Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 70 / 4 / c

Varying the canonical scalar action gives $\square \varphi +V_{,\varphi }=0$. In a homogeneous flat Friedmann-Lemaître-Robertson-Walker metric written in conformal time, $\sqrt{-g}=a^4$ and $g^{00}=a^{-2}$, soThe quadratic potential therefore givesEquivalently $\ddot{\varphi }+3H\dot{\varphi }+m^2\varphi =0$ in cosmic time. The intrinsic roll/oscillation time is $m^{-1}$ and the damping time is $H^{-1}$. If $m\gg H_0$, the scalar has long since begun rapid oscillations: averaging over a quadratic oscillation gives equal kinetic and potential energies, $⟨p⟩\simeq 0$, so it behaves as matter rather than accelerating dark energy. To remain nearly frozen or only begin rolling today requires the mass scale of quadratic quintessenceIf its energy is predominantly potential, equating $m^2{\varphi }_0^2/2$ to ${\rho }_{X0}=3M^2{H}_0^2{\Omega }_X$, with $M$ the reduced Planck mass, givesTaking the natural onset estimate $m\sim H_0$ and ${\Omega }_X\sim 0.7$ gives $|{\varphi }_0|\sim 2M\sim 5\times 10^{18}\mathrm{GeV}$, of order the Planck mass. More strongly frozen, negative-pressure evolution requires a smaller mass and correspondingly larger field. The density fixes the product $m|{\varphi }_0|$ at this accuracy, rather than a unique mass and field separately.

There are two normalization problems in the printed energy identity. In conformal time, the physical density of a homogeneous scalar is $T_{00}/a^2$, and its kinetic contribution contains ${\varphi }^{\prime 2}$, not one power of ${\varphi }^{\prime }$. The conformal density of a canonical scalar field follows directly from the stress-energy tensor:Writing $K={\varphi }^{\prime 2}/(2a^2)$, the condition $K-V=-(K+V)/3$ gives $V=2K$, henceThe reconstruction is consistent with this corrected identity; the unsquared printed identity cannot hold as an energy relation.

The constant-equation-of-state density scaling for this coasting fluid gives ${\rho }_{\varphi }={\rho }_{X0}{a}^{-2}$, where ${\rho }_{X0}=3M^2{H}_0^2{\Omega }_X$. Choose the branch on which $\varphi$ increases. ThenKeeping all three density parameters explicitly, the Friedmann equations in conformal time giveFor ${\Omega }_X>0$, defineThe square root is $\sqrt{(a+b{)}^2-A^2}$. Integrating on the expanding branch giveswhere $a({\varphi }_0)=1$. Thus the scalar-field reconstruction of a coasting component isSubstitution into the potential gives the requested present-field parametrization:The domain is the branch where $a>0$ and the hyperbolic argument is positive; reversing the direction of the field changes the sign of $B$. The actual discriminant condition is ${\Omega }_m^2/4>{\Omega }_r{\Omega }_X$. The supplied stronger bound suffices for physical density fractions $0<{\Omega }_X\le 1$. No density parameter has been eliminated using the flatness sum.

As a check, ${\varphi }^{\prime }$ is constant and $V_{,\varphi }=-2V(a^{\prime }/a)/{\varphi }^{\prime }$. Since $a^{2}V={\varphi }^{\prime 2}$ along this solution, ${\varphi }^{\prime \prime }+2(a^{\prime }/a){\varphi }^{\prime }+a^2{V}_{,\varphi }=0$ identically. The reconstructed field therefore satisfies its equation of motion as well as the Friedmann equations and required equation of state.