Chapter 11 · Section 8 of 11
Let us consider the Einstein field equations \[R_{\mu\nu}-frac12Rg_{\mu\nu}=8\pi G\left(T_{\mu\nu}+\frac\Lambda{8\pi G}g_{\mu\nu}\right).\] According to the Bianchi identities, (i) vacuum decay is possible only from a previous existence of some sort of non-vanishing matter and/or radiation, and (ii) the presence of a time-varying cosmological term results in a coupling between $T_{\mu\nu}$ and $\Lambda$. We will assume (unless stated otherwise) coupling only between vacuum and CDM particles, so that \[u_\mu,T_{;\nu}^{(CDM)\mu\nu}=-u_\mu\left(\frac{\Lambda g^{\mu\nu}}{8\pi G}\right)_{;\nu}= -u_\mu\left(\rho_\Lambda g^{\mu\nu}\right)_{;\nu}\] where $T_{\mu\nu}^{(CDM)}=\rho_{dm}u^\mu u^\nu$ is the energy-momentum tensor of the CDM matter and $\rho_\Lambda$ is the vacuum energy density. It immediately follows that \[\dot\rho_{dm}+3H\rho_{dm}=-\dot\rho_\Lambda.\] Note that although the vacuum is decaying, $w_\Lambda=-1$ is still constant, the physical equation of state (EoS) of the vacuum $w_\Lambda\equiv=p_\Lambda/\rho_\Lambda$ is still equal to constant $-1$, which follows from the definition of the cosmological constant.
(see Can vacuum decay in our Universe?, Interpreting Cosmological Vacuum Decay)
Since vacuum energy is constantly decaying into CDM, CDM will dilute in a smaller rate compared with the standard relation $\rho_{dm}\propto a^{-3}$. Thus we assume that $\rho_{dm}=\rho_{dm0}a^{-3+\varepsilon}$, where $\varepsilon$ is a small positive small constant. Find the dependence $\rho_\Lambda(a)$ in this model.
Conservation equation for CDM yields \[\rho_\Lambda=\bar\rho_{\Lambda0}+\frac{\varepsilon\rho_{dm0}}{3-\varepsilon}a^{-3+\varepsilon}.\]
Solve the previous problem for the case when vacuum energy is constantly decaying into radiation.
By considering that radiation will dilute more slowly compared to its standard evolution, $\rho_r\propto a^{-4}$, and that such a deviation is characterized by a positive constant $\alpha$ one finds \[\rho_r=\rho_{r0}a^{-4+\alpha}.\] By inserting this expression into radiation conservation equation \[\dot\rho_r+4H\rho_r=-\dot\rho_\Lambda,\] one obtains \[\rho_\Lambda=\bar\rho_{\Lambda0}+\frac{\alpha\rho_{r0}}{4-\alpha}a^{-4+\alpha}.\]
Show that existence of a radiation dominated stage is always guaranteed in scenarios, considered in the previous problem.
The ratio between the vacuum and radiation energy densities reads \[\frac{\rho_\Lambda}{\rho_r}=\frac{\bar\rho_{\Lambda0}}{\rho_{r0}}a^{4-\alpha}+\frac\alpha{4-\alpha}.\] The first term is asymptotically vanishing at early times whereas the second one is smaller than unity. Therefore, a radiation dominated stage is always guaranteed in this kind of scenarios.
Find how the new temperature law scales with redshift in the case of vacuum energy decaying into radiation.
For an adiabatic vacuum decay the equilibrium relations are preserved, as happens with the Stefan law, $\rho_r\propto T^4$. As a consequence, one may check that the $Ta^{1-\alpha/4}$. This implies that the new temperature law scales with redshift as \[T=T_0(1+z)^{1-\alpha/4}.\]
Since the energy density of the cold dark matter is $\rho_{dm}=nm$, there are two possibilities for storage of the energy received from the vacuum decay process:
(i) the equation describing concentration, $n$, has a source term while the proper mass of CDM particles remains constant;
(ii) the mass $m$ of the CDM particles is itself a time-dependent quantity, while the total number of CDM particles, $N=na^3$, remains constant.
Let us consider both the possibilities.
Find dependence of total particle number on the scale factor in the model considered in problem #IDE_78.
Since $\rho_{dm}=nm$, we find, using $\rho_{dm}=\rho_{dm0}a^{-3+\varepsilon}$, that \[n=n_0a^{-3+\varepsilon}.\] In the considered case, there is necessarily a source term in the current of CDM particles, that is, $N^\alpha_{;\alpha}$, where $N^\alpha=nu^\alpha$. In terms of the concentration it can be written as \[\dot n +3Hn=\psi.\] Inserting $n=n_0a^{-3+\varepsilon}$, one obtains \[\psi=\varepsilon\frac{\dot a}an\equiv n\Gamma.\] We have written particle source $\psi$ in terms of a decay rate, $\Gamma\equiv\dot N/N$.
Find time dependence of CDM particle mass in the case when there is no creation of CDM particles in the model considered in problem #IDE_78.
In this case the concentration of CDM particles satisfies the equation \[\dot n +3Hn=0,\] whose solution is $n=n_0a^{-3}$, which implies that $N(t)=const$. Naturally, if CDM particles are not being created, the unique possibility is an increasing in the proper mass of CDM particles. Actually, since $\rho_{dm0}=nm$, expressions \[\rho_{dm}=\rho_{dm0}a^{-3+\varepsilon}\] and \[n=n_0a^{-3}\] imply that the mass of the CDM particles scales as \[m(t)=m_0a(t)^\varepsilon,\] where $m_0$ is the present day mass of CDM particles.
Consider a model where the cosmological constant $\Lambda$ depends on time as $\Lambda=\sigma H$. Let a flat Universe be filled by the time-dependent cosmological constant and a component with the state equation $p_\gamma= (\gamma-1)\rho_\gamma$. Find solutions of Friedman equations for this system [1].
System of equations describing dynamics of the model reads \begin{align} \nonumber \rho_\gamma+\Lambda & = 3H^2,\\ \nonumber \dot\rho_\gamma + 3H\gamma\rho_\gamma & =-\dot\Lambda,\\ \nonumber \Lambda & =\sigma H. \end{align} Combine this equations to obtain \[2\dot H+3\gamma H^2-\sigma\gamma H=0.\] For $H>0$ (expanding Universe) the equation has the following solution \[a(t)=C\left[\exp\left(\frac{\sigma\gamma t}2\right)-1\right]^{3\gamma/2}.\] Here $C$ is one of the two integration constants (the equation for scale factor is the second-order one). The second integration constant is determined from the condition $a(t=0)=0$. Using the obtained solution, one finds \begin{align} \nonumber \rho_\gamma & =\frac{\sigma^2}3\left[1+\left(\frac C a\right)^{3\gamma/2}\right]\left(\frac C a\right)^{3\gamma/2},\\ \nonumber \Lambda & =\frac{\sigma^2}3\left[1+\left(\frac C a\right)^{3\gamma/2}\right]. \end{align}
Show that the model considered in the previous problem correctly reproduces the scale factor evolution both in the radiation-dominated and non-relativistic matter (dust) dominated cases.
In the radiation-dominated epoch $\gamma=w+1=4/3$. Therefore \[a(t)=C\left[\exp\left(\frac{2\sigma t}3\right)-1\right]^{1/2}.\] For small times $\sigma t\ll1$ the formula reproduces correct dependence $a(t)\propto t^{1/2}$. In the matter dominated case $\gamma=w+1=1$ and \[a(t)=C\left[\exp\left(\frac{2\sigma t}3\right)-1\right]^{2/3}.\] At $\sigma t\ll1$ we recover the standard law of non-relativistic matter evolution $a(t)\propto t^{2/3}$. Note that in the limit of long times $\sigma t\gg1$ the scale factor grows exponentially for all values of the parameter $\gamma$.
Find the dependencies $\rho_\gamma(a)$ and $\Lambda(a)$ both in the radiation-dominated and non-relativistic matter dominated cases in the model considered in problem #IDE_84.
Using the expressions for $\rho_\gamma(a)$ and $\Lambda(a)$ obtained in the problem #IDE_84, one finds that in the radiation-dominated epoch $(\gamma=4/3)$ \begin{align} \nonumber \rho_\gamma & =\frac{\sigma^2C^4}3\frac1{a^4}+\frac{\sigma^2C^2}3\frac1{a^2},\\ \nonumber \Lambda & =\frac{\sigma^2}3+\frac{\sigma^2C^2}3\frac1{a^2}. \end{align} In the limit $a\to0$ ($t\to0$) \begin{align} \nonumber \rho_\gamma & =\frac{\sigma^2C^4}3\frac1{a^4}=\frac{3}{4t^2},\\ \nonumber \Lambda & =\frac{\sigma^2C^2}3\frac1{a^2}=\frac\sigma{2t}. \end{align} At the matter-dominated epoch ($\gamma=1$) \begin{align} \nonumber \rho_\gamma & =\frac{\sigma^2C^3}3\frac1{a^3}+\frac{\sigma^2C^{3/2}}3\frac1{a^{3/2}},\\ \nonumber \Lambda & =\frac{\sigma^2}3+\frac{\sigma^2C^{3/2}}3\frac1{a^{3/2}}. \end{align}
Show that for the $\Lambda(t)$ models \[T\frac{dS}{dt}=-\dot\rho_\Lambda a^3.\]
Consider a two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant and rewrite the second Friedman equation in the following form \begin{equation}\label{mainDE} \frac{\ddot{a}}{a} = \frac{1}{2}\left( 1 + 3w\right) \left( \frac{\dot{a}^2}{a^2} + \frac{k}{a^2} \right) + \frac{1-3w}{6} \Lambda . \end{equation}
Consider a two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant with quadratic time dependence $\Lambda(\tau)=\mathcal{A}\tau^2$ and find the time dependence for of the scale factor.
Substitute the dependence $\Lambda(t)=Bt^2$ into the equation (\ref{mainDE}) and multiply both sides by $a^2$ to obtain \begin{equation} \ddot{a} a = \frac 12 (1+3w)\left(\dot{a}^2 + k\right) +\frac{1-3w}{6} a^2\mathcal{A}t^2\Lambda \label{mainDE1} \end{equation} Seek the solution in the form $a(\tau)=a_0\tau^\beta.$ Substitute this dependence into (\ref{mainDE1}) to show that the equality takes place only under the following condition $$\mathcal{A} = \frac{6\beta}{1-3w}\left(1 -\frac 1\beta +\frac 32 w \right)$$
Consider a flat two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant with quadratic time dependence $\Lambda(\tau)=\mathcal{A}\tau^{\ell}$. Obtain the differential equation for Hubble parameter in this model and classify it.
Recall that the definition of the Hubble parameter reads \begin{equation} H \equiv \frac{\dot{a}}{a} = H_0 \left( \frac{da}{a \, d\tau} \right) , \label{Hdefn} \end{equation} and use \eqref{mainDE} to obtain: \begin{equation} \frac{dH}{d\tau} = \left( \frac{-3\gamma}{2H_0} \right) H^2 + \left( \frac{\gamma}{2H_0} \right) \Lambda + \left( 1-\frac{3\gamma}{2} \right) \frac{k}{H_0 \, a^2}, \label{1stDE} \end{equation} where $\gamma = 1+w.$ This equation can presented in the following form \begin{equation} \frac{dH}{d\tau} = {\mathcal P}(\tau) \, H^2 + {\mathcal Q}(\tau) \, H + {\mathcal R}(\tau) , \label{RiccatiDE} \end{equation} where ${\mathcal P} \equiv - 3\gamma/2$, ${\mathcal Q} \equiv 0$ and ${\mathcal R}(\tau) \equiv (\gamma/2) \Lambda(\tau).$ The obtained equation \eqref{RiccatiDE} is of Riccati type. For further convenience we transit to dimensionless variables $H \to x$: \begin{equation} H \equiv -\frac{1}{{\mathcal P} x} \, \frac{dx}{d\tau} = \left( \frac{2H_0}{3\gamma} \right) \frac{dx}{x \, d\tau} , \label{xDefn} \end{equation} According to the condition of the problem one should consider the spatially flat Universe $k=0,$ so substitute into the equation \eqref{1stDE} the following dependence $\Lambda(\tau)=A\tau^{\ell}:$ \begin{equation} \tau^{\ell} \; \frac{d\,^2x}{d\tau^2} - \alpha \, x = 0 , \label{mainDE-t} \end{equation} where \begin{equation} \alpha \equiv \frac{3\gamma^2 {\mathcal A}}{4} . \label{alphaDef1} \end{equation} It is easy now to show that the scale factor $a(\tau)$ and the function $x(\tau)$ are linked by the relation \begin{equation} a(\tau) = [ x(\tau) ]^{2/3\gamma} . \label{xTOa} \end{equation} The constant $\alpha$ is determined by the relation \eqref{alphaDef1}.
Find solution of the equation obtained in the previous problem in the case ${\ell =1}$. Analyze the obtained solution.
In the considered case the equation \eqref{mainDE-t} takes the form \begin{equation} \tau \; \frac{d\,^2x}{d\tau^2} - \alpha \, x = 0 , \label{mainDE-t-1} \end{equation} The constant $\alpha$ is determined by the relation \eqref{alphaDef1}: \begin{equation} \alpha = (3\gamma/2)^2 \lambda_0 \tau_0 . \label{alphaDefn-1} \end{equation} Change the independent variable from $\tau$ to $z \equiv 2\sqrt{-\alpha\tau}$ to obtain \begin{equation} z^2 \, \frac{d\,^2x}{dz^2} - z \, \frac{dx}{dz} + z^2 \, x = 0 . \label{mainDE-t-1b} \end{equation} This equation can be reduced to the Bessel one $x(z) = c_1 z J_1(z) + c_2 z Y_1 (z),$ where $J_1(z)$ and $Y_1(z)$ are respectively Bessel and Neumann functions of first order. Use the relation (\ref{xTOa}) to obtain the explicit dependence for the scale factor: \begin{equation} a(\tau) = \tau^{1/3\gamma} \left[ c_1 J_1(z) + c_2 Y_1 (z) \right]^{\; 2/3\gamma} , \label{aSoln-1} \end{equation} where the factor $2\sqrt{-\alpha}$ enters the coefficients $c_1,c_2$. The Hubble parameter can be determined by substitution $x(z)$ into the expression \eqref{xDefn}: \begin{equation} H(\tau) = H_0 \, \sqrt{-\lambda_0 \left( \frac{\tau_0}{\tau} \right)} \, \left[ \frac{c_1 J_0(z) + c_2 Y_0(z)} {c_1 J_1(z) + c_2 Y_1(z)} \right] , \label{HSoln-1} \end{equation} where $J_0(z)$ and $Y_0(z)$ are respectively Bessel and Neumann functions of zero order. Note that from the definition (\ref{alphaDefn-1}) it follows that $z(\tau),$ $a(\tau)$ and $H(\tau)$, can be real-valued (for $\tau >0 $) if $\lambda_0 \leq 0.$ Though such possibility exists in theory, it contradicts the observational data. Thus the case $\ell=1$ is likely to be impossible in real Universe.
Solve the equation obtained in the problem #tauell for ${\ell =2}.$ Consider the following cases
a) $\lambda_0 > -1/(3\gamma\tau_0)^2,$
b) $\lambda_0 = -1/(3\gamma\tau_0)^2,$
c) $\lambda_0 < -1/(3\gamma\tau_0)^2$
(see the previous problem). Analyze the obtained solution.
In the case ${\ell =2}$ the equation (\ref{mainDE-t}) takes on the form
\begin{equation}
\tau^2 \; \frac{d\,^2x}{d\tau^2} - \alpha \, x = 0 ,
\label{mainDE-t-2}
\end{equation}
where $\alpha$ is determined by the equation (\ref{alphaDefn-2}):
\begin{equation}
\alpha = (3\gamma/2)^2 \lambda_0 \tau_0^2 .
\label{alphaDefn-2}
\end{equation}
This equation belongs to Euler type. Make the transformation of variables $y \equiv \ln\tau$ in the equation (\ref{mainDE-t-2}) to obtain
\begin{equation}
\frac{d\,^2x}{dy^2} - \frac{dx}{dy} - \alpha \, x = 0 .
\label{EulerDE}
\end{equation}
a) $\lambda_0 > -1/(3\gamma\tau_0)^2.$
As the astronomical observations give evidence in favor of positive value of $\lambda_0,$ thais model makes certain interest.
Solution of the equation (\ref{EulerDE}) for $x(y)$, where $x(\tau)$, is easy to obtain.
The scale factor and the Hubble parameter can be found using the relations (\ref{xTOa}) and (\ref{xDefn}), resulting in the following
\begin{equation}
a(\tau) = \tau^{1/3\gamma} \, \left( c_1 \tau^{m_0} + c_2
\tau^{-m_0} \right)^{2/3\gamma}
\label{aSoln-2a-2}
\end{equation}
\begin{equation}
H(\tau) = \frac{2H_0}{3\gamma} \, \left[ \frac
{m_1 c_1 \tau^{m_0} + m_2 c_2 \tau^{m_0}}
{\tau \, ( c_1 \tau^{m_0} + c_2 \tau^{-m_0})} \right] ,
\label{HSoln-2a-2}
\end{equation}
where $m_0 \equiv \frac{1}{2} \sqrt{1+(3\gamma\tau_0)^2 \lambda_0}$.
It is easy to see that $a(\tau)$ diverges at $\tau=0$ for $\lambda_0>0$ if
$c_2 \ne 0$. Thus the expressions (\ref{aSoln-2a-2}) and (\ref{HSoln-2a-2}) take on the simplified form
\begin{equation}
a(\tau) = \left( \frac{\tau}{\tau_0} \right)^{2m_1/3\gamma}
\label{aSoln-2a-3}
\end{equation}
\begin{equation}
H(\tau) = \frac{2H_0}{3\gamma} \, \left[ \frac{m_1}{\tau} \right] ,
\label{HSoln-2a-3}
\end{equation}
where $m_1 \equiv 1/2+m_0,$ à $\tau_0 = 2m_1/3\gamma.$
b) $\lambda_0 = -1/(3\gamma\tau_0)^2.$
Proceed in analogous way to obtain
\begin{equation}
a(\tau) = \tau^{1/3\gamma} \, (c_3 + c_4 \ln \tau)^{2/3\gamma}
\label{aSoln-2b-1}
\end{equation}
\begin{equation}
H(\tau) = \frac{2H_0}{3\gamma} \, \left[ \frac
{(c_3+2c_4) + c_4 \ln\tau}
{2\tau ( c_3 + c_4 \ln\tau)} \right] ,
\label{HSoln-2b-1}
\end{equation}
where $c_3,c_4$ are arbitrary constants.
In order to make $a(\tau)$ finite at $\tau=0$ set $c_4=0$, then
\begin{equation}
a(\tau) = (c_3 \sqrt{\tau})^{2/3\gamma}
\label{aSoln-2b-2}
\end{equation}
\begin{equation}
H(\tau) = \frac{H_0}{3\gamma\tau} .
\label{HSoln-2b-2}
\end{equation}
Substitute $H(\tau_0)=H_0$ into (\ref{HSoln-2b-2}) to find age of the Universe in the considered model
\begin{equation}
\tau_0 = 1/3\gamma.
\label{Age-2b}
\end{equation}
Substitute the expression (\ref{Age-2b}) into (\ref{aSoln-2b-2})
and $a(\tau_0)=1$ to find $c_3 = \sqrt{3\gamma}$. Then
\begin{equation}
a(\tau) = \left( \frac{\tau}{\tau_0} \right)^{1/3\gamma}.
\end{equation}
c) $\lambda_0 < -1/(3\gamma\tau_0)^2$
The dependencies $a(\tau)$ and $H(\tau)$ now take on the following form
\begin{equation}
a(\tau) = \tau^{1/3\gamma} \, \left[ c_5 \sin (m_3 \ln\tau) +
c_6 \cos (m_3 \ln\tau) \right]^{2/3\gamma}
\label{aSoln-2c-1}
\end{equation}
\begin{equation}
H(\tau) = \frac{2H_0}{3\gamma} \, \left\{ \frac
{(c_5-2m_3c_6)\sin(m_3\ln\tau)+(c_6+2m_3c_5)\cos(m_3\ln\tau)}
{2\tau[c_5\sin(m_3\ln\tau)+c_6\cos(m_3\ln\tau)]} \right\} ,
\label{HSoln-2c-1}
\end{equation}
where $m_3 \equiv \frac{1}{2} \sqrt{-(3\gamma\tau_0)^2 \lambda_0 - 1}$
and $c_5,c_6$ are arbitrary constants. Proceeding as usual, express $c_5$ and $c_6$ in terms of $\tau_0$:
\begin{equation}\label{c5defn-2}
c_5 = \frac{1}{\sqrt{\tau_0}} \sin(m_3\ln\tau_0) + \frac{1}{m_3}
\left[ \left( \frac{3\gamma}{2} \right) \sqrt{\tau_0} -
\frac{1}{2\sqrt{\tau_0}} \right] \cos(m_3\ln\tau_0)
\end{equation}
\begin{equation}
c_6 = \frac{1}{\sqrt{\tau_0}} \cos(m_3\ln\tau_0) - \frac{1}{m_3}
\left[ \left( \frac{3\gamma}{2} \right) \sqrt{\tau_0} -
\frac{1}{2\sqrt{\tau_0}} \right] \sin(m_3\ln\tau_0) .
\label{c6defn-2}
\end{equation}
Unlike the previous cases, it is now impossible to keep finite values $a(\tau)$ at $\tau=0$ turning one of the constants $c_5,c_6$ to zero.
Consider a flat two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant with the following scale factor dependence \begin{equation} \Lambda = {\cal B} \, a^{-m}. \label{Bam} \end{equation} Find dependence of energy density of matter on the scale factor in this model.
Use the conservation equation and the first Friedman equation to obtain \begin{equation} \frac{d}{da} \left( \rho a^{3\gamma} \right) = \left( \frac{m{\cal B}}{8\pi G} \right) a^{3\gamma-(m+1)} . \end{equation} Its integration leads to the required dependence of energy density of matter on the scale factor \begin{equation} \rho (a) = \rho_0 a^{-3\gamma} f(a) , \label{NewRho} \end{equation} where as usual $a(t_0)=a_0=1,\rho(a_0)=\rho_0$, and the function $f(a)$ takes on the form \begin{equation} f(a) \equiv 1 + \kappa_0 \times \left\{ \begin{array}{ll} \frac{ {\textstyle m(a^{3\gamma-m}-1)} } { {\textstyle 3\gamma-m} } & \mbox{ if } m \neq 3\gamma \\ 3\gamma \ln(a) & \mbox{ if } m=3\gamma \end{array} \right. \label{fDefn} \end{equation} \begin{equation} \kappa_0 \equiv {\cal B}/8\pi G \rho_0 . \label{kappaDef1} \end{equation} If $m=0$ then $f(a)=1$ and the equation (\ref{NewRho}) recovers the usual result $\rho\sim a^{-3}$ for non-relativistic matter ($\gamma=1$) and $a^{-4}$ for radiation ($\gamma=4/3$). The new parameter $\kappa_0$ can be determined by the decay law (\ref{Bam}) and astronomical observations implying ${\cal B} = \Lambda_0 = 3H_0^2\lambda_0$. Substitute this result into (\ref{kappaDef1}) to find: \begin{equation} \kappa_0 = \lambda_0/\Omega_0. \label{kappaDefn} \end{equation}
Find dependence of deceleration parameter on the scale factor for the model of previous problem.
Substitute (\ref{Bam}) and (\ref{NewRho}) into the Friedman equation to obtain \begin{equation} \frac{da}{d\tau} = a \left[ \Omega_0 a^{-3\gamma} f(a) - (\Omega_0+\lambda_0-1) a^{-2} + \lambda_0 a^{-m} \right]^{1/2} , \label{Expansion} \end{equation} where the function $f(a)$ is determined in the previous problem (see the formula \eqref{fDefn}). The second Friedman equation can be rewritten in the form \begin{equation} \frac{d^{\, 2}a}{d\tau^2} = \left( 1 - \frac{3\gamma}{2} \right) \Omega_0 a^{1-3\gamma} f(a) + \lambda_0 a^{1-m} . \label{Deceleration} \end{equation} Substitute this derivatives into the definition of the deceleration parameter to obtain the required dependence \begin{equation} q\equiv -\frac{\ddot{a}a}{\dot{a}^2}=-\frac{\left( 1 - \frac{3\gamma}{2} \right) \Omega_0 a^{1-3\gamma} f(a) + \lambda_0 a^{1-m}}{a\left[ \Omega_0 a^{-3\gamma} f(a) - (\Omega_0+\lambda_0-1) a^{-2} + \lambda_0 a^{-m} \right]}\end{equation}