Time-like Killing vector in FRW metric?

The spatially flat FRW metric in cartesian co-ordinates is given by: $$ds^2 = -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2)$$ As I understand it there are Killing vectors in the $x$, $y$, $z$ directions implying that momentum is conserved but there is no Killing vector in the $t$-direction which implies that energy is not conserved. I don't know much about Killing vectors. Is this correct?

If I transform the time co-ordinate $t$ to conformal time $\tau$ using the relationship: $$d\tau = \frac{dt}{a(t)}$$ the metric now becomes: $$ds^2 = a^2(\tau)(-d\tau^2 + dx^2 + dy^2 + dz^2)$$ Does this metric now have a Killing vector in the $\tau$-direction as well as in the $x$, $y$ and $z$ directions?

First of all the existence of Killing vector does not depend on the particular representation of the metric. However, given a coordinate system the vector tangent to a coordinate, say $\partial_{x^1}$, is a Killing vector of a given metric if and only if, referring to these coordinates the components of the metric do not depend on that coordinate, i.e. $x^1$. The proof is elementary. The Killing equation for a vector field $X$, namely $\nabla_a X_b + \nabla_b X_a =0$, where $\nabla$ is the Levi-Civita connection of the metric $g$, is equivalent to ${\cal L}_X g=0$. If $X= \partial_{x^1}$, the last identity can be explicitly written as $$\partial_{x^1} g_{ab}=0$$ referring to coordinates $x^1,x^2,\ldots, x^n$.

Therefore it is evident that $\partial_{x}$, $\partial_y$ and $\partial_z$ are Killing vectors of your considered metric. Conversely $\partial_\tau$ is not. However it is a conformal Killing vector (the Lie derivative of the metric with respect to that vector field is proportional to the metric itself). It implies in particular that, conformally invariant fields (like the electromagnetic field) "see" $\partial_\tau$ as a Killing vector. For instance it makes sense to define thermal states of such fields with respect to that notion of time.

• So if the Universe consisted of just photons and we measure conformal time then the energy of those photons would be conserved? – John Eastmond Jun 20 '14 at 11:58
• Yes, but it is not so interesting since you cannot use that notion when photons interact with massive particles. Instead, that notion of time is relevant to describe in what sense the cosmological background radiation state is stationary, i.e. an equilibrium state: it is such referring to $\tau$. There are two notions of temperature, the statistical mechanics one, which is constant in time ($\tau$), and that measured by thermometers which varies in time (proper time of thermometers). – Valter Moretti Jun 20 '14 at 12:36
• So would a photon confined inside a rigid reflective box lose energy with respect to conformal time or not? – John Eastmond Jun 20 '14 at 13:35
• I use the photon-in-a-box example as it could be considered as a model for a massive particle. – John Eastmond Jun 20 '14 at 13:47
• I think that if the box expands with the same rate of the universe referring to $\tau$ it should preserve its conformal energy content. However its proper energy decreases (the temperature, measured by a thermometer, of the cosmic radiation is decreasing indeed). – Valter Moretti Jun 20 '14 at 14:29

A vector field $\chi$ is conformal Killing vector field whenever $\mathcal{L}_\chi g= \phi g$ for some smooth real-valued $\phi$, its conformal factor. In the particular case of $\phi = 0$, $\chi$ is a Killing vector field.

Theorem: If $\chi$ is a conformal Killing vector field of $g_{\mu\nu}$, then it is also a conformal Killing vector field of $\tilde{g}_{\mu\nu} = e^{2\Omega}g_{\mu\nu}$, with conformal factor $\tilde{\phi} = \phi + \Omega_{,\alpha}\chi^\alpha$.

In your case, $\partial_\tau$ is a Killing vector field of the Minkowski metric we'd get if we ignore the front multiplier, but including it makes $\partial_\tau$ a conformal Killing vector field that is not itself a Killing vector field, except in the trivial case of static spacetime.