Cosmo All at 01:24, 23 May 2014

2014-05-23T01:24:06Z

Cosmo All: Created page with "B In a row of the problems below [following the paper Marco A. Reyes, On exact solutions to the scalar field equations in standard cosmology, arXiv:..."

2014-05-23T01:23:26Z

Created page with "B In a row of the problems below [following the paper Marco A. Reyes, On exact solutions to the scalar field equations in standard cosmology, arXiv:..."

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[[Category:Dark Energy|B]]

In a row of the problems below [following the paper Marco A. Reyes, On exact solutions to the scalar field equations in standard cosmology, arXiv: 0806.2292] we present a simple algebraic method to find exact solutions for a wide variety of scalar field potentials. Let us consider the function $V_a(\varphi)$ defined as
\begin{equation}\label{es_1}
V_a(\varphi)\equiv V(\varphi)+\frac12\dot\varphi^2.
\end{equation}
Derivative of this function reads
\[\frac{dV_a}{d\varphi}=\frac{dV}{d\varphi}+\ddot\varphi.\]
Hence, equations
\[H^2=\frac12\left(\frac12\dot\varphi^2+V(\varphi)\right),\]
\[\ddot\varphi+3H\dot\varphi+\frac{dV}{d\varphi}=0\]
can now be rewritten as
\begin{align}
\label{es_4}3H^2 & =V_a,\\
\label{es_5}3H\dot\varphi & =-\frac{dV_a}{d\varphi}.
\end{align}
To solve them, note that eq.(\ref{es_4}) defines $H$ as a function of $\varphi$, which when inserted into eq.(\ref{es_5}), gives the scalar field $\varphi(t)$ as a function of $t$, at least in quadratures
\[-3H(\varphi)\left(\frac{dV_a}{d\varphi}\right)^{-1}d\varphi=dt.\]
Finally, inserting $\varphi(t)$ into eqs.(\ref{es_1}) and (\ref{es_4}) gives $V(\varphi)$ and $a(t)$, respectively, and the solution is completed.

One could also use $H(t)$ to determine $\varphi(t)$, since
\[\dot H=-\frac12\dot\varphi^2.\]
implies that
\[\Delta\varphi(t)=\pm\int\sqrt{-2\dot H}dt.\]
Since $V_a(t)=3H^2(t)$, a complete knowledge of $H(t)$ fully determines the solution to the problem.

<div id=""></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 1 ===
<p style= "color: #999;font-size: 11px">problem id: </p>
For $H(t)=\alpha/t$ find $V(\varphi)$ and $\Delta\varphi(t)$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">\[V(\varphi)=(3\alpha^2-\alpha)\exp[-2\Delta\varphi/\sqrt{2\alpha}];\]
\[\Delta\varphi(t)=\sqrt{2\alpha}\ln t.\]</p>
</div>
</div></div>

<div id=""></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
<p style= "color: #999;font-size: 11px">problem id: </p>
Reconstruct $V(\varphi)$ and find $\varphi(t)$ and $a(t)$ for $V_a=\lambda\varphi^2$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In this case
\[H^2=\frac13\lambda\varphi^2,\quad \frac{dV_a}{d\varphi}=2\lambda\varphi.\]
Therefore, from
\[-3H(\varphi)\left(\frac{dV_a}{d\varphi}\right)^{-1}d\varphi=dt.\]
one finds that
\[\Delta\varphi(t)=\pm2\sqrt{\frac\lambda3}\Delta t.\]
Hence, $\dot\varphi$ is constant. Letting $\varphi(t_0=0)=0$, and using eqs. (\ref{es_1}) and (\ref{es_4}) we get
\begin{align}
\nonumber V(\varphi) & = \lambda\varphi^2-\frac23\lambda,\\
\nonumber a(t) & =a_0e^{-\frac\lambda3t^2}.
\end{align}
Obviously, one would be tempted to pick $\lambda<0$ in order to make $a(t)$ a growing function of $t$, but that would make $\varphi(t)$ an imaginary function of $t$.</p>
</div>
</div></div>

<div id=""></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
<p style= "color: #999;font-size: 11px">problem id: </p>
Reconstruct $V(\varphi)$ and find $\varphi(t)$ and $a(t)$ for $V_a=\lambda\varphi^4$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Proceeding the same way as in the previous problem one finds
\begin{align}
\nonumber \varphi(t) & =\varphi_0e^{\pm4\sqrt{\frac\lambda3}t},\\
\nonumber V(\varphi) & = \lambda\varphi^4-\frac83\lambda\varphi^2,\\
\nonumber a(t) & =a_0\exp\left[-\frac{\varphi_0^2}8e^{\pm8\sqrt{\frac\lambda3}(t-t_0)}\right].
\end{align}</p>
</div>
</div></div>

<div id=""></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 4 ===
<p style= "color: #999;font-size: 11px">problem id: </p>
Reconstruct $V(\varphi)$ and find $\varphi(t)$ and $a(t)$ for $V_a=\lambda\varphi^n$, $n>2$.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Proceeding the same way as above one finds
\begin{align}
\nonumber \varphi(t) & =\left[\varphi_0^{2-n}\pm2n(n-2)\sqrt{\frac\lambda3}(t-t_0)\right]^{-\frac1{n-2}};\\
\nonumber V(\varphi) & = \lambda\varphi^{2n}-\frac23\lambda n^2\varphi^{2(n-1)};\\
\nonumber a(t) & =a_0\exp\left\{-\frac1{4n}\left[\varphi_0^{2-n}\pm2n(n-2)\sqrt{\frac\lambda3}(t-t_0)\right]^{-\frac2{n-2}}\right\}.
\end{align}</p>
</div>
</div></div>

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 [[Category:Dark Energy|B]]
+__NOTOC__
 In a row of the problems below [following the paper Marco A. Reyes, On exact solutions to the scalar field equations in standard cosmology, arXiv: 0806.2292] we present a simple algebraic method to find exact solutions for a wide variety of scalar field potentials. Let us consider the function $V_a(\varphi)$ defined as
 \begin{equation}\label{es_1}

Exact Solutions - Revision history

Cosmo All at 01:24, 23 May 2014

Cosmo All: Created page with "B In a row of the problems below [following the paper Marco A. Reyes, On exact solutions to the scalar field equations in standard cosmology, arXiv:..."