Cosmological Inflation: The Canonic Theory - Revision history

Cherkaskiy: /* Problem 27 */

2013-08-27T10:06:26Z

‎Problem 27

Cherkaskiy: /* Problem 16 */

2013-08-27T09:52:00Z

‎Problem 16

Cherkaskiy: /* Problem 16 */

2013-08-27T09:51:08Z

‎Problem 16

Cherkaskiy: /* Problem 16 */

2013-08-27T09:50:32Z

‎Problem 16

Cherkaskiy: /* Problem 16 */

2013-08-27T09:49:49Z

‎Problem 16

Cherkaskiy: /* Problem 16 */

2013-08-27T09:42:11Z

‎Problem 16

Cherkaskiy: /* Problem 16 */

2013-08-27T09:40:27Z

‎Problem 16

Cherkaskiy: /* Problem 16 */

2013-08-27T09:39:33Z

‎Problem 16

Cherkaskiy: /* Scalar Field In Cosmology */

2013-08-27T09:36:45Z

‎Scalar Field In Cosmology

Cherkaskiy: /* Scalar Field In Cosmology */

2013-08-27T09:31:03Z

‎Scalar Field In Cosmology

@@ Line 708: / Line 708: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">For the case of homogeneous and isotropic Universe with scalar field coordinates one gets
+     <p style="text-align: left;">For the case of homogeneous and isotropic Universe with scalar field one gets
 $$
 \begin{array}{l}

@@ Line 503: / Line 503: @@
 \end{array}
 $$
-Substitute it into \ref{entropy_2} to obtain the equation for scalar field ïîëÿ
+Substitute it into \ref{entropy_2} to obtain the equation for scalar field
 $$
 \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.

@@ Line 493: / Line 493: @@
 \begin{array}{l}
 dE = Vd\rho + \rho dV,\\
-Vd\rho + (\rho + p)dV = 0.\label{entropy2}
+Vd\rho + (\rho + p)dV = 0.\label{entropy_2}
 \end{array}
 \end{equation}
@@ Line 503: / Line 503: @@
 \end{array}
 $$
-Substitute it into \ref{entropy2} to obtain the equation for scalar field ïîëÿ
+Substitute it into \ref{entropy_2} to obtain the equation for scalar field ïîëÿ
 $$
 \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.

@@ Line 493: / Line 493: @@
 \begin{array}{l}
 dE = Vd\rho + \rho dV,\\
-Vd\rho + (\rho + p)dV = 0.\label{entropy}
+Vd\rho + (\rho + p)dV = 0.\label{entropy2}
 \end{array}
 \end{equation}
@@ Line 503: / Line 503: @@
 \end{array}
 $$
-Substitute it into \ref{entropy} to obtain the equation for scalar field ïîëÿ
+Substitute it into \ref{entropy2} to obtain the equation for scalar field ïîëÿ
 $$
 \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.

@@ Line 490: / Line 490: @@
 $$
 one obtains
-\begin{eqnarray}{l}\nonumber
+\begin{equation}
 dE = Vd\rho + \rho dV,\\
 Vd\rho + (\rho + p)dV = 0.\label{entropy}
 \end{array}
 Use the definition for $\rho $ and $p$ to obtain
 $$

@@ Line 364: / Line 364: @@
 \end{array}
 $$
-Direct using of these expressions enables us to transform the initial system into the following
+Then
 $$
 \begin{array}{l}
@@ Line 373: / Line 373: @@
 \end{array}
 $$
-The prime denotes differentiation with respect to conformal time.</p>
+where the prime denotes differentiation with respect to conformal time.</p>
    </div>
 </div></div>
@@ Line 389: / Line 389: @@
      <p style="text-align: left;">$$
 \begin{array}{l}
-\displaystyle H = \frac{\dot a}{a};~\dot H = \frac{\ddot aa - \dot a^2 }{a^2 } = \frac{\ddot a}{a} - H^2,\\
+\displaystyle H = \frac{\dot a}{a};\quad \dot H = \frac{\ddot aa - \dot a^2 }{a^2 } = \frac{\ddot a}{a} - H^2,\\
-\displaystyle H^2 = \frac{8\pi G}{3}\rho ;\;\frac{\ddot a}{a} = - \frac{4\pi G}{3}(\rho + 3p), \\
+\displaystyle H^2 = \frac{8\pi G}{3}\rho ;\quad \frac{\ddot a}{a} = - \frac{4\pi G}{3}(\rho + 3p), \\
-\displaystyle \dot H > 0 \to p < - \rho. \\
+\displaystyle \dot H > 0 \Rightarrow p < - \rho. \\
 \end{array}
 $$
-Satisfaction of the latter condition is impossible for a scalar field with positively defined kinetic energy.
+The latter condition is impossible to satisfy for a scalar field with positively defined kinetic energy.
 </p>
    </div>
@@ Line 442: / Line 442: @@
 $$
 x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}},
-~w_{\varphi}=\frac{\dot{\varphi}^2+2V(\varphi)}{\dot{\varphi}^2-2V(\varphi)},
+\quad w_{\varphi}=\frac{\dot{\varphi}^2+2V(\varphi)}{\dot{\varphi}^2-2V(\varphi)},
 $$ in the system of units such that $8\pi G=1.$
 <div class="NavFrame collapsed">
@@ Line 462: / Line 462: @@
   \frac{V_{,\varphi}}{V} =-6\frac{H}{\dot{\varphi}}\left(\frac{x_{\varphi}}{1+ x_{\varphi}}\right)\left[1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right].
 $$
-As \[\frac{x_{\varphi}}{1+ x_{\varphi}} = \frac{1+ w_{\varphi}}{2}\] then
+As \[\frac{x_{\varphi}}{1+ x_{\varphi}} = \frac{1+ w_{\varphi}}{2},\] we have
 $$
   \frac{V_{,\varphi}}{V} =-3(1+w_{\varphi})\frac{H}{\dot{\varphi}}\left[1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right].
@@ Line 468: / Line 468: @@
 Note that $\rho_\varphi=\frac{\dot{\varphi}^2}{2}+V(\varphi)= (1+x_\varphi)=\frac{2V(\varphi)}{1-w_\varphi};~ H^2=\frac 13 \rho_{cr},~\dot{\varphi} = \pm\sqrt{2Vx_\varphi}$ and therefore
 $$
-\Omega_\varphi = \frac{\rho_\varphi}{\rho_{cr}} = \frac{2}{3}\frac{V}{H^2(1-w_\varphi)},
+\begin{array}{l}
-$$
+\Omega_\varphi = \frac{\rho_\varphi}{\rho_{cr}} = \frac{2}{3}\frac{V}{H^2(1-w_\varphi)},\\
-$$
 \pm \frac{V_{,\varphi}}{V} = \sqrt {\frac{3(1 + w)}{\Omega _\varphi(a)}} \left[ 1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right].
 $$</p>
    </div>

@@ Line 246: / Line 246: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">From the equation $\dot{H}=-4\pi G(p+\rho),$ obtained in the problem \ref{equ35} of Chapter 2, one finds that $\dot{H}=-4\pi G\dot{\varphi}^2$. Then one obtains
+     <p style="text-align: left;">From the equation $\dot{H}=-4\pi G(p+\rho),$ obtained in the problem \ref{equ35} of Chapter 2, one finds that $\dot{H}=-4\pi G\dot{\varphi}^2$. Then
 $$V=\frac{3H^2}{8\pi G}\left(1+\frac{\dot{H}}{3H^2}\right)$$ $$\varphi = \int dt \left( -\frac{\dot{H}}{4\pi G}\right)^{1/2}.$$</p>
    </div>
 </div></div>
 <div id="inf14"></div>
@@ Line 263: / Line 260: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">The first equation is precisely the first Friedman equation with inserted energy density of scalar field. To obtain the second equation differentiate with respect to time the definition of hubble parameter:
+     <p style="text-align: left;">The first equation is precisely the first Friedman equation with the energy density of scalar field. To obtain the second equation differentiate with respect to time the definition of Hubble parameter:
 $$\displaystyle \begin{array}{l}
 \displaystyle H = \frac{\dot a}{a};\quad \dot H = \frac{\ddot aa - \dot a^2 }{a^2 } = \frac{\ddot a}{a} - \frac{\dot a^2 }{a^2 } = \\
@@ Line 318: / Line 315: @@
 Substitute it into the first equation of the system to obtain
 $$
-H'^2 - 12\pi GH^2 = - 32\pi ^2 G^2 V\left( \varphi \right)
+H'^2 - 12\pi GH^2 = - 32\pi ^2 G^2 V\left( \varphi \right).
 $$
 The system of first-order equations
@@ Line 350: / Line 347: @@
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 12 ===
-Express the equations for the scalar field in terms of conformal time.
+Rewrite the equations for the scalar field in terms of conformal time.
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
@@ Line 359: / Line 356: @@
 $$
 $$
 \begin{array}{l}
 \displaystyle dt = ad\eta;~\frac{d}{dt} = \frac{d\eta }{dt}\frac{1}{d\eta } = \frac{1}{a}\frac{1}{d\eta },\\
 \displaystyle \dot \varphi = \frac{1}{a}\frac{d\varphi }{d\eta } = \frac{1}{a}\varphi'. \\