I'm considering a "classical" model of scalar field cosmology: A simple real scalar field minimally coupled to gravity, with a quartic Higgs-like field potential: \begin{equation}\tag{1} \mathcal{V}(\phi) = \frac{\lambda}{4} (\phi^2 - v^2)^2, \end{equation} where $v$ is the "true vacuum" field. We can derive this formula: \begin{equation}\tag{2} v = \pm \: \frac{m}{\sqrt{2 \lambda}}, \end{equation} where $m$ is the mass of the scalar field $\phi$, and $\lambda$ is the auto-coupling of the field. Take note that $\phi \sim \mathrm{L}^{-1}$, $m \sim \mathrm{L}^{-1}$ and $\lambda$ is a dimensionless number.

Now, the Friedmann-Lemaître equations and the scalar field equation are the following ($a(t) \sim \mathrm{L}$ is the cosmological scale factor, and dots are the usual cosmological time derivatives. $G \equiv \ell_P^2 \sim \mathrm{L}^2$): \begin{gather}\tag{3} \frac{\dot{a}^2}{a^2} + \frac{k}{a^2} = \frac{8 \pi G}{3} \big( \frac{1}{2} \; \dot{\phi}^2 + \mathcal{V}(\phi) \Big), \\[12pt] \tag{4} \frac{\ddot{a}}{a} = -\: \frac{8 \pi G}{3} \Big( \dot{\phi}^2 - \mathcal{V}(\phi) \Big), \\[12pt] \tag{5} \ddot{\phi} + 3 \frac{\dot{a}}{a} \dot{\phi} + \mathcal{V}^{\prime} = 0. \end{gather} The initial conditions that I need to use are these ($t = 0$ is the present time, i.e. our human epoc): \begin{align} a(0) &= a_0, & \dot{a}(0) &= H_0, \; \text{(Hubble's constant)} \\[12pt] \phi(0) &= \phi_0, \; \text{(any value)} & \dot{\phi}(0) &= \psi_0. \; \text{(any value)} \end{align} I can numerically solve these equations (after a scale transformation to remove all units), and get some very nice graphical output. I solve the second order equations (4) and (5) using the initial conditions defined above. Equation (3) is only used to find the curvature parameter $k$.

So the question is the following. I have 4 parameters as input to the numerical simulation:

  1. The field mass $m$.
  2. The field coupling constant $\lambda$.
  3. The field initial value (at present time): $\phi_0$.
  4. The field initial derivative (at present time): $\psi_0$.

I use an initial slow roll condition for simplicity: $\psi_0 = 0$. But what should be the "typical" realistic values of the three remaining parameters?

Currently, to get my nice graphical output, I had to use some very fantaisist and extravagant input:

  1. $m \approx 10^{- 68} \text{kg}$ (wow!)
  2. $\lambda \approx 10^{-121}$ (wowee!)
  3. $\phi(0) \sim \frac{1}{\ell_P}$, i.e. inverse Planck length (a very large field, to compensate the small values above).

EDIT : To numerically solve equations (4) and (5), we need to make a scale transformation to remove all units. I use these new variables : \begin{align} \tau &= H_0 \, t, \text{(dimensionless time, in units of the Hubble's time)} \tag{6} \\[12pt] \Phi &= \sqrt{\frac{8 \pi G}{3}} \; \phi, \text{(dimensionless scalar field, in units of the Planck lenght since $G \equiv \ell_P^2$)} \tag{7} \\[12pt] \tilde{m} &= \frac{m}{H_0}, \text{(dimensionless mass)} \tag{8} \\[12pt] \tilde{\lambda} &= \frac{3 \lambda}{8 \pi G H_0^2}, \text{(dimensionless coupling)} \tag{9} \end{align} Then, equations (4) and (5) become these (the prime is the derivative with respect to dimensionless time $\tau$) : \begin{align} \frac{a^{\prime \prime}}{a} = - \Phi^{\prime \, 2} + \mathcal{V}(\Phi), \tag{10} \\[12pt] \Phi^{\prime \prime} + 3 \frac{a^{\prime}}{a} \Phi^{\prime} + \frac{d \mathcal{V}}{d\Phi} = 0. \tag{11} \end{align} The typical input I used for my numerical simulation are $\tilde{m} \sim 1$, $\tilde{\lambda} \sim 1$ and $\Phi_0 \sim 1$ (not too small or too large numbers). From the scale transformation (6)-(9), this gives the fantaisist values cited above, for $m$, $\lambda$ and $\phi_0$.

  • $\begingroup$ The relevant time scale for cosmological evolution is the Hubble parameter $H_0\sim 10^{-60} $ in Planck units. The cosmological constant is $10^{-120}$. Thus if this model is to mimic dark energy then you should expect highly fine tuned parameters like the ones you found. $\endgroup$
    – Winther
    Jan 12, 2017 at 12:18
  • $\begingroup$ @Winther, there's nothing strange not specifying the curvature parameter. You start with the field values ($\phi(0)$ and $\dot{\phi}(0)$), to get the initial energy. If there's a lot of energy, you'll get a closed universe ($k = 1$). If you set the initial conditions to have much less field energy, then you get an open universe ($k = -1$). This is like throwing vertically a rock. You don't need to specify the "openeness" or "closeness" of its trajectory, before giving the rock an initial velocity. The initial conditions will define the closure of the rock's trajectory. $\endgroup$
    – Cham
    Jan 12, 2017 at 13:34
  • $\begingroup$ Yes my comment is not correct as written, I'll remove it. I was thinking of a different point. $\endgroup$
    – Winther
    Jan 12, 2017 at 13:47
  • 1
    $\begingroup$ btw this type of model have been studied in the literature before. Some papers that might be useful: as dark energy model (but coupled to matter) see e.g. Symmetron Fields ; Symmetron cosmology and as an inflation candidate see e.g. The Standard Model Higgs boson as the inflaton $\endgroup$
    – Winther
    Jan 12, 2017 at 13:53

1 Answer 1


The coupling constant $\lambda$ gives the overall energy scale of inflation, but otherwise does not effect any of the current observables directly.

As soon as we find Primordial gravity waves we will have a clear indication for the actual energy scale. Until that time you can set $\lambda$ to whatever is convenient in the sense of numerical evaluation. Physically people kind of guess it should be in the GUT energy scale, for numerous reasons.

$\phi_0$ should be very close to whatever value you have for $\phi$ at the end of inflation, since we see a very subtle accelerated expansion these days. So I think you have to set $\phi_0 \simeq \pm v$.

Now as for the value of $v$: the combination $\sqrt{\lambda} v^2$ should give an energy scale which is at the very least higher than the Electroweak energy scale (or 14TeV -current LHC energy scales), since we know physics up to theses scales and haven't found the inflaton yet. That is unless you think the Higgs is the inflaton, but it is probably not because of many different reasons.

EDIT: you can work out the actual units you need to use by reintroducing $\hbar,c,K_B$, etc.

  • $\begingroup$ I agree that $\phi_0 \sim \pm \; v$. But how is $\lambda$ related to an inflation energy scale ? $\lambda$ is dimensionless. Is it $\lambda v^2 \equiv \frac{1}{2} \; m^2$, thus $m$ ? And does that mean that $m$ should be higher than the Higgs boson mass ? $\endgroup$
    – Cham
    Jan 11, 2017 at 18:10
  • $\begingroup$ I'm sorry' you're right, in the model $\frac{1}{2}m^2 \phi^2$, the quantity $m$ does not affect the overall shape of H. but DOES change the actual value range of $H$, thus it will affect the number of efolds in this inflation. It is customary to take $N\sim 50-60$. This model (that you can later shift and renormalize) the number of efolds is roughly given by $N\sim \frac{\phi_{end}^2-\phi_i^2}{4}$, so you can make the calculation to find N, and then roughly have $N= H\cdot \Delta t$ where $t$ is the time it takes inflation to end, alternatively $N=\frac{1}{\sqrt{2\epsilon}}*\Delta \phi$ $\endgroup$
    – BeastRaban
    Jan 12, 2017 at 7:57
  • $\begingroup$ What is $H$ in your previous comment ? The Hubble expansion function $H(t) = \dot{a} / a$ ? $\endgroup$
    – Cham
    Jan 12, 2017 at 13:40
  • $\begingroup$ Yes, $H\equiv\frac{\dot{a}}{a}$ is the hubble parameter, $\epsilon_H$ is the first slow roll parameter $\epsilon_H\equiv -\frac{\dot{H}}{H^2}$, and $\epsilon_V\equiv\frac{1}{2}\left(\frac{V'}{V}\right)^2$ is the potential and derivative style definition for the first slow roll parameter. These two do not coincide perfectly but in quadratic inflation this is good enough. $\endgroup$
    – BeastRaban
    Jan 13, 2017 at 8:45

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