# I Hubble term versus inflaton

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1. Jan 10, 2017

### shinobi20

From cosmology,

$H^2 = \frac{ρ}{3M_p^2} = \frac{1}{3M_p^2}(½\dot φ^2 + ½m^2φ^2)$

Suppose $V(φ) = ½m^2φ^2$

where
$ρ$ = density
$M_p$ = planck mass

I want to graph $H$ vs. $φ$ but there is a $\dot φ$ and I know this is a differential equation, can somebody help me what to do here?

2. Jan 10, 2017

### Chalnoth

The only way is to solve the differential equation so that you know $\phi(t)$. For that, you'll need more than just this equation, as you have an unknown function of time on both sides of the equation. You'd need to make use of a second equation, possibly the second Friedmann equation, to resolve the discrepancy.

3. Jan 11, 2017

### shinobi20

I was hoping to get rid of $\dot φ$ but it seems I can't find any relationship for that. If I'm to use the second Friedmann equation,
$\frac{\ddot a}{a} = -\frac{1}{6M_p^2}(ρ+3p)~~~~~~~~~ ^*~H = \frac{\dot a}{a}~~→~~\dot H = \frac{\ddot a}{a} - (\frac{\dot a}{a})^2~~→~~\dot H = \frac{\ddot a}{a} - H^2$

$\dot H + H = -\frac{1}{6M_p^2}(ρ+3p)$

The problem is the form of $ρ$ and $p$. For warm inflation, should I consider $ρ = ρ_λ + ρ_r$ and $p = p_λ + p_r$? Given that $p_λ = -ρ_λ$ and $p_r = \frac{1}{3}ρ_r$

4. Jan 11, 2017

### Orodruin

Staff Emeritus
You cannot get rid of the time derivative in favour of other quantities. That would remove the dynamics of the field itself.

In some cases, like slow-roll inflation, you can neglect the kinetic term in the energy, but it is still there.

5. Jan 11, 2017

### shinobi20

Yes, that's why I'd like to solve the DE exactly but there is an $H^2$ in front which is also a variable.

6. Jan 11, 2017

### shinobi20

If I define $t_H = H^{-1}$ (Hubble time) then it would be just an ODE so I could use the typical numerical calculation in Mathematica?

Last edited: Jan 11, 2017
7. Jan 11, 2017

Have you considered the Klein-Gordon equation?
$$\ddot{\phi}+3H\dot{\phi}+\dfrac{dV}{d\phi}=0$$

8. Jan 11, 2017

### shinobi20

That would be the case in typical inflationary scenario but in warm inflation KG equation would be modified to

$\ddot{\phi}+(3H + Γ)\dot{\phi}+\dfrac{dV}{d\phi}=0$

There is an extra dissipation term $Γ$, which I would also need later, so that is also a problem.
Basically, I want to find the relationship of the tensor to scalar ratio $r$ with the $Γ$ dissipation term by the theoretical result $r = 16ε$ where $ε$ is the Hubble slow roll parameter, but from the equations I can see, $H$, $φ$, and $\dot φ$ are in the way since $ε = -\frac{\dot H}{H^2}$ so I think I can numerically calculate $H$ in terms of $φ$ in order to get different values of H to again numerically calculate $r$ in terms of $H$.

Last edited: Jan 11, 2017
9. Jan 13, 2017

### shinobi20

I can only think of two equations that I can use to get the behavior of $H$ in terms of $\phi$ for various dissipation term $\Gamma$,

$$\ddot{\phi}+3H\dot{\phi}+\dfrac{dV}{d\phi}=0 ,\quad H^2 = \frac{1}{6M_p^2}(\dot φ^2 + m^2φ^2)$$

Can anyone help me figure out how can I use mathematica to solve $H$ for different $\Gamma$? I only have basic knowledge of mathematica.