Why a Killing vector field is needed to judge positive/negative frequencies?

Question

Let $(M,g)$ be spacetime and consider the space of solutions to the Klein-Gordon equation endowed with the inner product:

$$(\phi_1,\phi_2)=i\int_\Sigma (\phi_1\nabla_\mu \phi_2^\ast-\phi_2^\ast\nabla_\mu \phi_1) n^\mu \sqrt{|\gamma|}d^{n-1}x$$

Let $\{\phi_i\}$ be a complete set of solutions on this space and $K$ a timelike Killing vector field on $(M,g)$.

If I understood well (and I might have not), in his book Spacetime and Geometry, Sean Carroll says that $\phi_i$ is a positive frequency solution to the KG equation if there is $\omega \in(0,\infty)$ such that

$$\mathscr{L}_K \phi_i=K\phi_i=-i\omega \phi_i,$$

and it is a negative-frequency solution if

$$\mathscr{L}_K \phi_i=K\phi_i=i\omega \phi_i.$$

Now why does one need a Killing field to judge this? Why this can't be done with an arbitrary timelike vector field?

The Lie derivative of a $C^\infty(M)$ function is just the directional derivative of the function along the integral lines of the field.

Why this condition says that a function is a positive or negative frequency, and why we need the Killing field?

Edit: I know a Killing field is a generator of isometries, so that the metric tensor is constant along its flow. Nevertheless, I really can't see why to define positive/negative frequency solutions we need a Killing field instead of just a timelike field. I don't see why any symmetry of spacetime should be involved.

Actually, a timelike future-directed vector field defines a family of observers and is a reference frame. I thought that this was enough to define positive/negative frequency: it is what this reference frame sees. I honestly don't know where the Killing part comes into play.

Answer 1

Killing vector field is needed in order to well define the vacuum state.

More precisely, positive and negative frequency solutions have meaning if the spacetime has a global timelike Killing vector field (it is stationary). If $K^\mu$ is globally timelike one can find a coordinate system where the metric is time independent and $K^\mu$ takes the form $(K^\mu)=(1,0,0,0)$ and the equation $\mathscr{L}_K\phi_i=-i\omega_i\phi_i$ takes the form

$$\frac{\partial\phi_i}{\partial t}=-i\omega_i\phi_i.$$

Now according to Noether's theorem, the energy is conserved and, since one has a well defined notion of conserved energy, one can well define the state of minimal energy, the vacuum.

So having a global timelike Killing vector ensures that your vacuum state stays as a vacuum state as time evolves. If you do not have a Killing vector but some timelike vector, then the vacuum state would would not evolve into the same vacuum state.

Answer 2

The significance of a Killing vector field, $\xi^a$, is that it indicates the presence of a spacetime symmetry. i.e., it is an "isometry" which means a symmetry of the metric: $\mathcal{L}_{\xi}g_{\mu\nu}=0$. Hence, defining the frequency with respect to a timelike Killing vector field gives us a notion of frequency which is related to a symmetry of the spacetime itself.

You could, of course, choose to pick an arbitrary timelike vector field instead but that choice would be, well, arbitrary with no relation to the underlying structure of spacetime. With such an arbitrary choice, you'd be defining the "frequency" in a way which may have no natural relation to anyone else's arbitrary definition. By defining it with respect to a spacetime symmetry, we have selected a definition which is not entirely arbitrary.

Possibly relevant comments on Killing vector fields:

In Appendix B of Carroll's book, he describes how one can always choose coordinates such that one of the coordinates (let's say the $x^0$-coordinate) lies along the integral curves of the Killing vector field, $\xi^a$. In such coordinates, the Lie derivative of a tensor field takes the especially simple form,

\begin{equation} \mathcal{L}_\xi T^{\mu_1\cdots \mu_k}_{\phantom{\mu_1\cdots \mu_k}\nu_1\cdots \nu_l}=\frac{\partial}{\partial x^0}T^{\mu_1\cdots \mu_k}_{\phantom{\mu_1\cdots \mu_k}\nu_1\cdots \nu_l}. \end{equation}

Now, the definition of the Killing vector field implies in these coordinates one has

\begin{equation} \frac{\partial}{\partial x^0} g_{\mu\nu}=0. \end{equation}

Thus we find that the presence of a Killing vector field allows us to always choose coordinates such that the metric is independent of a coordinate associated with the symmetry.