Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 62 / 2

Let $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$. In the metric variation specified here, the covariant metric tensor $g_{\mu \nu }$ is varied while the scalar field is held fixed. Differentiating the inverse and the determinant givesThus $\delta \mathcal{L}=\frac{1}{2}{\partial }^{\mu }\varphi {\partial }^{\nu }\varphi \delta g_{\mu \nu }$, and the action varies asThe functional derivative therefore gives $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi +g^{\mu \nu }\mathcal{L}$. Lowering both indices yields the stress-energy tensor of a canonical scalar field:The determinant variation has a positive sign because the variable is $g_{\mu \nu }$; varying the inverse metric tensor instead would reverse that intermediate sign.

For the flat FRW metric, $g^{00}=-1$, $g^{0i}=0$, and $g^{ij}=a^{-2}{\delta }^{ij}$. Consequently the kinetic contraction of the scalar field isInserting this in $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi +g^{\mu \nu }\mathcal{L}$, with ${\partial }^0\varphi =-\dot{\varphi }$ and ${\partial }^i\varphi =a^{-2}{\partial }_i\varphi$, givesThese are contravariant coordinate components of the stress-energy tensor; spatial components in an orthonormal frame carry different powers of $a$. In particular $T^{00}$ is the energy density seen by the comoving observer.

Vary the scalar field with compactly supported variation, leaving the metric tensor fixed. The action variation isIntegration by parts moves the derivative from $\delta \varphi$ and gives the Euler-Lagrange equationFor the FRW metric, $\sqrt{-g}=a^3$, so the covariant wave operator becomesRearranging gives the requested scalar field evolution:The damping term is Hubble friction; it comes from the expanding volume factor $a^3$, not from a microscopic dissipative force.

For a homogeneous scalar field, every spatial derivative vanishes. The stress-energy tensor consequently has no momentum density or directional stress:The only spatial tensor left is ${\delta }^{ij}$, so the stress is isotropic. This is the perfect fluid form in its comoving frame. Now expand the time component of stress-energy conservation using the covariant derivative:The first term is $\dot{\rho }$. In the second, only $\lambda =0$ contributes, and ${\sum }_i{\Gamma }^i{}_{i0}=3H$, giving $3H\rho$. The third isHere the quoted Christoffel symbols include their lower-index symmetry ${\Gamma }^i{}_{j0}={\Gamma }^i{}_{0j}$. Therefore the cosmological continuity equation isAs a check using the scalar field equation, $\dot{\rho }=\dot{\varphi }(\ddot{\varphi }+V_{,\varphi })=-3H{\dot{\varphi }}^2=-3H(\rho +P)$, with exactly the same sign.

With zero scalar potential and a homogeneous scalar field, the preceding perfect fluid identification givesThis equation of state is that of a stiff fluid. The field equation integrates to $a^3\dot{\varphi }=K$, since ${\partial }_t(a^3\dot{\varphi })=0$. Hence $\rho =K^2/(2a^6)$, agreeing with the cosmological continuity equation for $w=1$. If this nonzero component dominates a flat universe, the Friedmann equation giveson the expanding branch. Integrating once more, $a^3=3\sqrt{4\pi G/3}|K|(t-t_B)$. Choosing the Big Bang at $t_B=0$ yieldsThe homogeneous free scalar as a stiff fluid construction requires $K\ne 0$; a constant zero-potential field has zero energy density and cannot dominate this expansion.