Single Scalar Cosmology - Revision history

Cosmo All at 21:56, 18 June 2015

2015-06-18T21:56:24Z

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Cosmo All at 01:44, 23 April 2014

2014-04-23T01:44:55Z

Cosmo All: /* Problem 19 */

2014-04-23T01:43:49Z

‎Problem 19

Cosmo All: /* Problem 19 */

2014-04-23T01:43:21Z

‎Problem 19

Cosmo All: /* Problem 32 */

2014-04-23T01:41:00Z

‎Problem 32

Cosmo All: /* Problem 32 */

2014-04-23T01:38:55Z

‎Problem 32

Cosmo All: /* Problem 31 */

2014-04-23T01:38:29Z

‎Problem 31

Cosmo All at 01:36, 23 April 2014

2014-04-23T01:36:26Z

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Cosmo All: /* Problem 22 */

2014-04-23T01:12:47Z

‎Problem 22

Cosmo All: /* Problem 22 */

2014-04-23T01:08:26Z

‎Problem 22

← Older revision		Revision as of 01:44, 23 April 2014
Line 1:		Line 1:
	[[Category:Dark Energy\|7]]		[[Category:Dark Energy\|7]]

−	~~__NOTOC__~~	+	__TOC__

@@ Line 513: / Line 513: @@
 \end{equation}
 \begin{equation}
-\dot{H}+3H^{2} &=&-p_{\phi }=-\frac{\dot{\phi}^{2}}{2}+V\left( \phi \right),
+\dot{H}+3H^{2}=-p_{\phi }=-\frac{\dot{\phi}^{2}}{2}+V\left( \phi \right),
 \label{H1}
 \end{equation}
 \begin{equation}
-\ddot{\phi}+3H\dot{\phi}+V^{\prime }\left( \phi \right) &=& 0,  \label{phi}
+\ddot{\phi}+3H\dot{\phi}+V^{\prime }\left( \phi \right) = 0,  \label{phi}
 \end{equation}
 in terms of the parameter $G(\phi)$ introduced as

@@ Line 508: / Line 508: @@
 <p style= "color: #999;font-size: 11px">problem id: ES_0</p>
 Rewrite the equations of the single scalar cosmology
-\begin{eqnarray}
+\begin{equation}
-H^{2} &=&\rho _{\phi }=\frac{\dot{\phi}^{2}}{2}+V\left( \phi \right) ,
+H^{2} =\rho _{\phi }=\frac{\dot{\phi}^{2}}{2}+V\left( \phi \right) ,
-\label{H} \\
+\label{H}
 \dot{H}+3H^{2} &=&-p_{\phi }=-\frac{\dot{\phi}^{2}}{2}+V\left( \phi \right),
-\label{H1} \\
+\label{H1}
 \ddot{\phi}+3H\dot{\phi}+V^{\prime }\left( \phi \right) &=& 0,  \label{phi}
-\end{eqnarray}%
+\end{equation}
 in terms of the parameter $G(\phi)$ introduced as
 \[\dot\phi^2=2V(\phi)\sinh^2 G(\phi).\]
@@ Line 531: / Line 535: @@
 <div id="ES_1"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 18 ===
 <p style= "color: #999;font-size: 11px">problem id: ES_1</p>

@@ Line 1,044: / Line 1,044: @@
   w(\phi )=\left[ \frac{e^{-\alpha
 _{3}\phi }}{\sqrt{V(\phi )}}\right] \left( C_{1}+\frac{e^{3\alpha _{3}\phi }}{\alpha
-_{3}}\right) ^{1/3}.</p>
+_{3}}\right) ^{1/3}.
    </div>
 </div></div>

@@ Line 1,004: / Line 1,004: @@
 === Problem 32===
 <p style= "color: #999;font-size: 11px">problem id: ES_15</p>
-Obtain exact solution of the equation (\ref{fin}) in the case $M\left( \protect\phi \right) =V^{-3/2}$.
+Obtain exact solution of the equation (\ref{fin}) in the case $M\left( \phi \right) =V^{-3/2}$.
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>

@@ Line 948: / Line 948: @@
 === Problem 31===
 <p style= "color: #999;font-size: 11px">problem id: ES_14</p>
-Obtain exact solution of the equation (\ref{fin}) in the case $M\left( \protect\phi \right) =\protect\sqrt{V}$.
+Obtain exact solution of the equation (\ref{fin}) in the case $M\left( \phi \right) =\sqrt{V}$.
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
@@ Line 1,001: / Line 1,001: @@
 <div id="ES_15"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 32===
 <p style= "color: #999;font-size: 11px">problem id: ES_15</p>

@@ Line 606: / Line 606: @@
 _{0}^{2}-1\right) }.
 \end{equation}
-<div style="border: 1px solid #AAA; padding:5px;">
 [[File:exp1.jpg|center|thumb|400px|]]
-[[File:exp2.jpg|center|thumb|400px|Depicted is the time variation of the scale factor (in arbitrary units), in the first plot, and the time variation of the cosmological  scalar field, in the second plot,  with an exponential  potential for different values of $\alpha _0$: $\alpha _0=1.5 $ (solid curve), $\alpha _0=2.5$ (dotted curve), $\alpha _0=3.5$ (short dashed curve), $\alpha _0=4.5$ (dashed curve), and $\alpha _0=5.5$ (long dashed curve), respectively. The arbitrary integration constants $\phi _0$ and $V_0$ have been normalized so that $\exp \left(-\sqrt{3/2}\alpha _0\phi _0\right)=\sqrt{3V_0}$.]]</p>
+[[File:exp2.jpg|center|thumb|400px|Depicted is the time variation of the scale factor (in arbitrary units), in the first plot, and the time variation of the cosmological  scalar field, in the second plot,  with an exponential  potential for different values of $\alpha _0$: $\alpha _0=1.5 $ (solid curve), $\alpha _0=2.5$ (dotted curve), $\alpha _0=3.5$ (short dashed curve), $\alpha _0=4.5$ (dashed curve), and $\alpha _0=5.5$ (long dashed curve), respectively. The arbitrary integration constants $\phi _0$ and $V_0$ have been normalized so that $\exp \left(-\sqrt{3/2}\alpha _0\phi _0\right)=\sqrt{3V_0}$.]]
-</div>
 <br/>
 <br/>
-<div style="border: 1px solid #AAA; padding:5px;">
 [[File:exp3.jpg|center|thumb|400px|]]
-[[File:exp4.jpg|center|thumb|400px|Plots of the time variation of the exponential  scalar field potential, depicted in the fist figure, and the time variation of the deceleration parameter of the Universe filled with an exponential potential scalar field, depicted in the second figure,  for different values of $\alpha _0$: $\alpha _0=1.5 $ (solid curve), $\alpha _0=2.5$ (dotted curve), $\alpha _0=3.5$ (short dashed curve), $\alpha _0=4.5$ (dashed curve), and $\alpha _0=5.5$ (long dashed curve), respectively. The arbitrary integration constants $\phi _0$ and $V_0$ have been normalized so that $\exp \left(-\sqrt{3/2}\alpha _0\phi _0\right)=\sqrt{3V_0}$.]]
+[[File:exp4.jpg|center|thumb|400px|Plots of the time variation of the exponential  scalar field potential, depicted in the fist figure, and the time variation of the deceleration parameter of the Universe filled with an exponential potential scalar field, depicted in the second figure,  for different values of $\alpha _0$: $\alpha _0=1.5 $ (solid curve), $\alpha _0=2.5$ (dotted curve), $\alpha _0=3.5$ (short dashed curve), $\alpha _0=4.5$ (dashed curve), and $\alpha _0=5.5$ (long dashed curve), respectively. The arbitrary integration constants $\phi _0$ and $V_0$ have been normalized so that $\exp \left(-\sqrt{3/2}\alpha _0\phi _0\right)=\sqrt{3V_0}$.]]</p>
-</div>
 </p>
    </div>
 </div></div>