Difference between revisions of "Different Models of Inflation"

Use the result of the previous problem to obtain for the power-law potential $V\left( \varphi \right) = \lambda \varphi ^n $ and $$ \varphi _i \gg \varphi _e $$ $$ N \sim \int_{\varphi _e }^{\varphi _i } \frac{\varphi d\varphi }{M_{Pl}^2 } \sim \frac{\varphi _i^2 }{M_{Pl}^2 }. $$ As $V(\varphi _i ) \sim M_{Pl}^4 $, in the considered model $\varphi _i \sim \lambda ^{ - 1/4} M_{Pl} $ and therefore $$ N \sim 10^5. $$

$$ V\left( {\varphi _0 } \right) \sim M_{Pl}^4 $$ For the considered potential $$ \phi _b \sim \lambda ^{ - 1/4} M_{Pl} \gg M_{Pl} $$ inflation will continue until the scalar field's amplitude decreases to a value of the order of Planck mass (see previous problem). Thus inflation period corresponds to the following interval of the scalar field's values $$ \lambda ^{ - 1/4} M_{Pl} < \phi < M_{Pl}. $$

For great values of scalar field $\varphi $ ($\varphi > M_{Pl} $), when the inflation process is possible, the energy density can still remain less than the Planck value, $M_{Pl}^4 $. Consider for example the potential $V\left( \varphi \right) = \lambda \varphi ^4 $, where $\lambda $ is a dimensionless coupling constant. Under the condition $\lambda \ll 1$ the energy density at $\varphi \sim M_{Pl} $ is less than the Planck one $V(\phi \sim M_{Pl} ) \sim \lambda M_{Pl}^4 \ll M_{Pl}^4.$

The condition $\left| \varphi \right| \gg M_{Pl} $ guarantees realization of the slow-roll regime, then $$ a(\varphi ) \simeq a_0 \exp \left( {8\pi G\int_\varphi ^{\varphi _0 } {\frac{V}[[:Template:V'(\varphi )]]d\varphi } } \right). $$ Therefore for any potential $V(\varphi ) = \left( {1/n} \right)\lambda \varphi ^n $ one obtains $$ a(\varphi (t)) \simeq a_0 \exp \left( {4\pi G/n\left( {\varphi _0^2 - \varphi ^2 (t)} \right)} \right). $$

Neglecting the expansion, represent the equation for scalar field $$ \ddot \varphi + 3H\dot \varphi + V'_\varphi = 0 $$ in the form $$ \frac{d}{dt}\left( {\varphi \dot \varphi } \right) - \dot \varphi ^2 + \varphi V'_\varphi = 0. $$ After averaging over the period of oscillations the first term turns to zero and therefore $$ \left\langle {\dot \varphi ^2 } \right\rangle \simeq \left\langle {\varphi V'_\varphi } \right\rangle . $$ The effective (averaged) equation of state reads $$ w \equiv \frac{p}{\rho } \simeq \frac{{\left\langle {\varphi V'_\varphi } \right\rangle - \left\langle {2V} \right\rangle }}{{\left\langle {\varphi V'_\varphi } \right\rangle + \left\langle {2V} \right\rangle }}. $$

Using the result of previous problem, for $V \propto \varphi ^n$ one obtains $$ w \simeq \frac{n - 2}{n + 2}. $$ The case $n = 2$ corresponds to $w = 0$ (non-relativistic matter), and $w = 1/3$ - to $n = 4$ (radiation).

Let the potential have a minimum at $\varphi = 0$ and in the vicinity of the minimum $$ V(\varphi ) = \frac{m^2 \varphi ^2 }{2}. $$ The equation of motion for the scalar field in this potential reads $$ \ddot \varphi + 3H\dot \varphi + m^2 \varphi = 0. $$ Substitution $\varphi (t) = a^{ - 3/2} \chi (t)$ makes transition to an oscillator with variable frequancy and enables us to get rid of the first derivative $$ \ddot \chi + \left( {m^2 - \frac{3}{2}\frac{\ddot a}{a} - \frac{3}{4}\frac{\dot a^2 }{a^2 }} \right)\chi = 0. $$ In the regime of fast oscillations ($m^2 \gg H^2$) $$ \chi (t) = A\cos \left( {mt + \alpha } \right) $$ and the scalar field approaches the minimum as $$ \varphi(t) = Ca^{ - 3/2}\cos (mt + \alpha ). $$

Use the result of the previous problem to obtain for the power-law potential $V\left( \varphi \right) = \lambda \varphi ^n $ and $$ \varphi _i \gg \varphi _e $$ $$ N \sim \int_{\varphi _e }^{\varphi _i } \frac{\varphi d\varphi }{M_{Pl}^2 } \sim \frac{\varphi _i^2 }{M_{Pl}^2 }. $$ As $V(\varphi _i ) \sim M_{Pl}^4 $, in the considered model $\varphi _i \sim \lambda ^{ - 1/4} M_{Pl} $ and therefore $$ N \sim 10^5. $$

$$ V\left( {\varphi _0 } \right) \sim M_{Pl}^4 $$ For the considered potential $$ \phi _b \sim \lambda ^{ - 1/4} M_{Pl} \gg M_{Pl} $$ inflation will continue until the scalar field's amplitude decreases to a value of the order of Planck mass (see previous problem). Thus inflation period corresponds to the following interval of the scalar field's values $$ \lambda ^{ - 1/4} M_{Pl} < \phi < M_{Pl}. $$

For great values of scalar field $\varphi $ ($\varphi > M_{Pl} $), when the inflation process is possible, the energy density can still remain less than the Planck value, $M_{Pl}^4 $. Consider for example the potential $V\left( \varphi \right) = \lambda \varphi ^4 $, where $\lambda $ is a dimensionless coupling constant. Under the condition $\lambda \ll 1$ the energy density at $\varphi \sim M_{Pl} $ is less than the Planck one $V(\phi \sim M_{Pl} ) \sim \lambda M_{Pl}^4 \ll M_{Pl}^4.$

The condition $\left| \varphi \right| \gg M_{Pl} $ guarantees realization of the slow-roll regime, then $$ a(\varphi ) \simeq a_0 \exp \left( {8\pi G\int_\varphi ^{\varphi _0 } {\frac{V}[[:Template:V'(\varphi )]]d\varphi } } \right). $$ Therefore for any potential $V(\varphi ) = \left( {1/n} \right)\lambda \varphi ^n $ one obtains $$ a(\varphi (t)) \simeq a_0 \exp \left( {4\pi G/n\left( {\varphi _0^2 - \varphi ^2 (t)} \right)} \right). $$

Neglecting the expansion, represent the equation for scalar field $$ \ddot \varphi + 3H\dot \varphi + V'_\varphi = 0 $$ in the form $$ \frac{d}{dt}\left( {\varphi \dot \varphi } \right) - \dot \varphi ^2 + \varphi V'_\varphi = 0. $$ After averaging over the period of oscillations the first term turns to zero and therefore $$ \left\langle {\dot \varphi ^2 } \right\rangle \simeq \left\langle {\varphi V'_\varphi } \right\rangle . $$ The effective (averaged) equation of state reads $$ w \equiv \frac{p}{\rho } \simeq \frac{{\left\langle {\varphi V'_\varphi } \right\rangle - \left\langle {2V} \right\rangle }}{{\left\langle {\varphi V'_\varphi } \right\rangle + \left\langle {2V} \right\rangle }}. $$

Using the result of previous problem, for $V \propto \varphi ^n$ one obtains $$ w \simeq \frac{n - 2}{n + 2}. $$ The case $n = 2$ corresponds to $w = 0$ (non-relativistic matter), and $w = 1/3$ - to $n = 4$ (radiation).

Let the potential have a minimum at $\varphi = 0$ and in the vicinity of the minimum $$ V(\varphi ) = \frac{m^2 \varphi ^2 }{2}. $$ The equation of motion for the scalar field in this potential reads $$ \ddot \varphi + 3H\dot \varphi + m^2 \varphi = 0. $$ Substitution $\varphi (t) = a^{ - 3/2} \chi (t)$ makes transition to an oscillator with variable frequancy and enables us to get rid of the first derivative $$ \ddot \chi + \left( {m^2 - \frac{3}{2}\frac{\ddot a}{a} - \frac{3}{4}\frac{\dot a^2 }{a^2 }} \right)\chi = 0. $$ In the regime of fast oscillations ($m^2 \gg H^2$) $$ \chi (t) = A\cos \left( {mt + \alpha } \right) $$ and the scalar field approaches the minimum as $$ \varphi(t) = Ca^{ - 3/2}\cos (mt + \alpha ). $$