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Acceleration a Function of Density, yet Density is Constant?

  1. Mar 1, 2015 #1
    I have used the search feature and tried very hard to find a thread where someone has asked this question and received an answer that is at the [low] level of understanding that I have.

    I'm sorry if this question seems ignorant -- which it may be, because I am not a physics student -- but I am very tripped up by the above statement. It seems me like it makes a contradiction! If acceleration is a function of dark energy density (let's call this a(x)), and the acceleration of the universe has changed over time, then how can dark energy density be constant? That is, if the acceleration function a(x) is changing as a function of density, and acceleration is in fact changing, is it not necessary that density is changing as well? Am I missing something?

    Again, my apologies if this question has been asked before, and that my searching skills are just poor :wink:
     
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  3. Mar 1, 2015 #2

    PeterDonis

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    First, a comment: if you are quoting something (which it looks like you are), please give a reference for what you are quoting from. It helps to see the context.

    The acceleration has changed over time because the universe has more in it than just dark energy. If the universe had nothing in it but dark energy, the acceleration would be constant.

    Up until a few billion years ago, the density of ordinary matter in the universe was large enough to overcome the effect of dark energy, so the universe's expansion was decelerating. But unlike the dark energy density, the density of ordinary matter decreases as the universe expands, so its effect on the dynamics decreases too. A few billion years ago, the density of ordinary matter became small enough that its effect no longer overcame the effect of dark energy; that's when the expansion started accelerating.

    At some time in the far future, when the density of ordinary matter has decreased to the point where it is negligible compared to the dark energy density, the acceleration will approach a constant value.
     
  4. Mar 1, 2015 #3
    This isn't like me. I haven't been able to sleep lately, and I made this post at 4 in the morning. I borrowed the quote from the Wikipedia page on dark energy: https://en.m.wikipedia.org/wiki/Dark_energy. It can be found as the last paragraph of the subsection entitled "Effect of dark energy: a small constant negative pressure of vacuum."

    Your post is very insightful, but it again arouses my original question. I can tell that you made it a point to bring to my attention that dark energy plays its role on the universe's expansion by "repelling" ordinary matter. I understand that at the early stages deceleration occurred, and that this is due to the high density of ordinary matter w.r.t. the density of dark energy. The "repulsive" strength of dark energy was small in comparison to the attractive strength of ordinary matter, and that caused deceleration. I also understand that at its current stage acceleration is occurring because ordinary matter has become rare enough that the "repulsive" affect of dark energy is dominating.

    But it still seems to me that you are explaining that the acceleration is a function of ordinary matter density -- not dark energy density -- and at the very most, relative dark energy density (relative to ordinary matter density, that is). I still do not understand, from a mathematical standpoint, how a changing value of acceleration is dependent on the constant value of dark energy density.

    Thanks,
    HeavyMetal
     
  5. Mar 1, 2015 #4

    PeterDonis

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    Where did I say that? That's not really a very fruitful way of viewing what dark energy does. But it's not really germane to the present topic.

    Um, if it's a function of dark energy density relative to ordinary matter density, then it is a function of dark energy density, not just ordinary matter density. I think you are making this more complicated than it needs to be. See below.

    Acceleration is a function of both dark energy density and ordinary matter density. Dark energy density is constant. Ordinary matter density is changing. Therefore acceleration is changing, since one of the things it's a function of (ordinary matter density) is changing.

    Mathematically, the equation to look at is the second Friedmann equation, as given, for example, here:

    http://en.wikipedia.org/wiki/Friedmann_equations

    The acceleration is ##\ddot{a} / a##. It depends on ##\Lambda## (dark energy density), which is constant. But it also depends on ##\rho## (ordinary matter density), which isn't.

    (It also depends on the pressure of ordinary matter, ##p##; but the pressure of ordinary matter is negligible compared to its energy density, and has been for almost all of the universe's history. The pressure of radiation is not negligible; it's 1/3 of the energy density of radiation. So in the early universe, when the energy density of radiation was significant compared to the others, the ##\rho## term had a significant contribution from the energy density of radiation, and the ##p## term was significant.)
     
  6. Mar 1, 2015 #5
    Great! That's all I needed to know. I am aware that if you multiply a function by a constant, then that function is a function of both the value AND that constant, i.e. [tex]k*f(x)=f(k*x)[/tex]I was just confused by the wording of the wikipedia article that made it seem like the dependency was ONLY on dark energy density. Again, I am not a physics major, and I don't have any good textbooks on this subject. I must resort to you guys to understand better what I can find freely online.

    This is exactly what I was asking about. You da man! Thanks :)
     
  7. Mar 1, 2015 #6

    PeterDonis

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    Well, it is if you consider the constant to be a variable. :wink:

    However, this doesn't apply here because the equation for the acceleration (the Friedmann equation) does not have dark energy multiplying anything. The RHS of the equation adds the dark energy density term to the ordinary matter density (and pressure) term.

    I went back and looked at the article, and I agree the wording is confusing. For one thing, now that I read it in context, it is using the word "acceleration" in a different way than I have been using it.

    The way I have been using the term "acceleration", it simply refers to the term ##\ddot{a} / a## in the second Friedmann equation. This term is the second time derivative of the "scale factor" ##a##, which can be roughly thought of as the "size" of the universe. The first time derivative of ##a## is then the "speed" at which the size of the universe is changing, and the second time derivative is the "acceleration" of that change, i.e., the rate at which the "speed" is changing.

    The way the Wikipedia article was using the term "acceleration", it refers to the specific case in which ##\ddot{a} / a## is positive--i.e., the "speed" at which the universe is expanding is increasing. (If the speed were decreasing, using this terminology, you would call it "deceleration".) If you look at the second Friedmann equation, you will see that the only positive term on the RHS is the dark energy term; the ordinary matter term is negative. So if ##\ddot{a} / a## is positive, there must be dark energy present; it's impossible for ##\ddot{a} / a## to be positive if only ordinary matter (or radiation) is present.

    So using the article's terminology, the only way "acceleration" can happen is if dark energy is present; if you only have ordinary matter present, you can only get "deceleration", not "acceleration". That's what the confusing wording was really getting at. However, in either case, it's still true that ##\ddot{a} / a## depends on both the ordinary matter density and the dark energy density; the dependence on ordinary matter density doesn't go away just because ##\ddot{a} / a## happens to be positive. The article does not do a good job of making that clear.
     
  8. Mar 2, 2015 #7
    So the negative and the positive signs are what represent the respective attractive and repulsive forces, correct? I've been told that it's technically incorrect to call this a repulsive force, but I'm just trying to get things straight in my head. Ordinary matter attracts, dark energy "repells."

    So what is currently thought to be causing the repulsion? I know that this is one of the unsolved problems in physics; I'm just wondering, is there a consensus in the community on a generally accepted hypothetical force carrier that mediates this repulsion?
     
  9. Mar 2, 2015 #8

    PeterDonis

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    I wouldn't call them "forces", because all of the objects in this model (the galaxies, etc. in the universe) are in free fall, feeling no force; their relative motion is solely due to spacetime curvature. The negative sign can be thought of as "attractive" gravity and the positive sign can be thought of as "repulsive" gravity, but only in the GR sense of "gravity", where gravity is not a force but a manifestation of spacetime curvature.

    Dark energy. A better question is, why is the dark energy density what it is? That is one of the major unsolved problems in cosmology.

    There isn't a "force carrier", because there isn't a "force". See above.
     
  10. Mar 3, 2015 #9
    I am well aware of this, but I'm still glad that you made sure to nail this one into my head. It is an important lesson to learn early on. I first made this mistake while wondering how photons can change their path in the proximity of a massive object; I thought that the reason was attraction, and I wanted to know how these two interacted. Obviously, I was wrong, and so I was introduced to the concept of GR as a geometric representation of spacetime.

    Anyways, if we were to adopt the model of the universe that corresponds to a homogeneous, isotropic local geometry, could we say that ordinary matter causes positive (spherical) curvature and that dark energy causes negative (hyperbolic) curvature? Or do I have this backwards...?


    I'll just reiterate that I was cognizant of this when I asked my question. However, I've never quite been able to understand the necessity of the graviton; on the other hand, it is well accepted in the community and even predicted by the standard model, and so this is why I thought there might be a "force carrier" for dark energy too.

    About the graviton: does belief in its existence just simply stem from QFT in the sense that we need a particle to represent the smallest excitation of the gravitational field? Also, is the graviton considered a "force carrier" at all?

    I'll just take this time to state that learning physics as a non-physics major is rather challenging -- I am very motivated to learn more, but when people are rude to me about my ignorance I become discouraged. You've been relaxed and I appreciate your patience!
     
    Last edited: Mar 3, 2015
  11. Mar 3, 2015 #10

    PeterDonis

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    Not really. "Attractive gravity" does not correspond to "positive (spherical) curvature", and "repulsive gravity" does not correspond to "negative (hyperbolic) curvature". It's more complicated than that.

    First, you need to distinguish spacetime curvature from spatial curvature. Spacetime curvature is what is related to the kind of "stuff" that is present (ordinary matter or dark energy). But spacetime curvature is a fourth-rank tensor, the Riemann tensor; and even the part of it that is directly linked to matter and energy via the Einstein Field Equation (this part is called the Einstein tensor) is a second-rank tensor with ten independent components in the general case. So you can't really describe spacetime curvature by just one number or one adjective ("positive" vs. "negative").

    Spatial curvature is what is usually talked about in books or articles about cosmology, and is what is usually described as positive (spherical), negative (hyperbolic), or zero ("flat"). But spatial curvature depends on your choice of coordinates. The coordinates in which the spatial curvature of the universe is usually described are natural coordinates to use in cosmology (basically they're coordinates in which observers at rest see the universe as homogeneous and isotropic), but they're still just one particular system of coordinates. And the spatial curvature in these coordinates is not directly linked to the kind of "stuff" that is present (ordinary matter vs. dark energy). You can have positive spatial curvature in a universe that only has dark energy present, and you can have negative spatial curvature in a universe that only has ordinary matter present. Or you can have zero spatial curvature in a universe with both present--that's what our current best-fit model says our universe is like.

    Yes--at least, there seems to be a pretty general consensus among physicists that it exists. But there are still a significant minority of physicists who aren't so sure. See below.

    No, it isn't. The Standard Model does not include gravity at all, and it makes no predictions about gravity or the graviton. Any model of the graviton is outside the Standard Model.

    That's one key reason, yes. Basically, the idea is that you can't mix quantum and non-quantum things in the same theory; if everything else is a quantum field, gravity has to be a quantum field too. However, a number of physicists, including Freeman Dyson (who helped to develop quantum electrodynamics), have questioned whether this is really true. The issue is not closed.

    In theories that include it in an "expanded" sort of Standard Model, yes; it's considered just like the other "force carrier" particles (the photon, W and Z bosons, and gluon).

    Thanks! I'm glad our discussion is helping.
     
  12. Mar 4, 2015 #11
    I was actually talking about spacetime curvature, which as I learned from your above comment, does not use the naming conventions that I was using. I made the mistake of mixing up these two different concepts.

    This part is over my head, but I'd love to learn more. I think I might have a decent mathematical foundation to begin learning about [the basics of] this stuff; I've taken calc 1 and 2, vector calc, differential equations (mostly ordinary, some partial), and linear algebra. I need to start reading a solid introductory book that allows me to teach myself, and has practice problems. Can you suggest a GR book fitting this description?

    Regarding the part I italicized: I think I read the other day -- I don't remember where, and it's Murphy's Law that I cannot find it now -- that this is a safe assumption at scales ≥100 Mpc. Can you confirm this?

    Well, this is awkward...I seem to have been thinking of the Higgs boson. I guess today is the day to be mixing up my concepts o0)
     
  13. Mar 4, 2015 #12

    PeterDonis

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  14. Mar 4, 2015 #13

    PeterDonis

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    It all looks italicized. :eek: Unfortunately that's the way the quote feature works. Try quoting just the part you're asking about.
     
  15. Mar 4, 2015 #14
    It was the above quote. Sorry, just noticed that!

    Thank you so much for the book list and the suggestion on the lecture notes! Also, do you think that GR will require much more math than I currently have for at least the basics?
     
  16. Mar 4, 2015 #15

    PeterDonis

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    Ah, ok. I'm still not sure what you're asking, though; you said:

    What assumption are you talking about? The zero spatial curvature in the best-fit model isn't an assumption; it's a conclusion from the data.
     
  17. Mar 4, 2015 #16
    Okay, I see what you are saying now. What I was thinking about was unrelated. I actually just found where I read it, which was the Wikipedia page for the Friedmann equations. It stated:

    Thank you for all of your help! All of my questions were answered. I think I'm done with this thread for now :)
     
  18. Mar 4, 2015 #17

    Chronos

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    Yeah, all this stuff requires a stiff dose of math to wrap your head around it. That makes it suspect to all of us math challenged monkeys. I struggled with differential equations in college, so I sympathize.
     
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