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What does a Cosmological Constant mean?

  1. Feb 18, 2015 #1
    Simplistically, the GR equation is

    G = k T + l g

    G represents the curvature of the fabric of space-time

    T is the stress energy tensor, representing the fluxes and densities of matter and energy

    g is the metric tensor

    So... In another thread, someone said that the cosmological constant belongs in the equation, because it is admissible, and experiments and observations should reveal the coefficient (l). Is the proper interpretation of the term, about the matter and energy free form of the fabric of space-time? I.e. if (l=0) then mass and energy free space-time is Minkowskian... Whereas otherwise the fabric has an "intrinsic" curvature, in the sense of "not lying flat" even in the absence of matter and energy? ( I'm picturing a flatbed trailer which bows up in the middle, so as to be arched unloaded, so having an "intrinsic" curvature. )

    And observations suggest, that at large size scales, the fabric of space-time has such an "intrinsic" curvature?

    And, the curvature is such, that to impute the same into the fabric of space-time, using matter and energy, would require a fictitious fluid, having positive energy density but negative pressure density? I.e. the cosmological constant does not describe the universe as filled with such a fluid... Only that the fabric of space-time is "intrinsically" curved as if filled with such seeming strangeness.

    ( a flatbed trailer which bows up in the middle, does not actually imply the presence of antigravity. )
     
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  3. Feb 18, 2015 #2

    PeterDonis

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    Please give a link.

    Correct. It belongs in the equation because it is consistent with all of the constraints that went into deriving the equation. (Roughly, the constraint is that we want tensors involving the metric and its first and second derivatives, but no higher derivatives, whose covariant divergence is zero. The Einstein tensor is often described as the "only" tensor meeting those requirements, but the correct statement is that it's the only tensor that does, other than the metric itself, multiplied by any constant.)

    That's one way of interpreting it, yes.

    That's another way of interpreting it, yes.

    Not quite. The correct statement is that these are two different interpretations of the cosmological constant; but both lead to exactly the same predictions for all observables, so there is no way to distinguish between them as a matter of physics. The physics is equally consistent with either interpretation.
     
  4. Feb 18, 2015 #3

    Chronos

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    The Einstein field equations invoke the cosmological constant. It is effectively a constant of integration - re: http://arxiv.org/abs/1309.6590
     
  5. Feb 18, 2015 #4

    Chalnoth

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    It depends. You can put it in. But you can just as easily wrap the cosmological constant into the stress-energy tensor if you like. The two ways of writing down the equations are mathematically identical.
     
  6. Mar 13, 2015 #5
    As you will know the theory behind the cosmological constant is that it may not be zero, though there is no explanation as to what the constant may be, and whether there even is a constant in the first place.
    I believe that the working that leads to a possible figure is much simpler than what is being looked into, and it all revolves around gravity.
    In short, as the universe expands and new stars, planets etc. are created they gain a gravitational force that causes an effect, this effect that the gravitational pull has on each and everything around it in turn speeds up and slows down the expansion process.

    If you look at our own planet as an example, it has most directive force on our moon, which in turn also has a force on the planet, as well as the sun and the other planets, if you cancel all other gravitational forces out and you are just left with the force from our planet on the moon the moon would instantly be drawn to the planet. That idea represents incredible speed, the speed that would be acting on the expansion of the universe if gravity was selective.
    As gravity is not selective, the speed and force exerted depends on the size and composition of an object. This represents the unbalance in the universe, which is why there is an expansion, but in trying to measure the universe as a whole as one single expansion constant represented by a single number, the rate can not be quantified, or utilized without knowing the precise force of everything with in the universe which is impossible, considering we are discovering new objects, planets, galaxies constantly and do not know the true value of the universe itself.
    Even if that wasn't the case, it would have to be taken into account that the gravitational force exerted by an object has a different impact on any other objects depending on their, size, distance away from the object, composition etc.

    I believe we can only understand the rate of expansion via two possible ideas put into practice as a starting point:

    1. Follow the expansion of 2 solar systems, giving their gravitational effect on each of the planets and stars with in it, and seeing what the rate of expansion is between the two systems, and if there is a constant value even with the difference in composition and figures.

    2. Base the idea on the possibility of selectivity of different atoms and their difference in gravitational force upon one another, causing no constant rate of expansion, and in fact a continuously changing rate of expansion due to different rates of force causing constant changes.

    I am not an expert in this, though i believe it has allowed me to see it a little more simply, and i may be incorrect though i hope that there is some usefulness in this idea.
     
  7. Mar 13, 2015 #6

    PeterDonis

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    No, they don't. The new stars and planets are not "created" out of nothing; they are "created" out of matter that was already there, just not clumped into stars and planets. The gravity associated with that matter was already there; to someone far away from the star or planet being created, nothing changes gravitationally.

    No, it wouldn't. The moon has a sideways velocity that keeps it in orbit. If you eliminated all the forces from the other planets, the moon's orbit would change, yes, but it wouldn't just fall into the Earth.

    This doesn't make sense.

    The effect of gravity on an object is independent of its composition; and while the force of gravity on the object depends on its mass, its acceleration due to that force does not, because of the equivalence between inertial and gravitational mass.

    The rest of your post appears to be based on your incorrect understanding of the above, so it looks invalid to me.
     
  8. Mar 14, 2015 #7

    Garth

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    Nicholas Corso, welcome to these Forums, but you have to follow the Forum Guidelines. If you are not expert in the field you are posting in you have to qualify your statements by referring to refereed published papers, or at least Wikipedia articles that themselves cite refereed published papers. Of course we really like you to ask questions as a way of learning or starting a discussion.

    Getting back to a decent discussion of the subject in hand......
    Agreed.

    But I would like to point out that if left as a cosmological constant (on the 'left hand side') as an 'integration' constant', which is a necessary (but perhaps zero) component of the Einstein tensor Guv describing the curvature of space-time, then it simply means gravitation (as described by that Einstein tensor) acts as an attractive force (in a 'Newtonian' type of way) at short ranges (~ <109 pcs) and it acts as a repulsive force (in an 'anti-gravity' type of way) at cosmological ranges.

    If treated in this way it must be constant over all space and all time - otherwise it would violate the Bianchi identities, the covariant conservation properties of the Einstein tensor (Guv;u = 0).

    If treated as a component of the stress-energy tensor (on the 'right hand side') then it is this very mysterious substance called Dark Energy that has an equation of state of [itex]p = \omega \rho c^2[/itex] where [itex]\omega[/itex] = -1.

    In this case the parameter [itex]\omega[/itex] may be allowed to vary from -1 and whether it is actually a constant or not is the subject of much research, as in one example here: Is dark energy evolving?.

    At present there is no substantial evidence to say that [itex]\omega[/itex] is other than -1.

    Garth
     
    Last edited: Mar 14, 2015
  9. Mar 14, 2015 #8

    marcus

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    Entia non sunt multiplicanda praeter necessitatem.
    "Don't make things unnecessarily complicated."

    Nice stained-glass picture of William here:
    http://en.wikipedia.org/wiki/William_of_Ockham

    I think he would have approved of leaving the Lambda curvature constant on the lefthand side with the rest of the curvature terms.

    It has to be in the equation, as you point out. And it takes care of the observations of a slight acceleration after year 8 billion, without having to invent some exotic energy field with negative pressure. Plus it seems a bit arbitrary to yank it out of its original context and convert it to an equivalent energy density term on the righthand side.
    Especially without evidence suggesting that it arises from an energy.
     
  10. Mar 14, 2015 #9

    marcus

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    Some additional background to what Garth said: if you leave Lambda (the cosmological curvature constant) on the left side, where Einstein put it in 1917, then its value, as accurately as we have been able to measure so far,
    is very close to 10-35 seconds-2.
    As I recall it is 1.007 x 10-35 s-2 plus or minus some small uncertainty.

    This corresponds to a residual expansion rate of 1/173 of a percent per million years. That is built in to the dynamic geometry---so that the expansion rate can't decline to zero but instead has to level out at that fractional rate.

    TEFL asked a good question: what does this intrinsic curvature constant mean? Here is one meaning. It plays a remarkable role at microscopic level in an approach to quantum geometry, where the micoscopic geometric elements in the 4d case are the 4d analogs of 2d triangles and 3d tetrahedra. Let's imagine building quantum spacetime geometry out of these "4-simplex" building blocks. Suppose we include the curvature Lambda in the model by having this constant curvature built in to all the simplexes. Instead of being made out of flat material they are all cut out of slightly curved material.

    This has been worked out in 3d and still has to be extended to 4d. It turns out that instead of being infinite, the phase space of the evolving geometry is compact, and the quantum state space is finite dimensional and time is discrete. This could be mathematically convenient, it's a new result, and it's rather interesting. Potentially (if it goes through in 3+1 dimensions) it gives a new meaning to Lambda. If you introduce the intrinsic curvature into the materials you are working with, then something geometrical "closes on itself" and becomes finite. You can put a uniform probability measure on a finite thing (but not on an infinite), you may find extra symmetries. It's just nice. So that's a possible way of thinking about Lambda that doesn't involve imagining it arises from some made-up kind of "energy".
    That came up in a recent paper called "Compact phase space, cosmological constant, discrete time"

    You can get it by googling [compact phase discrete] without the brackets :smile:

    or click http://arxiv.org/abs/1502.00278
     
    Last edited: Mar 14, 2015
  11. Mar 14, 2015 #10

    Garth

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    Hi marcus! You tracked down my avatar, from the stained window in All Saint's Church, Ockham, near where I used to live in Surrey. (Ockham is the Anglo-Saxon spelling and Occam the Latin spelling)

    Yes indeed, keep things as simple as possible unless you have good reason (necessitatem).

    Garth
     
  12. Mar 15, 2015 #11

    Chronos

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    Not that it matters, but, I prefer the curvature thing [cosmological constant] over dark energy. It seems so much cleaner that way.
     
  13. Mar 15, 2015 #12
    Nicholas Corso,
    "If you look at our own planet as an example, it has most directive force on our moon..."

    If by "directive force" you mean gravitational acceleration, that of the Sun on the Moon is about .0059m/s^2 which is about twice that of the Earth on the Moon (about .0028m/s^2)... so maybe the Sun has most directive force on our Moon?
     
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