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Cruithne3753
2007-Jan-06, 08:28 PM
Hello, newbie here...

Ignoring the relatavistic effects of gravity, the absurdly short life of such a star, and a few other things besides, at 3.7 million Solar masses, just how bright would the galactic core black hole be if it shone as an ordinary star instead?

grant hutchison
2007-Jan-06, 08:38 PM
Ignoring the relatavistic effects of gravity, the absurdly short life of such a star, and a few other things besides, at 3.7 million Solar masses, just how bright would the galactic core black hole be if it shone as an ordinary star instead?It wouldn't be a star. It would be a black hole. It has too much mass to be a star.

Grant Hutchison

Cruithne3753
2007-Jan-06, 08:50 PM
Yes, I know it's a black hole (relatavistic effects -> gravity preventing the escape of light)... ignoring that. I did say it was a silly question!

grant hutchison
2007-Jan-06, 09:03 PM
Yes, I know it's a black hole (relatavistic effects -> gravity preventing the escape of light)... ignoring that!Yes, you told me to ignore that.
Black holes don't require relativity: Newtonian gravity will do (in fact, the Schwarzschild radius for a black hole is the same as would be predicted from simple Newtonian gravity). Laplace predicted the effect of escape velocity exceeding that of light for a sufficiently dense object in the 1700s, long before relativity.
So the question is whether any force exists that can prevent a single object from collapsing under its own weight to the point at which the escape velocity exceeds light. For something as massive as you want, there's no such force.
A cloud of gas of the mass you suggest would blow itself apart into smaller objects before forming a single star: there's something called the Eddington limit, which sets an upper boundary on the mass of stable stars, and it kicks in around a (very) few hundred solar masses.

Grant Hutchison

Disinfo Agent
2007-Jan-06, 09:45 PM
I suppose a reply more to the point might be that, even assuming there were some way to keep all that mass fusing in one place without collapsing on itself, the escape velocity of such an object would still exceed light speed, and it would not be visible from afar. In other words, it would still be a black whole for an outside observer.

I'm not an expert. Please be lenient with me. :o

P.S. Wait, I'm thinking in circles, aren't I? Escape velocity doesn't reach light speed until the object becomes dense enough to be a real black hole, I imagine. So I guess it really is a question of how much mass can hold on together before that, as Grant said.

Tim Thompson
2007-Jan-06, 10:22 PM
The mass - luminosity relationship for main sequence stars is roughly L/Lsun = (M/Msun)4. So, a 3,700,000 solar mass main sequence star would shine 1.87x1026 times brighter than the sun. Of course, such an object is impossible, so the answer really does not mean much. It goes without saying, I hope, that even at a distance of roughly 8000 parsec, such an object would be rather brighter than simply "naked eye".

Cruithne3753
2007-Jan-06, 11:06 PM
Well, what prompted this question was the short story Finis by Frank Pollack (text here (http://www.pulp-mag.net/pd/vI/POLLACKfinis.html)).
The science may seem ropey but it was written 100 years ago.

Spaceman Spiff
2007-Jan-06, 11:42 PM
The mass - luminosity relationship for main sequence stars is roughly L/Lsun = (M/Msun)4. So, a 3,700,000 solar mass main sequence star would shine 1.87x1026 times brighter than the sun. Of course, such an object is impossible, so the answer really does not mean much. It goes without saying, I hope, that even at a distance of roughly 8000 parsec, such an object would be rather brighter than simply "naked eye".

Not to split hairs over something clearly impossible, but the L-M relationship flattens out for very massive stars: L ~ M.
This is due to (a) radiation pressure gradient dominating the support of the star and (b) electron scattering (which doesn't depend much on density or T, except at extremes of either) dominating the star's opacity.

Several bad things happen which prevents such objects from existing (or forming, for that matter), and radiation pressure (the Eddington luminosity) is just one of them.

Ken G
2007-Jan-07, 03:39 AM
Yes, the L~M happens precisely to avoid the Eddington limit (and that would also avoid the black-hole problem: the radius would have to be quite large). So if those other 'bad things' didn't happen, I suppose one could argue that the luminosity would be about 10^10 solar luminosities, or in the ball park of the rest of the galaxy combined. Since it is about 10^9 times farther away than the Sun, we would still see it as billions of times less bright than our Sun. Indeed, since black hole accretion disks also tend to try to stay below the Eddington limit, that isn't a bad estimate of the maximum possible luminosity of our galactic core if the Milky Way were an active galaxy.

Tobin Dax
2007-Jan-07, 08:02 AM
Yes, the L~M happens precisely to avoid the Eddington limit (and that would also avoid the black-hole problem: the radius would have to be quite large). So if those other 'bad things' didn't happen, I suppose one could argue that the luminosity would be about 10^10 solar luminosities, or in the ball park of the rest of the galaxy combined. Since it is about 10^9 times farther away than the Sun, we would still see it as billions of times less bright than our Sun. Indeed, since black hole accretion disks also tend to try to stay below the Eddington limit, that isn't a bad estimate of the maximum possible luminosity of our galactic core if the Milky Way were an active galaxy.

I feel like doing a quick back-of-the-envelope calculation: 10^10 solar luminosities means 25 magnitudes brighter. Let's put it at 10 kpc for simplicity, and say the sun has M=5 for the same reason. That means the Sgr A* would have M=-20. m-M+5 = 5*log(d) => m = 20-5-5*4 = -5 So Sgr A* would appear 100 times brighter than the brightest star in the nighttime sky, but the sun would appear about 10^8 times brighter to us.

Kullat Nunu
2007-Jan-07, 12:24 PM
Except that the intervening dust prevents us seeing anything.

Ken G
2007-Jan-07, 03:33 PM
Good point-- the UV won't even get through the hydrogen gas, and the visible light (which would be somewhat dimmer than the UV to begin with) would get creamed by the dust. We'd have to see it in other galaxies.

Spaceman Spiff
2007-Jan-10, 03:08 PM
I feel like doing a quick back-of-the-envelope calculation: 10^10 solar luminosities means 25 magnitudes brighter. Let's put it at 10 kpc for simplicity, and say the sun has M=5 for the same reason. That means the Sgr A* would have M=-20. m-M+5 = 5*log(d) => m = 20-5-5*4 = -5 So Sgr A* would appear 100 times brighter than the brightest star in the nighttime sky, but the sun would appear about 10^8 times brighter to us.

And you're dealing with a hot star that emits most of its light in the UV (which doesn't penetrate dust or our atmosphere, or produce a useful image in our eye). So you've got an enormous bolometric de-correction to make to predict say its apparent visual magnitude.

Spaceman Spiff
2007-Jan-10, 03:09 PM
Yes, the L~M happens precisely to avoid the Eddington limit (and that would also avoid the black-hole problem: the radius would have to be quite large).....

Actually, L directly proportional to M is realized at the Eddington Limit. My meaning was that the power on M becomes smaller and approaches 1 for very high mass stars, due to the effects I mentioned.

Ken G
2007-Jan-10, 03:46 PM
Actually, L directly proportional to M is realized at the Eddington Limit.

Much more than that-- it is enforced by the Eddington limit. The star must find a way to make it true-- or the star ceases to be. The issue of what is the driver (here mass and the Eddington limit) and what is the passenger (here the luminosity) seems to keep cropping up in our discussions!

Nereid
2007-Jan-10, 03:49 PM
Yes, I know it's a black hole (relatavistic effects -> gravity preventing the escape of light)... ignoring that. I did say it was a silly question!As I think has become clear, it's not only relativistic effects that you have to ignore, in order to get an answer to your question!

In addition to the examples given above, there's a stability question - do you require, for it to be considered a 'star', that it is in hydrostatic equilibrium, even if only for a month (say)?

However, if we ignore the mass, and all the problems that creates, and simply ask "what is the luminosity of a spherical object of radius r, which emits EM as a blackbody, isotropically, and has a temperature of T?" then we could get a straight-forward answer.

Then we could move on to how to relate an object of mass 3 millions sols to the spheres we had just dealt with, making appropriate limiting assumptions as we went along ...

Ken G
2007-Jan-10, 04:16 PM
However, if we ignore the mass, and all the problems that creates, and simply ask "what is the luminosity of a spherical object of radius r, which emits EM as a blackbody, isotropically, and has a temperature of T?" then we could get a straight-forward answer. Then we could move on to how to relate an object of mass 3 millions sols to the spheres we had just dealt with, making appropriate limiting assumptions as we went along ...
And we did just that-- the answer is the Eddington-limit answer I and Spaceman Spiff gave above! That's how it would have to shake out-- it could be less luminous than that, but basically there are processes encouraging it to be more luminous, and others saying it can't, so there you have it.

Tobin Dax
2007-Jan-10, 05:24 PM
And you're dealing with a hot star that emits most of its light in the UV (which doesn't penetrate dust or our atmosphere, or produce a useful image in our eye). So you've got an enormous bolometric de-correction to make to predict say its apparent visual magnitude.

I stand corrected.

Nereid
2007-Jan-10, 05:37 PM
However, if we ignore the mass, and all the problems that creates, and simply ask "what is the luminosity of a spherical object of radius r, which emits EM as a blackbody, isotropically, and has a temperature of T?" then we could get a straight-forward answer. Then we could move on to how to relate an object of mass 3 millions sols to the spheres we had just dealt with, making appropriate limiting assumptions as we went along ...And we did just that-- the answer is the Eddington-limit answer I and Spaceman Spiff gave above! That's how it would have to shake out-- it could be less luminous than that, but basically there are processes encouraging it to be more luminous, and others saying it can't, so there you have it.I think it would be useful to take this a little slower - for starters, I expect that many readers of this thread, who are very interested in the answer, have little or no idea of what the Eddington limit is, let alone how it is relevant.

For example, the Eddington limit depends upon several factors, including the composition of the gas which makes up the 'star', so rolling this limit out already includes several of the limiting assumptions that I was proposing we explicitly examine ...

Ken G
2007-Jan-10, 08:10 PM
Indeed, you are right-- so far it's just jargon and the claim that L~M. The idea is that when the luminosity gets really high, eventually the outward "push" of the light on the matter can exceed gravity. That is called the Eddington limit, and in spherical symmetry, if it is exceeded the star explodes. So there's a maximum luminosity, and Nereid is correct that it depends on the opacity, the ability of light to interact with matter. So once you think you know what kind of matter you have, that gives you an opacity, and that in turn gives you a maximum luminosity that is proportional to the amount of mass. You don't know that you will necessarily achieve that luminosity, but objects that massive tend to find ways. Then if they try to exceed it, the loss of force balance will tend to disconnect whatever process was going too far. That's why very massive objects tend to emit at the Eddington limit, though not always, they still need some kind of energy source to maintain equilibrium.

The point about needing to include opacity in the luminosity is so well taken that it brings out an important point-- if much of the mass is in the form of dark matter, then that component will have no opacity, and the luminosity can rise proportionately.

Spaceman Spiff
2007-Jan-11, 03:57 AM
To be a bit more precise, the star doesn't explode (well, exploding stars do temporarily greatly exceed their Eddington luminosities, but stars at or near their Eddington luminosities aren't necessarily exploding in the sense of a supernova) - it just unbinds itself (the total mechanical energy goes to 0), until as KenG mentioned, the mass loss results in changing the dynamical structure of the star that then puts to a stop whatever process pushed it near or over the limit. Actually, wikipedia (http://en.wikipedia.org/wiki/Eddington_luminosity) has reasonable and brief overview.

Spaceman Spiff
2007-Jan-11, 04:04 AM
Much more than that-- it is enforced by the Eddington limit. The star must find a way to make it true-- or

if for some reason it cannot...


...the star ceases to be.

i.e., is most likely in the process of exploding.


The issue of what is the driver (here mass and the Eddington limit) and what is the passenger (here the luminosity) seems to keep cropping up in our discussions!

Well, it has, but only apparently. I haven't said anything here to the contrary (nor in the Neutron star thread (http://www.bautforum.com/showthread.php?p=892742#post892742), particularly here (http://www.bautforum.com/showpost.php?p=900734&postcount=86)...). But I still beg to differ from you somewhat re. MS star evolution. :)

Ken G
2007-Jan-11, 06:45 AM
Fair enough, we can't agree on everything, what fun would that be?

Nereid
2007-Jan-11, 09:02 PM
One of my reasons for taking this slowly was to allow us to explore all kinds of 'stars' (remember, the OP didn't restrict them, other than "Ignoring the relatavistic effects of gravity" ...

So what about a 3.7 million sol 'neutron star'? or a 'white dwarf'?

Or is it quite impossible to consider such 'stars', in the absence of the relativistic effects of gravity?

Ken G
2007-Jan-12, 12:48 AM
So what about a 3.7 million sol 'neutron star'? or a 'white dwarf'?

Or is it quite impossible to consider such 'stars', in the absence of the relativistic effects of gravity?It's not impossible to consider them without relativistic gravity. All you need is the Pauli exclusion principle, the Heisenberg uncertainty principle, and the meaning of pressure, to show that both those type of stars would collapse into black holes. Just write a crude force balance as P*R^2 = G*M^2/R^2, which is the force balance of virtually all astronomical objects in equilibrium. If you are talking white dwarf or neutron star, then their P is known to be purely a function of density, because it depends purely on the Heisenberg uncertainty principle-- the characteristic momentum is inversely proportional to the interparticle distance. You will find when you do that that if M exceeds a few solar masses, then the momentum must be greatly in excess of m*c, so you have highly relativistic particles and it's not going to matter much at all if they are neutrons or electrons. Then comes the real kicker-- relativistic degeneracy pressure is proportional to density to the 4/3 power. This is where the Heisenberg uncertainty principle comes in-- the interparticle separation scales as density to the minus 1/3 power, so the momentum goes like density to the plus 1/3 power, so the kinetic energy per particle does too (it's relativistic). The pressure will scale like the kinetic energy over the volume, so that's the kinetic energy per particle times the particle density, so voila, you get particle density to the 4/3 power. OK, so now go back to
P*R^2 = G*M^2/R^2 and plug in P proportional to M/R^3 to the 4/3 power (replace particle density with mass density because they are simply proportional to each other-- electrons pair with protons and the mass is essentially the same as a neutron), and you find that R drops out and you can solve for M! It comes out about 1.4 solar masses for electrons, and similarly for neutrons because they are essentially quite similar (except a lot of more complicated forces that don't change things much and are not that well understood). The interpretation of being able to solve for M when you assume the particles are relativistic is that there is no solution for a larger M at any radius or any density-- there's only one mass that can be supported by highly relativistic degeneracy pressure, and less relativistic will support even less mass.

Relativistic effects mean, among other things, that the gravitational binding energy lowers the gravitational mass below the baryonic mass, so this helps the star stay "afloat", but not nearly enough to save this puppy. It's black hole city-- the ultimate relativistic correction.

Note that the Eddington limit suffers from no such mass limit, without considering very complicated stability questions. In principle, without those considerations, you can get a force balance for any mass using high enough L, but the pressure you are using is radiation pressure, not degeneracy pressure.

publius
2007-Jan-12, 02:05 AM
Note that the Eddington limit suffers from no such mass limit, without considering very complicated stability questions. In principle, without those considerations, you can get a force balance for any mass using high enough L, but the pressure you are using is radiation pressure, not degeneracy pressure.

This is one of the key arguments of Mitra's ECO theory.

The radiation cannot escape as gravity gets stronger and stronger -- you essentially have "gravitational opacity", and it's that radiation pressure that prevents collapse. The result (according to him and his "converts") is a dense, radiation dominated soup -- no event horizon but a surface very close to it.

I forget what it is, but for mass M, the traditional cold collapse and degeneracy and all that places a lower limit on radius as some r = k*R, where R is the Schwarzschild radius and k > 1, I think k = 2 or 3 or something.

The ECO business says k can get very close to 1.

-Richard

Ken G
2007-Jan-12, 04:47 AM
That does make it easier to picture the idea, I have no idea if it holds water, or light.