Markus Hanke Posted December 26, 2021 Share Posted December 26, 2021 11 hours ago, geordief said: I did a search on "gravitational fluid" I made up this term, to describe a fluid the constituents of which interact gravitationally, instead of via EM. It’s not an official concept (I think). 11 hours ago, geordief said: BTW ,is there any possibility that DM could be clouds of mini black holes formed in the very early days after BB? If this were so, then ordinary matter (interstellar dust etc) should be detectable as it falls into these micro-BH, due to friction. DM clouds should thus be faintly luminous in the X-ray, radio, and gamma spectra. Also, colliding DM clouds should lead to BH mergers, which again should be detectable. I’m not aware of any data supporting this. 10 hours ago, QuantumT said: The suggestion I like the most, about the nature of DM, is also the most simple, and least mysterious: Ground state hydrogen My thinking is that this is incompatible with high-speed mergers of DM regions, such as the Bullet cluster. Hydrogen would behave like a gas, and thus generate friction and X-ray luminosity; however, observation tells us that in the Bullet cluster the DM region is separate and ahead of the gaseous baryonic region, and not luminous in any spectrum. It behaves differently than baryonic matter, and thus can’t be hydrogen. It’s interaction really is purely gravitational. Link to comment Share on other sites More sharing options...
geordief Posted December 26, 2021 Share Posted December 26, 2021 1 hour ago, Markus Hanke said: I made up this term, to describe a fluid the constituents of which interact gravitationally, instead of via EM. It’s not an official concept (I think No,I knew that Markus .You told us that but I still did a search on it.That is how I learn things sometimes. Sometimes,I may even think of a concept and just look it up to see if anyone has thought of it before(nothing of consequence, maybe even just a sequence of words) Link to comment Share on other sites More sharing options...
Danijel Gorupec Posted December 26, 2021 Share Posted December 26, 2021 On 12/24/2021 at 9:25 PM, Genady said: An easy derivation, with answers to the audience's questions relevant to the above discussion, can be found here, starting about 30 minutes into the lecture: Thanks for the video, I liked it. So, without the dark energy nuisance, we should be able to estimate the portion of uniformly distributed dark mater. But because of the poorly understood dark energy, I guess we cannot do much of an estimation? [Looking at equations derived by Susskind, purely mathematically speaking and putting things upside down, if the dark mater has a negative mass it might cause accelerating universe... supposing large majority of it being actually distributed uniformly, and being only slightly less concentrated around galaxies; as we can detect from its gravitational influence.] Link to comment Share on other sites More sharing options...
Bufofrog Posted December 26, 2021 Share Posted December 26, 2021 12 minutes ago, Danijel Gorupec said: if the dark mater has a negative mass it might cause accelerating universe... supposing large majority of it being actually distributed uniformly, and being only slightly less concentrated around galaxies; as we can detect from its gravitational influence. We have theorized dark matter based on it's gravitational attraction, so saying dark matter might have negative mass effects makes no sense. Link to comment Share on other sites More sharing options...
Danijel Gorupec Posted December 26, 2021 Share Posted December 26, 2021 14 minutes ago, Bufofrog said: We have theorized dark matter based on it's gravitational attraction, so saying dark matter might have negative mass effects makes no sense. Why? It seems the same to me - you can have positive-mass around galaxies, or you can have negative-mass everywhere else. How could you tell the difference? Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 14 minutes ago, Danijel Gorupec said: Why? It seems the same to me - you can have positive-mass around galaxies, or you can have negative-mass everywhere else. How could you tell the difference? By the Newton's theorem, a particle inside an empty massive sphere doesn't feel any gravity from the sphere. It holds the same for positive and for negative gravity. Neither positive nor negative mass homogeneously distributed outside galaxy have any gravitational effect on the galaxy. Link to comment Share on other sites More sharing options...
swansont Posted December 26, 2021 Share Posted December 26, 2021 8 minutes ago, Genady said: By the Newton's theorem, a particle inside an empty massive sphere doesn't feel any gravity from the sphere. It holds the same for positive and for negative gravity. Neither positive nor negative mass homogeneously distributed outside galaxy have any gravitational effect on the galaxy. Just a few days ago you claimed the opposite (“Yes, it would cause a net gravitational attraction and that would cause slowing of the universe expansion.”) Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 53 minutes ago, Danijel Gorupec said: So, without the dark energy nuisance, we should be able to estimate the portion of uniformly distributed dark mater. But because of the poorly understood dark energy, I guess we cannot do much of an estimation? Dark matter and dark energy are separable in a variety of ways. Dark matter attracts, dark energy repulses. Dark matter density falls off with a cube of expansion, dark energy density doesn't change with expansion. Dark matter non-uniformities accelerate clamping of regular matter, dark energy doesn't have such effect. ... Link to comment Share on other sites More sharing options...
Danijel Gorupec Posted December 26, 2021 Share Posted December 26, 2021 6 minutes ago, Genady said: By the Newton's theorem, a particle inside an empty massive sphere doesn't feel any gravity from the sphere. It holds the same for positive and for negative gravity. Neither positive nor negative mass homogeneously distributed outside galaxy have any gravitational effect on the galaxy. Sure, but there can be negative mass within the galaxy that will produce an effect on the galaxy. I suppose it would be possible to came out with some negative-mass-density profile (as a function of the distance from the galaxy center) that would give the observed gravitational effect. Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 2 minutes ago, swansont said: Just a few days ago you claimed the opposite (“Yes, it would cause a net gravitational attraction and that would cause slowing of the universe expansion.”) Right, it doesn't have an effect inside the sphere, it does have the effect outside it. Link to comment Share on other sites More sharing options...
swansont Posted December 26, 2021 Share Posted December 26, 2021 2 minutes ago, Genady said: Right, it doesn't have an effect inside the sphere, it does have the effect outside it. But there is no “outside” when considering the universe. You just need to make R bigger, and there is no effect. Link to comment Share on other sites More sharing options...
Danijel Gorupec Posted December 26, 2021 Share Posted December 26, 2021 5 minutes ago, Genady said: Dark matter attracts... This is exactly what I am talking about -> is this claim a jumping in conclusion? Because of effects like dark energy I cannot dare to claim that dark matter attracts. It might as well repulse... But I know very little, and maybe I am missing an important bit of information that indeed shows it is attractive. Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 4 minutes ago, swansont said: But there is no “outside” when considering the universe. You just need to make R bigger, and there is no effect. Correct again. All effect for the smaller R's add up. All effect s for the larger R's cancel. Link to comment Share on other sites More sharing options...
swansont Posted December 26, 2021 Share Posted December 26, 2021 10 minutes ago, Genady said: Correct again. All effect for the smaller R's add up. All effect s for the larger R's cancel. This makes no sense. There’s nothing to “add up” and there is only one R, which encompasses the universe. Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 14 minutes ago, swansont said: This makes no sense. There’s nothing to “add up” and there is only one R, which encompasses the universe. R in this calculation stands for distance from origin. Which of course is arbitrary. As Susskind has shown in his derivation, the final result does not depend on R. 40 minutes ago, Danijel Gorupec said: I suppose it would be possible to came out with some negative-mass-density profile (as a function of the distance from the galaxy center) that would give the observed gravitational effect. No, that calculation doesn't have such flexibility. The observable effects of dark matter lead to calculation of not only it's mass - positive - but also of its amount and distribution. The only observable effect of dark energy is acceleration of the universe expansion. Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 1 hour ago, Danijel Gorupec said: This is exactly what I am talking about -> is this claim a jumping in conclusion? Because of effects like dark energy I cannot dare to claim that dark matter attracts. It might as well repulse... But I know very little, and maybe I am missing an important bit of information that indeed shows it is attractive. I hope you will enjoy this video as well. Susskind presents an example of dark matter mass calculation starting at about 31st minute. (Only stuff inside an orbit has gravitational effect on the orbit.) Link to comment Share on other sites More sharing options...
swansont Posted December 26, 2021 Share Posted December 26, 2021 2 hours ago, Genady said: R in this calculation stands for distance from origin. Which of course is arbitrary. As Susskind has shown in his derivation, the final result does not depend on R. You’re mis-applying the principle here. The shell theorem applies to spherical symmetry, but you don’t have a finite sphere in this case. You have to have all the mass enclosed within R to apply it. You can look at translational symmetry, too. If you have a uniform distribution of mass, you can choose any origin you want and get the same answer, so there can’t be an acceleration toward any point. 1 Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 1 minute ago, swansont said: You’re mis-applying the principle here. The shell theorem applies to spherical symmetry, but you don’t have a finite sphere in this case. You have to have all the mass enclosed within R to apply it. You can look at translational symmetry, too. If you have a uniform distribution of mass, you can choose any origin you want and get the same answer, so there can’t be an acceleration toward any point. With all due respect, we are going circles for some time already. The answers to all these objections are in the Susskind lecture 1, I've linked earlier. The math is simple and the result is calculable. The outcome is that the uniform infinite distribution of mass causes it to shrink. All points toward all points. If it started with expansion, this distribution causes the expansion to decelerate. Link to comment Share on other sites More sharing options...
swansont Posted December 26, 2021 Share Posted December 26, 2021 1 minute ago, Genady said: With all due respect, we are going circles for some time already. The answers to all these objections are in the Susskind lecture 1, I've linked earlier. The math is simple and the result is calculable. The outcome is that the uniform infinite distribution of mass causes it to shrink. All points toward all points. If it started with expansion, this distribution causes the expansion to decelerate. Then show the math that demonstrates a net acceleration on a particular mass if you have a uniform distribution of mass in the background. I did not make any claim about expansion. Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 1 minute ago, swansont said: Then show the math that demonstrates a net acceleration on a particular mass if you have a uniform distribution of mass in the background. I did not make any claim about expansion. The acceleration is relative to an observer. If an observer is attached to the mass, it does not accelerate anywhere. To any other observer, it accelerates - free falls - toward them with G*M/(R^2), where R is the distance from the observer and M is the mass of the stuff inside a ball of radius R. Link to comment Share on other sites More sharing options...
joigus Posted December 26, 2021 Share Posted December 26, 2021 As Swansont is trying to tell you, the sphere is only for calculational purposes. There's no spherical symmetry. You place a sphere like that any place you want in the universe and calculate acceleration as due to the effect of gravitational attraction from all mass contained in that sphere. Resulting acceleration (which is an overall property of the whole universe) can be positive, negative, or zero, depending on relation between mass density and acceleration parameter, similar to escape velocity possible situations. Susskind makes that very clear in those lectures, as far as I can remember. It may be a bit puzzling that you can get the result from a calculation that involves an imaginary sphere, but that's the magic of Newton's theorem. It's not a sphere that's causing that acceleration. Or, perhaps, if you like a sphere of infinite radius of which any other sphere is at its centre. Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 Yes, the actual calculation should be done from the Friedman equations. The result is the same - a uniform infinite mass density causes deceleration. It is one of the components. The total result - positive, negative, or zero - depends on other components, specifically, on the current rate of expansion and on cosmological constant (or, dark energy, if you prefer.) Link to comment Share on other sites More sharing options...
Danijel Gorupec Posted December 26, 2021 Share Posted December 26, 2021 2 hours ago, Genady said: I hope you will enjoy this video as well. Susskind presents an example of dark matter mass calculation starting at about 31st minute. (Only stuff inside an orbit has gravitational effect on the orbit.) [I always enjoyed Susskind's lectures - did watch quite a few of them, but didn't know he has the cosmology series too.] Are you only considering the case of uniformly distributed mass density? Because if I put a black hole just outside Earth's orbit, it will affect the Earth trajectory. But even with uniformly distributed mass I find your claim confusing... Suppose you only have Earth orbiting the Sun. Now you add uniform mass density to the whole universe - you say that the Earth will only 'feel' the part that is inside its orbit? Does it mean the Earth will go into a tighter orbit with greater velocity? Ok... but remove the Sun now - will Earth go into some other orbit (because it feels the gravitational attraction of the mass inside its orbit) or will the Earth continue straight? Link to comment Share on other sites More sharing options...
Genady Posted December 26, 2021 Share Posted December 26, 2021 (edited) 33 minutes ago, Danijel Gorupec said: Suppose you only have Earth orbiting the Sun. Now you add uniform mass density to the whole universe - you say that the Earth will only 'feel' the part that is inside its orbit? Does it mean the Earth will go into a tighter orbit with greater velocity? Ok... but remove the Sun now - will Earth go into some other orbit (because it feels the gravitational attraction of the mass inside its orbit) or will the Earth continue straight? The answer depends on your frame of reference, i.e. an observer. In the first scenario, as we consider an orbit around the Sun, yes it will get tighter. In the second scenario, wherever you put an observer, the Earth will be free falling toward it (plus the initial velocity.) There is no absolute frame of reference to give an absolute answer. Edited December 26, 2021 by Genady Link to comment Share on other sites More sharing options...
swansont Posted December 26, 2021 Share Posted December 26, 2021 2 hours ago, Genady said: The acceleration is relative to an observer. If an observer is attached to the mass, it does not accelerate anywhere. To any other observer, it accelerates - free falls - toward them with G*M/(R^2), where R is the distance from the observer and M is the mass of the stuff inside a ball of radius R. Acceleration is not relative. Link to comment Share on other sites More sharing options...
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