaxy's stelliferous period represents just the first 0.001% of the timescale for galactic evaporation.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2020-02-28T18:01:26.620Z
@ChappoSaysSEDuddedMonica I think a galaxy of stellar remnants evaporate roughly with the same intensity. More interesting question: after this $10^{20}$ years, the stars will be dispersed in the Universe. Where is their gravitational potential energy is coming from? (What is "dudded"?)
peterh
CC BY-SA 4.0
2020-02-29T05:16:26.917Z
@peterh-ReinstateMonica *Dudded* is Australian slang for *tricked, robbed, left in the lurch*.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2020-02-23T01:28:03.203Z
2020-02-23T01:28:03.203Z
Anders Sandberg
<p>For a star, to leave its galaxy, it requires probably a 3-body near-collision at the edge of the galaxy. It is unlikely, but possible. And, if a star once somehow got the required velocity to escape the galaxy, then it will escape and never comes back.</p>

<p>Thus, galaxies should have a <em>characteristic time of their evaporation</em>, what could be calculated. I suspect, this time is probably longer, than other related processes (expansion of the Universe, end of the age of the stars, etc), but it exists can it can be probably calculated (most likely, by numeric simulations).</p>

<p>It probably also depends on the size and star density of the galaxy.</p>

<p>Was it calculated? How big is it?</p>

CC BY-SA 4.0
2020-02-22T18:37:25.537Z
Supermassive blackholes and "puny" blackholes can eject stars (not at the edge) from a galaxy (https://www.newscientist.com/article/dn10020-puny-black-holes-can-eject-milky-ways-stars/), so doesn't that need to be figured into the answer?  Also, what are you asking with the question "Did it happen?"  Are you asking if a galaxy has already evaporated?
Bob516
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2020-02-22T18:58:26.850Z
@Bob516 I understand this on the "characteristic time of the evaporation of a galaxy": how long would the galaxy disappear, if the evaporation rate (in star/year) would not change due to other processes.
peterh
<galactic-dynamics>
What is the characteristic time of the evaporation of the galaxies?
peterh
# What is the characteristic time of the evaporation of the galaxies?

(Asked by: username_0 on 2020-02-22)

For a star, to leave its galaxy, it requires probably a 3-body near-collision at the edge of the galaxy. It is unlikely, but possible. And, if a star once somehow got the required velocity to escape the galaxy, then it will escape and never comes back.

Thus, galaxies should have a *characteristic time of their evaporation*, what could be calculated. I suspect, this time is probably longer, than other related processes (expansion of the Universe, end of the age of the stars, etc), but it exists can it can be probably calculated (most likely, by numeric simulations).

It probably also depends on the size and star density of the galaxy.

Was it calculated? How big is it?

**Tags:** galactic-dynamics
> **Comments:**
> **username_1**: Supermassive blackholes and "puny" blackholes can eject stars (not at the edge) from a galaxy (https://www.newscientist.com/article/dn10020-puny-black-holes-can-eject-milky-ways-stars/), so doesn't that need to be figured into the answer?  Also, what are you asking with the question "Did it happen?"  Are you asking if a galaxy has already evaporated?  
> **username_0**: @Bob516 I understand this on the "characteristic time of the evaporation of a galaxy": how long would the galaxy disappear, if the evaporation rate (in star/year) would not change due to other processes.  

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## 1 answers

### Answer by username_2 on 2020-02-23. Score: 7 | Accepted answer

The standard treatment can be found in ([Binney & Tremaine 2008](https://press.princeton.edu/books/paperback/9780691130279/galactic-dynamics)), but see also ([Adams & Laughlin 1997](https://arxiv.org/abs/astro-ph/9701131)) for a good treatment.

The overall timescale for galactic evaporation is $$\tau\_{evap}= 100\tau\_{relax}\sim 10^{19}$$ years.

The relaxation timescale $$\tau\_{relax}=\frac{R}{v}\frac{N}{12 \ln(N/2)},$$ where $R$ is the size of the system, $v$ is the typical random velocity, and $N$ is the total number of stars. This corresponds to the time it takes to completely randomize the velocity of a star by interactions with other stars.

> **Comments:**
> **username_3**: This matches very closely the estimated time given here: https://en.wikipedia.org/wiki/Timeline_of_the_far_future  
> **username_4**: Does this assume that \$N\$ is constant (i.e. the same number of stars are born as die)?  
> **username_2**: @Graipher - Yes, it assumes $N$ stays constant. In practice this does not matter much since star formation declines from the early stelliferous era, so the error is at most a small multiplicative factor totally swamped by other uncertainties like the velocity dispersion.  
> **username_0**: Funny thing is, the evaporation of the galaxy is surprisingly low. We have $\approx 10^{11}$ stars in our galaxy, leading to an evaporation rate of 1 star per 100million years. WIth other words, $\approx$ 140 stars ejected our galaxy until now.  
> **username_2**: @peterh-ReinstateMonica - ~140 ejected by this kind of random event. There are a few more hypervelocity stars due to close encounters with the central black hole.  
> **username_0**: @AndersSandberg It looks the central black hole is a far more effective "evaporator". (Sorry for the unaccept, it looks this question is popular and I would like to attract attention. I will again accept in some days.)  
> **username_2**: @peterh-ReinstateMonica - There are several processes discussed in Binney & Tremaine, with different efficiencies over time (could bring them into the answer, but it would not be very concise). The eventual dissolution is more due to the slow statistical effects than "hard encounters", but there is a fair bit of calculation involved.  
> **username_5**: @AndersSandberg $10^{19}$ years is about a billion times the current age of the Universe. Does that evaporation timescale take into account the accelerating expansion of space due to dark energy?  
> **username_2**: @ChappoSaysSEDuddedMonica - Accelerated expansion does not strongly affect dynamics inside galactic systems (lots of other answers about that here). Long before the evaporation happens each galaxy will be in its own little island universe, and ejected stars soon (after a few billion years) will be alone in intergalactic space with nothing visible.  
> **username_5**: @AndersSandberg (after much more reading) yes, it's a dark and lonely future! In fact all the stars in our merged Milkomeda galaxy are expected to have exhausted their fuels after ~$10^{14}$ years, leaving just stellar remnants and brown dwarfs. If that's the case, our galaxy's stelliferous period represents just the first 0.001% of the timescale for galactic evaporation.  
> **username_0**: @ChappoSaysSEDuddedMonica I think a galaxy of stellar remnants evaporate roughly with the same intensity. More interesting question: after this $10^{20}$ years, the stars will be dispersed in the Universe. Where is their gravitational potential energy is coming from? (What is "dudded"?)  
> **username_5**: @peterh-ReinstateMonica *Dudded* is Australian slang for *tricked, robbed, left in the lurch*.  

---

