In QFT in curved spacetime, there is an indeterminacy in the local energy density (because of the indeterminacy in defining annihilation/creation operator) if the spacetime is not stationary.

Is it equivalent (due) to vacuum particle-antiparticle creation?


1 Answer 1


No. Notice you'd have a similar problem even in the classical theory. If spacetime is not stationary, you can't really say what is $T_{00}$ in the stress tensor to define what is the energy density, unless you are choosing to define it with respect to some observer, for example. Even in this case, you won't have a global conservation law due to the lack of a timelike Killing field. The fact it is happening due to the indeterminacy in the choice of creation and annihilation operators is pretty much the same problem with extra steps. Note it goes roughly like this:

  • particles are modes of positive-define energy;
  • there is no preferred notion of energy in a non-stationary spacetime;
  • hence, there is no preferred notion of particles in a non-stationary spacetime.
  • $\begingroup$ I see, it doesn't even make sense to talk about particle/antiparticle then. $\endgroup$ Apr 18, 2022 at 18:03
  • $\begingroup$ @Displayname Precisely. I believe this question might be related to your previous question concerning the Hawking paper. In there, he manages to define an approximate notion of particles within a certain region, in which he can use an approximate notion of energy (e.g. by picking $T_{00}$ with respect to an observer) but outside of this region the approximation fails more and more and you end up with the problem of not having a global notion of energy. $\endgroup$ Apr 18, 2022 at 18:13
  • $\begingroup$ In other words, Hawking manages to use an approximate notion of particles/antiparticles, but this notion eventually crumbles down if you go too far away from where you defined it. $\endgroup$ Apr 18, 2022 at 18:14
  • $\begingroup$ Exactly, since this arises near the horizon where we don't have a nice time-like Killing vector field. So can't we use this approximate notion of particles near the horizon to understand that a "negative-energy particle" falls into the BH and the other one escapes to infinity? $\endgroup$ Apr 18, 2022 at 18:24
  • $\begingroup$ @Displayname I must admit I've been thinking about your question for a while and I'm still not fully comfortable with it, but I'm quite sure the answer is no. Hawking himself states in his paper that the particle creation process is global, and mentions that an infalling observer would see no particles with frequency $\omega < M$ and just a few particles with $\omega > M$ (I'm not really comfortable because I thought they would see none at all, as opposed to a few). However, your argument would lead someone to expect the observer to see infinitely many, since there are infinitely many + $\endgroup$ Apr 21, 2022 at 23:52

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