# I Timelike Killing vectors & defining a vacuum state

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1. Jul 28, 2017

### Frank Castle

I've read that if a given spacetime possesses a timelike Killing vector, then it is possible to define a unique vacuum state by constructing positive and negative frequency modes with respect to this timelike Killing vector.

My question is, what does this mean? Explicitly, how does one use a timelike Killing vector to construct positive and negative frequency modes?

Is it something to do with the fact that if the spacetime possesses a timelike Killing vector, then one can find a coordinate frame in which the metric is time-independent, with the time coordinate defined with respect to the Killing vector $\frac{\partial}{\partial t}$. The equation of motion of a given quantum field (say a scalar for simplicity) is then separable (in this coordinate system) into a temporal and a spatial differential equation. One can then define positive and negative frequency modes as solutions to the temporal differential equation? If there is no timelike Killing vector then the equation of motion will not be separable and hence the solutions cannot be separated into a temporal and a spatial component, thus preventing one from classifying modes as positive 0r negative frequency?

Last edited: Jul 28, 2017
2. Jul 28, 2017

### jambaugh

You are correct. The time like Killing vector generates an isometry corresponding to a time-like translation. It defines a (curvilinear) coordinate $t$ such that the generator manifests as the $\partial /\partial t$ operator. This means, defining the flow generated by that killing field as time evolution establishes the corresponding conserved quantity (via Noether's Theorem), as energy $E \sim -i\hbar \frac{\partial}{\partial t}$. Having a well defined, conserved energy means you can define a minimal energy ground state a.k.a. vacuum.

The positive/negative frequency modes are the eigen-modes corresponding to positive (imaginary) and negative (imaginary) eigen-values, which in turn correspond to positive (real) and negative (real) energies.

Absent the existence of such a killing vector the dynamics is not symmetric under any time-like evolution and there is no definable conserved energy. If you define an arbitrary non-conserved energy corresponding to an arbitrary time-like Lie derivative (what would be a killing field if it generated a symmetry) then the vacuum at one point would not evolve to a vacuum at another point. There would be physical sources everywhere/when and it's rather difficult to justify calling that a "vacuum".

3. Jul 28, 2017

### Frank Castle

Does the space-time also have to be static in order to unambiguously define positive and negative frequency modes?

If so, would it be correct then to say, that for a static spacetime $\mathcal{M}$ with metric $g_{\mu\nu}$, a timelike Killing vector exists, and as such there exists a coordinate frame in which the metric is time-independent, i.e. we can choose coordinates such that $t$ is defined with respect to a corresponding time-like vector, $\partial_{t}:=\frac{\partial}{\partial t}$ and $\partial_{t}g_{\mu\nu}=0$, furthermore, there is no off-diagonal terms, i.e. $g_{0i}=0$. Given this, the d'Alembertian operator, $\Box$ is diagonalised, such that the frequency modes $f(t,\mathbf{x})$ satisfy $$\partial^{2}_{t}f(t,\mathbf{x})=-(g^{00})^{-1}\left(g^{ij}\partial_{i}\partial_{j}-m\right)f(t,\mathbf{x})$$ for which the solutions are separable and have the form $$f_{\omega}(t,\mathbf{x})=e^{-i\omega t}\bar{f}_{\omega}(\mathbf{x})$$, such that we can define the positive frequency modes with respect to the Killing vector $\partial_{t}$ as those that satisfy $$\partial_{t}f_{\omega}(t,\mathbf{x})=-i \omega f_{\omega}(t,\mathbf{x})\qquad\omega >0$$ The negative frequency modes can then be analogously defined in terms of the complex conjugate solutions $f^{\ast}_{\omega}(t,\mathbf{x})=e^{i\omega t}\bar{f}^{\ast}_{\omega}(\mathbf{x})$ satisfying, $$\partial_{t}f^{\ast}_{\omega}(t,\mathbf{x})=i \omega f^{\ast}_{\omega}(t,\mathbf{x})\qquad\omega >0$$ Is the point that the existence of a timelike Killing vector allows for a separable solution, in terms of a temporal and a spatial part, for the frequency modes such that they can be unambiguously classified as positive and negative. Furthermore, the timelike Killing vector ensures that the theory is time translation invariant meaning that the modes don't flip sign?!

Do you know of any books/notes that explain this stuff in detail?

4. Jul 29, 2017

### vanhees71

B. S. de Witt, Quantum field theory in curved space times, Phys. Rept. 19, 295 (1975)

5. Jul 29, 2017

### Frank Castle

Thanks for the recommendation

Would any of what I put in my last post be correct at all?

6. Jul 29, 2017

### Staff: Mentor

No, it just has to be stationary. "Stationary" means the spacetime has a timelike Killing vector field. "Static" means the timelike Killing vector field is hypersurface orthogonal--i.e., the spacetime can be "sliced" into a family of spacelike hypersurfaces that are everywhere orthogonal to the timelike Killing vector field. Heuristically, the spacetime around a spherically symmetric, non-rotating gravitating body is static; but the spacetime around a rotating gravitating body is only stationary.

This can be done for any stationary spacetime, as defined above. Note that the presence of a timelike Killing vector field is what defines a stationary spacetime.

7. Jul 29, 2017

### Frank Castle

So does the existence of a timelike Killing vector enable the d'Alembertian operator to be diagonalised, thus permitting solutions that can be separated into temporal and spatial components. One can then use the Killing vector to define sets of positive and negative frequency solutions, and since the Killing vector generates time translations under which the energy is conserved, this ensures that such a distinction is unambiguous?

8. Jul 29, 2017

### Staff: Mentor

Actually, I was not quite correct here: for a spacetime that is only stationary, not static, you can't find a global coordinate chart in which there are no nonzero $g_{0i}$ terms. Only in a static spacetime can all those terms be made to vanish.

As above, the d'Alembertian can only be diagonalized in a static spacetime (not in a spacetime that is stationary but not static). However, that is not necessary to separate solutions into positive and negative frequency parts. You can do that in any stationary spacetime.

9. Jul 29, 2017

### Frank Castle

Is this because the $g_{0i}$ components will be time independent (in an appropriately chosen coordinate system) and so the PDE will still be separable?

Also, I'm still not quite sure I understand to logic completely. Is the point that the timelike Killing vector ensures that there are solutions of the form $f_{\omega}(t,\mathbf{x})=e^{i\omega t}\bar{f}_{\omega}(\mathbf{x})$ to the equation $(\Box+m^{2})f=0$. One then uses the timelike Killing vector to distinguish positive and negative frequency modes $f$ and $f^{\ast}$ by defining them as solutions that satisfy $$K^{\mu}\partial_{\mu}f_{\omega}(t,\mathbf{x})=-i\omega f_{\omega}(t,\mathbf{x})$$ and $$K^{\mu}\partial_{\mu}f^{\ast}_{\omega}(t,\mathbf{x})=i\omega f^{\ast}_{\omega}(t,\mathbf{x})$$ respectively (with $\omega$>0)?

10. Jul 30, 2017

### Staff: Mentor

More or less. The point of a Killing vector field is that all of the metric coefficients are constant along its integral curves.

I would recommend working through the proof of Noether's Theorem, which is the theorem that shows why there is a conserved quantity associated with any Killing vector field. Having a conserved energy is just a special case of this where the Killing vector field is timelike.

11. Jul 30, 2017

### Frank Castle

Ok, I'll give that a go.
Is the takeaway point that the equations defining the positive and negative frequency modes with respect to a timelike vector are conserved along the integral curves of this timelike Killing vector field?

12. Jul 30, 2017

### Staff: Mentor

Equations aren't "conserved", so I'm not sure what you're trying to say here.

What sources (textbooks or peer-reviewed papers) are you looking at?

13. Jul 30, 2017

### Frank Castle

Sorry, I phrased this incorrectly. I was meaning that the sign on the right hand side of the equations (defining the positive and negative frequency modes) are invariant under time translations.

I've been reading Sean Carroll's book "Spacetime & Geometry: An Introduction". He discusses this issue briefly in his chapter near the end of the book on qft in curved spacetime.

14. Jul 30, 2017

### Staff: Mentor

They have to be, because if they aren't then no solutions of the form you wrote down exist.

That's a good source in general, although AFAIK it only gets into this topic at a very introductory level.

15. Jul 30, 2017

### Frank Castle

Ah ok, is this is what is meant by the fact that one can unambiguously define a notion of positive and negative frequency modes, which we can't do if such solutions as the ones I wrote down don't exist (i.e. the metric is time translation invariant [along the integral curves of a timelike Killing vector])?

Do you have any other texts that you would suggest? I've had a look at De Witt's book, but it hasn't helped me much unfortunately.

16. Jul 30, 2017

### Staff: Mentor

Yes, since the expressions you wrote down are the definitions of those terms.

Wald's monograph Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics is good, but definitely advanced; and even it does not go into too much mathematical detail about how the decomposition into positive and negative frequency parts is done. It does have a good set of references, though. I suspect that the mathematical detail is of more interest to mathematicians than physicists.

17. Jul 31, 2017

### vanhees71

Another textbook on the subject is

S. A. Fulling, Aspects of quantum field theory in curved spacetime, Cambridge University Press (1989)

18. Jul 31, 2017

### Frank Castle

Ok, thanks for the recommendation.
I think my problem is that I'm struggling to see the importance of defining the positive and negative frequency modes in such a manner. Is it simply because the Killing vector allows one to diagonalise the d'Alembertian such that one can solve the PDE via a separable solution, i.e. permitting solutions of the form $e^{-i\omega t}\bar{f}_{\omega}(\mathbf{x})$. Given this, one can then use the timelike Killing vector to distinguish a set of positive and negative frequency modes (as defined earlier). This is well-defined since the metric, and hence the sign of frequency modes, is invariant under time translation along the integral curves of the timelike Killing vector.

19. Jul 31, 2017

### Staff: Mentor

If so, continuing to repeat things you've already said isn't going to help.

Go back and read what @jambaugh said in post #2, particularly the second paragraph.

20. Aug 1, 2017

### Frank Castle

Having re-read @jambaugh 's post I think I might understand it now.

So the existence of a timelike Killing vector ensures that energy is conserved along the integral curves of this Killing vector, thus it is possible to define a well-defined vacuum state with respect this Killing vector since it will remain the state of lowest energy along its integral curves. Furthermore, one can find a coordinate frame in which the time coordinate is defined with respect to the timelike Killing vector and the metric in this frame is time-independent, thus the solutions to the equations of motion (EOM) are separable into a temporal and a spatial component. One can then define a natural set of creation and annihilation operators with respect to this timelike killing vector, since the separability of the solutions to the EOM makes it possible to classify them in terms of positive and negative frequency modes, and these distinctions are well-defined since their definitions are time-translation invariant along the integral curves of the Killing vector.