In a spacetime with a bifurcate Killing horizon, one can use the Killing vector on one side of the horizon to prepare a smooth global state across it, by performing a Euclidean path integral to the Hawking temperature. Viewed in terms of the Hamiltonian that Lie drags fields between slices of constant Killing time, this state will thus be a thermal state at the Hawking temperature.

In a maximally symmetric spacetime such as Minkowski spacetime, this state will also be a stationary state at zero temperature of the Hamiltonian of the global timelike Killing vector. I.e. the Minkowski vacuum is a thermal state at the Unruh temperature of a Rindler Hamiltonian.

However in general there will be no second timelike Killing vector. For example, in the most interesting case of Schwarzschild spacetime, the Hawking Hartle vacuum is a thermal state at the Hawking temperature of the Schwarzschild Hamiltonian. This state is supposed to be "positive frequency with respect to the Kruskal observers"; e.g. their appropriately-renormalized Unruh-DeWitt detectors are not excited.

However, the Kruskal Hamiltonian has an explicit time dependence. It is therefore not clear what it would mean for the Hawking-Hartle vacuum to be its "ground state". How does the Hawking-Hartle vacuum relate to it?


1 Answer 1


Let us denote the Kruskal coordinates as $u$ and $v$. Hence in the Kruskal frame, the light cone coordinates can be denoted by,

$$ \begin{align} U &= u+v\\ V &= u-v \end{align} $$

Now, you are right that the Kruskal frame metric has an explicit time dependence. So we can't use $\frac{\partial}{\partial t}$ as a timelike Killing vector field to define positive frequency modes like we usually do in Minkowski metric or any conformally flat metric. We have to use some other affine parameter for this matter. The Hawking-Hartle vacuum dictates exactly what timelike affine parameter we choose. In this particular case, we take V as the timelike vector to determine positiveness of the frequency of incoming modes and $U$ for the positiveness of the outgoing modes. Hope this helps!

For further reference on the choice of the vacuum state, you can have a look at this paper: Vacuum polarization in Schwarzschild spacetime.

  • $\begingroup$ Nice! Isn't V a null vector, though? $\endgroup$
    – AGML
    Jul 18, 2019 at 17:19
  • $\begingroup$ $V$ is null only at the horizons (u=v), like the proper time is null at the light cones. In that case we use a time-like affine parameter. I guess we can do the same thing here as well. $\endgroup$
    – abhijit975
    Jul 18, 2019 at 21:39

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