5 ms·

The Gravo-Thermal Catastrophe

105 points by terryf 2y ago

phkahler 2y ago
>> Also suppose they’re ‘gravitationally bound’. This means their total energy, kinetic and potential, is negative. That means they couldn’t all shoot off to infinity even if the sphere wasn’t there holding them in. This seems like an invalid assumption. We know that clusters of stars can eject some of their members. Lot of hand waving in this one.
- ISL 2y ago
  That's only an initial condition -- that requirement states only that the total energy is negative. We are gravitationally bound to Earth, but the Voyagers have left the solar system.
- pavel_lishin 2y ago
  That sentence does say "couldn't all shoot off to infinity".
keskival 2y ago
I suppose in the real world such stars would collide in the center of the sphere and possibly form a black hole before achieving the required density approaching infinity, and also catapult stars out so that they leave the system by exceeding the escape velocity without encountering an elastic wall returning them to the system.
- pdonis 2y ago
  > in the real world such stars would collide in the center of the sphere and possibly form a black hole Yes, the article mentions that towards the end.
  - xbar 2y ago
    “Things only get messier from here.”
AnotherGoodName 2y ago
My favourite along these lines is that the mass vs diameter relation for black holes scales in such a way that we are absolutely in a black hole right now according to current theory. As in the current mass of the universe is enough for a black hole with an event horizon diameter that extends beyond the universe.
- Nesco 2y ago
  This mass to event horizon radius relationship is a property of a Schwarzschild spacetime geometry, globally the universe has a FLRW spacetime geometry
  - trhway 2y ago
    The light has no chances of getting out of the 13.7B ly bubble due to Hubble expansion. Sounds a lot like black hole.
    - Nesco 2y ago
      The universe has no center, a black hole has one. The limits of the visible universe is an horizon on your frame of reference In nerdspeak, the geometries are not the same, one is isotropic the other anisotropic
      - hnuser123456 2y ago
        When you are in a black hole, you are always heading towards the center. The universe is expanding faster than the speed of light beyond a distance of 13.7b ly away, and the comoving diameter of the universe is already something like 93b ly. If you could teleport to another part of the universe, you would have a whole new local environment, but it would still be flying apart faster than you could ever hope to catch up to the furthest objects (without another teleport), and the furthest objects you can reach are getting more and more sparse as they are always flying outside of your observable universe. I.e., inside a black hole, where all possible directions only point further inward, and yet also "away from everything else", as anything that fell in before you, gets pulled in faster and faster "away from you, towards the singularity", and anything that falls in after you has increasingly no hope of catching up to you.
jessriedel 2y ago
Is this right?: * Although you can make the enveloping sphere as large as you want, the (anti-)equilibration process requires a sphere of some finite radius because if you wait long enough a few stars eventually get launched at escape velocity, and if these actually escaped they would effectively cool the remaining stars. * Therefore, the characteristic time scale for this process (i.e., the timescale on which the average kinetic energy rises substantially) gets longer and longer as the sphere gets larger. * In order for the pressure and average speed of the stars to keep rising, the gravitational potential needs to keep falling, so at least some stars need to get and stay very close. In real life, these turn into black holes, which cuts off the process by limiting the amount of gravitational potential energy that can be unlocked in any given volume with a given mass.
- pavel_lishin 2y ago
  > In real life, these turn into black holes I think this is right, and I think he explicitly calls out that these calculations were done with Newtonian physics modeling point particles - and we know that those two factors severely limit the application of this to the real-world.
  - jessriedel 2y ago
    Right, it wasn't criticism, but the point I added (that I think was not explicit in the article) is this: black holes provide a lower bound on the potential energy, not just an indication that the model is breaking down.
  - hnuser123456 2y ago
    Yes, so some of the gravitational potential energy gets converted into gravitational waves instead of flinging some stars further and further out, enabling at least something to stay behind.
- drudd 2y ago
  I was asked this question on my grad candidacy exam (which was quite a while ago, so my memory is hazy), and I believe stars tend to form binary systems which can halt the runaway gravitational collapse (the potential energy in a hard binary can be a very significant fraction of the total energy of a relatively loosely bound globular cluster).
pfdietz 2y ago
As I understand it, those simulations did not include three-body interactions that could leave particle pairs bound. If this happens, those binaries can now inject energy into the cluster as a whole, keeping it inflated and preventing collapse. Of course, the binaries' orbits shrink over time, so this doesn't go on forever.
- leephillips 2y ago
  What is a three-body interaction in classical gravity? If you calculate the force on each particle from every other particle, what’s left out?
  - pfdietz 2y ago
    I mean, they did not calculate interactions on a sufficiently short time scale that cases where three bodies come together and two come out bound would occur.
trhway 2y ago
Application of the 1/R2 gravity formula to the pointwise mass with R->0 can easily power your Romulan ships. In similar vein applying that classical gravity formula - which is valid only to spherical masses or masses at such large distances that they can be treated as such - to the stars inside disk galaxies gets you the "dark matter", and thus not surprisingly the flatter the disk galaxy the more "dark matter" :)
- meindnoch 2y ago
  >In similar vein applying that classical gravity formula - which is valid only to spherical masses What? Newtonian gravity is defined for point masses. Anything else you derive from that by integrating a mass density over a region.
  - trhway 2y ago
    it is equivalent formulations - the point masses case is obtained from the spherical in the limit. The spherical case is just more illustrative to show where the fantom of the "dark matter" in the disk galaxies comes from. >by integrating a mass density over a region. exactly. When you do that for a disk galaxy you get much flatter curves that the 1/R the proponents of the dark matter insist on (that 1/R is exactly what one would get if the galaxy was spherical or the star was far outside of the disk)
bbor 2y ago
A) …why? What makes this interesting to physicists? I understand this as “if stars weren’t stars but instead rigid spheres, and if they were in an impossibly-impervious giant sphere, then weird stuff happens”. And…? B) “since stars rather rarely collide” still blows my mind. I did some napkin math on Reddit a while back on why there will be very few stellar collisions (really, one star falling into another’s orbit?) when andromeda collides with the Milky Way, and the answer is that space is just mind-bogglingly huge. Even the most dense clusters in our galaxy are akin to ~70 1cm diameter spheres per olympic swimming pool. If god is real, he is surely a giant.
- LegionMammal978 2y ago
  For A), if you have a bunch of tiny atoms bouncing around within a regular-sized sphere, then thermodynamics predicts that the sphere will experience some constant amount of pressure, with tiny fluctuations up and down. This result is interesting, since it just takes the ordinary system and asks, "What if we scale it up so that the atoms (stars) interact gravitationally?" Then, there is no equilibrium pressure experienced by the sphere, since the gravitational potential of the stars keeps increasing.
  - trhway 2y ago
    > Then, there is no equilibrium pressure experienced by the sphere, since the gravitational potential of the stars keeps increasing. And GR fixes that by kind of moving the sphere walls farther away, ie. the space geometry changing by the changing gravitational potential.
mikhailfranco 2y ago
This idea has been explored by Julian Barbour in his book The Janus Point. http://www.platonia.com/books.html http://www.platonia.com/books.html The related math and modelling goes under the name Shape Dynamics: https://en.wikipedia.org/wiki/Shape_dynamics https://en.wikipedia.org/wiki/Shape_dynamics Shape Dynamics - An Introduction https://arxiv.org/abs/1105.0183 https://arxiv.org/abs/1105.0183
kazinator 2y ago
Something doesn't add up, obviously. For one thing, squeezing the sphere smaller against pressure requires work. That's an external energy input. The system is not closed if some agent is available that can squeeze the sphere smaller.
- hoseja 2y ago
  It's not getting squeezed, it's just gravitationally collapsing. The same thing happens when forming stars, except obviously there's all the electromagnetic repulsion, then nuclear repulsion, then black hole formation. The article speaking about "stars" as analogy for pointlike masses is kinda misleading. If it talked instead about gas in a nebula forming into a hot star, it would make much more sense.
  - kazinator 2y ago
    In this gedanken experiment, the stars are placed into a spherical container. Particles within this container bounce off its walls. It is like a cylinder full of gas. The container does not collapse with the stars; its dimensions are somehow controlled by the experimenter. Well, if there is pressure inside this sphere, then you cannot shrink it without working against the pressure, and doing so will add energy into the system because work is being done on it.

105 points by terryf 2y ago

>> Also suppose they’re ‘gravitationally bound’. This means their total energy, kinetic and potential, is negative. That means they couldn’t all shoot off to infinity even if the sphere wasn’t there holding them in. This seems like an invalid assumption. We know that clusters of stars can eject some of their members. Lot of hand waving in this one.

ISL 2y ago
That's only an initial condition -- that requirement states only that the total energy is negative. We are gravitationally bound to Earth, but the Voyagers have left the solar system.
pavel_lishin 2y ago
That sentence does say "couldn't all shoot off to infinity".

keskival 2y ago

I suppose in the real world such stars would collide in the center of the sphere and possibly form a black hole before achieving the required density approaching infinity, and also catapult stars out so that they leave the system by exceeding the escape velocity without encountering an elastic wall returning them to the system.

pdonis 2y ago
> in the real world such stars would collide in the center of the sphere and possibly form a black hole Yes, the article mentions that towards the end.
- xbar 2y ago
  “Things only get messier from here.”

AnotherGoodName 2y ago

My favourite along these lines is that the mass vs diameter relation for black holes scales in such a way that we are absolutely in a black hole right now according to current theory. As in the current mass of the universe is enough for a black hole with an event horizon diameter that extends beyond the universe.

Nesco 2y ago
This mass to event horizon radius relationship is a property of a Schwarzschild spacetime geometry, globally the universe has a FLRW spacetime geometry
- trhway 2y ago
  The light has no chances of getting out of the 13.7B ly bubble due to Hubble expansion. Sounds a lot like black hole.
  - Nesco 2y ago
    The universe has no center, a black hole has one. The limits of the visible universe is an horizon on your frame of reference In nerdspeak, the geometries are not the same, one is isotropic the other anisotropic
    - hnuser123456 2y ago
      When you are in a black hole, you are always heading towards the center. The universe is expanding faster than the speed of light beyond a distance of 13.7b ly away, and the comoving diameter of the universe is already something like 93b ly. If you could teleport to another part of the universe, you would have a whole new local environment, but it would still be flying apart faster than you could ever hope to catch up to the furthest objects (without another teleport), and the furthest objects you can reach are getting more and more sparse as they are always flying outside of your observable universe. I.e., inside a black hole, where all possible directions only point further inward, and yet also "away from everything else", as anything that fell in before you, gets pulled in faster and faster "away from you, towards the singularity", and anything that falls in after you has increasingly no hope of catching up to you.

jessriedel 2y ago

Is this right?: * Although you can make the enveloping sphere as large as you want, the (anti-)equilibration process requires a sphere of some finite radius because if you wait long enough a few stars eventually get launched at escape velocity, and if these actually escaped they would effectively cool the remaining stars. * Therefore, the characteristic time scale for this process (i.e., the timescale on which the average kinetic energy rises substantially) gets longer and longer as the sphere gets larger. * In order for the pressure and average speed of the stars to keep rising, the gravitational potential needs to keep falling, so at least some stars need to get and stay very close. In real life, these turn into black holes, which cuts off the process by limiting the amount of gravitational potential energy that can be unlocked in any given volume with a given mass.

pavel_lishin 2y ago
> In real life, these turn into black holes I think this is right, and I think he explicitly calls out that these calculations were done with Newtonian physics modeling point particles - and we know that those two factors severely limit the application of this to the real-world.
- jessriedel 2y ago
  Right, it wasn't criticism, but the point I added (that I think was not explicit in the article) is this: black holes provide a lower bound on the potential energy, not just an indication that the model is breaking down.
- hnuser123456 2y ago
  Yes, so some of the gravitational potential energy gets converted into gravitational waves instead of flinging some stars further and further out, enabling at least something to stay behind.
drudd 2y ago
I was asked this question on my grad candidacy exam (which was quite a while ago, so my memory is hazy), and I believe stars tend to form binary systems which can halt the runaway gravitational collapse (the potential energy in a hard binary can be a very significant fraction of the total energy of a relatively loosely bound globular cluster).

pfdietz 2y ago

As I understand it, those simulations did not include three-body interactions that could leave particle pairs bound. If this happens, those binaries can now inject energy into the cluster as a whole, keeping it inflated and preventing collapse. Of course, the binaries' orbits shrink over time, so this doesn't go on forever.

leephillips 2y ago
What is a three-body interaction in classical gravity? If you calculate the force on each particle from every other particle, what’s left out?
- pfdietz 2y ago
  I mean, they did not calculate interactions on a sufficiently short time scale that cases where three bodies come together and two come out bound would occur.

trhway 2y ago

Application of the 1/R2 gravity formula to the pointwise mass with R->0 can easily power your Romulan ships. In similar vein applying that classical gravity formula - which is valid only to spherical masses or masses at such large distances that they can be treated as such - to the stars inside disk galaxies gets you the "dark matter", and thus not surprisingly the flatter the disk galaxy the more "dark matter" :)

meindnoch 2y ago
>In similar vein applying that classical gravity formula - which is valid only to spherical masses What? Newtonian gravity is defined for point masses. Anything else you derive from that by integrating a mass density over a region.
- trhway 2y ago
  it is equivalent formulations - the point masses case is obtained from the spherical in the limit. The spherical case is just more illustrative to show where the fantom of the "dark matter" in the disk galaxies comes from. >by integrating a mass density over a region. exactly. When you do that for a disk galaxy you get much flatter curves that the 1/R the proponents of the dark matter insist on (that 1/R is exactly what one would get if the galaxy was spherical or the star was far outside of the disk)

bbor 2y ago

A) …why? What makes this interesting to physicists? I understand this as “if stars weren’t stars but instead rigid spheres, and if they were in an impossibly-impervious giant sphere, then weird stuff happens”. And…? B) “since stars rather rarely collide” still blows my mind. I did some napkin math on Reddit a while back on why there will be very few stellar collisions (really, one star falling into another’s orbit?) when andromeda collides with the Milky Way, and the answer is that space is just mind-bogglingly huge. Even the most dense clusters in our galaxy are akin to ~70 1cm diameter spheres per olympic swimming pool. If god is real, he is surely a giant.

LegionMammal978 2y ago
For A), if you have a bunch of tiny atoms bouncing around within a regular-sized sphere, then thermodynamics predicts that the sphere will experience some constant amount of pressure, with tiny fluctuations up and down. This result is interesting, since it just takes the ordinary system and asks, "What if we scale it up so that the atoms (stars) interact gravitationally?" Then, there is no equilibrium pressure experienced by the sphere, since the gravitational potential of the stars keeps increasing.
- trhway 2y ago
  > Then, there is no equilibrium pressure experienced by the sphere, since the gravitational potential of the stars keeps increasing. And GR fixes that by kind of moving the sphere walls farther away, ie. the space geometry changing by the changing gravitational potential.

mikhailfranco 2y ago

This idea has been explored by Julian Barbour in his book The Janus Point. http://www.platonia.com/books.html http://www.platonia.com/books.html The related math and modelling goes under the name Shape Dynamics: https://en.wikipedia.org/wiki/Shape_dynamics https://en.wikipedia.org/wiki/Shape_dynamics Shape Dynamics - An Introduction https://arxiv.org/abs/1105.0183 https://arxiv.org/abs/1105.0183

kazinator 2y ago

Something doesn't add up, obviously. For one thing, squeezing the sphere smaller against pressure requires work. That's an external energy input. The system is not closed if some agent is available that can squeeze the sphere smaller.

hoseja 2y ago
It's not getting squeezed, it's just gravitationally collapsing. The same thing happens when forming stars, except obviously there's all the electromagnetic repulsion, then nuclear repulsion, then black hole formation. The article speaking about "stars" as analogy for pointlike masses is kinda misleading. If it talked instead about gas in a nebula forming into a hot star, it would make much more sense.
- kazinator 2y ago
  In this gedanken experiment, the stars are placed into a spherical container. Particles within this container bounce off its walls. It is like a cylinder full of gas. The container does not collapse with the stars; its dimensions are somehow controlled by the experimenter. Well, if there is pressure inside this sphere, then you cannot shrink it without working against the pressure, and doing so will add energy into the system because work is being done on it.