Why is the expansion of a relativistic field into positive and negative frequency components $$\phi({\vec x},t)=\int\frac{d^3{k}}{~~(2\pi)^{3/2}}\left[A\left({\vec k}\right)e^{-i\omega_{\vec k}t}+A^*\left(-{\vec k}\right)e^{+i\omega_{\vec k}t}\right]e^{+i{\vec k}\cdot{\vec x}}$$ unambiguous in flat spacetime but not so in curved spacetime? I am very interested in knowing this.


In curved space-time "$t$" itself is ambiguous. You need some "time" translation symmetry to be able to use $e^{i\omega t}$ as a set for expanding something out. Technically this involves finding a timelike Killing vector field. There may be more than one of these fields and what is positive frequency with respect to the $t$ of one of them, may not be positive frequency with respect to the other. This problem even occurs in flat space. The metrics $$ d\tau^2 = dt^2-dx^2 \quad \hbox{Minkowski} $$ and $$ d\tau^2 = X^2 dT^2-dX^2 \quad \hbox{Rindler} $$ both describe flat space, and both are "time" ($t$ or $T$) independent --- but what is positive frequency with respect to $T$ is not necessarily positive with respect to $t$.

The mapping between $T$ and $t$ is implicit in the change of coordinates formulae $$ x= X\cosh T\\ t= X\sinh T $$

  • $\begingroup$ Why do you say that we need time-translation symmetry to expand in terms of $e^{i\omega t}$? $\endgroup$ – mithusengupta123 Nov 18 '20 at 14:45
  • $\begingroup$ I mean that the field equations for whatever we are expanding need to be $t$ independent for their normal modes to have an $e^{i\omega t}$ $t$ dependence. $\endgroup$ – mike stone Nov 18 '20 at 14:57
  • $\begingroup$ If the spacetime is flat and Minkowskian, why are the mode functions unique? Why two different observers will agree on the mode functions? Under Lorentz transformations, $t,{\vec x}$ and $\omega, {\vec k}$ of one observer is different from $t',{\vec x}'$ and $\omega', {\vec k}'$ of another observer. Why will they agree on mode functions? $\endgroup$ – mithusengupta123 Nov 18 '20 at 16:31
  • $\begingroup$ They won't agree on the modes. See this question and answer:physics.stackexchange.com/questions/418926/… $\endgroup$ – mike stone Nov 18 '20 at 16:41

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