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

Each separately conserved component obeys the constant-equation-of-state density scaling $\rho \propto a^{-3(1+w)}$. For the flat Friedmann-Lemaître-Robertson-Walker metric, the Friedmann equations therefore giveWrite $H(z)/H_0=1+h_{1}z+O(z^2)$, where $h_1=[3{\Omega }_m+4{\Omega }_r+3(1+w){\Omega }_X]/2$. Expanding the reciprocal inside the luminosity distance integral givesMatching its quadratic coefficient gives the deceleration parameter $q_0=h_1-1$. Using flatness in this part,A negative deceleration parameter signifies accelerated expansion; negative pressure must overcome the positive matter and radiation terms. Restoring units replaces $H_0{d}_L$ by $H_0{d}_L/c$ in the expansion, leaving $q_0$ unchanged.

In the observational setting of this paper, the cosmic microwave background favours nearly flat geometry, while cluster dynamics and large-scale structure favour a matter density near ${\Omega }_m\sim 0.3$. The missing contribution to the critical density is then roughly ${\Omega }_X\sim 0.7$. The supernova luminosity distance measurements require recent accelerated expansion. With present radiation negligible, $q_0<0$ requiresFor an X-dominated universe the threshold would instead be $w<-1/3$. Thus X cannot be ordinary pressureless matter or radiation. It must be comparatively smooth on cluster scales, so that it contributes to the background expansion without being counted as clustered dynamical matter. Its negative pressure makes its density dilute more slowly than matter, allowing late dominance. A cosmological constant, with $w=-1$, or sufficiently potential-dominated quintessence can have these properties.

The comparison with a perfect fluid in general relativity needs a distinction between background stress and perturbations. A homogeneous canonical scalar has canonical scalar stress as a perfect fluid:It has isotropic pressure and no background shear stress, just as a perfect fluid in general relativity does. Negative pressure therefore does not, by itself, prevent a perfect-fluid description. However, an adiabatic barotropic fluid with the constant closure $p=w\rho$ would have $c_s^2=dp/d\rho =w$. For negative $w$, a short-wavelength density mode then has ${\omega }^2\simeq w{k}_{\mathrm{phys}}^2<0$, the negative-w barotropic-fluid gradient instability. Such a closure cannot give a stable, smooth propagating X component.

A canonical scalar instead has rest-frame sound speed squared equal to one: its spatial and temporal kinetic terms have the same coefficient. Its pressure perturbation need not obey the background relation $\delta p=w\delta \rho$. This nonadiabatic response suppresses small-scale clustering without the negative-sound-speed instability. A cosmological constant is a separate limiting case with no independent propagating fluid density mode. Thus the issue is the microphysics and perturbation closure, not whether the background stress-energy tensor has perfect-fluid form.

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.