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The Cosmological Constant with Inflation is not Constant

  1. Sep 21, 2014 #1
    The title says a lot about my doubt. I dont know a lot about inflation maths but the idea is that in Inflation, Im seeing that the Cosmological Constant (ie Vacuum Energy) is not Constant in two ways:
    1 With time; the Vacuum Energy was larger in the beginning and that triggered Inflation. Some time later, the Vacuum Energy got smaller and then the rate of expansion got smaller too.
    2 With space: The Vacuum Energy got smaller in some places first and in some others later, and that produced inhomogeneities.
    But in GR the Cosmological Constant must be Constant because, if it is not constant, then the left hand of the Einstein Equation is not a Tensor (or something like that).
    So, I'm seeing a trouble trying to reconcile Inflation with GR. Obviously, I must be thinking something wrong, but I cant see where! I will find your thoughts very helpful as always!!!

    Thanks in advance
  2. jcsd
  3. Sep 22, 2014 #2


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    The vacuum energy responsible for inflation is not caused by the cosmological constant, but more likely a dynamical field that evolves as the universe expands. The field evolves in such a way that its vacuum energy decreases in time. If the cosmological constant is nonzero, it must have a magnitude smaller than the inflationary energy density.
  4. Sep 22, 2014 #3
    Thanks for your answer but that does not answer my doubt. When I saw (Susskind) / read about inflation, the idea is that the vacum energy plays the role of Cosmological Constant in GR and that is why space expansion accelerates. But in GR that Cosmological Constant should be a constant, however, the vacuum energy is not constant. So, what is added in GR is not a constant and the equations of motion don't remain equations between tensors and everything tears appart. So, there is something in the rough idea that I grasped from what I heard that is not working well. Am I understanding well the logic behind inflation and behind GR.

    Thanks and sorry if you have already answered and I could not get your idea.
  5. Sep 22, 2014 #4


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    I think you have misunderstood Susskind. Space will expand at an accelerated rate if it is filled with any homogeneous fluid satisfying the following equation of state: [itex]p < -\rho/3[/itex], where [itex]p[/itex] is the pressure and [itex]\rho[/itex] the density of the fluid (generally written [itex]p = w\rho[/itex]). The cosmological constant is just one example of such a "fluid", having [itex]p = -\rho[/itex] (i.e. [itex]w=-1[/itex]). But inflation is importantly a dynamic process: there is initially a high density of vacuum energy that very closely approximates a cosmological constant [itex]p \approx -\rho[/itex]. As inflation progresses, [itex]w[/itex] varies, changing the relationship between [itex]p[/itex] and [itex]\rho[/itex]. So, as inflation progress [itex]w[/itex] ranges from close to [itex]-1[/itex] up to [itex]w=-1/3[/itex] and inflation ends.

    Note that there is no need to have a cosmological constant at all for inflation to work or to get accelerated expansion in general. In fact, remove [itex]\Lambda[/itex] entirely from your equations!! Instead, the vacuum energy that powers inflation is due to the dynamics of a scalar field and is part of the matter side of the Einstein Equations. Look up the Friedmann Equations (http://en.wikipedia.org/wiki/Friedmann_equations) and notice that you still get accelerated expansion even when [itex]\Lambda = 0[/itex].
    Last edited: Sep 25, 2014
  6. Sep 22, 2014 #5
    Ok, I will go back to Susskind, I see what you are saying.

  7. Sep 22, 2014 #6


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    No, the equations still work just fine; but, as bapowell says, if you have a "cosmological constant" type term where the "constant" isn't really constant (i.e., you have a term ##\phi g_{\mu \nu}##, a scalar multiple of the metric where the scalar can change in space or time), it's more usual think of it as part of the RHS of the field equation, not the LHS.

    The original supposed "problem" with having a non-constant cosmological constant was that it was thought of as belonging on the LHS of the field equation, and that side isn't supposed to include any "fields" except the metric and its derivatives. A constant times the metric would be OK, but a scalar that can vary times the metric wouldn't be OK because the scalar would count as "another field" and would belong on the RHS. However, this is more of a matter of how you conceptualize the terms in the field equation; it doesn't indicate any actual breakdown of the math.

    In fact, the current tendency appears to be to consider even a truly constant "cosmological constant" term (i.e., a constant times the metric tensor) to belong on the RHS of the field equation--i.e., to consider it as just another type of "fluid", as bapowell notes. Again, this is perfectly fine mathematically; it's just a question of how you conceptualize the terms.
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