Phase diagram method

Question

I was trying to find the famous attractor solution of the inflaton field which follows the equation

$$\frac{d\dot{\phi}}{d\phi}=-\frac{\sqrt{12\pi}(\dot{\phi}^2+m^2\phi^2)^{1/2}\dot{\phi}+m^2\phi}{\dot{\phi}}$$

in ''Physical Foundations of Cosmology by Viatcheslav Mukhanov'' the author claims the equation can be studied using the phase diagram method and the behavior of the solutions in the $\phi$-$\dot{\phi}$ plane is shown in Figure 5.3

How does one reached such a plot? I tried solving the differential equation using Wolfram Mathematica but it couldn't generate a single point.

Answer 1

Just write it as a vector field instead of a line-field: \begin{align} &\frac{d\phi}{dt} = \dot{\phi}\\ &\frac{d\dot{\phi}}{dt} = - \,m^2\phi \, - \, \sqrt{12\pi(\dot{\phi}^2+m^2\phi^2)\,}\, \dot{\phi} \end{align} and run something simple like Runge-Kutta. I basically, renamed the variables $x = \phi$ and $y = \dot{\phi}$ and rewrote the renamed vector field: \begin{align} &\frac{dx}{dt} = y\\ &\frac{dy}{dt} = - \,m^2\,x \, - \, \sqrt{12\pi(y^2+m^2x^2)\,}\, y \end{align} In Wolfram alpha (online Mathematica), I set up m = 3, then I just typed

StreamPlot[{y; - 3^2*x - sqrt(12*pi*(y^2 + 3^2*x^2))*y}, {x, - pi/8, pi/8}, {y, -5/sqrt(12*pi), 5/sqrt(12*pi)}]

and got the following picture:

Answer 2

with :

$\frac{d}{d\varphi}\left(\frac{d\varphi}{dt}\right)=\frac{d}{dt}$

thus:

$\frac{d}{dt}\left(\frac{d\varphi}{dt}\right)=\frac{d^2\varphi}{dt^2}$

you get this differential equation:

$${\frac {d^{2}}{d{t}^{2}}}\varphi \left( t \right)+ \sqrt {12\,\pi }\,\pm\,\sqrt { \left( {\frac {d}{dt}}\varphi \left( t \right) \right) ^{2}+{m}^{2} \left( \varphi \left( t \right) \right) ^{2}}{ \frac {d}{dt}}\varphi \left( t \right) +{m}^{2}\varphi \left( t \right) =0$$ because the square-root function you have two solution, one for plus square-root and one for minus square-root.

there is no analytical solution but you can solve it numerically

Maple code