Dark energy creates a constant acceleration between any two objects in the universe regardless of their masses.
Nice. I submit, however, that kg & km may not be the most convenient units of mass & distance for this problem. :-) Perhaps solar masses & LY, or Milky Way masses & Mpc?So, it depends upon the mass we're accelerating away from. Dark energy creates a constant acceleration between any two objects in the universe regardless of their masses. Plugging in some numbers, the distance is given by:
$$r = (260,000 km) \left({M \over 1kg}\right)^{1 \over 3}$$
Haha, sure :)Nice. I submit, however, that kg & km may not be the most convenient units of mass & distance for this problem. :-) Perhaps solar masses & LY, or Milky Way masses & Mpc?
Haha, sure :)
If I'm doing my unit conversions correctly:
$$r = (350ly) \left({M \over M_\odot}\right)^{1 \over 3}$$
For practical purposes, it's the lighter one that is, or isn't, bound. So the relevant mass is 10 solar. The cube root of 10 is 2 (close enough!), so that means 700 LY.Thank you for that. But as my math is pretty crap coukd I ask you do an example so I can see how it works. Say how far would an object of 2 solar masses have to be from an object of say 10 solar masses before they were no longer gravitationally bound.
Or just about (1 Mpc) * (mass/Milky Way's mass)1/3 -- convenient! [Yes, if you did the conversions correctly, which I didn't check -- but doesn't this seem a little too small?]Haha, sure :)
If I'm doing my unit conversions correctly:
$$r = (350ly) \left({M \over M_\odot}\right)^{1 \over 3}$$
For practical purposes, it's the lighter one that is, or isn't, bound. So the relevant mass is 10 solar. The cube root of 10 is 2 (close enough!), so that means 700 LY.
Wolfram Alpha is your friend. The region bound to the local galactic group would be of a size of order 1/100000 of the diameter of the observable Universe. That works out to a volume that is about ##1/10^{15}## of the volume of the observable Universe. ##10^{15} \gg 10\cdot 10^9##.
This calculation isn't enough to state that something is gravitationally-bound. What the calculation suggests is that nothing which is much lighter than the Milky Way can possibly be gravitationally-bound to it if it is further than 3.1 million light years away. This number is, on its face, not absurd, given that all of the Milky Way's satellite dwarf galaxies are substantially closer than this:Ok great thanks.
So taking that the Milky Way is about 700 billion solar masses, then that works out that anything within 3.1 million light years would be gravitational bound to it.
This calculation isn't enough to state that something is gravitationally-bound. What the calculation suggests is that nothing which is much lighter than the Milky Way can possibly be gravitationally-bound to it if it is further than 3.1 million light years away. This number is, on its face, not absurd, given that all of the Milky Way's satellite dwarf galaxies are substantially closer than this:
https://en.wikipedia.org/wiki/Satellite_galaxies_of_the_Milky_Way
Things that are closer can still be unbound if they are moving quickly enough. Things that are further might become bound in the future if they're on a trajectory towards the Milky Way and will slow down due to interactions with the galaxy (e.g. when the Andromeda galaxy collides with us in a few billion years).
Oh, yeah. Dark energy is only a late-time effect. The early expansion was vastly faster than it is today. Dark energy puts a floor on that rate of expansion, preventing it from decreasing further. In the far future, the expectation is that the rate of a expansion will be a constant.Yes of course. My thinking was along the lines.... if we say for the sake of simplicity that the universe is 90 billion ly across and there are around 90 billion galaxies. And assuming the cosmological principle then that would give a grid of galaxies around 300,000 ly apart in any direction. So if they were all of similar mass to the Milky Way then they would, without any prior momentum, all be gravitationally bound.
Of course the universe is a bit more lumpy than that but the outcome should still be the same. So there must have been something that gave early mass momentum away from each other.
I always thought of inflation as just the grid of co-moving objects getting bigger. But obviously it must be more than that.
Yes of course. My thinking was along the lines.... if we say for the sake of simplicity that the universe is 90 billion ly across and there are around 90 billion galaxies. And assuming the cosmological principle then that would give a grid of galaxies around 300,000 ly apart in any direction. So if they were all of similar mass to the Milky Way then they would, without any prior momentum, all be gravitationally bound.
Of course the universe is a bit more lumpy than that but the outcome should still be the same. So there must have been something that gave early mass momentum away from each other.
Accelerated expansion is a late time effect. What you are computing is the distance over which the cosmological constant is relevant. It was irrelevant in the early Universe. Also, you should probably not use the word "inflation" for the accelerated expansion due to our current cosmological constant. The connotation of the word is to the hypothetical period of inflation in the early Universe before the onset of the standard Big Bang.I always thought of inflation as just the grid of co-moving objects getting bigger. But obviously it must be more than that.
I think it should be stressed that this is only true in a universe that grows to be large enough for the cosmological constant to become relevant (i.e., where it becomes a significant portion of the contribution to the energy density). It is perfectly possible to imagine a universe where the initial matter density is large enough for the universe to collapse and contract before the cosmological constant has had any relevance whatsoever.Dark energy puts a floor on that rate of expansion, preventing it from decreasing further. In the far future, the expectation is that the rate of a expansion will be a constant.
Yes, almost exactly one Mpc for one "galactic mass". :-)What the calculation suggests is that nothing which is much lighter than the Milky Way can possibly be gravitationally-bound to it if it is further than 3.1 million light years away. This number is, on its face, not absurd, given that all of the Milky Way's satellite dwarf galaxies are substantially closer than this:
https://en.wikipedia.org/wiki/Satellite_galaxies_of_the_Milky_Way
I have seen the conclusion that the Norma Cluster at the core of Laniakea is beyond the critical distance, but that Virgo is not.
The implication is that the Local Group is just barely bound against DE
I'll take your word for it that it's ~ 5x. (I'd have guessed 3x.) But if @kimbyd's equation is right, then r ~ 1 Mpc x ("galactic masses")1/3, right? So that's a distance r ~ 51/3 Mpc, somewhat < 2 Mpc.No, it isn't. The relevant mass ##M## in the formula is not the mass of one galaxy, but the mass of the entire Local Group, which includes two Milky-Way sized galaxies plus many smaller ones, totalling 54 galaxies in all. The total mass of the group seems to be about 5 times the mass of the Milky Way, which gives an ##r## of about 15 Mpc.
"Specific": can't. Almost certainly either Tyson's Cosmos or History Channel's Universe, in one of their episodes on the long(er)-term fate of the universe. A practicing astrophysicist, yes, but not a scholarly presentation.Please give a specific reference.
I'll take your word for it that it's ~ 5x.
if @kimbyd's equation is right, then r ~ 1 Mpc x ("galactic masses")1/3, right? So that's a distance r ~ 51/3 Mpc
I'd be very curious to see if I've made an error in the above calculations.Yes, almost exactly one Mpc for one "galactic mass". :-)
But this distance does seem a bit too small to my eyes. The implication is that the Local Group is just barely bound against DE, a conclusion that I would have expected to read long before now if true, and have not. I have seen the conclusion that the Norma Cluster at the core of Laniakea is beyond the critical distance, but that Virgo is not. (More mass in both cases, of course, but tens of thousands of times more for Virgo?)
Yes, not only a confound, but a potential observation bias, as the population of small but numerous dwarfs, low- vs. high-gas spirals, and stellar streams --not to mention DM -- become increasingly hard to weigh.I'd be very curious to see if I've made an error in the above calculations.
One confounding factor is that as you go further away, the total mass of the system grows because more mass is included.
That's the best estimate I can come up with from what I can find online; the most detailed paper I can find is here:
https://arxiv.org/abs/1312.2587
This seems to be giving a total mass of stars of about 2.4 trillion Suns, which is roughly 5 times the Milky Way. However, it also gives a total mass of dark matter (which should probably be included) of about 4.2 trillion Suns; the sum of the two, which would be the total mass overall, is 6.6 trillion Suns, or about 13 times the Milky Way.
Ah, yes, you're right, r goes like the cube root of the mass.
I cannot resolve the paradox between M31 ~ 1 T Suns (DM+LM), MW ~ 1 T, and your numbers ~ 6 T for the whole LG.
My understanding is that the DM/LM ratio isn't actually relevant to the calculations. The way the mass is measured comes from measuring the velocities of objects as far as possible from the center of the galaxy. If we assume the objects are gravitationally-bound, then they will have a velocity dispersion which depends upon the mass of the galaxy closer than the object in question.Ah. The paradox (to my mind) that MW plus M31 plus, say, twice M33 only accounts for ~ 3 T Suns. I use an extra M33 as a proxy for all the dwarfs, even though the other 51 would have to be about as large as the LMC on average, whereas not even M32 seems to be that big.
Of course, this factor of 2 in total mass may just show how variable the DM/LM ratio is, both from galaxy to galaxy and from galaxies to inter-galaxy space in the LG.
The problem there (in terms of my question about the LG's mass) is that I am not aware of any objects that can be said to orbit the LG as a whole: They are either too close to one of MW or M31, or else they are too far to be measured, or too far to be dominated by the LG's gravity instead of some other small cluster's gravity.My understanding is that the DM/LM ratio isn't actually relevant to the calculations. The way the mass is measured comes from measuring the velocities of objects as far as possible from the center of the galaxy. If we assume the objects are gravitationally-bound, then they will have a velocity dispersion which depends upon the mass of the galaxy closer than the object in question.
The accuracy of this mass estimate, then, depends upon measuring the velocities of as many objects as possible. As you go further, you include more mass, but the number of objects drops. There's also the problem that we can only accurately measure the velocities along the line-of-sight, which means that geometry has to be taken into account.
You can get smaller error bars by assuming a mass profile (often the NFW profile is used), and including nearer objects into the fit. But then you add a systematic error that stems from how well that assumed mass profile matches the true mass profile.
Haha, sure :)
If I'm doing my unit conversions correctly:
$$r = (350ly) \left({M \over M_\odot}\right)^{1 \over 3}$$
Unfortunately that's a bit of a tricky question, and I'd have to do some work to figure out the answer.Just a quick question, if I am comparing the distance of two galaxies equal in mass do I include both masses in the formula?
This seems like the correct answer, though I would defer to some of our GR aficionados for the definitive judgment. (Notice that the reduced mass has half the mass of the pair, rather than their sum.)Unfortunately that's a bit of a tricky question, and I'd have to do some work to figure out the answer.
When working with classical Newtonian gravity and no cosmological constant, the two-body problem is solved using the concept of "reduced mass":
$$\mu = {m_1 m_2 \over m_1 + m_2}$$
Using the reduced mass allows you to work with the system as if there was only one mass. I'm sure you can do the same sort of thing in the case where the cosmological constant is relevant, but some careful work would have to be done to make sure that the math was all done properly.
Using the reduced mass allows you to work with the system as if there was only one mass.
Quite right. The question, I suppose, is whether the hypothetical second object is external. E.g., after multiplying the masses by 10 or so, we could ask about the mutual attraction and binding of the LG and a nearby group of similar mass. It wouldn't make sense to ask about the MW and a similar-sized galaxy that isn't in the LG, since an external one would be attracted to the LG as a whole in that case, not only to us.More precisely, it allows you to solve for the internal motions of the system (i.e., the motion of each mass relative to the other) as if there was only one mass. But I don't think this method will give the correct gravitational force exerted by the system on an external object. And the latter is what is relevant for this discussion.
However, I suppose you could the same question of two separated galaxies near the edge of a void, where both would be effectively isolated, not members of groups.
