I am told that equation for a scalar field is:
$$\ddot{\phi}+3 H \dot{\phi}+V^{\prime}(\phi)=0$$
where $H$ is defined as: $H=\dot{R} / R$
For a potential $V(\phi)=V_{0} \exp (-\lambda \phi)$, I am supposed to show that:
$$R(t)=R_{0}\left(\frac{t}{t_{0}}\right)^{\frac{2}{\lambda^{2}}}$$
and that
$$\phi(t)=\frac{2}{\lambda} \ln \left[\left(\frac{V_{0}}{2\left(6-\lambda^{2}\right)}\right)^{1 / 2} \lambda^{2} t\right]$$
I can verify that these solutions indeed satisfy the equation. How would I go about solving it directly though? (I am happy with complete solutions, but hints are also appreciated.)
Verification of the provided solution
Consider:
$$\dot{\phi}(t)=\frac{2}{\lambda} t^{-1}$$
$$\ddot{\phi}(t)=-\frac{2}{\lambda} t^{-2}$$
$$\dot{R}(t)=R_{0}\left(\frac{1}{t_{0}}\right)^{\frac{2}{\lambda^{2}}} \frac{2}{\lambda^{2}} t^{\frac{2}{\lambda^{2}}-1}$$
$$\frac{\dot{R}}{R} \dot{\phi}=\frac{R_{0}\left(\frac{1}{t_{0}}\right)^{\frac{2}{\lambda^2}} \frac{2}{\lambda^{2}} t^{\frac{2}{\lambda^{2}}-1}}{R_{0}\left(\frac{t}{t_{0}}\right)^{\frac{2}{\lambda^{2}}}} \frac{2}{\lambda} t^{-1}=\frac{4}{\lambda^{3}} t^{-2}$$
Then we can substitute into original equation:
$$\ddot{\phi}+\frac{3 \dot{R}}{R} \dot{\phi}-\lambda V_{0} e^{-\lambda \phi}=$$
$$=-\frac{2}{\lambda} t^{-2}+3 \frac{4}{\lambda^{3}} t^{-2}-\lambda \operatorname{V_0exp}\left(-\lambda \frac{2}{\lambda} \ln \left[\left(\frac{V_{0}}{2\left(6-\lambda^{2}\right)}\right)^{1 / 2} \lambda^{2} t\right]\right)$$
$$=-\frac{2}{\lambda} t^{-2}+\frac{12}{\lambda^{3}} t^{-2}-\lambda V_{0}\left[\left(\frac{V_{0}}{2\left(6-\lambda^{2}\right)}
\right)^{\frac{1}{2}} \lambda^{2} t\right]^{-2}$$
$$=-\frac{2}{\lambda} t^{-2}+\frac{12}{\lambda^{3}} t^{-2}-\lambda V_{0}\left(
\frac{V_{0}}{2\left(6-\lambda^{2}\right)}
\right)^{-1} \lambda^{-4} t^{-2}$$
$$=-\frac{2}{\lambda} t^{-2}+\frac{2 \lambda^{2}}{\lambda^{3}} t^{-2}=0$$
as wanted.
My thoughts:
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I think I would be able to solve for $\phi$ if $H$ was constant. Would probably resort to writing $\phi$ as $\phi=\sum a_n x^n$ & then find $a_n$ coefficients, which is probably not the best way but I think it would work.
-
I am not used to solve ODEs where I have two functions to worry about ($\phi$ & $R$). Is this ODE a certain class of ODE which I should look up?
Best Answer
You are missing one key equation to close the system.
A scalar field carries energy and its this energy that determines how fast the Universe expands. For a canonical scalar field the kinetic energy is $\frac{1}{2}\partial_\mu \phi\partial^\mu\phi$ which is just $\frac{1}{2}\dot{\phi}^2$ in a smooth Friedmann universe. The potential energy is $V(\phi)$ so the Friedmann equation (in Planck units: $8\pi G = 1$) is $$H^2 = \left(\frac{d\log R}{dt}\right)^2 = \frac{1}{3} \rho_\phi = \frac{1}{3}\left(\frac{1}{2}\dot{\phi}^2 + V(\phi)\right)$$ when the scalar field is the only source of energy. Thus the full system reads $$H^2 = \frac{1}{3} \rho_\phi = \frac{1}{3}\left(\frac{1}{2}\dot{\phi}^2 + V(\phi)\right)$$ $$\ddot{\phi} + 3H\dot{\phi} + V'(\phi) = 0$$ Only for very special classes of potentials do we have analytical solutions. A general method of studying such systems is to write them as a coupled set of first order equations - a so-called dynamical system.
If we introduce $x_1 = \frac{\dot{\phi}}{\sqrt{6}H}$ and $x_2 = \frac{\sqrt{V}}{\sqrt{3}H}$ then the Hubble equation says that $x_1^2 + x_2^2 = 1$. We further find with $x = \log R$ as our time-coordinate: $$\frac{d x_1}{dx} = -3x_1 + \frac{\sqrt{6}}{2}\lambda_\phi x_2^2 - x_1 \frac{d\log H}{dx}$$ $$\frac{d x_2}{dx} = -\frac{\sqrt{6}}{2}\lambda_\phi x_1x_2 - x_2 \frac{d\log H}{dx}$$ $$\frac{d\lambda_\phi}{dx} = -\sqrt{6}\lambda_\phi^2(\Gamma-1)x_1$$ where $$\frac{d\log H}{dx} = -\frac{3}{2}(1+x_1^2 - x_2^2)$$ $$\lambda_\phi = -\frac{V'}{V},\,\,\,\Gamma = \frac{V''V}{V'^2}$$ Now for an exponential potential $\lambda_\phi = \lambda$ stays constant and $\Gamma = 1$ and the system simplifies greatly down to $$\frac{d x_1}{dx} = -3x_2^2[x_1 - \frac{\lambda}{\sqrt{6}}]$$ $$\frac{d x_2}{dx} = 3x_1x_2[x_1 - \frac{\lambda}{\sqrt{6}}]$$ or simply $$\frac{d x_1}{dx} = -3(1-x_1^2)[x_1 - \frac{\lambda}{\sqrt{6}}]$$ This is possible to integrate though its hard to invert the resulting function, but we see right away that if $x_1 \equiv \frac{\lambda}{\sqrt{6}}$ then we have one solution. This is a so-called fixpoint of this system (put derivatives to zero and solve the resulting algebraic system). Finding and classifying such fixpoints (i.e. are they attractive or repellent) is super useful (much more than trying to find analytical solutions that might not exist) for understanding the dynamics of the system and the given analytical solution is such a fixpoint. We can also see that the fixpoint is attractive so even if we did not start with this initial value of $x_1$ we would be driven towards it and the system would settle close to it.
If you now unwrap this solution you will find the given solution. For example $\frac{d\log H}{dx} = -3x_1^2 = -\frac{\lambda^2}{2}$ gives us $H \propto R^{-\frac{\lambda^2}{2}}$ and $H = \frac{d\log R}{dt}$ gives us $R(t)$ and from the definition of $x_1$ we get $\dot{\phi} \propto H \implies \phi \propto \log R$.