Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 64 / 1

The spatial coordinates are cyclic because the flat Friedmann-Lemaître-Robertson-Walker metric depends only on time. Differentiating the printed affine Lagrangian gives the canonical momentumThe Euler-Lagrange equations give $d{P}_i^c/d\tau =\partial L/\partial x^i=0$, soThis is the conserved quantity associated with spatial translation invariance. The overall minus sign in the stated action makes this canonical covector opposite to the usual future-directed mechanical momentum; it does not alter the geodesic trajectories or the conservation result. The canonical momentum sign in the affine particle action will be kept explicit below.

The mass-shell condition and $P_{\mu }^c=-m{g}_{\mu \nu }{\dot{x}}^{\nu }$ imply $g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }=-1$, so the affine parameter is proper time up to orientation. Choose its future orientation. With conformal time $t$,In a local comoving orthonormal frame both time and spatial coordinate intervals acquire the same factor $a$, so the measured peculiar velocity is precisely $\mathbf{v}=d\mathbf{x}/dt$. Thus the physical momentum obeysThe fixed vector direction also gives $\mathbf{P}(t)=a(t_{\ast })\mathbf{P}(t_{\ast })/a(t)$. With the conventional opposite overall sign for the affine particle action, the familiar comoving momentum is $\mathbf{q}=a\mathbf{P}={\mathbf{P}}^c$ instead. The redshift law is identical under either sign convention.

For signature $(-+++)$, $\sqrt{-g}=a^4$ and $g^{\mu \nu }=a^{-2}{\eta }^{\mu \nu }$. The scalar action becomesVary $\varphi$ with compactly supported variations and integrate its derivative terms by parts:Setting this coefficient to zero gives the Klein-Gordon equation in the expanding background:The additional first-derivative term is the conformal-time expansion damping; the time coordinate here is not the proper cosmological time used in Question 2.

Put $\chi =a\varphi$. Direct differentiation of the previous equation yieldsIn radiation domination, $a\propto t$, and for a massless scalar field both extra terms vanish. Each nonzero Fourier mode therefore satisfies ${\chi }_{\mathbf{k}}^{\prime \prime }+k^2{\chi }_{\mathbf{k}}=0$. The general real radiative solution isSumming over both wavevector directions supplies the two traveling-wave solutions, with constant amplitudes. In infinite space the sum is replaced by the corresponding Fourier integral.

There is a qualification to calling the printed oscillatory expression completely general. At $k=0$, ${\chi }_0^{\prime \prime }=0$, so ${\chi }_0=C_1+C_{2}t$ rather than a single zero-frequency exponential. The zero mode of a radiation-era massless scalar isA constant nonzero field is an immediate counterexample to the completeness of the fixed-amplitude oscillatory formula at zero wavenumber. Include this homogeneous term, or explicitly restrict the displayed plane-wave expansion to the radiative modes with no homogeneous background.

The spatial momentum operator acts only on the plane wave:Thus the modes have fixed canonical/comoving wavevector momentum, not fixed physical momentum. In the usual positive-energy field-mode convention the local physical momentum and energy are $\hslash \mathbf{k}/a$ and $\hslash k/a$. This agrees with conserved spatial comoving momentum and cosmological redshift proportional to $a^{-1}$.

The overall particle-action sign noted in (i) reverses the literal worldline canonical label relative to this usual mechanical momentum: it is ${\mathbf{P}}^c=-a\mathbf{P}$ for a future-directed particle with that printed action. The conserved plane-wave label therefore has the opposite sign if it is identified with that literal worldline label. This sign convention does not change the wavenumber magnitude or either conservation/redshift conclusion.