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The force of dark energy

  1. Jul 12, 2010 #1
    Okay so dark energy repels mass, but why is that? The reason why masses repel other masses when brought together is the electromagnetic force, but dark energy also doesn't interact at all with electromagnetic waves, and if it exhibited electromagnetic force it would interact with electromagnetic waves, so then, what is the repelling force of dark energy and matter? Is it quantifiable? Where does it come from? etc...
     
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  3. Jul 12, 2010 #2

    mathman

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    Dark energy force is anti-gravity. It has nothing to do with electromagnetic. DE according to current ideas (subject to change) is related to the concept of a cosmological constant in general relativity.
     
  4. Jul 12, 2010 #3
    does that mean the repelling force of dark energy is equal to that of the force of gravity?
     
  5. Jul 12, 2010 #4
    Dark Energy and Dark Matter are a patch on physics theories based on mathematics that have failed--new epicycles.
     
  6. Jul 12, 2010 #5
    If I understand it correctly (and I don't think I do), dark energy is not related to any forces, such as gravity or the electromagnetic force. It's thought to be a property of the universe itself, in essence the "fabric" of space-time stretching itself out due to quantum fluctuations. So there isn't really a "repelling force" in the sense I think you mean, since its not a force that acts on matter, but rather its what drives the universe to expand.
     
  7. Jul 12, 2010 #6

    bapowell

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    The only thing dark matter and dark energy have in common is the word 'dark' and the fact that crackpots like to discount their existence without giving any tenable arguments.
     
  8. Jul 12, 2010 #7

    bapowell

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    The simplest way to model dark energy in general relativity is to hypothesize a smooth energy component of the universe that has a negative pressure (quantum vacuum energy has this property). When this type of energy is used as the source in Einstein's equations, one obtains an isotropic and homogeneous universe that expands at an accelerated pace. So, it's not that individual masses are repelling each other; it's that the spacetime itself is expanding (much like ordinary non-accelerated expansion, only that this rate of expansion is itself increasing), and individual masses 'repel' each other as a result of the the expanding space between them.

    So, accelerated expansion should be thought of as a gravitational effect of dark energy.
     
  9. Jul 13, 2010 #8

    Ich

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    I think it's absolutely valid to say that dark energy is the source of repulsive gravity and thus repels matter. It doesn't "catalyse" masses to repel each other.
     
  10. Jul 13, 2010 #9

    bapowell

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    ....so we agree?
     
  11. Jul 13, 2010 #10

    Ich

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    Do you think our statements are the same?
     
  12. Jul 13, 2010 #11

    bapowell

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    I have no idea. I agree that dark energy sources repulsive gravity; but test masses do not. The universe accelerates whether the test masses are present or not. If they are, then I guess it appears as if they are repelling each other, but really they are comoving with the expansion. I was trying to move away from the thinking that dark energy makes a kind of repulsive Newton's law possible for test masses -- it doesn't.
     
  13. Jul 13, 2010 #12

    Ich

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    Yes, we agree on that. That's why I didn't understand your "individual masses 'repel' each other as a result of the the expanding space between them". Not the test masses repel each other, nor does DE make them repel each other. DE repels them.
    But you agree DE acts like kind of a negative matter density in the Newtonian sense? Just like a positive matter density attracts test masses, a negative one repels them.
     
  14. Jul 13, 2010 #13
    So basically what you guys are saying is that this "dark energy" is actually the expansion of spacetime as a biproduct of quantum fluxuations? Does this mean that the amount of dark energy is uniform across all of space? And if it is just the stretching of spacetime, then how is it that we can put a value of "how much" dark energy there is in space?
     
  15. Jul 13, 2010 #14

    bapowell

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    Yes, that's one way to think of it. Quantum fluctuations are the right kind of stress energy for driving accelerated expansion, and are a candidate for dark energy. In this case, the dark energy would be perfectly uniform across space (a cosmological constant, if you will). In other models, in which the dark energy is the vacuum energy of a dynamical scalar field, the dark energy is time dependent and spatially homogeneous modulo quantum fluctuations of the field.

    Using the equations that govern the homogeneous and isotropic expansion of the universe, it is possible to relate the expansion rate (by way of the Hubble parameter, H) to the energy density of the universe. In this way, 'more' dark energy results in a more rapid expansion.
     
  16. Jul 14, 2010 #15
    I'm sorry but I understood basically every other word you said. according to how much I do understand, you say that quantum fluctuations create stress energy, wouldn't this energy actually cause matter to contract towards one another? (because of the whole gravity thing..)
     
  17. Jul 14, 2010 #16

    bapowell

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    No, this is precisely where the antigravity comes in. Quantum fluctuations have a negative pressure, as opposed to more ordinary items like radiation, which have positive pressure. The negative pressure leads to a repulsive, accelerated expansion of spacetime.
     
  18. Jul 14, 2010 #17

    George Jones

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    The weak-field limit of Einstein's equation without cosmological constant/dark energy leads to Poisson's equation,

    [tex]\nabla^2 \Phi = - \vec{\nabla} \cdot \vec{g}= 4 \pi G \rho,[/tex]

    where [itex]\Phi[/itex] is gravitational potential and [itex]\vec{g}= - \vec{\nabla} \Phi[/itex] is the acceleration of a small test mass.

    The weak-field limit of Einstein's equation with cosmological constant/dark energy [itex]\Lambda[/itex] leads to a modified "Poisson" equation,

    [tex]\nabla^2 \Phi = 4 \pi G \rho - \Lambda c^2.[/tex]

    For a spherical mass [itex]M[/itex], the divergence theorem applied to the above gives

    [tex]\vec{g} = \left(-\frac{GM}{r^2} + \frac{c^2 \Lambda}{3} r \right) \hat{r}.[/tex]

    The second term is a "springy" repulsive term for positive [itex]\Lambda[/itex].
     
  19. Jul 14, 2010 #18
    okay, why do quantum fluctuations have negative pressure? What forces them to have this characteristic, and why is it different?

    Okay sorry george jones, but I didn't understand any of that. My mathematics are extremely subpar to my level of understanding for the logical part of it, if that makes sense.
     
    Last edited: Jul 14, 2010
  20. Jul 14, 2010 #19

    George Jones

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    Here is a very rough, heuristic way to think about it.

    Think of the vacuum as being a fluid, and consider a fluid element. As the universe expands, the the volume [itex]V[/itex] of this fluid element expands. Event though the fluid element expands, its energy density [itex]\rho[/itex] remains constant, i.e., there is "more" of the same vacuum fluid. The energy [itex]E[/itex] of the fluid element is [itex]\rho V[/itex]. The first law of thermodynamics then gives that

    [tex]
    \begin{equation*}
    \begin{split}
    \Delta E &= - p \Delta V \\
    \rho \Delta V &= - p \Delta V.
    \end{split}
    \end{equation*}
    [/tex]

    Consequently, [itex]p = - \rho[/itex], so, if the energy density [itex]\rho[/itex] is postive, pressure [itex]p[/itex] is negative.
     
  21. Jul 14, 2010 #20

    Ich

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    Great explanation, george!

    A little addendum to your previous post:
    More generally, and with c=1, it is
    [tex]
    \nabla^2 \Phi = 4 \pi G (\rho - 3p).
    [/tex]
    And for a homogeneous source,
    [tex]
    \vec{g} =-\frac{4}{3}\pi G (\rho - 3p) \vec{r},
    [/tex]
    which is the second Friedmann equation in disguise.
     
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