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Do all galaxies with dark matter halo have flat rotation curve?
Since flat rotation curve is usually mentioned in relation to spiral galaxies, I was wondering if it’s true of all types of galaxies with a dark matter halo.Vanadium 50 said:What have you looked into and what part of it didn't you get?
Further, if you tell that there is extra matter by the flat(ter) rotation curve, aren't you asking a tautological question? And if it's not tautological, maybe you should explain what you mean more.
I think you are asking if also non-spiral galaxies "need" dark matter as a halo to explain their properties?Ranku said:I was wonder if it’s true of all types of galaxies with a dark matter halo.
That would be also another way of putting it.malawi_glenn said:I think you are asking if also non-spiral galaxies "need" dark matter as a halo to explain their properties?
Because "having a dark matter halo" is not something we know that any galaxy have.Ranku said:That would be also another way of putting it.
Then that goes back to my original question as to is dark matter halo unique to spiral galaxies, and if so why. Since dark matter is present in much greater proportion to matter, we’d expect it to accumulate around all types of galaxies as well.malawi_glenn said:Because "having a dark matter halo" is not something we know that any galaxy have.
It is something we have thrown into explain the galaxies seemingly anomalous behavior from Newtonian and Einsteinian gravity.
Yeah there is no equivalence between dark mater haloes and flat galactic rotation curves.Maarten Havinga said:I warn against this way of naming asymptotically flat rotation curves tautological with dark matter haloes.
I would not call it "discredits". MOND researchers discredits Particle DM research so.Maarten Havinga said:Because it simply ignores and discredits all the work done on MOND,
malawi_glenn said:Because "having a dark matter halo" is not something we know that any galaxy have.
It is something we have thrown into explain the galaxies seemingly anomalous behavior from Newtonian and Einsteinian gravity.
I am probably asking a question which has been asked before, but I have only just joined the forum.malawi_glenn said:Because "having a dark matter halo" is not something we know that any galaxy have.
It is something we have thrown into explain the galaxies seemingly anomalous behavior from Newtonian and Einsteinian gravity.
They see the stars, which tells how much matter is there.davicle said:My question is - on what basis do astronomers expect the curve that they say they expect.
You can measure luminosity of different parts of the galaxy and that gets you an estimate of the mass distribution. Then you plug that into Newtonian gravity and equate the gravitational force to the centripetal force to get a predicted rotation rate as a function of distance from galactic center.davicle said:My question is - on what basis do astronomers expect the curve that they say they expect.
davicle said:My question is - on what basis do astronomers expect the curve that they say they expect. And what theoretical justification is there for this basis? What equation are they using to calculate the orbital velocity as a function of distance?
You can easily generalise Newton's equations for diffuse masses. It's not usually done in introductory physics courses because it formally requires vector calculus, but the concept isn't difficult. Essentially you divide the diffuse mass up into a lot of small masses and calculate the total force from treating each small mass as a point source. That's only an approximation, but the approximation gets better if you divide the mass up into more smaller masses. And you can consider the limiting case as you divide the mass up into infinitely many infinitely small sources (that's where calculus comes in) to make the error in the approximation go to zero.davicle said:Can anyone shed any further light on theoretical justificationfor the above "assumptions"?
In a dark matter particle paradigm there are dark matter halos that are inferred in galaxies (but not all galaxies), galaxy clusters, and also in intergalactic media.Ranku said:Do all galaxies with dark matter halo have flat rotation curve?
Does the matter outside the star's orbit have to be homogenous in density for it to sum to net zero? Or at least cylindrically symmetrical? Or does that matter?Ibix said:Once you've done that, it does indeed turn out that the only mass that matters is the mass inside the star's orbit because the gravitational effect of matter outside its orbit turns out to add up to a net zero. And the matter inside (assuming cylindrical symmetry, which is reasonable for a galaxy) turns out to have the same gravitational effect as it would if it were all concentrated at a point at the center.
Good point - it needs to be cylindrically symmetric (or close enough for jazz, in practice).DaveC426913 said:Does the matter outside the star's orbit have to be homogenous in density for it to sum to net zero?
The expected curve is based upon visible matter only using Newtonian gravity (i.e. F=GM/r^{2}) and a reasonable model of the distribution of the visible matter in the galaxy based upon observations made (usually described in a formula that is discussed in the methods section of the paper doing the calculation). There is some calculus that goes into it, but that is the basic concept.davicle said:I am probably asking a question which has been asked before, but I have only just joined the forum.
I have looked at sketch curves showing the actual and "expected" galaxy rotation curves. My question is - on what basis do astronomers expect the curve that they say they expect. And what theoretical justification is there for this basis? What equation are they using to calculate the orbital velocity as a function of distance?
Thanks very much for this. The calculus involved in dealing with a graded density circular disc must be complicated but I would like to see it and hopefully understand it. Could you give me a link to a paper where the derivation is explained step by step?Ibix said:You can easily generalise Newton's equations for diffuse masses. It's not usually done in introductory physics courses because it formally requires vector calculus, but the concept isn't difficult. Essentially you divide the diffuse mass up into a lot of small masses and calculate the total force from treating each small mass as a point source. That's only an approximation, but the approximation gets better if you divide the mass up into more smaller masses. And you can consider the limiting case as you divide the mass up into infinitely many infinitely small sources (that's where calculus comes in) to make the error in the approximation go to zero.
Once you've done that, it does indeed turn out that the only mass that matters is the mass inside the star's orbit because the gravitational effect of matter outside its orbit turns out to add up to a net zero. And the matter inside (assuming cylindrical symmetry, which is reasonable for a galaxy) turns out to have the same gravitational effect as it would if it were all concentrated at a point at the center. So the assumptions you're talking about aren't assumptions - they are actually rigorously derived results.
Google 'shell theorem'. It's a one liner if you know Gauss' Theorem, and with high school calculus I don't think it's particularly difficult, just messy.davicle said:Thanks very much for this. The calculus involved in dealing with a graded density circular disc must be complicated but I would like to see it and hopefully understand it. Could you give me a link to a paper where the derivation is explained step by step?
davicle said:Newton's equation strictly applies to two point masses M and m separated by a distance d, which in theory is accurately known. But when one of the masses M is diffuse e.g. a galaxy, you run into a problem of just what is d. The above text attempts to get round this problem by assuming (I can only put it like this as no justification or reference is given in the text) that only the mass of the galaxy inside the orbiting star's orbit gravitationally acts on the star. Surely the whole mass of the galaxy acts on the star, not just the mass inside the star's orbit.
A further assumption is made that the galaxy's gravitationally active mass is always placed at the centre of the galaxy. This has to be done in order to deduce d. The only case I know where this can be shown to be true is for a point mass outside a sphere, where the gravitational centre of the sphere acts at the centre of mass of the sphere regardless of distance. But it is not true for any other shape I have looked at - ring, rod, planar rectangle or circular disc.
Can anyone shed any further light on theoretical justificationfor the above "assumptions"?
I am inclined to support davicle here, not Ibix.Ibix said:You can easily generalise Newton's equations for diffuse masses. It's not usually done in introductory physics courses because it formally requires vector calculus, but the concept isn't difficult. Essentially you divide the diffuse mass up into a lot of small masses and calculate the total force from treating each small mass as a point source. That's only an approximation, but the approximation gets better if you divide the mass up into more smaller masses. And you can consider the limiting case as you divide the mass up into infinitely many infinitely small sources (that's where calculus comes in) to make the error in the approximation go to zero.
Once you've done that, it does indeed turn out that the only mass that matters is the mass inside the star's orbit because the gravitational effect of matter outside its orbit turns out to add up to a net zero. And the matter inside (assuming cylindrical symmetry, which is reasonable for a galaxy) turns out to have the same gravitational effect as it would if it were all concentrated at a point at the center. So the assumptions you're talking about aren't assumptions - they are actually rigorously derived results.
If you were to model DM as having the same type of mass distribution as luminous matter, sure. But collisionless matter cannot form a disc, as the particles have a hard time losing energy or exchanging momenta.davicle said:This is in direct conflict with what the effect of dark matter is supposed to be - speeding up the star's velocity from the "expected" slow velocity.
That is basically the NFW distribution. But, the NFW distribution is far from the best fit to the data for halo distribution models. The best fit DM halo mass distributions are not spherically symmetric, although they aren't as compressed as the luminous matter distribution.Bandersnatch said:If you were to model DM as having the same type of mass distribution as luminous matter, sure. But collisionless matter cannot form a disc, as the particles have a hard time losing energy or exchanging momenta.
So it has to remain in a roughly spherical distribution, with the vast majority of particles streaming through the central area on nearly degenerate orbits. That's why it's referred to as dark mater 'halo'.
I.e. to make a quick-and-dirty model of the effect of DM on the orbital velocities of your luminous matter, you should assume spherical distribution with density increasing with decreasing radius.
Oh, sure. But it's not terrible if one just wants to get an idea of what's going with their doodling. I wasn't even thinking of the NFW profile for this purpose. More like, assume ##\rho (r)## is inversely proportional to r and see what happens to the velocities. Then assume ##1/{r^2}##, see what happens. Then maybe constant density, and so on. Spherical symmetry makes this easy to do.ohwilleke said:far from the best fit to the data
Why should vast majority of particles stream through the central area, i. e. have low periapse requiring low individual angular momentum? In case of no collisions, would there be any mechanism to clear out high periapse orbits?Bandersnatch said:If you were to model DM as having the same type of mass distribution as luminous matter, sure. But collisionless matter cannot form a disc, as the particles have a hard time losing energy or exchanging momenta.
So it has to remain in a roughly spherical distribution, with the vast majority of particles streaming through the central area on nearly degenerate orbits.
FYI, you longer need to "doodle" with these sorts of nontrivial computations...davicle said:[...] my own doodlings have shown that this certainly does not happen with a disc. [...]
Hey, thanks for this question. I've been trying to recall the argument for that, and I'm increasingly certain I've just pulled it out of my lower back at some point in the past and then started believing it.snorkack said:Why should vast majority of particles stream through the central area, i. e. have low periapse requiring low individual angular momentum? In case of no collisions, would there be any mechanism to clear out high periapse orbits?
The entire mass is in the halo for the simple reason that the mass is infinite - 1/r density obviously means that the density diverges to infinity at r=0, while the mass diverges to infinity with r^{2}.Bandersnatch said:Because as you say, it shouldn't be the case, should it? And in any case, the density being inversely dependent on the radius doesn't require 'the vast majority' of DM to be going through the central region. The vast majority of the mass should actually be in the halo, given the approx. 1/r density profile, since the volume obviously grows faster with r.