Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

In what way is "dark energy" more than "none of the above"?

  1. Jan 30, 2015 #1
    Hello,

    A bit of a long post - still trying to get clear for myself what I really want to ask. You can perhaps skip the background info below and go straight to 'My question is'.

    Background

    On p300 of my Penguin edition of "The Fabric of the Cosmos", Brian Greene writes:

    The recession speed of a supernova depends on the difference between the gravitational pull of ordinary matter and the gravitational push of the "dark energy" supplied by the cosmological constant. Taking the amount of matter, both visible and dark, to be about 30 percent of the critical density, the supernova researchers concluded that the accelerated expansion they had observed required an outward push of a cosmological constant whose dark energy contributes about 70 percent of the critical density.​

    He then stresses how, satisfyingly, results from two different empirical investigations - measuring/estimating the density of the universe (not discussed in detail how it was done, cf p291), and measuring/estimating the expansion rate of the universe (via supernova study) - mutually confirm the conclusion that

    [...] the outward push demonstrated by the supernova data can be explained by just the right amount of dark energy to account for the unseen 70 percent of the universe['s density] that inflationary cosmologists had been scratching their heads over. The supernova experiments and inflationary cosmology are wonderfully complementary. They confirm each other. Each provides a corroborating second opinion for the other. [emphasis added]
    I don't get how one gets from the information about expansion rate to the numbers about the proportions of stuff that make up the universe (or perhaps vice versa).

    What do we have (forgive my no doubt very clumsy equations - just trying to get a rough idea across):
    1. There is a universe expansion rate e (empirically found via supernovae). I may be dead wrong, but I'm guessing that, simplifying, something like this holds [eq1]: e = eunknown - edm - evm, where we know, or at least guess, the magnitudes of the impacts of visual matter/energy and dark matter/energy (negative terms since their gravitational impact is the 'regular' contractive one). There used to be an unknown eunknown, the value of which has sprung from the equation after establishing e through the supernova measurements. Let's call whatever is causing the eunknown term "dark energy", and we must assume it has negative pressure (for it causes repulsive gravity); but otherwise we have no clue what it is (is that so?).
    2. There is a universe density d (empirically found via means not quite detailed on p291), for which something like this holds [eq2]: d = dunknown + ddm + dvm. Expressed in proportions, d corresponds to 100%, and our empirical guesses for ddm and dvm are 25% and 5% respectively. So dunknown must be 70%.
    My question is: if we have no clue what the "dark energy" (from eq1, expansion rate related) is (i.e. what its properties are apart from what it does to e), how can we say anything about its density from equation 1? How is this a case of accounting for the missing 70% in eq2, rather than just attributing dunknown to this something ('dark energy') which we decided to use as a term to mean that what accounts for eunknown?

    Here is an analogy which is imperfect on purpose - see next paragraph. Suppose that we find we have 100 apples in a basket (corresponding to e). We know Don took out 10, and Val took out 10 too. We conclude that something or other, "dark energy" must have put in 120 apples. We also happen to know the total number of chihuahuas that all apple-handling entities have; let's say it's 100 too (corresponding to d). And we know that Don has 5 chihuahuas and Val 25. Sure, it follows - trivially - that "dark energy" must have 70 chihuahuas. But we could have said that without knowing all the stuff about how many apples are in the basket. So it seems we are really just using the same term ("dark energy") for the things behind the up to that point unknown terms in our equations... but that doesn't seem to say much. You might as well call it "previously unknown remainder". And I don't see how one previously unknown remainder confirms another previously unknown remainder.

    I figure the answer may be in the actual relationship between expansion rate and density. In my analogy this is missing: I implicitly assume there is no particular relationship between number of apples put in/taken out and number of chihuahuas owned. In contrast, density seems to have a lot to do with expansion rate, at least to my feeble intuition on this. Maybe for simplification, Greene has skipped the details? Or no less likely, I'm missing something. Enlightenment most welcome! Thanks.

    Just for good measure:
    • Post not meant as criticism of Greene's book. It's not faultless I think, but I'm grateful for all I'm learning from it.
    • I don't mean to sound like "you (physicists) don't know what you're talking about". I bet you do. To the contrary, I don't understand what you're talking about. (I hope you understood what I'm asking :^)
     
  2. jcsd
  3. Jan 30, 2015 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No. The kinds of things that can have negative pressure are very limited: the only candidates are a cosmological constant or a scalar field that basically acts like a cosmological constant (a scalar field can vary in space and time, but it can't vary very much because the acceleration of the universe's expansion is extremely uniform). Both of these things have well-established relationships between their pressure and their density, so once we know their pressure (from the effect on expansion rate), we know their density. That's how we can determine the contribution to equation 2 once we know the contribution to equation 1.
     
  4. Jan 30, 2015 #3

    marcus

    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    "we have no clue what it is (is that so?)."
    Peter says no, and I agree. He mentions two possibilities: cosmo constant and scalar field
    We have very strong clues that we are looking at the cosmological curvature constant that occurs naturally in Einstein's GR and was already in the equation as he wrote it in 1917 or earlier. The curvature constant Lambda belongs in the equation because it is one of the two physical constants allowed by general covariance, i.e. by the underlying symmetry of the theory. Traditionally in physics if a term is allowed by a theory's symmetry you put it in the equation, and then you check experimentally to see if it is zero or has some non-zero value that you can measure.

    Einstein noticed that the curvature constant belonged in the equation and he put it in and even used the Greek letter lambda for it. (A trivial detail, in 1917 he used the lower case λ and we now use the uppercase Λ).
    Spacetime curvature is associated with the percentage rate of distance growth. the effect of the positive curvature constant Λ in the equation is to place a limit on how low the percentage distance growth rate can go, as expansion continues and the matter content thins out.

    the curvature constant can be expressed as either inverse length squared or inverse time squared. Intuitively curvature is a number per unit area, a reciprocal area, one over length squared, but in relativity you can switch time or length units more or less freely so second-2 which is "square Hz" is OK. Some people like that for Λ.
    Percentage growth rate is also a "per unit time" unit---reciprocal time---and the lower limit on distance growth rate is essentially just the square root of the Λ curvature constant.

    Naturally if the percentage growth rate of distances can't get below a certain positive (albeit very small) rate then eventually we have what amounts to exponential growth. Growth at a fixed or nearly constant percentage rate.

    Probably "energy" is the wrong thing to think of in this connection. distance growth is not like ordinary motion (nobody gets anywhere by it, everybody just becomes farther apart) it is geometry itself changing. GR is about dynamic geometry. Geometry changing over time = curvature.
     
  5. Jan 30, 2015 #4

    marcus

    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    The way I remember the cosmological curvature constant Λ is to remember the longterm distant growth rate (as measured by the Planck mission most recently) namely 1/173 percent per million years.
    the growth rate has been declining since the BB start of expansion and it is now 1/144% per million years and destined, we think, to continue declining to a limit of 1/173%.

    So let's square that and see what it is in "square Hz". there is a conventional factor of 3, and then that's Λ.

    If you put this:
    3 ((1/173) percent per million years)^2
    into google you get the answer:
    1.00656 × 10-35 s-2

    As far as we know this is a universal eternal constant, so that appears to suggest that whatever graduate students created the universe for fun in their departmental laboratory had TEN fingers and wore wristwatches that counted seconds rather than some other unit. Because the value of the constant comes out essentially to be 10-35 s-2
    or as some people like to say "10 square atto Hertz"
    because "atto" is a billionth of a billionth namely 10-18
    so "attoHertz" is 10-18 s-1
    and "square attoHz" is 10-36 s-2

    So that "ten square attoHz" is what we said: 10-35 s-2

    I find it easier to remember it by the two main parameters in cosmology, the present and longterm Hubble expansion rates: which you see whenever you use the standard cosmic model built into, for example, lightcone calculator. Link in my sig:
    http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html
    The present and longterm Hubble radiuses are 14.4 billion LY, and 17.3 billion LY (the two most important lengths in standard cosmology)
    and that means the present and longterm expansion rates are
    1/144% and 1/173% per million years.
    those are two quantities everything in cosmology depends on in a very immediate direct way, so I can't forget 1/173% per million years. And I can always square that an multiply by the conventional 3 to get the vacuum curvature of "ten square attoHertz" (really just an arbitrary historical accident it should come out round numbers)
     
    Last edited: Jan 30, 2015
  6. Feb 2, 2015 #5
    Thank you, both -- those are very helpful answers! :)

    Right... I didn't realize that 'negative pressure' is special in that sense -- it's not expanded upon much in the book -- "meaning that the pressure sucks inward instead of pushing outward" -- that sounds like an everyday occurrence, but I'm guessing the everyday phenomenon relates to relatively negative pressure, whereas the negative pressure relevant to dark energy is absolute?

    Is it still unknown what the cosmological constant or the cosmological-constant-like-but-not-constant-field is constituted by? Is that even a sensible question? Or are the properties like the pressure, expansion rate, and density, really all there is to know about this 'thing'?

    Thanks for working through this relationship step by step -- it's making things more concrete for me! That last identity is quite an Aha!-statement. Suddenly 'curvature of 3D space" becomes a lot less impossible to picture :) But... to make sure, so we always speak about the curvature of space, not spacetime, right?

    This surprises me. So the growth rate is expected to reach a stationary value? But Greene writes (p301): "About 7 billion years ago, ordinary gravitational attraction became weak enough for the gravitational repulsion of the universe's cosmological constant to become dominant, and since then the rate of spatial expansion has been continually increasing." Is there a contradiction here, or am I missing something?

    What is the exact relationship between Hubble radius ([distance] beyond which objects recede from that observer at a rate greater than the speed of light, says Wikipedia) and expansion rate? Can't be hard, given the numbers, but I somehow fail to see it... :oops:

    Thanks very much again, both!
     
  7. Feb 2, 2015 #6

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Not really. The everyday occurrence that you are describing is that if you have a vacuum, matter will rush in to fill it. That's not negative pressure; it's just zero pressure in some region, which therefore can't stop positive pressure in other regions from pushing matter into the zero pressure region.

    Negative pressure is something different. A region of space with negative pressure will suck things into it even from a region with zero pressure--i.e., there is nothing pushing matter from the zero pressure region into the negative pressure region.

    The following two articles give more information on this topic and are worth reading:

    http://www.astro.ucla.edu/~wright/cosmo_constant.html

    http://math.ucr.edu/home/baez/vacuum.html

    Basically, yes. We know the basic properties that it has to have (density and pressure), but we don't have a good explanation for why the candidate fundamental fields in question (a cosmological constant, or a scalar field that acts like one) produce the values for those properties that we actually observe. The second link I gave above goes into this.
     
  8. Feb 2, 2015 #7

    marcus

    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    To continue with part of Diploria's discussion where there were questions about couple of things I said (shown indented) that I could expand on.
    ==quote Diploria==
    https://www.physicsforums.com/threads/in-what-way-is-dark-energy-more-than-none-of-the-above.795083/goto/post?id=4993255#post-4993255 [Broken]
    the growth rate has been declining since the BB start of expansion and it is now 1/144% per million years and destined, we think, to continue declining to a limit of 1/173%​

    This surprises me. So the growth rate is expected to reach a stationary value? But Greene writes (p301): "About 7 billion years ago, ordinary gravitational attraction became weak enough for the gravitational repulsion of the universe's cosmological constant to become dominant, and since then the rate of spatial expansion has been continually increasing." Is there a contradiction here, or am I missing something?

    https://www.physicsforums.com/threads/in-what-way-is-dark-energy-more-than-none-of-the-above.795083/goto/post?id=4993255#post-4993255 [Broken]
    The present and longterm Hubble radiuses are 14.4 billion LY, and 17.3 billion LY (the two most important lengths in standard cosmology) and that means the present and longterm expansion rates are 1/144% and 1/173% per million years.​

    What is the exact relationship between Hubble radius ([distance] beyond which objects recede from that observer at a rate greater than the speed of light, says Wikipedia) and expansion rate? Can't be hard, given the numbers, but I somehow fail to see it...
    =======endquote======
    Diploria good follow-on discussion! When I say "rate" I try always to have, in the same paragraph, already said "percentage rate" and/or "growth rate" so that the reader knows what sort of rate is meant. That is different from SPEED. If distances grow at a constant percentage rate, then if you watch a particular distance the size will get larger at an increasing speed. That is simply exponential growth at a given fixed percent rate. So there are two quite different concepts. The percentage growth rate can even be DECLINING and if it falls off gradually enough there can be nearly exponential growth.

    imagine having a savings account at a bank where the bank very gradually reduces the percent interest it pays, your savings could still be growing almost exponentially by an increasing annual DOLLAR AMOUNT because the principal is growing.

    I think Greene used confusing language in that quote because he did not explain the difference between rate (like growth rate and interest rate) that many people normally think of, and annual dollar amount or kilometer amount increments---a kind of "speed". But it is probably a bit misleading to think of distance expansion as ordinary motion, characterized by speed, in any case. The Hubble rate is a percentage growth rate, or if you want to sound more academic, a "fractional growth rate" The fraction by which something increases per unit time. Percentage sounds more colloquial and is less off-putting, but fractional sounds more academic/technical.
    =================
    Diploria we estimate that at present cosmological distances (not between objects gravitationally bound within a galaxy, but large-scale unbound) are all growing at 1/144% per million years.

    So you tell me, ASSUMING THAT growth rate, look at a distance of size 14.4 billion LY, what speed is that distance growing?
    Well, in a million years its size will have grown by 1/144%. OK. What is 1/144% of 14.4 billion LY?

    You asked what is the exact relationship between the Hubble radius and the percent growth rate of distance. This is it. You are right to imagine it is very simple.

    Suppose instead of the present measured 1/144% the distance growth rate were 1/10 % per million years. Invert the fraction and put decimal point in and get 1.0.
    Look at a distance of size 1.0 billion LY. How much longer is it in a million years? Well, 1/10 of a percent is a thousandth. what is a thousandth of 1.0 billion LY?
    It is a million LY. So it grows at a speed of one million LY per million Y. That is the speed of light. So 1.0 billion LY is the Hubble radius
     
    Last edited by a moderator: May 7, 2017
  9. Feb 2, 2015 #8

    Jorrie

    User Avatar
    Science Advisor
    Gold Member

    We have to be specific, because the curvature of space may be zero, while the curvature of spacetime is non-zero in our present cosmos. I think of "flat space" as if you "freeze-frame" expansion, parallel lines 'here' remain parallel 'there' in flat space (on large scale). "Un-freeze" the expansion and parallel lines 'here' will diverge 'there', due to our present negative spacetime curvature, i.e. accelerated expansion.

    If this picture is not true, will someone please correct?
     
  10. Feb 3, 2015 #9

    marcus

    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    Jorrie, I think my understanding of it is along similar lines, so if someone can correct our notions I will learn something too. I imagine that to talk about spatial curvature and spatial flatness probably presupposes you have a solution, not too different from the Friedmann model with some initially even distribution of matter. so there is a preferred time-slicing and we know what space we are talking about the curvature of. So there is a semi-realistic model of the universe with a preferred "universe standard time" like what we are familiar with in usual cosmology.
    And then you can say look! the spatial slices are exactly or at leas approximately flat! Triangles add up to 180 degrees! and so on. Zero spatial curvature. But if you lay all those 3d slices on top of each otther to make a 4d block you find that parallel lines are diverging and it's NOT "flat". There is non-zero spacetime curvature.

    Diploria, good questions! BTW you have a good nickname for a scuba diver or marine biologist. Our family has done a bit of scuba in the Caribbean and we've seen some wonderful coral, including the large roundish whiteish brain coral, with a maze of folds on its surface.
     
  11. Feb 4, 2015 #10
    Thanks everyone, you really help me understand (a bit more) and your efforts are much appreciated!

    Right, I get it... a bit better. Baez's article was enlightening. I found Wright's a little harder for my current understanding. I'll keep the "Observational Limits" details for some possible future; but I am also a bit unclear on the initial stuff about negative pressure. He writes:

    «The vacuum energy density must be associated with a negative pressure because:
    • The vacuum energy density must be constant because there is nothing for it to depend on.
    A 'true vacuum' provides nothing for anything to depend on; but in a 'false vacuum' are there not those quantum fluctuations...?
    • If a piston capping a cylinder of vacuum is pulled out, producing more vacuum, the vacuum within the cylinder then has more energy which must have been supplied by a force pulling on the piston.
    • If the vacuum is trying to pull the piston back into the cylinder, it must have a negative pressure, since a positive pressure would tend to push the piston out.»
    Well, "if". Do we know whether it does? Has this experiment been done, can it even be done?

    Also, if you can assume that P = -u = -ρ*c2, then yes, it follows that 'Einstein needed P < 0'. But that equation for P seems really weird: more matter/energy means lower, indeed negative, pressure? I suppose that's precisely the surprising revelation, that this is true for the (false) vacuum. So, is P = -u then a relationship which must be right from the equations even though it is unintuitive?

    Thanks for that, and for working out the Hubble radius / expansion rate relationship. All clear now! (my bank account feels optimistic too now ;-) ).

    Okay, so for space, 'curvature' is geometry changing over space, whereas for spacetime, it is geometry changing over time and/or space?

    Glad you enjoyed that! I'm neither a scuba diver nor a marine biologist, but I do like snorkeling and admiring lifeform variety... Your comment inspired me to ask a question about the Diploria that live in seas rather than post on boards (-:
     
  12. Feb 4, 2015 #11

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Sure, those are what make the (false) vacuum have positive energy (more precisely, positive energy relative to the true vacuum). But those fluctuations have to be Lorentz invariant, because they are vacuum fluctuations; that is what he means by "there is nothing for anything to depend on". If the energy density due to the vacuum fluctuations varied in space or time, they would pick out some particular direction in space or time and would not be Lorentz invariant.

    Logically, it follows from his previous statement, that it took a force to pull the piston out.

    No, because we have no way of making a false vacuum to put inside the piston chamber.

    Sure, that was the whole point of his argument in what you quoted. Positive vacuum energy means negative pressure, and more positive vacuum energy means more negative pressure.

    Yes.
     
  13. Feb 12, 2015 #12
    Thanks again, Peter.

    Right. But that statement ("If a piston capping a cylinder of vacuum is pulled out, producing more vacuum, the vacuum within the cylinder then has more energy which must have been supplied by a force pulling on the piston. ") seems odd too. Why should it be so? Sure, pulling the piston must have required energy. But I don't readily see how/why that alters something about the energy of the system -- unlike in the case of a piston pulled in a non-vacuum system.

    If
    there were a true vacuum (no quantum fluctuations, absolutely nothing) both inside the cylinder and outside, it just can't change at all -- nothing can only stay nothing. But apparently the situation is not like that: the vacuum inside the cylinder is in fact a false vacuum, which can become more energetic.

    I think I'm confused about what is being postulated.

    Is it (case 1) "I pull the piston and find the vacuum in the cylinder must have changed energy, for I notice the piston then moving back, so the vacuum must be a false vacuum (and must have acquired negative pressure)"? Then there was no initial assumption of a false vacuum in the cylinder: it's a conclusion. But it is not one which has been found by actual experiment. So it must derive from equations. But how could that happen if you started out assuming true vacuum inside the cylinder? It seems like there is nothing to give rise to the term which says the piston is moving back.

    Or is it (case 2) "I know/postulate there is a false vacuum in the cylinder, and a true one outside. After I pull the piston, it moves back, so the false vacuum must have acquired negative pressure"? But in that case, if -again- this is not (and cannot be) found through experiment, it must come from equations - and part of these describe the postulated false vacuum. But if the P = -u property was already in those equations, we only got out what we put in. Presumably this is not the case.

    So then: is it case 2, with P = -u not part of the postulated equations, but deriving from a theoretical system representing 'only' the cylinder and the quantum fluctuations as modeled by QT?
     
  14. Feb 12, 2015 #13

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    I don't understand. If pulling the piston out required energy, that means it added energy to the system. That's what "pulling the piston out requires energy" means. If it didn't add energy to the system, it wouldn't require energy to do it.

    Another way of looking at it is that, if pulling the piston out requires energy, that means doing it does work on the system (the cylinder). Doing work on the system increases its energy. See further comments below.

    Yes. More precisely, its energy density (energy per unit volume) is constant, so adding volume means adding energy.

    The starting postulate is that there is false vacuum inside the cylinder, and that it has a constant positive energy density (energy per unit volume). That is what the term "false vacuum" means. (More precisely, as I said in a previous post, the starting postulate is that the false vacuum inside the cylinder has a constant positive energy density relative to the energy density of the true vacuum outside the cylinder, which is assumed to be zero.)

    Since the false vacuum inside the cylinder has a constant positive energy density, if the piston moves out, increasing the volume inside the cylinder, the energy (energy density times volume) of the false vacuum inside the cylinder must increase. That means it must require a force to move the piston out; if it didn't, then the piston could just move out spontaneously, but energy can't just spontaneously increase; work has to be done on the system for its total energy to increase. Exerting a force on the piston to move it out supplies the work that must be done to increase the system's energy.

    So, since it requires a positive force to move the piston out, it must require a negative force to move the piston in. (If you wonder about this, suppose we first exert positive force to move the piston out, and then reverse the process, moving the piston back to its original position. Doing that must also return the system's total energy back to its original value; but that means the excess energy must have gone somewhere. The only place it can go is the environment, by way of the piston--in other words, moving the piston back in means the system has done work on its environment, the reverse of the work the environment did on the system to move the piston out. The system doing work on the environment is what "negative force" means.) This means the false vacuum inside the cylinder must have negative pressure--if it had positive pressure, the pressure would push on the piston so that it required positive force to push the piston in, and if it had zero pressure, it would require no force at all to move the piston either in or out.

    Does that make the logical sequence clearer?
     
  15. Feb 13, 2015 #14
    Right... so I was imagining a true vacuum both inside and outside the cylinder. Then there is nothing to stop the piston once it's set in motion (I'm assuming there is no friction either). But to get it moving, and again to make it stop after a little while, you'd have to apply a (possibly infinitesimal?) force. So it would require some energy to displace the piston. Yet after it is in its new position, there would be no trace of that energy other than how the piston has been displaced? It's not like that gives it some potential energy, like when in Earth conditions you lift something on top of a table -- for there is no gravity... Hm well, actually, there is? Between the piston and the cylinder chamber? So is that what would cause an energy change even in the true-vacuum-everywhere setup?

    In any case, the true-vacuum-everywhere assumption clearly is not what Wright and you are talking about!

    Yes! Thank you for your patience - really appreciated.
     
  16. Feb 13, 2015 #15

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Not really. If there is literally zero pressure and zero friction on both sides of the piston, it takes zero force to move it, if the piston itself is idealized as having zero mass. We have been implicitly idealizing the piston that way in this discussion, but it hasn't been pointed out explicitly; sorry about that.

    (If the piston has nonzero mass, then yes, it takes nonzero force to get it moving, and therefore nonzero energy, even in the complete absence of pressure and friction. But if that is the case, the arguments we have been discussing need to be revised to account for the fact that the piston itself has nonzero energy. So really we've been idealizing the piston as having zero energy regardless of its state of motion, but that means we're idealizing it as having zero mass.)
     
  17. Feb 17, 2015 #16
    This may be getting too off-topic... but this struck me as bewildering:
    That sounds like an 'unmoved mover' -- not something I'd think has a place in modern physics.

    I can more or less see the sense in which it would require zero force. On the other hand: how would something accelerate without any force acting on it? Seems fundamentally incoherent.

    Is it merely due to the fact that we are talking about an idealization? Or is there more to it? (Is it perhaps that quantum fluctuations are conceived of as arising acausally?)
     
  18. Feb 17, 2015 #17

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No, it's just an unrealistic idealization.

    Because it is (unrealistically) idealized as having zero mass.

    Yes, of course. It's just a thought experiment to try and give a heuristic understanding of why positive vacuum energy goes with negative pressure.

    If you want to make it slightly more "realistic", instead of idealizing the piston as having zero mass and taking zero force to move in the absence of pressure, just say that its mass, and the consequent force required to move it in the absence of pressure, are "negligible", i.e., too small to matter for this analysis. You get the same answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: In what way is "dark energy" more than "none of the above"?
  1. More on dark energy (Replies: 31)

  2. What dark energy? (Replies: 4)

  3. What is dark energy? (Replies: 4)

Loading...