I How Can I Graph Hubble Term vs. Inflaton Using Mathematica?

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To graph Hubble term H versus inflaton φ using Mathematica, it is essential to solve the differential equation involving φ(t), which requires additional equations like the second Friedmann equation. The presence of the time derivative in the equations indicates that the dynamics of the field cannot be ignored. In warm inflation scenarios, the Klein-Gordon equation is modified to include a dissipation term Γ, complicating the relationship between H, φ, and their derivatives. To explore the tensor to scalar ratio r in relation to the dissipation term, numerical calculations of H in terms of φ for various Γ values are necessary. Assistance with Mathematica for these calculations is sought, particularly for those with basic knowledge of the software.
shinobi20
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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?
 
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shinobi20 said:
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?
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.
 
Chalnoth said:
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.

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##
 
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.
 
Orodruin said:
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.
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.
 
Chalnoth said:
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.
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?
 
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Have you considered the Klein-Gordon equation?
$$ \ddot{\phi}+3H\dot{\phi}+\dfrac{dV}{d\phi}=0$$
 
MinasKar said:
Have you considered the Klein-Gordon equation?
$$ \ddot{\phi}+3H\dot{\phi}+\dfrac{dV}{d\phi}=0$$
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##.
 
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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.
 
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