1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinite universe filled with uniform dust

  1. Jan 28, 2010 #1
    If you had an infinite universe filled with a static, homogeneous, electrically neutral dust, the same in all directions, so that there are no forces to consider other than gravity, what would happen? Would it remain in that static state? I have two ways of thinking about it that lead to different answers.

    First, the obvious: if the universe is the same in every direction, then the force of attraction on any dust particle is the same in all directions. So there is no net force on any particle, and the universe remains static.

    Second: Pick any arbitrary point in the universe (call it O). Then every dust particle is attracted to point O. The dust particle (call it P) lies on a sphere S centered at O. The dust particles inside the sphere attract P towards O as if their total mass were concentrated at O. On the other hand, the dust particles in any spherical shell outside the sphere S do not have any net attractive force on P. So there is no net force on P coming from particles outside S.

    Which one of these arguments is flawed? Or maybe both are? My apologies if this is an old puzzle that has been worked out many times. I just thought of it a few minutes ago.
     
  2. jcsd
  3. Jan 28, 2010 #2
    Gravity causes tidal forces, so that's a complication. Everything would experience infinite crushing tidal forces maybe, or none... Hard to imagine.
     
  4. Jan 28, 2010 #3
    Suppose that first paragraph is correct.

    Than the first must be correct as well.

    Then Second is flawed. For there is no way that the particles outside the sphere O would have NET attractive force of zero on P, since this is against the laws of first paragraph. They ARE interacting with the particles inside the sphere as well as with the particle P, creating a zero net force on ANY particle.

    I have a problem with the first paragraph though. :-)

    Universe is RANDOM. ( GOD plays DICE ! :-) )
    So even IF at the beginning of time universe was homogeneous... an infinitesimally short instant after that, random event would occur and homogeneous universe is gone... WIth the (gravitational) wind...
     
  5. Jan 28, 2010 #4
    Certainly the first argument and the second argument are mutually contradictory. But they both seem plausible to me. Maybe you are talking about this sentence of mine:

    I worded this sentence poorly. Each individual dust particle does indeed exert an attractive force on particle P, but the sum of the forces on P from all the particles in a spherical shell that is outside the sphere S is zero (isn't it?).

    Yes, of course. I'm talking about a hypothetical universe that is filled with (initially) stationary dust particles, not our actual universe. I am aware that the particles in our actual universe are in perpetual motion.
     
  6. Jan 28, 2010 #5
    Is this question in the context of Newtonian gravity?

    If so, you are quite correct; Newtonian gravity is inconsistant with an infinite universe filled with equally distributed matter for the very reason you've given.
     
  7. Jan 28, 2010 #6
    It seems to me that, even if the initial conditions are perfectly uniform, they would not remain so indefinitely. At some point, minute quantum effects would cause instability, resulting in this Universe to start "clumping". Once that began, there would no longer be a net force of zero on particle, which would result in a ripple of gravitational effects coming from the original disturbance. At least, this would happen in a Universe in which quantum effects exist.

    So, an infinite Universe seems to be inconsistent with idealized Newtonian gravitation - an imaginary, non-existent Universe. However, I don't see that this leaves out the possibility of an infinite Universe that includes quantum effects. At least, not on these grounds.
     
  8. Jan 29, 2010 #7

    Yes, I was thinking of Newtonian gravity. I was trying to understand why any small deviation from uniform density would cause matter to clump together, when I realized that Newtons theory predicts that, even in with a uniform distribution of matter, matter will clump around any point you choose. And that seemed weird.

    Yes, a uniform distribution could not be realized in the actual universe because of the instability you mention. It would be too improbable. I was thinking about matter "clumping" together when I stumbled on this puzzle.

    Still, I wonder if the first people to recognize this inconsisitency thought of it as a good argument against an infinite universe. If I were living in a pre-relativity, pre-QM world, I think I would throw out the infinite universe hypothesis before I would throw out Newtonian gravity, which works pretty well.
     
  9. Jan 29, 2010 #8
    In 'The Measure of the Universe', John North says that Newton toyed with the idea of some sort of cosmological constant, trying to force a universe behaving according to argument #2 to be static. Newton eventually settled for argument #1, but he must have seen the problem. So did others - by 1760 Boscovich was also trying to introduce some sort of cosmological constant.
    No, the problem is still there in a finite universe.
     
  10. Jan 29, 2010 #9

    Dale

    Staff: Mentor

    "Uniform" is not the same as "spherically symmetric". I don't think it is valid to replace a problem with uniform symmetry with one with spherical symmetry. I suspect that one of the mathematical simplifications that are made in deriving the spherical shell result is not valid as the radius tends to infinity.
     
  11. Jan 29, 2010 #10
    That is neat background info. Thanks you. I am not surprised that Newton, brillant as he was, was the first to notice things like this.

    Are you talking about the problem of a non-static universe? The matter in a finite Newtonian universe still falls in on itself, but at least there is no logical contradiction. It is just contrary to observation, right?

    I was wondering about this myself. Spherically symmetric does not imply uniform, but uniform does imply spherically symmetric from any arbitrary point, right? So I don't think my assumption of spherical symmetry about an arbitrary point in the second argument contradicts my assumption of uniformity in the first.

    I hadn't thought about the spherical shell result breaking down as the radius tends to infinity. I don't see why it would, but I also am not sure that it wouldn't.
     
  12. Jan 29, 2010 #11

    Dale

    Staff: Mentor

    There are a lot of simplifications involved in getting the integral to converge. One or more of them must be invalid.

    Regarding uniform vs. spherical symmetry. I don't think you can say that uniform implies spherically symmetric. In your example, we know immediately by the uniform symmetry that the force is zero. If you take any spherical section we know the force due to that mass and therefore the force due to the rest of the universe must be equal and opposite. This is not consistent with the spherical symmetry idea. The problem lies in the fact that for any R your spherically symmetric universe always has more mass on one side of the test point than on the other side. This is not consistent with uniform symmetry.
     
  13. Jan 29, 2010 #12

    Ah, okay, I think I see what you mean. Saying the universe is spherically symmetric about point O does seem to mean that it is not also uniformly symmetric about dust particle P, because in every spherical shell about point O, there is more mass to one side of P than to the other.

    But it doesn't pose a problem with the forces balancing, because even though there is more mass to one side of P (whatever it means to say there is more mass to one side in an infinite universe), it is also farther away from P and therefore exerts less force.

    EDIT: of course, I'm forgetting the mass inside the sphere S that particle P lies on. That doesn't get balanced out, and it does seem to break the uniformity about P.
     
  14. Jan 29, 2010 #13
    Awfully good argument, Dale. But is there still a contradiction... It's something worth pondering.

    Would forces still be equal and opposite?
     
  15. Jan 30, 2010 #14
    I've been thinking about it more, and I'm still inclined to think that, in an infinite universe, uniformity implies every other kind of symmetry. It was inaccurate of me to talk about the universe being uniform about the particle P. There is no need to refer to any point in space when talking about uniformity. The most complete symmetry about a point is spherical.

    For instance, wouldn't a uniform infinite universe also be translation invariant? If you move dust particle P over to the right one meter, would the net force on it change, now that there is infinitely more matter to the left of it than there used to be? I would say no. Please correct me if I'm missing the point. I get confused thinking about infinities.

    Aren't the assumptions that the universe is roughly translation invariant and spherically symmetric about any point the justifications for conservation of linear momentum and angular momentum?
     
  16. Jan 30, 2010 #15
    I think the problem is with both points, for the following reason:

    The 3D integral that calculates the amount of force on particle P from other dust particles outside S is not well-defined. In a well defined integral, you can always split the domain in two parts (arbitrarily), and the total value should be the sum of each contribution. But if you do this in this case, e.g. by dividing the universe in two pieces, you get two infinite results. Think about e.g. splitting the universe in two pieces, with P on the edge.

    These contributions are supposed to cancel each other, but that is not mathematically well defined. So the integral of the forces acting on P is not well-defined, at least in the usual mathematical definition of an integral.

    Both the points suffer from this problem. At least we are dealing with a problem that is not describable using ordinary mathematics, I think :-)

    Actually, I can't remember right now, but I seem to remember that there are some other "paradoxes" that appear like this.

    EDIT: One additional thing: in order to regulate the contributions from the integrals, the measure needs to dampen the contributions at large radius. The effect will be that the two viewpoints will be compatible, I think.

    Torquil
     
  17. Jan 31, 2010 #16
    Its going to be an unstable equilibrium so any deviations from uniformity will cause your dust filled toy universe to collapse. If the dust has a temperature other than absolute zero, it will cause a chaotic collapse and would probably be a decent model for our actual universe. Regions would congeal into spheres that would rotate about each other and probably form complex structures like galaxies.
     
  18. Jan 31, 2010 #17
    Okay, I guess I see that. But the first argument doesn't seem to require any careful reasoning about the cancellation of improper integrals. It's just a symmetry argument. It seems to me that symmetry such a fundamental principle that you can appeal to it without having to worry about mathematical analysis. Maybe I'm wrong.

    DaleSpam was also concerned about the second argument breaking down because of assumptions made in obtaining the result about spherical shells not exerting force on a test particle inside the shell. I worked out the integral for the case of arbitrary radius, but assumed that the units can be chosen so that the radius=1 and the density of the shell = 1. If the shell is centered at the origin, and the test particle is located at (a, 0, 0) where a < 1, then the force on the particle is along the x-direction (by symmetry).

    [tex]F = \int_0^\pi \frac {2\pi \sin \theta}{(1+a^2-2a\cos\theta)} \frac{(\cos \theta - a)}{\sqrt{1+a^2-2a\cos\theta}}d\theta[/tex]

    [tex]F = \pi(I_1 - I_2)[/tex]

    where

    [tex]I_1 = \int_0^\pi \frac {2 \sin\theta\cos\theta}{(1+a^2-2a\cos\theta)^{3/2}}d\theta[/tex]

    [tex]I_2 = \int_0^\pi \frac {2a \sin \theta}{(1+a^2-2a\cos\theta)^{3/2}}d\theta[/tex]

    After making the substitution [tex]x = 1+ a^2 - 2a\cos\theta[/tex], it turns out that both integrals are equal:

    [tex]I_1 = I_2 = \frac{4a}{1-a^2}[/tex]

    and so they cancel exactly. Why can't this be extended to infinite radius? Is it because I assumed that units could be chosen so that R = 1?
     
    Last edited: Jan 31, 2010
  19. Feb 1, 2010 #18
    Consider the opposite viewpoint: The gravity acting at one particular point P depends on the dust at all other points of the universe. But since the mathematical expression for this cannot be represented as a 3d integral over space with a 1/r gravitational law, you cannot express this as individual contributions coming from each particular point. I.e. you cannot define a "density" of gravity in this case.

    So I would say that the 1/r gravity law is inconsistent (ie. impossible) in an infinite dust filled universe, as long as physics is based on ordinary calculus. Physics depends on mathematics to be well-defined, so I don't think that this is a mere technicality.

    Btw, it reminds me of the sum 1-1+1-1+1-1+1-1....

    Its value is not well-defined because it doesn't converge, but if you use a symmetry cancellation argument:

    (1-1) + (1-1) + (1-1) + ...

    you can be fooled into believing that the sum is zero since each term is zero (similar to the statement "each shell contributes 0 gravity"). This is a more extreme example of course, but I think it illustrates the same problem.

    Torquil
     
  20. Feb 1, 2010 #19
    Thanks, Torquil. I think I finally see what you and DaleSpam are trying to get at. When integrating over an infinite space, you can't necessarily divide it up into chunks however you like. You may come up with different answers depending on how it is divided, which is the same thing as saying the integral is not well-defined. Your example of the non-convergent series made that clear to me. Now I remember learning an even more startling result in calculus: a conditionally convergent series that is not absolutely convergent can be rearranged to add up to anything you want. That one blew my mind.

    If I lived in the 19th century, I think these considerations would make me reject the infinite universe hypothesis. The explanatory power of Newton's theory of gravity is just too great. Even though the universe is evidently not filled with uniform matter, it seems odd to think that simply arranging things so that it was would lead to a contradiction.

    ........
    Yep, this is exactly the kind of stuff I was thinking about when I thought of the subject of this post. I figured that I wouldn't get very far if I couldn't even resolve what would happen in a universe at absolute zero with no deviations.
     
    Last edited: Feb 1, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook